Guess Paper XII Chemistry 2026 with complete solution

📘 Guess Paper XII Chemistry 2026

Prepared by Inam Jazbi – Learn Chemistry

• 🎯 Targeted Preparation: Covers key sections likely to appear in exams.
• 🔢 Important Reactions: Step-by-step solutions for scoring high marks.
• 🧪 Concept Clarity: Smart Short-answer questions explained in simple language.
• ⏰ Smart Time Management: Suggested timings for each part to maximize efficiency.
• ✅ Exam-Oriented Approach: Structured exactly according to the paper pattern.

📝 Instructions

📚 Attempt 14 questions in all ---- 9 from Section B, 4 from Section C and the compulsory question No. 1 (Section A) of MCQs.

🔢 Write down proper numbering according to question paper. Do not copy questions.

✍️ Section B

🔹 Attempt 9 Short Questions ➡️ 4 marks each 🎯 Total = 36 marks

⏰ Suggested Time for Solution

🕒 Total: 90 mins

✍️ Each part: 9 mins

⚗️ Inorganic–General Chemistry Section 🧪

➡️ Attempt Any 4 Questions 📝

Q2 (i)

Explain the group trend and irregularities of ionization energy and atomic radii in the periodic table.

OR

What is bleaching powder? How it is prepared? Give its reactions with water and dilute HCl.

Q2 (ii)

What is diagonal relationship? Give diagonal relationship of Li-Mg, Be-Al and B-Si.

OR

Why transition elements have the tendency to form alloy? Write the name of three alloys of transition elements along with their composition.

OR

What are transition elements and outer transition elements? How are they classified? Why outer transition elements are called d-block element?

OR

Define d-block elements, why do they form colored compounds? Explain it in term of Crystal Field Theory.

Q2 (iii)

Write the IUPAC names of the following complexes:

⚛️ Na₂[Pt(OH)₄] →✅ Sodium tetrahydroxoplatinate(II)
⚛️ [Zn(NH₃)₄]²⁺ →✅ tetraamminezinc(II) ion
⚛️ [Ni(SCN)₄]²⁻ →✅ tetrathiocyanatonickelate(II) ion
⚛️ K₂[Fe(CN)₅NO] →✅ Potassium pentacyanonitrosylferrate(III)
⚛️ [Ag(NH₃)₂]OH →✅ diamminesilver(I) hydroxide
⚛️ K₃[Fe(CN)₆] →✅ Potassium hexacyanoferrate(III)
⚛️ K₃[Cr(C₂O₄)₂Cl₂] →✅ Potassium dichlorobis(oxalato)chromate(III)
⚛️ [Pt(en)₂(NO₂)₂]SO₄ →✅ Bis(ethylenediamine)dinitroplatinum(IV) sulphate
⚛️ K₃[Cu(C₂O₄)₂(CN)₂] →✅ Potassium dicyanobis(oxalato)cuprate(III)
⚛️ [Ni(CO₃)₂(OH)₂]²⁻ →✅ Dihydroxobis(carbonato)nickelate(II) ion
⚛️ [Cr(NH₃)₄Cl₂] →✅ tetraamminedichlorochromium(II)
⚛️ [Pt(en)₂Br₂] →✅ dibromobis(ethylenediamine)platinum(II)
⚛️ Na₃[Co(NO₂)₆] →✅ Sodium hexanitritocobaltate(III)
⚛️ NH₄[Cr(SCN)₄(NH₃)₂] →✅ Ammonium diamminetetrathiocyanatochromate(III)
⚛️ [Cr(NH₃)₂(NSC)₄]⁻ →✅ diamminetetrathiocyanatochromate(III) ion
⚛️ Na₂[Fe(CN)₅NO] →✅ Sodium pentacyanonitrosylferrate(III)
⚛️ [Fe(CN)₅NO]²⁻ →✅ pentacyanonitrosylferrate(III) ion
⚛️ K₃[Cr(C₂O₄)₂Cl₂] →✅ Potassium dichloridobis(oxalato)chromate(III)
⚛️ [Pt(en)₂(NO₂)₂]Cl₂ →✅ Bis(ethylenediamine)dinitroplatinum(IV) chloride
⚛️ [Cu(NH₃)₄]SO₄ →✅ tetraamminecopper(II) sulfate

OR

Q2. Give the formulae of the following complexes.

🟠 tetraamminedichlorochromium(III) chloride →✅ [Cr(NH₃)₄Cl₂]Cl
🟠 tetracarbonylnickel(0) →✅ [Ni(CO)₄]
🟠 tetranitrochromate(III) ion →✅ [Cr(NO₂)₄]⁻
🟠 Tollen’s reagent →✅ [Ag(OH)₂]OH
🟠 potassium hexacyanoferrate(III) →✅ K₃[Fe(CN)₆]
🟠 Nessler’s Reagent →✅ K₂[HgI₄]

Q2 (iv)

Explain colour formation, magnetic properties and variable oxidation state of transition elements.

OR

What is meant by binding energy? Explain the trend of binding energy in 3d series of transitions elements.

Q2 (v)

Why Beryllium differs markedly from other members of II A group? Write down four properties of beryllium that show its unique behaviour in group II A. Also write the reason why beryllium does not react with cold water and steam?

OR

What is flame test? What is the basis of flame test? Mention the colour flame of s-block elements.

Q2 (vi)

Using electronic configuration, identify the block, period & group of the elements with the Z = 16, 24, 29, 35, 47 and 53.

OR

Define ligands and chelating agents with examples. Write down names and formulae of 5 neutral, 5 negative, 2 bidentate and 2 polydentate ligands.

Q2 (vii)

Give the general valence shell electronic configuration of the following:

➡️ Representative elements (ns¹⁻² to ns² np¹⁻⁶)
➡️ Chalcogens (ns² np⁴) and Halogens (ns² np⁵)
➡️ Outer (n–1)d¹⁻¹⁰ ns¹⁻² and Inner transition elements (n–2)f²⁻¹⁴ (n–1)d¹⁻² ns²
➡️ Alkali metals (ns¹) and coinage metals (n–1)d¹⁰ ns¹)

Q2 (viii)

What is Catalytic converter? What are the main pollution targeted by catalytic converter?

OR

How was ozone layer formed? Explain the causes of depletion of ozone layer.

OR

Define green house gases. How green house gases cause to global warming?

Q2 (ix)

What is industrial smog and how is it formed?

OR

Explain four fundamental methods for the testing of waste water.

OR

What information about the structure of a molecule we can get from mass spectroscopy? Give the applications of mass spectroscopy.

Q2 (x)

Give the scope of pharmaceutical industries in Pakistan. Write down names of five drugs with their uses.

OR

What is meant by acid rain? Describe its cuases and adverse effects on human life style and health. What measures can be taken to prevent acid rain?

OR

Describe the preparation and two properties of nylon, terylene and PVC.

Q2 (xi)

Write down complete balanced action of following reactions:

➡️ Magnesium is heated with nitrogen gas.
➡️ Potassium is put into ethyl alcohol.
➡️ Chlorine react with nitrogen.
➡️ Fluorine reacts with oxygen.
➡️ Aluminium reacts with water.
➡️ Silicon reacts with steam.
➡️ Bleaching powder is dissolved in water.
➡️ Phosphorus reacts vigorously with water.
➡️ Reaction of chromium with steam.
➡️ Reaction of dichromate with ferrous salt.
➡️ Reaction of manganese with air.
➡️ Sulphur reacts at high temperature with water.
➡️ Silicon is heated with nitrogen at high temperatures.
➡️ Phosphorus reacts with nitrogen at high temperature.
➡️ Nitrogen reacts with oxygen in the presence of catalyst.
➡️ A piece of aluminium is dropped into concentrated sulphuric acid.
➡️ Ferric chloride is mixed in an aqueous solution of caustic soda.
➡️ Chlorine gas is passed through an aqueous solution of caustic soda.
➡️ Reaction of conc. nitric acid with copper.
➡️ Reaction of conc. sulphuric acid with copper.
➡️ Reaction of permanganate with oxalic acid.
➡️ Reaction of manganese with dilute sulphuric acid.

Q2 (xii)

Complete and balance the following chemical equations:

➡️ 2Li₍ₛ₎ + H₂(g) → 2LiH
➡️ 2Na₍ₛ₎ + Cl₂(g) → 2NaCl₍ₛ₎
➡️ 6Na₍ₛ₎ + N₂(g) → 2Na₃N₍ₛ₎
➡️ 3Ca₍ₛ₎ + N₂(g) → Ca₃N₂₍ₛ₎
➡️ 6Li₍ₛ₎ + N₂(g) → 2Li₃N₍ₛ₎
➡️ 4Li₍ₛ₎ + O₂(g) → 2Li₂O₍ₛ₎
➡️ 4Na₍ₛ₎ + O₂(g) → 2Na₂O₍ₛ₎
➡️ 2Na₍ₛ₎ + O₂(g) —Excess O₂→ Na₂O₂₍ₛ₎
➡️ K₍ₛ₎ + O₂(g) → KO₂₍ₛ₎
➡️ Rb₍ₛ₎ + O₂(g) → RbO₂₍ₛ₎
➡️ 2Be₍ₛ₎ + O₂(g) → 2BeO₍ₛ₎
➡️ Sr₍ₛ₎ + O₂(g) → SrO₂₍ₛ₎
➡️ Si₍ₛ₎ + 2H₂O₍ₗ₎ → SiO₂₍ₛ₎ + 2H₂↑
➡️ 3Si₍ₛ₎ + 2N₂(g) → Si₃N₄₍ₛ₎
➡️ Si₍ₛ₎ + 2H₂(g) → SiH₄(g)
➡️ Si₍ₛ₎ + 2H₂O₍ₗ₎ → SO₂(aq) + 2H₂↑
➡️ 4P(g) + 5O₂(g) —Excess O₂→ 2P₂O₅(g)
➡️ 4P(g) + 3O₂(g) —Limited O₂→ 2P₂O₃(g)
➡️ 2N₂(g) + O₂(g) —Catalyst→ 2N₂O(g)
➡️ 2Na + 2C₂H₅OH(aq) → 2C₂H₅ONa(aq) + H₂(g)↑
➡️ 6OH⁻ + 3Cl₂ —Excess O₂→ 5Cl⁻ + ClO₃⁻ + 3H₂O
➡️ 6NaOH + 3Cl₂ —Excess O₂→ 5NaCl + NaClO₃ + 3H₂O
➡️ CaOCl₂ + 2H₂O₍ₗ₎ → Ca(OH)₂ + 2HOCl
➡️ 2K₍ₛ₎ + 2HCl(aq) + H₂O₍ₗ₎ → 2KCl(aq) + H₂(g)↑
➡️ 4B₍ₛ₎ + 3O₂(g) → 2B₂O₃₍ₛ₎
➡️ N₂(g) + O₂(g) → 2NO(g)
➡️ N₂(g) + 3Cl₂(g) → 2NCl₃(g)
➡️ 2P₄₍ₛ₎ + 12H₂O₍ₗ₎ → 3H₃PO₄(aq) + 5PH₃↑
➡️ 6P₍ₛ₎ + 5N₂(g) → 2P₃N₅₍ₛ₎
➡️ 2F₂(g) + O₂(g) → 2OF₂(g)
➡️ 2Al₍ₛ₎ + 6H₂O₍ₗ₎ → 2Al(OH)₃₍ₛ₎ + 3H₂↑
➡️ 2Al₍ₛ₎ + N₂(g) → 2AlN₍ₛ₎
➡️ Zn₍ₛ₎ + 2NaOH(g) → Na₂ZnO₍aq₎ + H₂↑
➡️ 2Na/Hg₍ₗ₎ + 2H₂O₍ₗ₎ → 2NaOH₍aq₎ + H₂↑ + 2Hg₍ₗ₎

Q2 (xiii)

Give reasons of any 4 of the following:

➡️ Ionization energy decreases from top to bottom in s-block elements.
➡️ H₂O and NH₃ act as ligands but H₃O⁺ and NH₄⁺ do not.
➡️ Fluorine is the strongest oxidizing agent and lithium is the strongest reducing agent.
➡️ Alkali metals are good conductor of electricity.
➡️ Melting and boiling points of zinc are exceptionally low.
➡️ Multidentate ligands are known as chelating agents.
➡️ The transition metals complexes are coloured.
➡️ Mn²⁺ show maximum paramagnetic character amongst the bivalent ions of first transition series.
➡️ Melting point of d-block elements increase up to middle of the series and then decrease why.
➡️ Configuration of ₂₄Cr is 3d⁵ 4s¹ instead of 3d⁴ 4s² & Configuration of ₂₉Cu is 3d¹⁰ 4s¹ instead of 3d⁹ 4s².
➡️ Melting point of d-block elements increase up to middle of the series and then decrease why.
➡️ Why Cu²⁺ ions is blue but Zn²⁺ is Colorless.
➡️ Acidity of hydrogen halide increases from HF to HI.
➡️ Gallium has smaller atomic radii than aluminum despite being below the aluminum in group IIIA.
➡️ Electronegativity decreases regularly from top to bottom in s-block elements.
➡️ Boiling point of halogens increases down the group in the periodic table.
➡️ Transition elements have ability to form alloys.
➡️ Why are the binding energy of Mn²⁺ and Fe³⁺ ions the highest and that of zinc is least in 3d series.
➡️ Why do transition elements show variable oxidation states.
➡️ Paramagnetic behaviour is the strongest for Fe³⁺ and Mn²⁺.
➡️ The maximum OS increase in each transition series up to the middle of series & then decreases afterward.
➡️ How the given reaction is avoided during the preparation of NaOH?
6OH⁻ + Cl₂ ➔ Cl⁻ + ClO₃⁻ + 3H₂O
➡️ Ligands are generally called Lewis bases.

⚗️ Organic Chemistry Section 🧪

➡️ Attempt Any 5 Questions 📝

Q2 (xiv)

Define any FOUR of the following:

Catenation, Reforming, homologous series, Functional group, chiral carbon, enantiomers, knocking, octane number, Polymerization (polymers), isomerism (isomers), metamerism, knock-inhibitor, carbonization, heterocyclics, Saponification, Glycosidic linkage, peptide bond, zwitterion, electrophile, nucleophile, Huckle rule, aromaticity.

OR

Write down Two differences between the following:

(a) Reducing and non-reducing sugars
(b) Aliphatic and aromatic compounds
(ii) Saturated and unsaturated compounds (BUT)
(d) Total & partial synthesis of organic compounds

OR

Define Bucky Ball. Explain its structure and mention its some properties and uses.

Q2 (xv)

We often use the term iso and neo in the common system of naming of alkanes. Explain with examples.

OR

If an organic compound contains both double and triple bond in the main carbon chain, what rules you follow to write its IUPAC names. Explain by giving an example.

Q2 (xvi)

How is coal produced under the earth crust? Write the name of four types of coal and mention the %age of carbon content in them. Explain destructive distillation of coal and various products obtained from it.

OR

Define homologous series and write its three general properties.

Q2 (xvii)

Draw the orbital structure of ethane or ethene and ethyne and explain how ethyne is distinguished from ethene by a simple chemical test.

OR

What is free radical? Give stepwise mechanism for the chlorination of methane.

Q2 (xviii)

Why benzene show stability towards addition reaction? Why benzene gives electrophilic substitution reaction? Write the chemical equations for the Friedel-Craft reaction and Sulphonation. Also write stepwise the mechanism of nitration or acylation of benzene.

OR

Identify each of the following with one laboratory test:

➡️ Alcohol
➡️ Phenol
➡️ Alkene
➡️ Aldehyde

Q2 (xix)

What is meant by stereo isomerism, chiral carbon and plane polarized light? Define optical and cis and trans isomers with examples. Explain optical isomerism briefly.

OR

What is meant by isomerism? Explain four different types of structural isomers and two types of stereoisomerism and give one example of each. Draw all possible isomers of pentyl alcohol (or pentyl chloride), pentene, butyne & compound with formula C₃H₆O and C₂H₄O₂.

OR

Write note on classification of organic compounds and Natural sources of Organic Compounds.

Q2 (xx)

Define organo-metallic compound with examples. What is Grignard’s reagent? Write the equation of its (methyl magnesium iodide) like reaction with:

➡️ Water    ➡️ Dry ice    ➡️ Formalin    ➡️ Methyl amine    ➡️ Carbonyl compounds

OR

Write the equation for the reaction of acetaldehyde with the following:

Chromic acid, lithium aluminium hydride, Zinc-mercury amalgam, hydroxylamine, ammonia, hydrogen cyanide.

Q2 (xxi)

Why are alkyl amines basic in nature? How a primary alkyl amine is converted into secondary & tertiary amine?

OR

Define primary, secondary and tertiary amines. Why are amines basic in nature? Explain why secondary amines are more basic than primary amines. How can we prepare ethyl amine from the following compounds?

* Ethyl iodide
* Methyl cyanide
* Ethanamide

Q2 (xxii)

Name four derivatives of carboxylic acids with their class formulae and write the equations of their preparation.

OR

Discuss the acidic nature of carboxylic acid. How is carboxylic acid prepared by:

➡️ Hydrolysis of alkyl nitrile (Methyl Nitrile)
➡️ Oxidation of primary alcohols (Ethanal)
➡️ Oxidation of ketone (Acetone)
➡️ Carbonation of Grignard’s reagent (Methyl magnesium chloride)

Q2 (xxiii)

Define nucleophilic substitution reaction. How can we prepare following compounds using CH₃Br?

➡️ CH₃SH
➡️ CH₃OCH₃
➡️ CH₃COOCH₃
➡️ CH₃CN

Q2 (xxiv)

Give the scope of pharmaceutical industries in Pakistan. Write down names of five drugs with their uses.

OR

What is antihistamine drug? Give the symptoms in which it is used.

Q2 (xxv)

Consider the following structures and answer the following questions:

(A) CH₂=CH₂    (B) C₆H₆

(a) Draw the hybrid structure of A
(b) Write equation for the conversion of B into acetophenone.
(c) Write the equation for the conversion of A into acetic acid.
(d) Write the equations for the conversion of B into phenol and TNT.

OR

The structure of two organic compounds ‘A’ and ‘B’ are shown below:

(A) CH₂=CH₂    (B) HC≡CH

(a) Draw and explain the orbital structure of A and specify hybridization and bond angle.
(b) Write the equations, when A and B react with ozone (ozonolysis).

Q2 (xxvi)

What are synthetic polymers? Write down the names of two synthetic and two natural polymers? Write down the preparation of two condensation polymers and one addition polymer with equations.

OR

Distinguish two types of polymers based on mode of polymerization and action of heat. Which polymer is obtained on polymerization of following monomers. Write complete reaction equation:

➡️ Vinyl chloride
➡️ Adipic acid and hexamethylene diamine
➡️ Ethylene glycol and Terephthalic acid

Q2 (xxvii)

Give the equation and write the name of final product in the following process. (write only equation)

➡️ Reaction of benzene diazonium chloride with water at high temperature
➡️ Reaction of propyl alcohol with thionyl chloride
➡️ Reaction of phenol with concentrated sulphuric acid at low and high temperature
➡️ Acetylene or propyne reacts with water in presence of H₂SO₄ and HgSO₄ at 75°C
➡️ 1,2-dibromoethane is heated with alcoholic KOH
➡️ Reduction of acetic acid with LiAlH₄
➡️ Reaction of sodium ethanoate in the presence of soda lime
➡️ Oxidation of acetone with acidified K₂Cr₂O₇
➡️ Ethyne is treated with hydrogen bromide
➡️ Reaction of propene with HBr
➡️ Ethanol in excess, is heated in presence of H₂SO₄
➡️ Ethene or 2-butene is ozonolyzed
➡️ Ethene is treated with oxygen in presence of peracetic acid at 100°C

OR

How can we prepare following compounds (any four):

(a) Ethanal from ethyne
(b) Phenyl hydrazone from formaldehyde
(c) Ethanol from organo-metallic compound
(d) Oxime from acetaldehyde
(e) tert-butyl alcohol from Grignard’s reagent
(f) Ethene from ethanol
(g) Bromohydrin from ethene
(h) Ethyne from ethene

Q2 (xxviii)

Write only the equations for the following reactions:

Williamson’s synthesis, Dow’s process, Esterification, Saponification, Clemmensen Reduction.

OR

What is Lucas reagent and Lucas test? How is this test used to distinguish three types of alcohols.

OR

Write down two methods of preparation of ethers. How is oxonium salt of ether formed?

Q2 (xxix)

Explain the following with scientific reasons:

(a) Ethanol is liquid but ethyl chloride is a gas at room temperature.
(b) Boiling point of alcohol is higher than ether and carbonyl compounds.
(c) Formaldehyde is highly soluble in water as compared to other aldehydes.
(d) Oxidation of aldehydes is faster than ketones.
(e) The boiling point of carboxylic acids are high than alcohol.
(f) The structure of carboxylic acid is trigonal planar.
(g) tert-alcohols cannot be oxidized.

OR

Differentiate between atomic absorption and emission spectroscopy. What is the purpose of UV-visible spectroscopy? What are its applications in chemistry and biology?

✍️ Section C

🔹 Attempt 4 Detailed Questions ➡️ 8 marks each 🎯 Total = 32 marks

⏰ Suggested Time for Solution

🕒 Total: 70 mins

✍️ Each part: 17.5 mins

Q3

Draw a flow diagram of contact process and describe various steps involved in the industrial manufacture of oil of vitriol. Write the reactions of concentrated Sulphuric acid with Al and sucrose.

OR

Describe various steps involved in the extraction of 99.99% pure copper from its chalcopyrite ore. (Draw diagram where necessary)

Q4

What is water pollution? Discuss different types of water pollution also explain any parameters of drinking water analysis.

OR

Define Troposphere and Stratosphere. Describe the chemistry involved due to the presence of oxides of carbon and nitrogen in the troposphere. (Write equations where necessary)

Q5

Explain with the help of a diagram of Castner Kellner cell, how caustic soda is obtained by the electrolysis of aqueous sodium chloride. Write down merits and demerits of Castner Kellner process. Write down action of sodium hydroxide solution on zinc and aluminium metal.

OR

Write the balanced chemical equations for the following chemical processes:

➡️ Bleaching powder is dissolved in water.
➡️ Fluorine reacts with oxygen.
➡️ Potassium is put into ethyl alcohol.
➡️ Carbon heated with nitrogen at high temperature.
➡️ Permanganate ion reacts with oxalic acid.
➡️ Manganese reacts with air.
➡️ Potassium dichromate dissolved in water at neutral pH.
➡️ Phosphorus is put in water.
➡️ Sodium burns in excess of air.
➡️ Silicon reacts with steam.
➡️ Sulphur reacts at high temperature with water.
➡️ Phosphorus reacts with nitrogen at high temperature.
➡️ Chlorine reacts with nitrogen.
➡️ A piece of aluminium is dropped into concentrated sulphuric acid.
➡️ Ferric chloride is mixed in an aqueous solution of caustic soda.
➡️ Chlorine gas is passed through hot and cold aqueous solution of caustic soda.
➡️ Reaction of cuprous oxide with cuprous sulphide.
➡️ Reaction of chromium with steam at high temperature.
➡️ A mixture of carbon and silicon is heated under elevated temperature.
➡️ Bleaching powder is treated with hydrochloric acid.
➡️ Copper is treated with concentrated nitric acid.
➡️ A piece of chromium is put into dilute hydrochloric acid.
➡️ Reaction between KMnO₄ and FeSO₄ in the presence of H₂SO₄ (write ionic equation).
➡️ Reaction between K₂Cr₂O₇ and FeSO₄ in the presence of H₂SO₄ (write ionic equation).

Q6

(a) What is the main cause of Global warming? How does it effect on weather pattern?

(b) Complete and balance the following chemical equations:

➡️ 2Cr₍ₛ₎ + 3H₂O₍ₗ₎ → Cr₂O₃₍ₛ₎ + 3H₂↑
➡️ 2Li₍ₛ₎ + H₂₍g₎ → 2LiH₍ₛ₎
➡️ 6Na₍ₛ₎ + N₂₍g₎ → 2Na₃N₍ₛ₎
➡️ 3Ca₍ₛ₎ + N₂₍g₎ → Ca₃N₂₍ₛ₎
➡️ 6Li₍ₛ₎ + N₂₍g₎ → 2Li₃N₍ₛ₎
➡️ K₍ₛ₎ + O₂₍g₎ → KO₂₍ₛ₎
➡️ 2Cr₍ₛ₎ + 6HCl₍aq₎ → 2CrCl₃₍aq₎ + 3H₂↑
➡️ 3Mn₍ₛ₎ + 2O₂₍aq₎ → Mn₃O₄₍ₛ₎
➡️ MnO₄⁻₍aq₎ + 5Fe²⁺₍aq₎ + 8H⁺₍aq₎ → Mn²⁺₍aq₎ + 5Fe³⁺₍aq₎ + 4H₂O
➡️ 2MnO₄⁻₍aq₎ + 5H₂C₂O₄₍aq₎ + 6H⁺₍aq₎ → 2Mn²⁺₍aq₎ + 10CO₂₍g₎ + 8H₂O
➡️ 2Cr₂O₇²⁻₍aq₎ + 3H₂C₂O₄₍aq₎ + 8H⁺₍aq₎ → 4Cr³⁺₍aq₎ + 6CO₂₍g₎ + 7H₂O

⚗️ Organic Chemistry Section (Long Questions)🧪

➡️ Attempt Any 2 Questions 📝

Q7

What is meant by nucleophile? Differentiate between Sɴ₁ and Sɴ₂ reactions. Explain the reaction mechanism of Sɴ₁ and Sɴ₂ reactions.

OR

Write down names, type formula and characteristic groups of three types of monohaloalkane. Explain the mechanism of following reactions: (No description is required).

(i) NaOH or KOH reacts with 2-bromo-2-methylpropane; (CH₃)₃CBr (3° alkyl halide)
(ii) NaSH/KSH reacts with 2-chlorobutane (2° alkyl halide) in the presence of aprotic or protic solvents
(iii) NaCN or KCN reacts with 1-chlorobutane; CH₃CH₂CH₂CH₂Cl (1° alkyl halide)

Q8

Write down structural formulae of any 8 of the following organic molecules:

TNT, Benzophenone, Picric acid, p-cresol, isobutyraldehyde, oxalic acid, ethylene glycol, neopentyl alcohol, di-isopropyl ketone, iso-valeric acid, Resorcinol, 2-ethoxyhexane, trifluroacetic acid, triphenyl amine, divinyl acetylene, carbinol, isobutyric acid, glyoxal, Resorcinol, isobutanoyl bromide, Adipic acid, Cyclopentane, diphenyl ether, 1,2,3-benzentriol (Pyrogallol), isopropyl butanoate, divinyl acetylene, 2-methoxy-2-methylbutane, α,β-dimethyl valerte, isopropyl propionate, Ethyl neo-pentyl ether, Neo Pentyl iodide, Benzamide, Benzene-1,4-dioic acid (Terephthalic Acid).

Q9

Define β-elimination reactions. Mention its two types and give difference between them and Outline the mechanism of reaction between sec-butyl chloride & alcoholic KOH in the presence of polar protic and polar aprotic solvents?

OR

What are proteins? Classify various types of proteins on the basis of their function and structures. Also give biological significance and properties of proteins.

Q10

Explain ONE simple laboratory test to distinguish between any 4 of the following pairs of compounds:

(i) Alkanes and alkenes
(ii) Propanal and propanone
(iii) Alcohol and phenol
(iv) n-hexane and benzene
(v) 1-butyne and 2-butyne
(vi) Alkenes and alkynes

Q11

What is meant by orientation of benzene? Explain ortho-para and meta directing groups. Write the equation for the preparation of the following compounds from benzene:

* TNT
* m-nitrotoluene
* m-nitrobenzoic acid
* o- and p-nitrotoluene

OR

Complete and balance any five of the following reactions:

OR

Write the equation for the nucleophilic addition reaction of formaldehyde with hydrogen cyanide, primary alcohol, ammonia, hydroxylamine, primary alcohol and Grignard’s reagent.

Q12

What are carbohydrates? Classify them on the basis of structure and give their biological significance. Explain glycosidic linkage and open chain and close chain structures of glucose and fructose.

OR

What is green house effect? How does human activities contribute to the enhancement of the green house effect?

OR

What is the main cause of Global warming? How does it effect on weather pattern?

OR

Explain structural isomerism and stereoisomerism with suitable examples.

Q13

Write the equation for the following reactions. (Any Eight)

➡️ Formation of amide from ethanoic acid
➡️ Reaction of acetaldehyde with hydroxyl amine
➡️ Oxidation of 1°-alcohol with PCC
➡️ Formation of acetic anhydride from acetic acid
➡️ Reaction of ethylene glycol with periodic acid
➡️ Formation of picric acid from phenol
➡️ Formation of 3° alcohol from ketone
➡️ Reduction of acetone by lithium aluminium hydride
➡️ Reduction of ethanal by mixture of zinc amalgam and concentrated HCl
➡️ Formation of benzoquinone from phenol
➡️ Formation of primary alcohol from carboxylic acid
➡️ Reduction of acetone by mixture of zinc amalgam and concentrated HCl
➡️ Reaction of phenol with H₂SO₄ at 20°C
➡️ Reaction of propanone with hydroxyl amine
➡️ Ethanol, in excess, is heated up to 140°C in presence of H₂SO₄
➡️ Oxidation of 2°-alcohol with K₂Cr₂O₇/H₂SO₄
➡️ Dehydration of ethyl alcohol at 170°C in conc. H₂SO₄
➡️ Reduction of acetone with LiAlH₄

OR

Write down the IUPAC names of any 8 of the following organic compounds:

◉ CH₂=CH−HC=CH−C≡CH
◉ CH₃−CH=CH−CH₂−COOH
◉ CH₃–CH₂–O–C(CH₃)₂C₂H₅
◉ (CH₃)₂CH−CO−CH(C₂H₅)₂
◉ CH₂=CH−CH−C≡CH
◉ CH₂=CH−C≡C−CH=CH₂
◉ CH₂=CH−(CH₂)₄−COOH
◉ (CH₃)₃C−CO−CH₂CH₂−CHO
◉ CH₃COCCl₃
◉ F₃C−COOH
◉ OHC−CHO
◉ (CH₃)₃C−CO−CH₂CH₃
◉ (C₆H₅)₃C−CHO
◉ CH₂=CH−CH(OH)−CH₂−COOH
◉ C₂H₅−O−CH₂−C(CH₃)₃
◉ C₂H₅−CH₂−COO−CH(CH₃)₂
◉ (CH₃)₂CH−CO−C(CH₃)₃
◉ CH₃−(CHOH)₂−CH₂OH
◉ CH₂(OH)₂
◉ Adipic acid (HOOC−(CH₂)₄−COOH)
◉ Cyclopentane (C₅H₁₀)
◉ Diphenyl ether (C₆H₅−O−C₆H₅)
◉ 1,2,3-benzenetriol (Pyrogallol)
◉ Benzamide (C₆H₅−CONH₂)
◉ Benzene-1,4-dioic acid (Terephthalic Acid, HOOC−C₆H₄−COOH)

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📝 Model Paper MCQs Karachi Board XII Chemistry 2026 (Click to reveal answer with reason.🎯📌

1. The oxidation states of the elements of group VIIA:

A) +1 and +3

B) -1, +5 and -5

C) -3, +3 and +5

D) -1, +1, +3, +5 and +7

✅ Correct Answer: D) -1, +1, +3, +5 and +7
Reason: Halogens (Group VIIA) commonly show oxidation states from -1 to +7 depending on the compound.

2. This s-block element forms superoxide when burnt in air:

A) Li

B) Na

C) K

D) Mg

✅ Correct Answer: C) K
Reason: Potassium reacts with oxygen to form potassium superoxide (KO₂).

3. The metal used in making jewelry and also serves as a catalyst:

A) Pt

B) Fe

C) Pb

D) Al

✅ Correct Answer: A) Pt
Reason: Platinum is precious, used in jewelry, and acts as a catalyst in many reactions (e.g., hydrogenation).

4. The coordination number of Ni in the complex compound [Ni(en)₂(NO₂)₂]²⁺:

A) 4

B) 6

C) 2

D) -2

✅ Correct Answer: B) 6
Reason: Ni is bonded to 2 ethylenediamine ligands (each bidentate = 4 sites) and 2 nitrite ions, giving coordination number 6.

5. The knocking of internal combustion engine can be reduced by the following petroleum process:

A) Reforming

B) Refining

C) Distillation

D) Condensation

✅ Correct Answer: A) Reforming
Reason: Reforming increases the octane number of petrol, reducing knocking in engines.

6. Welding gas among the following is:

A) Ethene

B) Ethane

C) Methane

D) Ethyne

✅ Correct Answer: D) Ethyne
Reason: Ethyne (acetylene) mixed with oxygen is used as a welding gas due to its very high flame temperature.

7. This pair of compounds represents functional group isomerism:

A) 1-butene and 2-butene

B) Ethanol and dimethyl ether

C) n-butane and iso-butane

D) Diethyl ketone and methyl propyl ketone

✅ Correct Answer: B) Ethanol and dimethyl ether
Reason: Ethanol (alcohol) and dimethyl ether (ether) have the same molecular formula (C₂H₆O) but different functional groups.

8. Benzene burns with a smoky flame because of its:

A) Inflammability

B) High carbon percentage

C) High resonance energy

D) Aromaticity

✅ Correct Answer: B) High carbon percentage
Reason: Benzene has a high carbon content, leading to incomplete combustion and a smoky flame.

9. Sɴ₂ reaction occurs most easily if the substrate molecule is:

A) Methyl iodide

B) Ethyl iodide

C) 2-iodopropane

D) 2-iodo-2-methylpropane

✅ Correct Answer: A) Methyl iodide
Reason: S_N2 reactions are fastest with least steric hindrance; methyl iodide has minimal hindrance.

10. Lucas reagent is a mixture of:

A) Zn and Hg

B) Zn and HCl

C) NaOH and CaO

D) ZnCl₂ and HCl

✅ Correct Answer: D) ZnCl₂ and HCl
Reason: Lucas reagent (conc. HCl + ZnCl₂) is used to distinguish primary, secondary, and tertiary alcohols.

11. Clemmensen reduction is the conversion of aldehydes and ketones into:

A) Alkanes

B) Alkenes

C) Alkyl halides

D) Alcohols

✅ Correct Answer: A) Alkanes
Reason: Clemmensen reduction uses Zn(Hg) and HCl to reduce carbonyl groups (aldehydes/ketones) to alkanes.

12. Formic acid is naturally found in:

A) Bee sting

B) Venom of ants

C) Vinegar

D) Butter

✅ Correct Answer: B) Venom of ants
Reason: Formic acid is present in ant venom and is responsible for the burning sensation of ant bites.

13. Saponification is the formation of soap by the reaction of fat and oil with:

A) Sugar

B) HCl

C) An alkali

D) Glycerol

✅ Correct Answer: C) An alkali
Reason: Saponification involves hydrolysis of fats/oils with NaOH or KOH to produce soap and glycerol.

14. Drugs that lower the body temperature to normal are known as:

A) Antibiotics

B) Antipyretic

C) Antiallergic

D) Antihistamines

✅ Correct Answer: B) Antipyretic
Reason: Antipyretics (like paracetamol) reduce fever by lowering body temperature to normal.

15. This element is used to build strong bones and teeth:

A) Zn

B) Ca

C) Fe

D) Mg

✅ Correct Answer: B) Ca
Reason: Calcium is the main mineral in bones and teeth, providing strength and structure.

16. Ozone is produced in the stratosphere region due to photochemical reaction of sun rays and:

A) CFCs

B) NO₂

C) NO

D) O₂

✅ Correct Answer: D) O₂
Reason: Ozone forms when UV rays split O₂ molecules into atomic oxygen, which then combines with O₂ to form O₃.

17. Infra-red spectroscopy is a technique used to determine:

A) Double and triple bonds

B) Functional group

C) Mass to charge ratio

D) Conjugated system

✅ Correct Answer: B) Functional group
Reason: IR spectroscopy identifies functional groups by their characteristic absorption frequencies.

📝 Model Paper MCQs Karachi Board XII Chemistry 2025 (Click to reveal answer with reason.🎯📌

1. In III A group of periodic table, the ionization of gallium (Ga) is higher than aluminium (Al) due to insufficient shielding of the nuclear charge in gallium (Ga) by:

A) 4s electrons

B) 4p electrons

C) 3d electrons

D) 3p electrons

✅ Correct Answer: C) 3d electrons
Reason: Poor shielding by 3d electrons increases effective nuclear charge in Ga, raising ionization energy compared to Al.

2. This metal is essential for growth of bones and teeth:

A) Calcium

B) Magnesium

C) Sodium

D) Potassium

✅ Correct Answer: A) Calcium
Reason: Calcium is the main mineral in bones and teeth, providing strength and structure.

3. The boiling point and viscosity of H₂SO₄ is high because:

A) It is strong oxidizing agent

B) It has strong hydrogen bonding

C) It is highly volatile

D) It forms stable sulfur-oxygen bonds

✅ Correct Answer: B) It has strong hydrogen bonding
Reason: Extensive hydrogen bonding in H₂SO₄ molecules leads to high boiling point and viscosity.

4. The coordination number of Pt in [Pt(en)₃]Cl₄ is:

A) 3

B) 4

C) 6

D) +4

✅ Correct Answer: C) 6
Reason: Each ethylenediamine (en) ligand is bidentate, so 3 ligands donate 6 coordination sites to Pt.

5. This metal has maximum number of unpaired electrons in 3d orbitals:

A) Copper (Cu)

B) Manganese (Mn)

C) Cobalt (Co)

D) Iron (Fe)

✅ Correct Answer: B) Manganese (Mn)
Reason: Mn (Z=25) has configuration [Ar] 3d⁵ 4s², with 5 unpaired electrons in 3d orbitals — the maximum.

6. The general formula of the homologous series of cyclo alkane is:

A) CₙH₂ₙ

B) CₙH₂ₙ₊₂

C) CₙH₂ₙ₋₂

D) CₙH₂ₙ₊₄

✅ Correct Answer: A) CₙH₂ₙ
Reason: Cycloalkanes have the general formula CₙH₂ₙ, similar to alkenes, due to ring closure reducing two hydrogens.

7. The final product obtained when hydrogen bromide (HBr) is added to an ethyne molecule:

A) 1,2–Dibromo ethane

B) 1,1–Dibromo ethane

C) Bromoethene

D) 1,1,2,2–Tetrabromo ethane

✅ Correct Answer: B) 1,1–Dibromo ethane
Reason: Addition of 2 moles of HBr to ethyne gives 1,1–dibromoethane as the final product.

8. The sooty flame observed when benzene burns is due to:

A) High percentage of carbon atoms

B) Low molecular weight

C) High percentage of hydrogen atoms

D) High density of benzene

✅ Correct Answer: A) High percentage of carbon atoms
Reason: Benzene has a high carbon content, leading to incomplete combustion and a smoky flame.

9. The carbonyl carbon of aldehyde is:

A) sp hybridized

B) sp² hybridized

C) sp³ hybridized

D) dsp³ hybridized

✅ Correct Answer: B) sp² hybridized
Reason: The carbonyl carbon forms a double bond with oxygen, giving it trigonal planar geometry and sp² hybridization.

10. Lucas reagent is a mixture of:

A) Zn and Hg

B) ZnCl₂ and HCl

C) Na and Hg

D) CaO and NaOH

✅ Correct Answer: B) ZnCl₂ and HCl
Reason: Lucas reagent (conc. HCl + ZnCl₂) is used to distinguish primary, secondary, and tertiary alcohols.

11. When a Grignard reagent reacts with Ketone it produces:

A) Primary alcohol

B) Secondary alcohol

C) Tertiary alcohol

D) Ester

✅ Correct Answer: C) Tertiary alcohol
Reason: Grignard reagents add to ketones forming tertiary alcohols after hydrolysis.

12. An example of transport protein is:

A) Albumin

B) Haemoglobin

C) Lipase

D) Casein

✅ Correct Answer: B) Haemoglobin
Reason: Haemoglobin transports oxygen in blood, making it a transport protein.

13. This compound among the following is known as oxime:

A) R–CH=N–NH

B) R–CO=NH

C) R–CH=N–OH

D) R–CH(CN)–OH

✅ Correct Answer: C) R–CH=N–OH
Reason: Oximes are formed by reaction of aldehydes/ketones with hydroxylamine, giving R–CH=N–OH.

14. Antimalarial drug among the following is:

A) Ibuprofen

B) Chloroquine

C) Paracetamol

D) Diphenyl hydramine

✅ Correct Answer: B) Chloroquine
Reason: Chloroquine is a well-known antimalarial drug used to treat malaria infections.

15. Ozone is destroyed by:

A) Sulphur dioxide (SO₂)

B) Carbon dioxide (CO₂)

C) Chlorofluorocarbon (CFCs)

D) Hydrofluorocarbons (HFCs)

✅ Correct Answer: C) Chlorofluorocarbon (CFCs)
Reason: CFCs release chlorine radicals that catalytically destroy ozone in the stratosphere.

16. The IUPAC name for formic acid is:

A) Acetic acid

B) Methanoic acid

C) Ethanoic acid

D) Benzoic acid

✅ Correct Answer: B) Methanoic acid
Reason: Formic acid is the simplest carboxylic acid, with IUPAC name methanoic acid (HCOOH).

17. The pH of acid rain is:

A) Between 7 to 8

B) Between 6 to 7

C) Below 5

D) Above 8

✅ Correct Answer: C) Below 5
Reason: Acid rain has a pH less than 5 due to dissolved SO₂ and NOₓ forming strong acids.

📝 Karachi Board MCQs XII Chemistry 2025 (Click to reveal answer with reason.🎯📌

i) This gives violet colour to the flame:

A) Li

B) Na

C) K

D) Be

✅ Correct Answer: C) K
Reason: Potassium salts impart a violet (lilac) colour to the flame in flame tests.

ii) The order of reducing power of halide ion is:

A) I⁻ > Br⁻ > Cl⁻ > F⁻

B) I⁻ > Cl⁻ > F⁻ > Br⁻

C) F⁻ > Cl⁻ > Br⁻ > I⁻

D) Br⁻ > Cl⁻ > I⁻ > F⁻

✅ Correct Answer: A) I⁻ > Br⁻ > Cl⁻ > F⁻
Reason: Reducing power increases down the group; iodide is the strongest reducing agent among halides.

iii) The coordination number of [Cr(en)₃]³⁺ is:

A) 3

B) 4

C) 5

D) 6

✅ Correct Answer: D) 6
Reason: Each ethylenediamine (en) ligand is bidentate, donating 2 sites. With 3 en ligands, total coordination number = 6.

iv) This fertilizer provides Nitrogen and Phosphorus to the plants:

A) Urea

B) Potassium Nitrate

C) Ammonium Nitrate

D) Diammonium Phosphate

✅ Correct Answer: D) Diammonium Phosphate
Reason: Diammonium phosphate (DAP) supplies both nitrogen and phosphorus nutrients to plants.

v) This is the hottest part of atmosphere:

A) Thermosphere

B) Stratosphere

C) Mesosphere

D) Troposphere

✅ Correct Answer: A) Thermosphere
Reason: The thermosphere absorbs high‑energy solar radiation, making it the hottest atmospheric layer.

vi) Cathode in Castner Kellner Cell is:

A) Titanium Blocks

B) Carbon rods

C) Iron Container

D) Mercury

✅ Correct Answer: D) Mercury
Reason: In the Castner Kellner process, mercury acts as the cathode to form sodium amalgam.

vii) Diagonal relationship of lithium with:

A) Be

B) Na

C) Ca

D) Mg

✅ Correct Answer: D) Mg
Reason: Lithium shows diagonal relationship with magnesium due to similar size, charge density, and properties.

viii) This is low quality coal:

A) Bituminous

B) Lignite

C) Anthracite

D) Peat

✅ Correct Answer: D) Peat
Reason: Peat is the lowest quality coal, with high moisture and low carbon content.

ix) This is used as an Ink preservative:

A) Benzene

B) Toluene

C) Phenol

D) Alcohol

✅ Correct Answer: C) Phenol
Reason: Phenol prevents microbial growth, making it useful as an ink preservative.

x) This one is more acidic:

A) Carboxylic acid

B) Phenol

C) Alcohol

D) Water

✅ Correct Answer: A) Carboxylic acid
Reason: Carboxylic acids are stronger acids due to resonance stabilization of their conjugate base.

xi) This is meta directing group:

A) –OH

B) –NH₂

C) –CN

D) –OR

✅ Correct Answer: C) –CN
Reason: –CN is an electron withdrawing group, directing substitution to the meta position in benzene.

xii) Lucas reagent is a mixture of:

A) Zn and Hg

B) ZnCl₂ and HCl

C) Zn and HCl

D) NaOH and CaO

✅ Correct Answer: B) ZnCl₂ and HCl
Reason: Lucas reagent (conc. HCl + ZnCl₂) is used to distinguish primary, secondary, and tertiary alcohols.

xiii) The hybridization in Carbonyl Carbon of aldehydes and ketones is:

A) sp²

B) sp

C) sp³

D) dsp³

✅ Correct Answer: A) sp²
Reason: The carbonyl carbon forms a double bond with oxygen, giving trigonal planar geometry and sp² hybridization.

xiv) This molecule contains Ketonic group:

A) Glucose

B) Galactose

C) Mannose

D) Fructose

✅ Correct Answer: D) Fructose
Reason: Fructose is a ketohexose, containing a ketonic group, unlike glucose, galactose, and mannose which are aldoses.

xv) Ozone layer is present in:

A) Troposphere

B) Stratosphere

C) Mesosphere

D) Thermosphere

✅ Correct Answer: B) Stratosphere
Reason: The ozone layer lies in the stratosphere, where it absorbs harmful UV radiation from the sun.

xvi) Carbolic acid contains this functional group:

A) –COOH

B) –NO₂

C) –OH

D) –SO₃H

✅ Correct Answer: C) –OH
Reason: Carbolic acid is another name for phenol, which contains the –OH functional group attached to an aromatic ring.

xvii) In s-block element this forms superoxide:

A) Mg

B) Li

C) Na

D) Cs

✅ Correct Answer: D) Cs
Reason: Cesium, being a heavier alkali metal, reacts with oxygen to form cesium superoxide (CsO₂).

📝 Karachi Board MCQs XII Chemistry 2024 (Click to reveal answer with reason.🎯📌

1. The colour of flame produced by Beryllium is:

A) Red

B) No colour

C) Silver white

D) Deep red

✅ Correct Answer: B) No colour
Reason: Beryllium does not impart any characteristic colour to the flame due to its high ionization energy.

2. The reason that benzene burns with soot is:

A) Inflammability

B) Higher carbon content

C) High resonance energy

D) Aromaticity

✅ Correct Answer: B) Higher carbon content
Reason: Benzene has a high carbon percentage, leading to incomplete combustion and a sooty flame.

3. Drugs that make body temperature normal are called:

A) Antibiotic

B) Antipyretic

C) Antiallergic

D) Antihistamin

✅ Correct Answer: B) Antipyretic
Reason: Antipyretics (like paracetamol) reduce fever and normalize body temperature.

4. This process is used to improve octane number of gasoline:

A) Reforming

B) Refining

C) Distillation

D) Condensation

✅ Correct Answer: A) Reforming
Reason: Reforming increases the octane number by converting straight-chain hydrocarbons into branched or aromatic hydrocarbons.

5. The general formula of nitrides for alkaline earth metals is:

A) MN₃

B) M₂N₃

C) M₃N

D) M₃N₂

✅ Correct Answer: D) M₃N₂
Reason: Alkaline earth metals (M²⁺) form nitrides with formula M₃N₂, balancing charges between M²⁺ and N³⁻ ions.

6. Formic acid is naturally found in:

A) Apple

B) Bee’s sting

C) Vinegar

D) Butter

✅ Correct Answer: B) Bee’s sting
Reason: Formic acid is present in ant and bee stings, causing the burning sensation.

7. An example of thermoplastic is:

A) Polythene

B) Nylon

C) PVC

D) Bakelite

✅ Correct Answer: A) Polythene
Reason: Polythene is a thermoplastic that softens on heating and hardens on cooling, unlike Bakelite which is a thermosetting plastic.

8. The general formula for cycloalkane is:

A) CₙH₂ₙ

B) CₙH₂ₙ₊₂

C) CₙH₂ₙ₋₂

D) CₙH₄ₙ

✅ Correct Answer: A) CₙH₂ₙ
Reason: Cycloalkanes have the general formula CₙH₂ₙ due to ring closure reducing two hydrogens compared to alkanes.

9. NMR spectroscopy is concerned with the interaction of molecules with:

A) Infrared radiation

B) Visible radiation

C) Ultraviolet radiation

D) Radio waves

✅ Correct Answer: D) Radio waves
Reason: NMR spectroscopy uses radio waves to probe the magnetic properties of nuclei in a magnetic field.

10. Oxidation state of manganese in KMnO₄ (potassium permanganate) is:

A) +4

B) +2

C) +7

D) +5

✅ Correct Answer: C) +7
Reason: In KMnO₄, K = +1 and O = -2 (×4 = -8). To balance, Mn must be +7.

11. The greenhouse gas is:

A) O₂

B) N₂

C) Ar

D) CO₂

✅ Correct Answer: D) CO₂
Reason: Carbon dioxide is a major greenhouse gas responsible for trapping heat in the atmosphere.

12. The extra glucose of bloodstream is converted into:

A) Fat

B) Glycogen

C) Gluconic acid

D) Triglyceride

✅ Correct Answer: B) Glycogen
Reason: Excess glucose in the bloodstream is stored in the liver and muscles as glycogen.

13. Beryllium is diagonally related to this:

A) Li

B) Si

C) Al

D) Mg

✅ Correct Answer: c) Al
Reason: Beryllium shows diagonal relationship with aluminium due to similar ionic size, charge density, and chemical properties.

14. On oxidation, a secondary alcohol gives:

A) Ether

B) Aldehyde

C) Ketone

D) Amine

✅ Correct Answer: C) Ketone
Reason: Secondary alcohols oxidize to ketones because the carbon bearing the –OH group is bonded to two other carbons.

15. The coordination number of cobalt in Na[Co(C₂O₄)₃] is:

A) 3

B) 4

C) 6

D) 2

✅ Correct Answer: C) 6
Reason: Each oxalate (C₂O₄²⁻) ligand is bidentate, donating 2 sites. With 3 oxalates, cobalt has coordination number 6.

16. This element is not used in electroplating:

A) Zn

B) Sn

C) Cr

D) Mn

✅ Correct Answer: D) Mn
Reason: Manganese is not commonly used in electroplating, unlike Zn, Sn, and Cr which are widely used for protective coatings.

17. Ozone depletion in upper atmosphere is mainly caused by:

A) SO₂

B) NO

C) CO

D) CFC

✅ Correct Answer: D) CFC
Reason: Chlorofluorocarbons release chlorine radicals in the stratosphere, which catalytically destroy ozone molecules.

📝 Chapter-wise Important MCQs XII Chemistry (Click to reveal answer with reason.🎯📌

Textbook MCQs on Chapter 1 ….. Group Trend of Representative Elements🎯📌

1. Melting & boiling points of which of the following group of representative elements decrease regularly down the group:

A) Group IA

B) Group IIIA

C) Group VIIA

D) Group VIIIA

✅ Correct Answer: A) Group IA
Reason: Alkali metals (Group IA) show decreasing melting and boiling points down the group due to weaker metallic bonding as atomic size increases.

2. Which of the following s-block element forms super oxide when burned in air?

A) Li

B) Na

C) K

D) Mg

✅ Correct Answer: C) K
Reason: Potassium reacts with oxygen to form potassium superoxide (KO₂).

3. Which of the following formula of nitrides for alkaline earth metal is possible?

A) MN₃

B) M₂N₃

C) M₃N

D) M₃N₂

✅ Correct Answer: D) M₃N₂
Reason: Alkaline earth metals (M²⁺) form nitrides with formula M₃N₂, balancing charges between M²⁺ and N³⁻ ions.

4. The flame colour of which of the following alkali metal is yellow:

A) Na

B) K

C) Rb

D) Cs

✅ Correct Answer: A) Na
Reason: Sodium salts impart a bright yellow colour to the flame in flame tests.

5. The chemical used in the fireworks is:

A) Sodium bicarbonate

B) Bleaching powder

C) Potassium nitrate

D) Potash alum

✅ Correct Answer: C) Potassium nitrate
Reason: Potassium nitrate is a strong oxidizer used in fireworks and gunpowder to produce bright flames and explosions.

6. Cathode in Castner Kellner cell is:

A) Titanium blocks

B) Carbon rods

C) Mercury

D) Iron container

✅ Correct Answer: C) Mercury
Reason: In the Castner Kellner process, mercury acts as the cathode to form sodium amalgam.

7. The diagonal member of beryllium is:

A) Mg

B) Al

C) Si

D) C

✅ Correct Answer: B) Al
Reason: Beryllium shows diagonal relationship with aluminium due to similarities in ionic size and properties.

8. Purification of Sulphur dioxide from arsenic oxide is an essential step in contact process to avoid:

A) Catalyst poisoning

B) Temperature elevation

C) Pressure controlling

D) Air mixing

✅ Correct Answer: A) Catalyst poisoning
Reason: Arsenic oxide poisons the vanadium pentoxide catalyst used in the contact process, so SO₂ must be purified.

9. Oil of vitriol refers to:

A) Borax

B) Sulphuric Acid

C) Alum

D) Caustic soda

✅ Correct Answer: B) Sulphuric Acid
Reason: Oil of vitriol is the old name for concentrated sulphuric acid due to its oily appearance.

10. The best oxidizing agent among halogens is:

A) F₂

B) Cl₂

C) Br₂

D) I₂

✅ Correct Answer: A) F₂
Reason: Fluorine is the strongest oxidizing agent among halogens due to its highest electronegativity and reactivity.

📝 Textbook MCQs on Chapter 2 ….. Outer Transition Elements (Click to reveal answer with reason)🎯📌

1. Zn²⁺ ion is colourless because:

A) Its 4s orbital is empty

B) Its 3d orbitals have all unpaired electrons

C) Its 3d orbitals have all paired electrons

D) Its d-orbitals cannot split up into t₂g and e_g

✅ Correct Answer: C) Its 3d orbitals have all paired electrons
Reason: Zn²⁺ has a 3d¹⁰ configuration (all paired), so no d–d transitions occur, making it colourless.

2. The coordination number of cobalt in Na₄[Co(C₂O₄)₃] is:

A) 3

B) 4

C) 6

D) 8

✅ Correct Answer: C) 6
Reason: Each oxalate (C₂O₄²⁻) ligand is bidentate, donating 2 sites. With 3 oxalates, cobalt has coordination number 6.

3. An example of a bidentate ligand among the following is:

A) OH⁻

B) C₂O₄²⁻

C) CO₃²⁻

D) CN⁻

✅ Correct Answer: B) C₂O₄²⁻
Reason: Oxalate ion is bidentate, coordinating through two oxygen atoms.

4. A highly paramagnetic ion among the following is:

A) Fe²⁺

B) Fe³⁺

C) Co²⁺

D) Cr³⁺

✅ Correct Answer: B) Fe³⁺
Reason: Fe³⁺ has 5 unpaired electrons (3d⁵), making it highly paramagnetic.

5. The highest oxidation state of chromium is:

A) +4

B) +5

C) +6

D) +7

✅ Correct Answer: C) +6
Reason: Chromium exhibits a maximum oxidation state of +6, as in CrO₃ and dichromates.

6. This element is not used for electroplating:

A) Zinc

B) Tin

C) Chromium

D) Manganese

✅ Correct Answer: D) Manganese
Reason: Manganese is not commonly used in electroplating, unlike Zn, Sn, and Cr which are widely used for protective coatings.

7. Which of the following steel is typically used in making Fry pans?

A) Carbon steel

B) Stainless steel

C) Tool steel

D) Alloy steel

✅ Correct Answer: B) Stainless steel
Reason: Stainless steel resists rust and corrosion, making it ideal for cookware like fry pans.

8. Which step in the extraction of copper from chalcopyrite is involved in the elimination of gangue impurities?

A) Concentration

B) Roasting

C) Smelting

D) Bessemerization

✅ Correct Answer: A) Concentration
Reason: Concentration removes gangue impurities from chalcopyrite ore before further processing.

9. 5d series of outer transition elements is:

A) Sc to Zn

B) Y to Cd

C) La to Hg

D) Ac to Cn

✅ Correct Answer: C) La to Hg
Reason: The 5d transition series extends from lanthanum (La) to mercury (Hg).

10. Oxidation of manganese in air gives the following oxide:

A) MnO

B) MnO₂

C) Mn₂O₃

D) Mn₃O₄

✅ Correct Answer: D) Mn₃O₄
Reason: On oxidation in air, manganese forms Mn₃O₄.

📝 Textbook MCQs on Chapter # 3, Organic Compounds (Click to reveal answer with reason)🎯📌

1. Which of the following functional group contains a nitrogen atom?

A) Alcohol

B) Ketone

C) Ester

D) Amide

✅ Correct Answer: D) Amide
Reason: Amides contain nitrogen in the –CONH₂ group, unlike alcohols, ketones, or esters.

2. A closed chain hydrocarbon with an alternate double bond is known as:

A) Aliphatic

B) Aromatic

C) Alicyclic

D) Carboxylic

✅ Correct Answer: B) Aromatic
Reason: Aromatic hydrocarbons have conjugated double bonds in a closed ring, e.g., benzene.

3. Fractional distillation of coal gives coke, coal tar, coal gas and:

A) Natural gas

B) Ammonia liquor

C) Naphtha

D) Petroleum

✅ Correct Answer: B) Ammonia liquor
Reason: Fractional distillation of coal produces coke, coal tar, coal gas, and ammonia liquor.

4. The nuisance of knocking can be reduced by the process:

A) Reforming

B) Refining

C) Distillation

D) Condensation

✅ Correct Answer: A) Reforming
Reason: Reforming increases the octane number of gasoline, reducing knocking in engines.

5. The general formula of the homologous series of alcohol is:

A) CₙH₂ₙO

B) CₙH₂ₙ₊₂O

C) CₙH₂ₙ₋₂O

D) CₙH₂ₙ₊₁O

✅ Correct Answer: B) CₙH₂ₙ₊₂O
Reason: Alcohols follow the general formula CₙH₂ₙ₊₂O, derived from alkanes with one –OH group.

6. The number of five membered and six membered rings in C₆₀ Bucky ball are respectively:

A) 12 and 12

B) 5 and 15

C) 12 and 20

D) 40 and 20

✅ Correct Answer: C) 12 and 20
Reason: The C₆₀ fullerene (Bucky ball) consists of 12 pentagons and 20 hexagons arranged like a soccer ball.

7. Which of the following pairs of compounds represents isomerism?

A) C₂H₅–OH & C₃H₇OH

B) CH₃–O–CH₃ & C₂H₅OH

C) C₂H₅CH₂Cl & C₃H₇CH₂Cl

D) CH₃NH₂ & CH₃CH₂NH₂

✅ Correct Answer: B) CH₃–O–CH₃ & C₂H₅OH
Reason: Dimethyl ether (CH₃–O–CH₃) and ethanol (C₂H₅OH) are functional group isomers.

8. Which of the following is hard and high ranked coal?

A) Peat

B) Lignite

C) Bituminous

D) Anthracite

✅ Correct Answer: D) Anthracite
Reason: Anthracite is the hardest, highest carbon content coal, producing the most heat.

9. Which of the following hydrocarbon is the chief constituent of natural gas?

A) CH₄

B) C₂H₆

C) C₃H₈

D) C₄H₁₀

✅ Correct Answer: A) CH₄
Reason: Methane (CH₄) is the main component of natural gas, making up about 70–90%.

10. Urea was first synthesized by Wohler from an inorganic material named as:

A) Ammonium nitrate

B) Ammonium chloride

C) Ammonium bicarbonate

D) Ammonium cyanate

✅ Correct Answer: D) Ammonium cyanate
Reason: Wohler synthesized urea in 1828 by heating ammonium cyanate, marking the first synthesis of an organic compound from inorganic material.

📝 Textbook MCQs on Chapter 4 ….. Nomenclature of Organic Compounds (Click to reveal answer with reason)🎯📌

1. Which of the following molecules possess acyl functional group?

A) R – CO – X

B) R – CONH₂

C) R – COO – CH₃

D) R – CO – R

✅ Correct Answer: A) R – CO – X
Reason: The acyl group is –CO– attached to another atom (like halogen or oxygen). R–CO–X represents an acyl compound.

2. A hydrocarbon with the molecular formula C₇H₁₂ is possibly:

A) Heptane

B) Heptene

C) Heptyne

D) Heptadiene

✅ Correct Answer: C) Heptyne
Reason: CₙH₂ₙ₋₂ is the general formula for both alkynes and dienes. For n=7, C₇H₁₂ fits both heptyne and heptadiene.But in our book it is marked as heptyne.

3. An organic compound possesses the structural formula CH₃CH=CH–C≡CH, its correct IUPAC name is:

A) 1-pentyne-3-ene

B) 3-pentene-1-pentyne

C) 3-penten-1-yne

D) 2-penten-3-yne

✅ Correct Answer: C) 3-penten-1-yne
Reason: The compound has both double and triple bonds. Correct numbering gives 3-penten-1-yne.

4. IUPAC name of isopropyl alcohol is:

A) 1-propanol

B) 2-propanol

C) 1-butanol

D) 2-butanol

✅ Correct Answer: B) 2-propanol
Reason: Isopropyl alcohol has the structure CH₃–CHOH–CH₃, named 2-propanol in IUPAC.

5. Resorcinol is a phenol, it contains two –OH groups at:

A) Position 1 and 2

B) Position 1 and 3

C) Position 1 and 4

D) Position 2 and 4

✅ Correct Answer: B) Position 1 and 3
Reason: Resorcinol is 1,3-dihydroxybenzene, with –OH groups at positions 1 and 3 of the benzene ring.

6. Formula of a saturated hydrocarbon is C₄H₈, it should be:

A) Butane

B) Butene

C) Butyne

D) Cyclobutane

✅ Correct Answer: D) Cyclobutane
Reason: C₄H₈ fits the general formula CₙH₂ₙ for cycloalkanes. Cyclobutane is a saturated cyclic hydrocarbon.

7. In the IUPAC naming of ketones, the ending –e of main carbon chain is replaced by:

A) –yl

B) –ol

C) –al

D) –one

✅ Correct Answer: D) –one
Reason: Ketones are named by replacing the –e of the parent alkane with –one (e.g., propane → propanone).

8. In the common system, carboxylic acid with six carbon atoms in straight chain is named as:

A) Propionic acid

B) Valeric acid

C) Caproic acid

D) Capric acid

✅ Correct Answer: C) Caproic acid
Reason: Hexanoic acid (six carbons) is commonly called caproic acid.

9. The correct structure of 1,3-pentadiene is:

A) CH₃–CH=CH–CH=CH₂–CH₃

B) CH₂=CH–CH₂–CH=CH₂

C) CH₂=CH–CH=CH–CH₃

D) CH₃–CH=C=CH–CH₃

✅ Correct Answer: C) CH₂=CH–CH=CH–CH₃
Reason: 1,3-pentadiene has two double bonds at positions 1 and 3 in a five-carbon chain.

10. The IUPAC name of CH₃COOCH(CH₃)₂ is:

A) Isopropyl ethanoate

B) Ethyl propanoate

C) Isopropyl acetate

D) Propyl ethanoate

✅ Correct Answer: A) Isopropyl ethanoate
Reason: The ester CH₃COOCH(CH₃)₂ is named as isopropyl ethanoate (common name: isopropyl acetate).

📝 Textbook MCQs on Chapter # 5, Hydrocarbons (Click to reveal answer with reason)🎯📌

1. The final product obtained when hydrogen bromide (HBr) is added to an ethyne molecule:

A) 1,1-dibromoethane

B) 1,2-dibromoethane

C) 1,1,2,2-tetrabromoethane

D) Bromoethane

✅ Correct Answer: A) 1,1-dibromoethane
Reason: Addition of HBr to ethyne occurs stepwise, giving vinyl bromide first, then 1,1-dibromoethane as the final product.

2. The formula of saturated hydrocarbon is C₃H₆, it should be:

A) Propane

B) Propene

C) Propyne

D) Cyclopropane

✅ Correct Answer: D) Cyclopropane
Reason: C₃H₆ fits the general formula CₙH₂ₙ for cycloalkanes. Cyclopropane is a saturated cyclic hydrocarbon.

3. Ozonide on heating with zinc dust produces:

A) Alcohol

B) Aldehyde

C) Alkene

D) Ether

✅ Correct Answer: B) Aldehyde
Reason: Ozonolysis of alkenes followed by reduction with zinc dust produces aldehydes or ketones.

4. Which of the following pairs of compounds represents functional group isomerism?

A) 1-butene and 2-butene

B) Ethanol and dimethyl ether

C) n-butane and isobutene

D) Diethyl ketone and methyl propyl ketone

✅ Correct Answer: B) Ethanol and dimethyl ether
Reason: Ethanol (alcohol) and dimethyl ether (ether) have the same molecular formula but different functional groups.

5. The substituent that can act as a meta director is:

A) –Cl

B) –CH₃

C) –OH

D) –COOH

✅ Correct Answer: D) –COOH
Reason: –COOH is an electron withdrawing group, directing substitution to the meta position in benzene.

6. Welding gas among the following is:

A) Ethylene

B) Acetylene

C) Ethane

D) Methane

✅ Correct Answer: B) Acetylene
Reason: Acetylene mixed with oxygen produces a very hot flame, widely used in welding and cutting metals.

7. Benzene burns with smoky flame because of its:

A) Inflammability

B) High carbon %age

C) High resonance energy

D) Aromaticity

✅ Correct Answer: B) High carbon %age
Reason: Benzene has a high carbon content, leading to incomplete combustion and a sooty flame.

8. Select the suitable chemical to distinguish between ethene and ethyne:

A) Alkaline KMnO₄

B) Acidified KMnO₄

C) Bromine water

D) Ammonical AgNO₃

✅ Correct Answer: D) Ammonical AgNO₃
Reason: Ethyne reacts with ammoniacal silver nitrate to form a precipitate of silver acetylide, while ethene does not.

9. Meta directing group among the following is:

A) –OH

B) –NH₂

C) –CH₃

D) –NO₂

✅ Correct Answer: D) –NO₂
Reason: –NO₂ is a strong electron withdrawing group, directing substitution to the meta position in benzene.

10. Acylation of benzene in the presence of AlCl₃ gives:

A) Toluene

B) Acetophenone

C) Phenol

D) Xylene

✅ Correct Answer: B) Acetophenone
Reason: Acylation of benzene with CH₃COCl in the presence of AlCl₃ (Friedel–Crafts acylation) produces acetophenone.

📝 Textbook Multiple Choice Questions on Chapter 6 alkyl halides and Amines (Click to reveal answer with reason)🎯📌

1. Which of the following composition justifies the secondary alkyl halide?

A) R₃CX

B) R₂CHX

C) RCH₂X

D) CH₃X

✅ Correct Answer: B) R₂CHX
Reason: In secondary alkyl halides, the carbon attached to halogen is bonded to two other alkyl groups (R₂CHX).

2. Which of the following alkyl halide cannot produce an alkene while treated with alcoholic potassium hydroxide?

A) Methyl bromide

B) Ethyl bromide

C) Propyl bromide

D) Butyl bromide

✅ Correct Answer: A) Methyl bromide
Reason: Methyl bromide has no β-hydrogen, so elimination cannot occur to form an alkene.

3. Ethyl magnesium bromide with carbon dioxide yields:

A) Methanoic acid

B) Ethanoic acid

C) Propanoic acid

D) Butanoic acid

✅ Correct Answer: C) Propanoic acid
Reason: Grignard reagent (C₂H₅MgBr) reacts with CO₂ to form magnesium salt of propanoic acid, which on hydrolysis gives propanoic acid.

4. Grignard’s reagent with ester produces:

A) Aldehyde

B) Carboxylic acid

C) Ketone

D) Ether

✅ Correct Answer: C) Ketone
Reason: Grignard reagents react with esters to form ketones after hydrolysis.

5. Amine act as bases because:

A) They accept OH⁻

B) They donate the OH⁻

C) They donate H⁺

D) They accept H⁺

✅ Correct Answer: D) They accept H⁺
Reason: Amines have a lone pair of electrons on nitrogen, allowing them to accept protons (H⁺), thus acting as bases.

6. The structure of primary amine is:

A) Planar trigonal

B) Regular tetrahedral

C) Tetrahedral pyramidal

D) Linear

✅ Correct Answer: C) Tetrahedral pyramidal
Reason: Primary amines have a nitrogen atom with one lone pair and three bonds, giving a pyramidal geometry.

7. Alkyl amine when reacts with nitrous acid in the presence of hydrochloric acid yields:

A) Diazonium salt

B) Aldehyde

C) Ketone

D) Alcohol

✅ Correct Answer: A) Diazonium salt
Reason: Primary aliphatic amines react with nitrous acid to form alcohols via diazotization followed by decomposition.But in our text book it is marked as Diazonium salt.

8. Sɴ₂ reaction occurs most easily if the substrate molecule is:

A) Methyl iodide

B) Ethyl iodide

C) 2-iodopropane

D) 2-iodobutane

✅ Correct Answer: A) Methyl iodide
Reason: Sɴ₂ reactions are fastest with least steric hindrance; methyl iodide has minimal hindrance.

9. Suitable reagent required for the synthesis of propane from methyl magnesium iodide is:

A) H₂O

B) NH₃

C) CH₃OH

D) CH₃NH₂

❌ Note: All given options are incorrect.
✅ Correct Answer: C₂H₅Cl
Reason: Methyl magnesium iodide reacts with ethyl chloride to form propane via Grignard reaction.

10. The rate of Sɴ₁ mechanism depends upon:

A) Conc. of substrate only

B) Conc. of attacking nucleophile only

C) Conc. of both substrate and attacking nucleophile

D) Polar solvent

✅ Correct Answer: A) Conc. of substrate only
Reason: Sɴ₁ is a unimolecular reaction; the rate depends only on the concentration of the substrate.

📝 Text Book Multiple Choice Questions on Chapter 7 Alcohols, Phenols and Ethers (Click to reveal answer with reason)🎯📌

1. In the molecule of phenol, the carbon atom which is attached to hydroxyl group is:

A) sp-hybridized

B) sp²-hybridized

C) sp³-hybridized

D) Unhybridized

✅ Correct Answer: B) sp²-hybridized
Reason: In phenol, the hydroxyl group is attached to a benzene ring carbon, which is sp²-hybridized.

2. Which of the following is a trihydric phenol?

A) Resorcinol

B) Cresol

C) Pyrogallol

D) Catechol

✅ Correct Answer: C) Pyrogallol
Reason: Pyrogallol (1,2,3-trihydroxybenzene) contains three –OH groups on the benzene ring.

3. Ethanol reacts with PCl₃ to form:

A) Diethyl ether

B) Ethene

C) Ethyl chloride

D) Ethanoic acid

✅ Correct Answer: C) Ethyl chloride
Reason: Alcohols react with phosphorus trichloride to form alkyl chlorides (here ethanol → ethyl chloride).

4. Which of the following alcohols has the highest boiling point?

A) Neo-pentyl alcohol

B) n-pentyl alcohol

C) Iso-pentyl alcohol

D) Ethyl alcohol

✅ Correct Answer: B) n-pentyl alcohol
Reason: Straight-chain alcohols have stronger van der Waals forces than branched ones, giving higher boiling points.

5. Which of the following products is mainly formed if ethanol is dehydrated with conc. sulphuric acid at 170°C?

A) Ethene

B) Ethyne

C) Ethanol

D) Diethyl ether

✅ Correct Answer: A) Ethene
Reason: At 170°C, concentrated H₂SO₄ dehydrates ethanol to ethene.
📌 Note: Some textbooks list diethyl ether (option D) as the product at lower temperatures (~140°C), but at 170°C the major product is ethene.

6. Lucas reagent is a mixture of:

A) Zn and Hg

B) NaOH and CaO

C) ZnCl₂ and HCl

D) Zn and HCl

✅ Correct Answer: C) ZnCl₂ and HCl
Reason: Lucas reagent (ZnCl₂ + conc. HCl) is used to distinguish primary, secondary, and tertiary alcohols by their reactivity.

7. Oxidative cleavage of 1,2-diol occurs in the presence of:

A) K₂Cr₂O₇

B) KMnO₄

C) HNO₃

D) HIO₄

✅ Correct Answer: D) HIO₄
Reason: Periodic acid (HIO₄) cleaves vicinal diols oxidatively, breaking the C–C bond between them.

8. Which of the following molecule cannot form hydrogen bonding with water molecule?

A) Phenol

B) Resorcinol

C) Ethyl chloride

D) Ethyl alcohol

✅ Correct Answer: C) Ethyl chloride
Reason: Ethyl chloride lacks –OH or –NH groups, so it cannot form hydrogen bonds with water.

9. Secondary alcohols undergo oxidation with potassium dichromate to produce carboxylic acid through an intermediate product known as:

A) Aldehyde

B) Ketone

C) Ether

D) Alkyl halide

✅ Correct Answer: B) Ketone
Reason: Secondary alcohols oxidize to ketones first; further oxidation can lead to acids under strong conditions.

10. Which of the following is an anaesthetic agent?

A) Phenol

B) Ethyl alcohol

C) Diethyl ether

D) Acetone

✅ Correct Answer: C) Diethyl ether
Reason: Diethyl ether was historically used as an anaesthetic agent due to its volatility and depressant effect on the CNS.

📝 Text Book MCQs on Chapter 8 Aldehyde and Ketone (Click to reveal answer with reason)🎯📌

1. Ketones when treated with LiAlH₄, they reduce to:

A) Primary alcohol

B) Secondary alcohol

C) Tertiary alcohol

D) Dihydric alcohol

✅ Correct Answer: B) Secondary alcohol
Reason: Ketones are reduced by LiAlH₄ to secondary alcohols because the carbonyl carbon is bonded to two alkyl groups.

2. The reagent used to oxidize ketones into carboxylic acid is:

A) Ammoniacal silver nitrate

B) Potassium dichromate

C) Fehling solution

D) Benedict solution

✅ Correct Answer: B) Potassium dichromate
Reason: Strong oxidizing agents like K₂Cr₂O₇ can oxidize ketones to carboxylic acids under vigorous conditions.

3. The carbonyl carbon of aldehydes and ketones is:

A) sp-hybridized

B) sp²-hybridized

C) sp³-hybridized

D) dsp³-hybridized

✅ Correct Answer: B) sp²-hybridized
Reason: The carbonyl carbon forms three sigma bonds and one pi bond, giving it sp² hybridization.

4. Acetophenone is the member of ketone family, it contains:

A) One aryl & one H atom

B) Two aryl groups

C) One alkyl & one aryl group

D) Two alkyl groups

✅ Correct Answer: C) One alkyl & one aryl group
Reason: Acetophenone has a phenyl group (aryl) and a methyl group (alkyl) attached to the carbonyl carbon.

5. The most reactive molecules towards nucleophilic addition in the following is:

A) Formaldehyde

B) Acetaldehyde

C) Diethyl ketone

D) Acetophenone

✅ Correct Answer: A) Formaldehyde
Reason: Formaldehyde is most reactive towards nucleophilic addition because it has no alkyl groups to donate electrons, making the carbonyl carbon highly electrophilic.

6. Clemmensen reduction is the conversion of aldehydes and ketones into:

A) Alkanes

B) Alkenes

C) Alkyl halides

D) Alcohols

✅ Correct Answer: A) Alkanes
Reason: Clemmensen reduction (Zn-Hg + conc. HCl) reduces carbonyl groups of aldehydes and ketones to alkanes.

7. Hydration of propyne in the presence of H₂SO₄ and HgSO₄ gives:

A) Formaldehyde

B) Methyl ethyl ketone

C) Acetone

D) Acetaldehyde

✅ Correct Answer: C) Acetone
Reason: Hydration of propyne gives initially an enol, which tautomerizes to acetone or dimethyl ketone (CH₃COCH₃).

8. Which of the following carbonyl compounds is most soluble in water?

A) Formaldehyde

B) Acetaldehyde

C) Benzaldehyde

D) Acetophenone

✅ Correct Answer: A) Formaldehyde
Reason: Formaldehyde is the smallest carbonyl compound, highly polar, and most soluble in water.

9. Which of the following gives silver mirror test with Tollen’s reagent?

A) HCHO

B) CH₃–O–CH₃

C) C₂H₅OH

D) CH₃COOH

✅ Correct Answer: A) HCHO
Reason: Formaldehyde (HCHO) is an aldehyde, which gives a positive silver mirror test with Tollen’s reagent.

10. On reduction of carbonyl compound by Zn-Hg and conc. HCl, it is converted to an alkane. This reaction is known as:

A) Dow reduction

B) Wolff–Kishner reduction

C) Clemmensen reduction

D) Cope reduction

✅ Correct Answer: C) Clemmensen reduction
Reason: Clemmensen reduction uses Zn-Hg and conc. HCl to reduce aldehydes and ketones to alkanes.

📝 Textbook Multiple Choice Questions on Chapter 9 Carboxylic acids and Functional Derivatives (Click to reveal answer with reason)🎯📌

1. The most common compound found in pineapple is:

A) Acetic acid

B) Ethanol

C) Acetone

D) Ethyl butanoate

✅ Correct Answer: D) Ethyl butanoate
Reason: Ethyl butanoate is the ester responsible for the characteristic pineapple aroma.

2. Two molecules of acetic acid on condensation gives:

A) Ethyl acetate

B) Acetamide

C) Acyl halide

D) Acetic anhydride

✅ Correct Answer: D) Acetic anhydride
Reason: Condensation of two acetic acid molecules produces acetic anhydride with elimination of water.

3. Carboxylic acid is stronger acid than:

A) HCl

B) HNO₃

C) C₂H₅OH

D) H₂SO₄

✅ Correct Answer: C) C₂H₅OH
Reason: Carboxylic acids are stronger acids than alcohols due to resonance stabilization of the carboxylate ion.

4. The reagent that cannot produce an acyl halide in reaction with a carboxylic acid is:

A) PCl₃

B) PCl₅

C) HCl

D) SOCl₂

✅ Correct Answer: C) HCl
Reason: HCl alone cannot convert carboxylic acids into acyl halides; reagents like PCl₃, PCl₅, and SOCl₂ are required.

5. Benzoic acid is the product of oxidation of:

A) Benzene

B) Ethyl benzene

C) Aniline

D) Phenol

✅ Correct Answer: B) Ethyl benzene
Reason: Side-chain oxidation of ethyl benzene produces benzoic acid.

6. Formation of acyl halide and amide by carboxylic acid involves:

A) Replacement of hydrogen

B) Replacement of carbonyl group

C) Replacement of hydroxyl group

D) Replacement of oxygen

✅ Correct Answer: C) Replacement of hydroxyl group
Reason: In carboxylic acids, the –OH group is replaced to form acyl halides and amides.

7. Formic acid is naturally found in:

A) Venom of ants

B) Bees sting

C) Vinegar

D) Butter

✅ Correct Answer: A) Venom of ants
Reason: Formic acid is present in ant venom and is responsible for its sting.

8. Among the following compounds, the one with the highest boiling point is:

A) Ethanol

B) Acetaldehyde

C) Acetic acid

D) Ethyl chloride

✅ Correct Answer: C) Acetic acid
Reason: Acetic acid forms strong hydrogen bonds and has the highest boiling point among the given compounds.

9. The formula of caproic acid is:

A) C₄H₉COOH

B) C₅H₁₁COOH

C) C₆H₁₃COOH

D) C₇H₁₅COOH

✅ Correct Answer: B) C₅H₁₁COOH
Reason: Caproic acid is hexanoic acid (contains total six carbons), with the formula C₅H₁₁COOH.

10. The reaction of acetic acid with ethanol in the presence of conc. sulphuric acid gives:

A) Ethyl acetate

B) Acetamide

C) Acetic anhydride

D) Ethane

✅ Correct Answer: A) Ethyl acetate
Reason: Acetic acid reacts with ethanol in the presence of conc. H₂SO₄ to form ethyl acetate (esterification).

📝 Text Book MCQs on Chapter 10 Biochemistry (Click to reveal answer with reason)🎯📌

1. Starch and sucrose are examples of:

A) Monosaccharides and disaccharides

B) Disaccharides and oligosaccharides

C) Polysaccharides and disaccharides

D) Monosaccharides and polysaccharides

✅ Correct Answer: C) Polysaccharides and disaccharides
Reason: Starch is a polysaccharide, while sucrose is a disaccharide.

2. Amino acid units bonded in protein molecule through:

A) Glycosidic linkage

B) Ether linkage

C) Peptide linkage

D) Hydrogen bridge

✅ Correct Answer: C) Peptide linkage
Reason: Proteins are formed by amino acids linked via peptide bonds (–CO–NH–).

3. Proteins are composed of:

A) Amino acids

B) Carbohydrates

C) Lipids

D) Nucleic acids

✅ Correct Answer: A) Amino acids
Reason: Proteins are polymers of amino acids joined by peptide bonds.

4. A condensation polymer of amino acid is:

A) Protein

B) Lipids

C) Glycogen

D) Starch

✅ Correct Answer: A) Protein
Reason: Proteins are condensation polymers of amino acids linked by peptide bonds.

5. Saponification is the formation of soap by the reaction of fat and oil with:

A) An alkali

B) An acid

C) Sugar

D) Glycerol

✅ Correct Answer: A) An alkali
Reason: Saponification involves hydrolysis of fats/oils with alkali (NaOH/KOH) to form soap and glycerol.

6. Which of the following mineral is considered to be essential for immune system?

A) Iron

B) Zinc

C) Magnesium

D) Calcium

✅ Correct Answer: D) Calcium
Reason: Calcium plays a vital role in immune function.

7. Rancidity is a chemical process involving:

A) Oxidation & hydrolysis

B) Condensation & reduction

C) Polymerization

D) Decarboxylation

✅ Correct Answer: A) Oxidation & hydrolysis
Reason: Rancidity of fats/oils occurs due to oxidation and hydrolysis, producing foul-smelling compounds.

8. Lipid which is a major component of cell membrane is:

A) Triglyceride

B) Phospholipid

C) Glycolipid

D) Steroid

✅ Correct Answer: B) Phospholipid
Reason: Phospholipids form bilayers, the fundamental structure of cell membranes.

9. Total numbers of alpha amino acids are:

A) 19

B) 22

C) 25

D) 28

✅ Correct Answer: B) 22
Reason: There are 22 known alpha amino acids, including the 20 standard ones plus selenocysteine and pyrrolysine.

10. Sugar molecules are classified as:

A) Fats

B) Proteins

C) Carbohydrates

D) Lipids

✅ Correct Answer: C) Carbohydrates
Reason: Sugars are carbohydrates, serving as primary energy sources in living organisms.

📝 Text Book MCQs on Chemical Industries Chapter 11 (Click to reveal answer with reason)🎯📌

1. Super glue is chemically named as:

A) Cyano acrylate

B) Polyvinyl acetate

C) Epoxy resins

D) Polyurethane

✅ Correct Answer: A) Cyano acrylate
Reason: Super glue is chemically cyanoacrylate, known for its strong adhesive properties.

2. DDT is a chemical which commercially known as:

A) Insecticide

B) Herbicide

C) Pesticide

D) Fungicide

✅ Correct Answer: A) Insecticide
Reason: DDT (dichlorodiphenyltrichloroethane) is widely known as an insecticide.

3. Nail polish removers mainly consist of:

A) Pigments

B) Acetone

C) Diethyl ether

D) Ethyl alcohol

✅ Correct Answer: B) Acetone
Reason: Acetone is the primary solvent used in nail polish removers.

4. Antimalarial drug among the following is:

A) Ibuprofen

B) Chloroquine

C) Diphenyl hydramine

D) Paracetamol

✅ Correct Answer: B) Chloroquine
Reason: Chloroquine is an antimalarial drug, effective against Plasmodium parasites.

5. Aspirin is a pain reliever, its chemical name is:

A) Ascorbic acid

B) Acetyl salicylic acid

C) Nicotinic acid

D) Benzoic acid

✅ Correct Answer: B) Acetyl salicylic acid
Reason: Aspirin is chemically acetyl salicylic acid, used as an analgesic and antipyretic.

6. Nylon 6,6 is a condensation polymer of hexamethylene diamine and:

A) Benzoic acid

B) Adipic acid

C) Phthalic acid

D) Valeric acid

✅ Correct Answer: B) Adipic acid
Reason: Nylon 6,6 is formed by condensation of hexamethylene diamine and adipic acid.

7. Which of the following is NOT a synthetic plastic?

A) Nylon

B) Teflon

C) Cellulose

D) Polyethene

✅ Correct Answer: C) Cellulose
Reason: Cellulose is a natural polymer, not a synthetic plastic.

8. Drugs that lower the body temperature to normal are known as:

A) Antibiotics

B) Antipyretic

C) Antihistamines

D) Antiallergic

✅ Correct Answer: B) Antipyretic
Reason: Antipyretics reduce fever by lowering body temperature to normal.

9. Which of the following is used as an oxidizing agent in permanent hair dyes?

A) Acetone

B) Hydrogen peroxide

C) Polyvinyl acetate

D) Resorcinol

✅ Correct Answer: B) Hydrogen peroxide
Reason: Hydrogen peroxide acts as an oxidizing agent in permanent hair dyes, helping color fixation.

10. An example of thermosetting plastic is:

A) Polyethene

B) PVC

C) Nylon

D) Bakelite

✅ Correct Answer: D) Bakelite
Reason: Bakelite is a thermosetting plastic, once set it cannot be remelted or reshaped.

📝 Text Book Multiple Choice Questions On Environmental Chemistry Chapter 12 (Click to reveal answer with reason)🎯📌

1. The greenhouse gas is:

A) Oxygen (O₂)

B) Nitrogen (N₂)

C) Carbon dioxide (CO₂)

D) Argon (Ar)

✅ Correct Answer: C) Carbon dioxide (CO₂)
Reason: CO₂ is a major greenhouse gas responsible for trapping heat in the atmosphere.

2. The pH of acid rain is:

A) Between 7 to 8

B) Between 6 to 7

C) Below 5

D) Above 8

✅ Correct Answer: C) Below 5
Reason: Acid rain has a pH below 5 due to dissolved SO₂ and NOₓ forming acids.

3. Ozone depletion in upper atmosphere is mainly caused by:

A) Sulphur dioxide

B) Nitrogen oxides (NOₓ)

C) Chlorofluorocarbons (CFCs)

D) CO

✅ Correct Answer: C) Chlorofluorocarbons (CFCs)
Reason: CFCs release chlorine radicals that destroy ozone molecules in the stratosphere.

4. The region of sphere which extends from 11 km to 50 km from our Earth is known as:

A) Stratosphere

B) Mesosphere

C) Troposphere

D) Thermosphere

✅ Correct Answer: A) Stratosphere
Reason: The stratosphere lies between 11–50 km above Earth and contains the ozone layer.

5. The increase in global average temperatures is primarily attributed to:

A) Solar radiations

B) Oxides of sulphur (SOₓ)

C) Greenhouse gases emission

D) Natural climate variations

✅ Correct Answer: C) Greenhouse gases emission
Reason: Rising CO₂ and other greenhouse gases are the main drivers of global warming.

6. The primary goal of green chemistry is:

A) Maximizing industrial production

B) Minimizing waste generation

C) Reducing energy consumption

D) Developing environmentally friendly chemicals

✅ Correct Answer: D) Developing environmentally friendly chemicals
Reason: Green chemistry focuses on designing safer, sustainable, and eco-friendly chemical processes.

7. Smog is word used for the combination of:

A) Water vapours and fog

B) Smoke and fog

C) Sediments and colloids

D) Oxides of nitrogen and sulphur

✅ Correct Answer: B) Smoke and fog
Reason: Smog is formed from smoke and fog, often worsened by pollutants like NOₓ and SO₂.

8. The catalyst used in automobile catalytic converter is a mixture of:

A) Pt and Pd

B) Ni and Fe

C) Cu and Cr

D) Pb and Hg

✅ Correct Answer: A) Pt and Pd
Reason: Catalytic converters use platinum and palladium to oxidize CO and hydrocarbons, and reduce NOₓ.

9. Ozone layer is present in which of the following regions of atmosphere?

A) Stratosphere

B) Mesosphere

C) Troposphere

D) Thermosphere

✅ Correct Answer: A) Stratosphere
Reason: The ozone layer lies in the stratosphere, protecting Earth from harmful UV radiation.

10. The major source of water pollution is:

A) Organic farming

B) Renewable energy sources

C) Industrial activities

D) Global warming

✅ Correct Answer: C) Industrial activities
Reason: Industrial discharge of chemicals and waste is the major contributor to water pollution.

📝 Textbook MCQs on Spectroscopy Chapter 13 (Click to reveal answer with reason)🎯📌

1. The energy of IR radiations is:

A) Higher than radio waves

B) Higher than UV waves

C) Lower than microwaves

D) Higher than x-rays

✅ Correct Answer: A) Higher than radio waves
Reason: IR radiation has higher energy than radio waves but lower than visible, UV, and X-rays.

2. The fingerprint region of an IR spectrum is typically found in the range of:

A) 1500–4000 cm⁻¹

B) 2000–4000 cm⁻¹

C) 1500–500 cm⁻¹

D) 4000–8000 cm⁻¹

✅ Correct Answer: C) 1500–500 cm⁻¹
Reason: The fingerprint region lies between 1500–500 cm⁻¹, unique for each molecule.

3. Which unit of measurement is used for the wave number of IR spectrum?

A) Nanometer (nm)

B) Angstrom (Å)

C) Centimeter⁻¹ (cm⁻¹)

D) Hertz (Hz)

✅ Correct Answer: C) Centimeter⁻¹ (cm⁻¹)
Reason: IR wave numbers are expressed in reciprocal centimeters (cm⁻¹).

4. Electronic excitation occurs in electromagnetic spectrum if the molecule absorbs:

A) IR radiations

B) UV-visible radiations

C) Radio waves

D) Microwaves

✅ Correct Answer: B) UV-visible radiations
Reason: UV-visible absorption promotes electrons to higher energy levels (electronic excitation).

5. In NMR spectroscopy, the hydroxyl proton of C₂H₅OH appears as a broad singlet around:

A) 1–2 ppm

B) 2–3 ppm

C) 4–5 ppm

D) 6–7 ppm

✅ Correct Answer: C) 4–5 ppm
Reason: Hydroxyl protons in alcohols typically appear as broad singlets around 4–5 ppm.

6. Infrared spectroscopy is a technique used to determine ………….. in the given organic molecule:

A) Double and triple bonds

B) Mass to charge ratio

C) Functional group

D) Conjugated system

✅ Correct Answer: C) Functional group
Reason: IR spectroscopy identifies functional groups by their characteristic absorption bands.

7. Highest UV-visible absorption energy is required for the transition of:

A) σ → σ*

B) π → π*

C) n → σ*

D) n → π*

✅ Correct Answer: A) σ → σ*
Reason: σ → σ* transitions require the highest energy among UV-visible transitions.

8. NMR spectroscopy is carried out if radiations interact with molecules in high magnetic field:

A) UV-visible

B) Radio waves

C) X-rays

D) Infrared rays

✅ Correct Answer: B) Radio waves
Reason: NMR spectroscopy uses radio waves in a strong magnetic field to excite nuclear spins.

9. Atomic absorption spectrum is represented by:

A) Dark lines against bright background

B) Bright lines against dark background

C) Bright lines against bright background

D) Dark lines against dark background

✅ Correct Answer: A) Dark lines against bright background
Reason: Atomic absorption spectra show dark lines where radiation is absorbed by atoms.

10. In which of the following spectroscopy technique, D₂O can be used as solvent?

A) Mass spectroscopy

B) NMR spectroscopy

C) Atomic absorption spectroscopy

D) IR spectroscopy

✅ Correct Answer: B) NMR spectroscopy
Reason: D₂O is used in NMR spectroscopy to avoid interference from hydrogen in solvents.

📝 Past Papers MCQs XII Chemistry (Click to reveal answer with reason)🎯📌

1. The atomic number of an element belonging to group VIA and 3rd period is:

A) 24

B) 8

C) 16

D) 34

✅ Correct Answer: C) 16
Reason: Group VIA (16) and 3rd period corresponds to Sulfur, atomic number 16.

2. Sodium amalgam is an alloy of:

A) Sodium and lead

B) Sodium and silver

C) Sodium and Hg

D) Sodium and Iron

✅ Correct Answer: C) Sodium and Hg
Reason: Sodium amalgam is an alloy of sodium and mercury, used in chemical reductions.

3. Elements of group IB are called:

A) Normal elements

B) Coinage metals

C) Chalcogens

D) Pnictogens

✅ Correct Answer: B) Coinage metals
Reason: Group IB elements (Cu, Ag, Au) are called coinage metals due to their historical use in coins.

4. Which lists species with the same electron configuration?

A) S²⁻, Cl⁻, K⁺

B) Mg²⁺, Ca²⁺, Sr²⁺

C) Mg, Ca, Sr

D) F⁻, S²⁻, As³⁻

✅ Correct Answer: A) S²⁻, Cl⁻, K⁺
Reason: All have 18 electrons, same as Argon (isoelectronic species).

5. The metal forms superoxide is:

A) Be

B) Li

C) Na

D) Cs

✅ Correct Answer: D) Cs
Reason: Cesium forms superoxide (CsO₂) due to its large size and low ionization energy.

6. The Outer Electronic Configuration of an element is 4s² 3d¹⁰ 4p¹, its group and period are respectively:

A) I-A & 3rd

B) II-A & 4th

C) III-A & 3rd

D) III-A & 4th

✅ Correct Answer: D) III-A & 4th
Reason: Configuration corresponds to Gallium (Ga), group III-A (13), period 4.

7. EDTA is the ______________ type of Ligand.

A) Bi-dentate

B) Tetra-dentate

C) Penta-dentate

D) Hexa-dentate

✅ Correct Answer: D) Hexa-dentate
Reason: EDTA (Ethylenediaminetetraacetic acid) can bind through six donor atoms.

8. Coinage metals are elements of group IB and these include:

A) Cu, Ag & Au

B) Zn, Cd & Hg

C) Fe, Co & Ni

D) Cu, Zn & Ni

✅ Correct Answer: A) Cu, Ag & Au
Reason: Coinage metals are copper, silver, and gold, all group IB elements.

9. The metal ion having highest number of unpaired electrons is:

A) Mn²⁺

B) Fe²⁺

C) Co²⁺

D) Ni²⁺

✅ Correct Answer: A) Mn²⁺
Reason: Mn²⁺ has 3d⁵ configuration with 5 unpaired electrons, maximum among the given.

10. On burning in excess of Oxygen Sodium forms its:

A) Superoxide

B) Peroxide

C) Monoxide

D) Dioxide

✅ Correct Answer: B) Peroxide
Reason: Sodium forms sodium peroxide (Na₂O₂) when burned in excess oxygen.

11. The process of covering iron sheets by a layer of Zinc is known as:

A) Tempering

B) Tin plating

C) Galvanizing

D) Annealing

✅ Correct Answer: C) Galvanizing
Reason: Galvanizing protects iron sheets from rusting by coating them with zinc.

12. The chemical name of laughing gas is:

A) Nitrogen pentaoxide

B) Nitrous oxide

C) Nitrogen trioxide

D) Nitric oxide

✅ Correct Answer: B) Nitrous oxide
Reason: Laughing gas is N₂O, commonly used as an anesthetic.

13. Which one of the following is the formula of superoxides of alkali metals (M)?

A) MO₂

B) MO

C) M₂O

D) M₂O₂

✅ Correct Answer: A) MO₂
Reason: Superoxides of alkali metals have the formula MO₂ (e.g., KO₂).

14. The oxide of beryllium is:

A) Basic

B) Acidic

C) Neutral

D) Amphoteric

✅ Correct Answer: D) Amphoteric
Reason: BeO reacts with both acids and bases, showing amphoteric behavior.

15. Which of the following form normal oxide?

A) Na

B) Li

C) Ca

D) K

✅ Correct Answer: B) Li
Reason: Lithium forms normal oxide (Li₂O), unlike heavier alkali metals which form peroxides or superoxides.

16. The elements of which one of the following pairs of group would be expected to form compound of formula MX:

A) IIIA and IVA

B) IA and VIA

C) IIA and VIA

D) IVA and VIIA

✅ Correct Answer: C) IIA and VIA
Reason: Group IIA (alkaline earth metals) and group VIA (chalcogens) are both divalent and hence form compounds of formula MX (e.g., MgS) in 1:1 ratio.

17. The electron configuration of atoms of element X is [Ar] 3d¹⁰ 4s². Which is the most likely formula for the compound of this element and oxygen?

A) X₂O₃

B) XO

C) XO₂

D) X₂O

✅ Correct Answer: B) XO
Reason: The configuration corresponds to Zn (atomic number 30) which is divalent and oxygen is also bivalent, hence they combine to form oxide with formula ZnO (XO type) in 1:1 ratio.

18. Which one of the following represents the pair of metalloids?

A) B and Al

B) Mg and Si

C) As and Sb

D) Ge and C

✅ Correct Answer: C) As and Sb
Reason: Arsenic and antimony are metalloids, showing both metallic and non-metallic properties.

19. Which one of the following is the formula of peroxides of alkaline earth metals (M)?

A) M₂O₂

B) MO

C) M₂O

D) MO₂

✅ Correct Answer: D) MO₂
Reason: Alkaline earth metals form peroxides (MO₂).

20. The element with atomic number 9 is closest to in chemical properties with element of Z:

A) 27

B) 37

C) 47

D) 53

✅ Correct Answer: D) 53
Reason: Atomic number 9 is Fluorine, a halogen. Element 53 is Iodine, also a halogen, sharing similar chemical properties.

21. The Hybridization in the Carbon atom of Carbonyl group is:

A) sp

B) d²sp³

C) sp³

D) sp²

✅ Correct Answer: D) sp²
Reason: The carbonyl carbon forms three sigma bonds and one pi bond, giving sp² hybridization.

22. The formula for Caproic acid is:

A) CH₃(CH₂)₂COOH

B) CH₃(CH₂)₃COOH

C) CH₃(CH₂)₅COOH

D) CH₃(CH₂)₄COOH

✅ Correct Answer: D) CH₃(CH₂)₄COOH
Reason: Caproic acid is hexanoic acid, with six carbons in the chain.

23. This is the type formula for Ketones:

A) R–CO–H

B) R–CO–R

C) R–CO–OH

D) R–O–R

✅ Correct Answer: B) R–CO–R
Reason: Ketones have a carbonyl group bonded to two alkyl/aryl groups.

24. Only this one of the compound given below obeys Markownikoff rule on reaction with HCl:

A) CH₂=CH₂

B) CH₃–CH=CH₂

C) Cl₂C=CCl₂

D) CH≡CH

✅ Correct Answer: B) CH₃–CH=CH₂
Reason: In propene, HCl adds according to Markownikoff’s rule, giving 2-chloropropane.

25. Which one of the following is NOT the general formula of carbonyl compounds?

A) CₙH₂ₙO₂

B) CₙH₂nO

C) (CₙH₂ₙ+1)₂CO

D) CₙH₂ₙ+2CO

✅ Correct Answer: A) CₙH₂ₙO₂
Reason: CₙH₂ₙO₂ is the formula of carboxylic acids/esters, not carbonyl compounds.

26. Another name for methane is:

A) Marsh gas

B) Mustard gas

C) Oil gas

D) Coal gas

✅ Correct Answer: A) Marsh gas
Reason: Methane is commonly called marsh gas, as it is produced in marshy places.

27. Which one of the following formula represents acid halides?

A) R–X

B) R–COR

C) RCOX

D) Ar–X

✅ Correct Answer: C) RCOX
Reason: Acid halides have the general formula RCOX, where X is a halogen.

28. The general formula of alkanoic acids is:

A) CₙH₂ₙO₂

B) CₙH₂ₙO

C) CₙH₂ₙ+2O₂

D) CₙH₂ₙX₂

✅ Correct Answer: A) CₙH₂ₙO₂
Reason: Alkanoic acids (carboxylic acids) follow the general formula CₙH₂ₙO₂.

29. A carbon atom having a positive charge is called:

A) Carbanion

B) Carbonium ion

C) Carbon ion

D) Arenium ion

✅ Correct Answer: B) Carbonium ion
Reason: A positively charged carbon atom is called a carbocation or carbonium ion.

30. Anti-knocking character of gasoline is improved by the addition of:

A) Ethyl lead

B) Triethyl lead

C) Diethyl lead

D) Tetraethyl lead

✅ Correct Answer: D) Tetraethyl lead
Reason: Tetraethyl lead was historically added to gasoline to reduce knocking.

31. The number of isomers of hexane are:

A) 3

B) 5

C) 9

D) 18

✅ Correct Answer: B) 5
Reason: Hexane has 5 structural isomers (n-hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane, 2,3-dimethylbutane).

32. Cycloalkanes have the general formula:

A) CₙH₂ₙ

B) CₙH₂ₙ+2

C) CₙH₂ₙ-2

D) CₙH₂ₙ+4

✅ Correct Answer: A) CₙH₂ₙ
Reason: Cycloalkanes follow the formula CₙH₂ₙ, same as alkenes, due to ring closure.

33. The functional group in RCN is:

A) Thioalcohol

B) Ether

C) Nitrile

D) Thioether

✅ Correct Answer: C) Nitrile
Reason: RCN represents nitriles, containing the –C≡N group.

34. Propanone and propanal are:

A) Functional isomers

B) Positional isomers

C) Tautomers

D) Metamers

✅ Correct Answer: A) Functional isomers
Reason: Propanone (ketone) and propanal (aldehyde) have same molecular formula but different functional groups.

35. The oxidation states of the elements of group VA is:

A) +1 and +3

B) -3 and -5

C) -3, +3 and +5

D) +1, -1 and +3

✅ Correct Answer: C) -3, +3 and +5
Reason: Group VA elements (N, P, As, Sb, Bi) exhibit oxidation states of -3, +3, and +5.

36. Which of the following s-block elements form superoxide when burn in air:

A) Li

B) Na

C) K

D) Mg

✅ Correct Answer: C) K
Reason: Potassium forms superoxide (KO₂) when burned in air.

37. In the equilibrium of dichromate ion, the colour of CrO₄²⁻ is:

A) Orange

B) Green

C) Yellow

D) Red

✅ Correct Answer: C) Yellow
Reason: Chromate ion (CrO₄²⁻) is yellow, while dichromate (Cr₂O₇²⁻) is orange.

38. The coordination number of Pt in [Pt(en)₃]Cl₄ is:

A) 3

B) 4

C) 6

D) 2

✅ Correct Answer: C) 6
Reason: Ethylenediamine (en) is bidentate; three en ligands give coordination number 6.

39. The knocking of internal combustion engine can be reduced by the following petroleum process:

A) Refining

B) Resonance

C) Distillation

D) Reforming

✅ Correct Answer: D) Reforming
Reason: Reforming increases the octane number of gasoline, reducing knocking.

40. How many optical isomers are possible for HOOC(Br)CH–CH(OH)COOH:

A) 2

B) 3

C) 4

D) 5

✅ Correct Answer: C) 4
Reason: The molecule has two chiral centers, giving 2² = 4 optical isomers.

41. Which of the following pairs of compounds represent functional group isomerism?

A) Propanone and propanal

B) Ethanol and dimethyl ether

C) Acetic acid & methyl formate

D) All of these

✅ Correct Answer: D) All of these
Reason: Each pair has same molecular formula but different functional groups, hence functional group isomers.

42. Which of the following pairs of compounds represent positional isomerism?

A) 1-butene and 2-butene

B) Iso-propyl alcohol & 1-propanol

C) n-butane and iso-butane

D) Both a and b

✅ Correct Answer: D) Both a and b
Reason: Positional isomers differ in the position of double bond or functional group.

43. Lucas Test is used to distinguish between:

A) Alcohol and phenol

B) Three types of alcohols

C) Alcohols and amines

D) None of these

✅ Correct Answer: B) Three types of alcohols
Reason: Lucas test distinguishes primary, secondary, and tertiary alcohols by their reactivity with Lucas reagent.

44. Lucas reagent is a mixture of:

A) Zn and Hg

B) Zn and HCl

C) ZnCl₂ and HCl

D) NaOH & CaO

✅ Correct Answer: C) ZnCl₂ and HCl
Reason: Lucas reagent is ZnCl₂ in concentrated HCl, used for alcohol classification.

45. The kinetics and molecularity of elimination bimolecular reaction are respectively:

A) Second order and 2

B) First order and 1

C) Zero order and 2

D) None of these

✅ Correct Answer: A) Second order and 2
Reason: E2 elimination is bimolecular, rate depends on both substrate and base concentration.

46. The region of sphere which extends from 11 km to 50 km from our Earth is known as:

A) Thermosphere

B) Mesosphere

C) Troposphere

D) Stratosphere

✅ Correct Answer: D) Stratosphere
Reason: Stratosphere lies between 11–50 km and contains the ozone layer.

47. Clemmensen reduction is the conversion of aldehydes and ketones into:

A) Alkyl halides

B) Alcohols

C) Alkenes

D) Alkanes

✅ Correct Answer: D) Alkanes
Reason: Clemmensen reduction (Zn-Hg/HCl) reduces carbonyl groups to alkanes.

48. Formic acid is naturally found in:

A) Vinegar

B) Valerian root

C) Butter

D) Bee’s sting

✅ Correct Answer: D) Bee’s sting
Reason: Formic acid occurs naturally in bee and ant stings.

49. An example of quaternary structure of protein is:

A) Globulin

B) Albumin

C) Myoglobin

D) Hemoglobin

✅ Correct Answer: D) Hemoglobin
Reason: Hemoglobin has quaternary structure with four polypeptide chains.

50. It is a three membered cyclic ether:

A) Diazonium salt

B) Epoxide

C) Benzoquinone

D) Ozonide

✅ Correct Answer: B) Epoxide
Reason: Epoxides are three-membered cyclic ethers formed by oxidation of alkenes.

51. They are used as a versatile building blocks in the synthesis of pharmaceuticals, agrochemicals, polymers, surfactants:

A) Diazonium salt

B) Epoxide

C) Benzoquinone

D) Ozonide

✅ Correct Answer: B) Epoxide
Reason: Epoxides are highly reactive three-membered cyclic ethers, widely used as intermediates in synthesis.

52. The final product of ozonolysis of an alkyne like ethyne is:

A) Diazonium salt

B) Epoxide

C) Glyoxal

D) Ozonide

✅ Correct Answer: C) Glyoxal
Reason: Ozonolysis of ethyne yields glyoxal (CHO–CHO) as the final product.

53. Primary amines react with aldehydes and ketones to produce:

A) Imine

B) Schiff’s base

C) Both of them

D) None of them

✅ Correct Answer: C) Both of them
Reason: Primary amines react with carbonyl compounds to form imines, also called Schiff’s bases.

54. When amines react with nitrous acid in the presence of hydrochloric acid at below 10°C gives:

A) Diazonium salt

B) Benzoquinone

C) Imines

D) Amide

✅ Correct Answer: A) Diazonium salt
Reason: Aromatic primary amines react with nitrous acid at low temperature to form stable diazonium salts.

55. Reduction of carboxylic acid by LiAlH₄ gives:

A) Primary alcohol

B) Alkanes

C) Aldehyde

D) Amine

✅ Correct Answer: A) Primary alcohol
Reason: LiAlH₄ reduces carboxylic acids directly to primary alcohols.

56. Reduction of esters by LiAlH₄ gives:

A) Primary alcohol

B) Alkanes

C) Aldehyde

D) Amine

✅ Correct Answer: A) Primary alcohol
Reason: Esters are reduced to two primary alcohols by LiAlH₄.

57. Reduction of methyl acetate by LiAlH₄ gives:

A) Ethyl alcohol

B) Methyl alcohol

C) Both of them

D) None of them

✅ Correct Answer: C) Both of them
Reason: Methyl acetate reduces to methanol and ethanol (two primary alcohols).

58. Reduction of acetic acid by LiAlH₄ gives:

A) Ethyl alcohol

B) Methyl alcohol

C) Both of them

D) None of them

✅ Correct Answer: A) Ethyl alcohol
Reason: Acetic acid (CH₃COOH) reduces to ethanol (CH₃CH₂OH).

59. Tertiary alcohols cannot be oxidized by any oxidizing agent due lack of:

A) Alpha H

B) Beta H

C) Gamma H

D) None of them

✅ Correct Answer: A) Alpha H
Reason: Oxidation requires α-hydrogen adjacent to the –OH group, absent in tertiary alcohols.

60. The oxidation of ketone involves the breaking of:

A) C–C bond

B) C–H bond

C) C=O bond

D) None of them

✅ Correct Answer: A) C–C bond
Reason: Oxidation of ketones involves cleavage of the C–C bond adjacent to the carbonyl group.

61. Vicinal diol are the types of dihydric alcohols containing two hydroxyl groups on position:

A) 1 and 2

B) 1 and 3

C) 1 and 1

D) None of them

✅ Correct Answer: A) 1 and 2
Reason: Vicinal diols have hydroxyl groups on adjacent carbons (1,2 positions).

62. Vicinal diol are also called:

A) 1,2-diols

B) 1,3-diols

C) Glycols

D) Both a and c

✅ Correct Answer: D) Both a and c
Reason: Vicinal diols are 1,2-diols, commonly called glycols (e.g., ethylene glycol).

63. The oxidative cleavage of 1,2-diols by periodic acid gives two molecules of same or different:

A) Carbonyl compounds

B) Carboxylic acids

C) Glycols

D) Both a and c

✅ Correct Answer: A) Carbonyl compounds
Reason: Periodic acid cleaves vicinal diols to yield aldehydes or ketones (carbonyl compounds).

64. During the oxidative cleavage of 1,2-diols, periodic acid is reduced into:

A) Iodic acid

B) Hypoiodous acid

C) Iodous acid

D) Both a and c

✅ Correct Answer: A) Iodic acid
Reason: Periodic acid (HIO₄) is reduced to iodic acid (HIO₃) during cleavage.

65. The oxidative cleavage of ethylene glycol with periodic acid gives:

A) Formaldehyde

B) Acetone

C) Ethyl alcohol

D) Both a and c

✅ Correct Answer: A) Formaldehyde
Reason: Ethylene glycol (HO–CH₂–CH₂–OH) cleaves to give two molecules of formaldehyde.

66. Reduction of aldehyde and ketones by catalytic hydrogenation by H₂ or chemical reduction with reducing agents like LiAlH₄ or NaBH₄ gives:

A) Alcohol

B) Amine

C) Carboxylic acids

D) Both a and c

✅ Correct Answer: A) Alcohol
Reason: Aldehydes reduce to primary alcohols, ketones to secondary alcohols.

67. Aldehydes on reduction gives:

A) Primary alcohol

B) Secondary alcohol

C) Carboxylic acids

D) Both a and c

✅ Correct Answer: A) Primary alcohol
Reason: Aldehydes reduce to primary alcohols (R–CHO → R–CH₂OH).

68. Ketones on reduction gives:

A) Primary alcohol

B) Secondary alcohol

C) Carboxylic acids

D) Both a and c

✅ Correct Answer: B) Secondary alcohol
Reason: Ketones reduce to secondary alcohols (R–CO–R → R–CHOH–R).

69. Which ion is resonance-stabilized?

A) Phenoxide ion

B) Carboxylate ion

C) Amide ion

D) Both a and b

✅ Correct Answer: D) Both a and b
Reason: Phenoxide and carboxylate ions are resonance-stabilized, delocalizing negative charge.

70. The preparation of benzene by the alkaline hydrolysis of chlorobenzene followed acidification is called:

A) Dow’s process

B) Down’s process

C) Cannizaro reaction

D) Both a and b

✅ Correct Answer: A) Dow’s process
Reason: Dow’s process involves alkaline hydrolysis of chlorobenzene to produce benzene.

71. Which gas is released when diazonium chloride is heated with water:

A) Phenol

B) Nitrogen

C) Oxygen

D) CO₂

✅ Correct Answer: B) Nitrogen
Reason: Diazonium salts decompose with water to give phenols and release nitrogen gas.

72. Due to its explosive nature in dry form, picric acid is stored in bottles under a layer of:

A) Water

B) Kerosene

C) Petrol

D) None of these

✅ Correct Answer: A) Water
Reason: Picric acid is kept moist under water to prevent explosion in dry state.

73. When phenol is treated with concentrated sulphuric acid at 15–20°C, the major product is:

A) Ortho-phenol sulphonic acid

B) Para-phenol sulphonic acid

C) Both of them

D) None of them

✅ Correct Answer: A) Ortho-phenol sulphonic acid
Reason: At low temperature, ortho product predominates in sulphonation of phenol.

74. When phenol is treated with concentrated sulphuric acid at 100°C, the major product is:

A) 2-hydroxy benzene sulphonic acid

B) 4-hydroxy benzene sulphonic acid

C) Both of them

D) None of them

✅ Correct Answer: B) 4-hydroxy benzene sulphonic acid
Reason: At higher temperature, para product predominates in sulphonation of phenol.

75. Which one is a conjugated diketone?

A) Benzoquinone

B) Diazonium salt

C) Acetophenone

D) Acetone

✅ Correct Answer: A) Benzoquinone
Reason: Benzoquinone is a conjugated diketone with alternating double bonds and two carbonyl groups.

76. The oxidation of phenol by oxidizing agents like acidified dichromate gives:

A) Benzoquinone

B) Diazonium salt

C) Acetophenone

D) Acetone

✅ Correct Answer: A) Benzoquinone
Reason: Phenol oxidizes to benzoquinone with strong oxidizing agents.

77. The colour of benzoquinone gives:

A) Yellow

B) Green

C) Blue

D) White

✅ Correct Answer: A) Yellow
Reason: Benzoquinone is a yellow crystalline solid.

78. The fruity smell of ester in ester test indicates the presence of ………. in the given organic compound:

A) Alcoholic group

B) Ester group

C) Amino group

D) Phenolic group

✅ Correct Answer: A) Alcoholic group
Reason: Ester test detects alcohols; fruity smell indicates ester formation from alcohol.

79. Appearance of violet, blue or purple colouration indicates the formation of complex during ferric chloride test and identifies the presence of:

A) Alcohols

B) Ester

C) Amines

D) Phenols

✅ Correct Answer: D) Phenols
Reason: Ferric chloride test gives violet/purple complex with phenols.

80. The violet or purple coloured complex formed in ferric chloride test for phenol identification is named as:

A) Ferric phenoxide complex

B) Ferric phenolate complex

C) Ferric phenate complex

D) All of them

✅ Correct Answer: D) All of them
Reason: The complex can be referred to as ferric phenoxide, phenolate, or phenate complex.

81. The IUPAC name of ferric phenoxide complex is:

A) Ferric phenoxide complex

B) Ferric phenate complex

C) Hydrogen hexaphenoxidoferrate(III)

D) None of them

✅ Correct Answer: C) Hydrogen hexaphenoxidoferrate(III)
Reason: The systematic IUPAC name for ferric phenoxide complex is hydrogen hexaphenoxidoferrate(III).

82. The formula of ferric phenoxide complex is:

A) H₃[Fe(C₆H₅O)₆]

B) H₂[Fe(C₆H₅O)₃]

C) H₃[Fe(C₆H₅O)₅]

D) None of them

✅ Correct Answer: A) H₃[Fe(C₆H₅O)₆]
Reason: The ferric phenoxide complex has the formula H₃[Fe(C₆H₅O)₆].

83. Disappearance of brown colour of bromine water and appearance of white precipitates of 2,4,6-tribromophenol during bromine water test identifies the presence of:

A) Alcohols

B) Ester

C) Amines

D) Phenols

✅ Correct Answer: D) Phenols
Reason: Phenols react with bromine water to form white precipitate of 2,4,6-tribromophenol.

84. The ozonolysis of ethene finally gives:

A) Formaldehyde

B) Acetone

C) Ozonide

D) None of them

✅ Correct Answer: A) Formaldehyde
Reason: Ethene undergoes ozonolysis to yield two molecules of formaldehyde.

85. The ozonolysis of 2,3-dimethyl-2-butene finally gives:

A) Formaldehyde

B) Acetone

C) Ozonide

D) None of them

✅ Correct Answer: B) Acetone
Reason: Ozonolysis of 2,3-dimethyl-2-butene produces two molecules of acetone.

86. In the oxidation of alcohols by pyridinium chlorochromate (PCC), the –OH functional group is converted to:

A) Carbonyl group (C=O)

B) Carboxylic group

C) Ester

D) None of them

✅ Correct Answer: A) Carbonyl group (C=O)
Reason: PCC oxidizes alcohols to carbonyl compounds without over-oxidation.

87. The selective oxidizing agent PCC oxidizes primary alcohols to:

A) Aldehydes

B) Ketones

C) Carboxylic acids

D) Esters

✅ Correct Answer: A) Aldehydes
Reason: PCC selectively oxidizes primary alcohols to aldehydes without further oxidation to acids.

88. This is the 37–40% aqueous solution of formaldehyde:

A) Formalin

B) Vinegar

C) Metaformaldehyde

D) Paraldehyde

✅ Correct Answer: A) Formalin
Reason: Formalin is a 37–40% aqueous solution of formaldehyde, widely used as preservative.

89. Formalin is used as a:

A) Disinfectant

B) Biological specimens preservative

C) Both of them

D) None of them

✅ Correct Answer: C) Both of them
Reason: Formalin is used both as disinfectant and preservative for biological specimens.

90. PCC oxidizes isopropyl alcohol to:

A) Acetone

B) Acetaldehyde

C) Acetic acid

D) Propionic acid

✅ Correct Answer: A) Acetone
Reason: PCC oxidizes secondary alcohols like isopropyl alcohol to ketones (acetone).

91. The addition of hydrazine to carbonyl compounds gives:

A) Hydrazone

B) Oxime

C) Cyanohydrin

D) Imine

✅ Correct Answer: A) Hydrazone
Reason: Hydrazine reacts with carbonyl groups to form hydrazones.

92. The reduction of carbonyl compounds results in the formation of:

A) Alkane

B) Alcohol

C) Both of them

D) None of them

✅ Correct Answer: B) Alcohol
Reason: Reduction of aldehydes gives primary alcohols, ketones give secondary alcohols.

93. The reduction of carbonyl compounds into saturated hydrocarbons by zinc amalgam and concentrated HCl is known as:

A) Clemmensen reduction

B) Wolff-Kishner reaction

C) Lucas reaction

D) None of them

✅ Correct Answer: A) Clemmensen reduction
Reason: Clemmensen reduction converts carbonyl compounds to alkanes using Zn-Hg/HCl.

94. Acetaldehyde is reduced by a mixture of zinc amalgam and concentrated HCl into:

A) Ethane

B) Ethene

C) Ethanol

D) Propane

✅ Correct Answer: A) Ethane
Reason: Clemmensen reduction of acetaldehyde (CH₃CHO) gives ethane (CH₃CH₃).

95. Acetone is reduced by a mixture of zinc amalgam and concentrated HCl into:

A) Ethane

B) Ethene

C) Ethanol

D) Propane

✅ Correct Answer: D) Propane
Reason: Clemmensen reduction of acetone (CH₃COCH₃) gives propane (CH₃CH₂CH₃).

96. The conversion of carbonyl compounds into alkane using hydrazine and potassium hydroxide through hydrazone intermediate is known as:

A) Clemmensen reduction

B) Wolff-Kishner reaction

C) Lucas reaction

D) None of them

✅ Correct Answer: B) Wolff-Kishner reaction
Reason: Wolff-Kishner reduction converts carbonyls to alkanes via hydrazone intermediate.

97. The addition of hydrazine to carbonyl compounds gives:

A) Hydrazone

B) Imine

C) Oxime

D) Diazonium salt

✅ Correct Answer: A) Hydrazone
Reason: Hydrazine reacts with carbonyl groups to form hydrazones.

98. Hydrazone is converted into alkane by heating with potassium hydroxide with the release of:

A) Nitrogen

B) Oxygen

C) Hydrogen

D) None of these

✅ Correct Answer: A) Nitrogen
Reason: Wolff-Kishner reduction releases nitrogen gas during conversion of hydrazone to alkane.

99. Carbonyl compounds on reduction with strong reducing agents such as NaBH₄ or LiAlH₄ gives:

A) Alcohols

B) Carboxylic acids

C) Oxime

D) Amines

✅ Correct Answer: A) Alcohols
Reason: NaBH₄ and LiAlH₄ reduce aldehydes and ketones to alcohols.

100. Acetaldehyde is reduced by strong reducing agents such as NaBH₄ or LiAlH₄ to give:

A) Ethyl alcohol

B) Acetone

C) Ethane

D) Acetic acid

✅ Correct Answer: A) Ethyl alcohol
Reason: Acetaldehyde (CH₃CHO) reduces to ethanol (CH₃CH₂OH).

✏️ Smart Answers of Short-Answer Questions of ⚗️ Inorganic–General Chemistry Section 🧪 ✏️

Q2. (i) Explain the group trend and irregularities of ionization energy and atomic radii in the periodic table.

✍️ Answer

🌟 Ionization Energy (I.E./ I.P. / I) – Group Trend

🔹 Definition of Ionization Energy?

Ionization energy is the minimum energy required to remove the most loosely bound valence electron from an isolated neutral gaseous atom (or ion), forming a positive ion.
👉 Higher ionization energy = harder to remove the electron.

📉 Group Trends of IE (Top ➝ Bottom) 🔽

⚛️ Group Trend of IE in Group IA & IIA (s-block metals)

📉 General Trend: ⬇️ Ionization energy decreases down the group due to increased atomic size and shielding effect.
📌 Detailed Reason (optional)
📏 Atomic Size ↑: Outer electron is farther → easier to remove.
🛡️ Shielding ↑: Inner shells reduce nuclear pull.
⚡ Effective Attraction ↓: Despite higher nuclear charge, nucleus holds valence electron more weakly.
🧪 Example: In Group 1 (Alkali metals): Li > Na > K > Rb > Cs (IE decreases from top to bottom).

⚛️ Group Trend of IE in Group IIIA (Boron Family)

📉 General Trend: ⬇️ IE generally decreases down the group, but shows irregular behavior.
⚠️ Irregularities Observed: Ga has higher I.E. than Al; Tl has higher I.E. than In.
🛡️ Reason: Poor shielding of nuclear charge by 3d electrons in Ga and 4f electrons in Tl.
⚡ This poor shielding increases effective nuclear attraction, making the outer electron harder to remove.

⚛️ Group Trend of IE in Group IVA (Carbon Family)

📉 General Trend: ⬇️ IE generally decreases down the group, but Sn and Pb show irregularity due to lanthanide contraction and stronger nuclear attraction.
⚠️ Irregularity (Sn & Pb): Tin (Sn) and Lead (Pb) show nearly similar atomic radii.
🔬 Reason: Due to lanthanide contraction, Pb does not increase much in size, so the nucleus attracts outer electrons more strongly.
⚡ As a result, more energy is required to remove the electron than expected.

⚛️ Group Trend of IE in Group VA to VIIIA (p-block metals)

📉 General Trend: ⬇️ Ionization energy decreases down the group due to increased atomic size and shielding effect.

🌟 Atomic Radii

📏 Definition of Atomic Radii

The atomic radius is the average distance between the nucleus and the outermost electron shell of an atom.

📉 Group Trends of Atomic Radii (Top ➝ Bottom) 🔼

⚛️ Group Trend of Atomic Radii in Group IA & IIA (s-block metals)

📉 General Trend: ⬆️ Radius increases down the group ⬇️ due to extra electron shells 🟡.
🟢 First element → smallest, last element → largest.
💡 Example: IA: Li (smallest) → Fr (largest); IIA: Be (smallest) → Ra (largest).
⚡ Alkali metals have largest radii in their periods (weaker nuclear attraction).
🟠 Alkaline earth metals have slightly smaller radii.
📊 Atomic Radii Values of Alkali Metals (Group IA) – in pm:
Li → 152 ➡️ Na → 186 ➡️ K → 227 ➡️ Rb → 248 ➡️ Cs → 265 ➡️ Fr → 348
📊 Atomic Radii Values of Alkaline Earth Metals (Group IIA) – in pm:
Be → 112 ➡️ Mg → 145 ➡️ Ca → 194 ➡️ Sr → 219 ➡️ Ba → 253 ➡️ Ra → 215

⚛️ Group Trend of Atomic Radii in Group IIIA (Boron family)

📉 General Trend: ⬆️ Radius increases down the group ⬇️ due to extra electron shells 🟡.
⚠️ Exception: Ga < Al due to poor d-electron shielding.

⚛️ Group Trend of Atomic Radii in Group IVA to VIIIA

📉 General Trend: ⬆️ Radius increases down the group ⬇️ due to extra electron shells 🟡.

OR
What is bleaching powder? How it is prepared? Give its reactions with water and dilute HCl.

✍️ Answer

🧪 Bleaching Powder

📌 Definition & Composition

Bleaching powder is chemically called Calcium chlorohypochlorite (Proposed by Prof. Odling).

🧾 Formula

Ca(OCl)Cl or CaOCl₂

⚙️ Preparation

🏭 Method

Prepared by passing chlorine gas (Cl₂) over dry slaked lime [Ca(OH)₂] in the Hasenclever plant.

🧪 Reaction

Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O

🔄 Process

➡️ Dry slaked lime is fed from the top.
➡️ Chlorine gas enters from below.
➡️ They react to form bleaching powder.

⚗️ Chemical Properties

💧 Reaction with Water

➡️ Forms calcium hydroxide and hypochlorous acid (bleaching agent).
➡️ Ca(OCl)Cl + H₂O → Ca(OH)₂ + HCl + HOCl (responsible for bleaching & disinfecting action) 👈

🧪 Reaction with Dilute HCl

➡️ Reacts with acids to liberate chlorine gas.
➡️ CaOCl₂ + 2HCl → CaCl₂ + Cl₂ + H₂O

🌟 Uses (extra)

💧 1. Sterilization of drinking water
🧵 2. Bleaching cotton, linen & paper
⚗️ 3. Preparation of chloroform & chlorine gas
🧴 4. Used as an oxidizing agent

(ii) What is diagonal relationship? Give diagonal relationship of Li-Mg, Be-Al and B-Si.

✍️ Answer

Definition of Diagonal Relationship

👉 Diagonal relationship is the similarity in properties shown by certain pairs of elements that are placed diagonally to each other in adjacent groups and periods of the periodic table.

🔗 Important Diagonal Pairs ↘️

Li → Mg (Group IA & IIA)
Be → Al (Group IIA & IIIA)
B → Si (Group IIIA & IVA)

✨ Similarities

🔹 Lithium–Magnesium (Li–Mg) Diagonal Relationship

📏 Similar atomic size & electronegativity (Li =152 pm and Mg = 160 pm | EN: Li =1.0 and Mg = 1.2).
⚖️ Harder & lighter metals in their groups
🧪 Oxides less soluble in water

🔹 Beryllium–Aluminium (Be–Al) Diagonal Relationship

⚡ Similar electronegativity (1.5).
🛡️ Show passivity with conc. HNO₃
🧲 Chlorides (BeCl₂, AlCl₃) act as Lewis acids

🔹 Boron–Silicon (B–Si) Diagonal Relationship

⚡ Nearly same electronegativity (B =2.0 and Si =1.8).
⚖️ Similar density (B = 2.35 g/cm³ and Si = 2.34 g/cm³).
🔷 Metalloids & do not form cations

OR
Why transition elements have the tendency to form alloy? Write the name of three alloys of transition elements along with their composition.

✍️ Answer

🔍 Reason for Alloy Formation

Transition elements tend to form alloys because they have similar atomic sizes and closely packed crystal structures, allowing their atoms to replace each other easily in the lattice without changing the overall structure.

Composition and Important Uses of Some Transition Metal Alloys

➡️ Stainless steel

(Fe, Cr and Ni), used in making cutlery and surgical instruments

➡️ Duralumin

(Al, Cu, Mg and Mn), used in making utensils, aero plane etc.

➡️ Brass

(Cu and Zn); used in plumbing and automotive parts etc.

➡️ Bronze

(Cu and Sn); used in making medals, statues, coins etc.

OR
What are transition elements and outer transition elements? How are they classified? Why outer transition elements are called d-block element?

✍️ Answer

📘 Definition of Transition Elements

Transition elements are those elements whose atoms or at least one of their ions contain partially filled d- or f-orbitals characterized by the gradual filling of (n–1)d orbitals and (n–2)f orbitals.

🗂️ Classification of Transition Elements

💠 1. d-block elements (Outer transition elements)

💠 2. f-block elements (Inner transition elements)

📘 Definition of Outer Transition Elements (d-Block Elements)

Outer transition elements or d-Block elements are those in which the (n–1)d orbitals are being filled and which have partially filled d-orbitals in their atomic state or in at least one oxidation state.

🌀 General Electronic Configuration

(n−1)d¹⁻¹⁰ ns¹⁻² (where n = 4 to 7)

🧠 Why are they called “Outer Transition Elements? 🤔

They are called outer transition elements because the differentiating electron enters the penultimate (n–1)d subshell (outer d-orbitals), which lies closer to the outermost shell than the inner (n–2)f subshell.

🧠 Why are they called “d-Block Elements? 🤔

They are called d-block elements because the differentiating electron enters a d-orbital, and their chemical properties are mainly governed by partially filled d-orbitals.

Four Series of d-block Elements (10 elements each)

🔶 First Transition Series (3d series)

4th period (₂₁Sc to ₃₀Zn)

🔶 Second Transition Series (4d series)

5th period (₃₉Y to ₄₈Cd)

🔶 Third Transition Series (5d series)

6th period (₅₇La, ₇₂Hf to ₈₀Hg)

🔶 Fourth Transition Series (6d series)

7th period (₈₉Ac, ₁₀₄Rf to ₁₁₂Cn)

⚠️ Special Case: Zinc, Cadmium, and Mercury / Non-typical transition elements

Zinc, Cadmium, and Mercury – filled d¹⁰ configuration elements with no variable oxidation states or colour; called non-typical transition elements.

OR
Define d-block elements, why do they form colored compounds? Explain it in term of Crystal Field Theory.

✍️ Answer

📘 Definition of d-Block Elements

Elements in which (n–1)d orbitals are being filled and have partially filled d-orbitals in atoms or ions.

🎨 Why d-Block Compounds Are Colored

Most d-block compounds are colored because they have partially filled d-orbitals.
Color arises from d–d electronic transitions, where electrons jump between split d-orbitals.

d-block elements form colored compounds because ligands split d-orbitals into t₂g and eg orbitals, and electrons absorb visible light to jump from t₂g → eg (d–d transition) and the complementary color is observed.

🔬 Crystal Field Theory (CFT) Explanation

In an octahedral field (e.g., [Cu(H₂O)₆]²⁺), the five d-orbitals split into two sets:

t₂g orbitals

(dxy, dxz, dyz) → lower energy

eg orbitals

(dz², dx²–y²) → higher energy

Electrons in t₂g orbitals can absorb visible light to jump to eg orbitals.
The energy difference (Δ₀) corresponds to the wavelength of light absorbed, and the complementary color is observed.

Example

[Cu(H₂O)₆]²⁺ absorbs red light → appears blue

(iii) Write the IUPAC names of the following complexes:

✍️ Answer

⚛️ Complexes and Their IUPAC Names

Na₂[Pt(OH)₄] ------------------→ ✅ Sodium tetrahydroxoplatinate(II)
[Zn(NH₃)₄]²⁺ ------------------→ ✅ tetraamminezinc(II) ion
[Ni(SCN)₄]²⁻ ------------------→ ✅ tetrathiocyanatonickelate(II) ion
K₂[Fe(CN)₅NO] ----------------→ ✅ Potassium pentacyanonitrosylferrate(III)
[Ag(NH₃)₂]OH -----------------→ ✅ diamminesilver(I) hydroxide
K₃[Fe(CN)₆] ------------------→ ✅ Potassium hexacyanoferrate(III)
K₃[Cr(C₂O₄)₂Cl₂] -------------→ ✅ Potassium dichlorobis(oxalato)chromate(III)
[Pt(en)₂(NO₂)₂]SO₄ ------------→ ✅ Bis(ethylenediamine)dinitroplatinum(IV) sulphate
K₃[Cu(C₂O₄)₂(CN)₂] -----------→ ✅ Potassium dicyanobis(oxalato)cuprate(III)
[Ni(CO₃)₂(OH)₂]²⁻ -------------→ ✅ Dihydroxobis(carbonato)nickelate(II) ion
[Cr(NH₃)₄Cl₂] -----------------→ ✅ tetraamminedichlorochromium(II)
[Pt(en)₂Br₂] ------------------→ ✅ dibromobis(ethylenediamine)platinum(II)
Na₃[Co(NO₂)₆] ----------------→ ✅ Sodium hexanitritocobaltate(III)
NH₄[Cr(SCN)₄(NH₃)₂] ----------→ ✅ Ammonium diamminetetrathiocyanatochromate(III)
[Cr(NH₃)₂(NSC)₄]⁻ -------------→ ✅ diamminetetrathiocyanatochromate(III) ion
Na₂[Fe(CN)₅NO] ----------------→ ✅ Sodium pentacyanonitrosylferrate(III)
[Fe(CN)₅NO]²⁻ -----------------→ ✅ pentacyanonitrosylferrate(III) ion
K₃[Cr(C₂O₄)₂Cl₂] --------------→ ✅ Potassium dichloridobis(oxalato)chromate(III)
[Pt(en)₂(NO₂)₂]Cl₂ -------------→ ✅ Bis(ethylenediamine)dinitroplatinum(IV) chloride
[Cu(NH₃)₄]SO₄ ----------------→ ✅ tetraamminecopper(II) sulfate

OR
Give the formulae of the following complexes:

🟠 tetraamminedichlorochromium(III) chloride

✅ [Cr(NH₃)₄Cl₂]Cl

🟠 tetracarbonylnickel(0)

✅ [Ni(CO)₄]

🟠 tetranitrochromate(III) ion

✅ [Cr(NO₂)₄]⁻

🟠 Tollen’s reagent

✅ [Ag(OH)₂]OH

🟠 potassium hexacyanoferrate(III)

✅ K₃[Fe(CN)₆]

🟠 Nessler’s Reagent

✅ K₂[HgI₄]

(iv) Explain colour formation, magnetic properties and variable oxidation state of transition elements.

✍️ Answer

🎨 Colour Formation

Transition elements often form coloured compounds because of the presence of partially filled d-orbitals.
Ligands surrounding the metal ion split the five d-orbitals into two energy levels (t₂g and eg).
When light falls on the compound, electrons absorb specific wavelengths to jump from lower t₂g orbitals to higher eg orbitals (d–d transitions).
The absorbed wavelength corresponds to a certain colour, and the compound appears in the complementary colour.
Example: [Cu(H₂O)₆]²⁺ absorbs red light and appears blue.

🧲 Magnetic Properties

The magnetic behaviour of transition elements depends on the number of unpaired d-electrons.
If d-orbitals contain unpaired electrons, the compound shows paramagnetism (attracted to a magnetic field).
The greater the number of unpaired electrons, the stronger the magnetic moment.
If all d-orbitals are paired, the compound becomes diamagnetic (repelled by a magnetic field).
Example: Mn²⁺ with five unpaired electrons is strongly paramagnetic, while Zn²⁺ with fully filled d-orbitals is diamagnetic.

⚖️ Variable Oxidation States

Transition elements exhibit variable oxidation states because both ns and (n–1)d electrons can participate in bonding.
This flexibility allows them to form multiple stable ions and compounds with different oxidation numbers.
Example: Iron shows +2 and +3 states, manganese shows +2, +4, +6, and +7 states.
The availability of multiple oxidation states makes transition elements useful in redox reactions and as catalysts.
This property also explains their ability to form complex compounds with ligands.

OR
What is meant by binding energy? Explain the trend of binding energy in 3d series of transition elements.

✍️ Answer

⚛️ Definition of Binding Energy

It is the energy required to separate constituents of a bound system; measures the strength of attractive forces.

The binding energy of d-block elements is higher than that of other elements due to the strong attraction to their outermost d-electrons.

📈 Trend in d-Block Elements

Across a Period (Sc → Zn)

Increases due to higher nuclear charge and smaller atomic radius.

Down a Group

Decreases due to larger atomic size and increased shielding, reducing effective nuclear charge.

Special Cases

Mn²⁺ and Fe³⁺ have highest binding energies due to maximum unpaired 3d electrons.

(v) Why Beryllium differs markedly from other members of II A group? Write down four properties of beryllium that show its unique behaviour in group II A. Also write the reason why beryllium does not react with cold water and steam?

✍️ Answer

Beryllium differs markedly

Beryllium differs markedly from other alkaline earth metals due to its small size, high ionization energy, and high charge density, which leads to covalent character in its compounds.

🌟 Four Unique Properties of Beryllium

🔥 High Melting & Boiling Points

Much higher than other Group II A metals.

🤝 Forms Covalent Compounds

Forms covalent bonds due to small atomic size.⚡

⚖️ Lower density & higher melting point

💪 Harder and more rigid

Than other group members.

🛡️ Protective Oxide Layer

Forms a protective oxide layer, giving chemical stability and preventing corrosion.

❌ Resists Attack by Water & Steam

Does not react with cold water or steam under normal conditions.

💧 Amphoteric Oxide & Hydroxide

BeO and Be(OH)₂ react with both acids and bases. 🧪

Reason for Not Reacting with Water/Steam

It does not react with water/steam because of its protective oxide layer. Beryllium has a strong Be–O bond in its oxide layer and high ionization energy, so it does not react readily with cold water or steam.

OR
What is flame test? What is the basis of flame test? Mention the colour flame of s-block elements.

✍️ Answer

🔥 Flame Test

👉 It is a preliminary qualitative test used to identify s-block metal (except Be & Mg) in their white salts by their characteristic flame colours 🌈
🔥 When heated, electrons get excited and on returning to lower energy levels, emit coloured light ✨
💡 Each alkali and alkaline earth metal has a unique flame colour (except Be & Mg), helping in quick identification.

🧪 Basis of Flame Test

When a metal is heated in a flame, its electrons absorb energy and jump to a higher energy level 🔼
When they fall back 🔽 to a lower level, they emit light ✨
The emitted light has a specific colour 🌈 unique to each metal
👉 Colour is due to electronic transitions between energy levels (excited electrons release energy as coloured light ✨).

🌟 Group IA Metals & Their Flame Colours 🔥

➡️ Li → 🔴 Crimson red
➡️ Na → 🟡 Golden yellow
➡️ K → 🟣 Violet
➡️ Rb → 🟥 Red-violet
➡️ Cs → 🔵 Blue-violet

🌟 Group IIA Metals & Their Flame Colours 🔥

➡️ Be → ❌ No colour
➡️ Mg → ⚪ Silver white
➡️ Ca → 🟠 Brick red (Orange red)
➡️ Sr → 🔴 Crimson red (bright deep red)
➡️ Ba → 🟢 Apple green (Pale green)

(vi) Using electronic configuration, identify the block, period & group of the elements with the Z = 16, 24, 29, 35, 47 and 53.

✍️ Answer

➡️ Z = 16 (S): 1s² 2s² 2p⁶ 3s² 3p⁴ or [Ne] 3s² 3p⁴ →
👉 p-block, Period 3 (highest n), Group VI A or 16 (total outer ē)

➡️ Z = 24 (Cr): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹ or [Ar] 3d⁵ 4s¹ →
👉 d-block, Period 4, Group VIB or 6

➡️ Z = 29 (Cu): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ or [Ar] 3d¹⁰ 4s¹ →
👉 d-block, Period 4, Group IB or 11

➡️ Z = 35 (Br): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁵ or [Ar] 3d¹⁰ 4s² 4p⁵ →
👉 p-block, Period 4, Group VII A or 17

➡️ Z = 47 (Ag): [Kr] 4d¹⁰ 5s¹ →
👉 d-block, Period 5, Group IB or 11

➡️ Z = 53 (I): [Kr] 4d¹⁰ 5s² 5p⁵ →
👉 p-block, Period 5, Group VII A or 17

OR
Define ligands and chelating agents with examples. Write down names and formulae of 5 neutral, 5 negative, 2 bidentate and 2 polydentate ligands.

✍️ Answer

⚛️ Ligands and Chelating Agents

Definition of Ligands

A ligand is an ion or molecule that donates a pair of electrons to a central metal atom/ion to form a coordination complex.
Example: NH₃, Cl⁻, H₂O

Definition of Chelating Agents

Chelating agents are ligands that have multiple donor atoms forming stable ring-like structures (chelate rings) or ring complexes with a central metal ion.
Example: Ethylenediamine (en), EDTA⁴⁻

📌 Examples of 5 Neutral Ligands

Ammine (NH₃), aquo (H₂O), Carbonyl (CO), nitrosyl (NO), en or ethylenediamine (NH₂CH₂CH₂NH₂), dinitrogen (N₂), pyridine (C₆H₅N), phosphine (PH₃)

📌 Examples of 5 Negative (Anionic) Ligands

Halo (X⁻ like Cl⁻), cyano (CN⁻), isocyano (NC⁻), hydroxo (OH⁻), Thiocyanate (SCN⁻), Nitrato (NO₃⁻), thiocyanato (SCN⁻), iso thiocyanato (NCS⁻)

📌 Examples of 2 Bidentate Ligands

Ethylene diamine (H₂N–CH₂–CH₂–NH₂ or en), oxalato (C₂O₄²⁻)

📌 Examples of 2 Polydentate or Multidentate Ligands

Ethylenediaminetetraacetato (EDTA⁴⁻ or [CH₂N(CH₂COO⁻)₂]₂⁴⁻ or C₁₀H₁₆N₂O₈⁴⁻)
Dimethyl glyoxime (DMG⁻ or CH₃C(NOH)C(NO)CH₃⁻)

(vii) Give the general valence shell electronic configuration of the following:

✍️ Answer

➡️ Representative elements ------------- ns¹⁻² to ns² np¹⁻⁶
➡️ Chalcogens ------------------- ns² np⁴ and Halogens ----------- ns² np⁵
➡️ Outer transition elements ---------- (n–1)d¹⁻¹⁰ ns¹⁻² and
Inner transition elements ------- (n–2)f²⁻¹⁴ (n–1)d¹⁻² ns²
➡️ Alkali metals ………… ns¹ and coinage metals ………….. (n–1)d¹⁰ ns¹

(viii) What is Catalytic converter? What are the main pollution targeted by catalytic converter?

✍️ Answer

🚗 Gasoline Combustion & Pollution

Burning gasoline produces:
🚫 Carbon Monoxide (CO)
🌬️ Nitric Oxide (NO)
🛢️ Unburnt Hydrocarbons (HC)
These pollutants harm the environment and living beings.

🌟 Catalytic Converter

A catalytic converter is a device in a car’s exhaust system that uses catalysts like platinum, palladium, rhodium to convert harmful gases into less or non-toxic substances before they are released into the air. (It reduces CO, NO, and HC emissions).

♻️ Role of Catalytic Converter

Convert harmful gases → less harmful/non-harmful gases.
Catalytic converters clean car exhaust by turning CO, HC, and NOₓ into safer gases, protecting air quality and health.

♻️ Main Pollutants Targeted 🌍

🚫 Carbon Monoxide (CO) → converted to CO₂
🌬️ Nitrogen Oxides (NOₓ) → converted to N₂ + O₂
🛢️ Unburnt Hydrocarbons (HC) → converted to CO₂ + H₂O

Quick Memory Tip:

CO, NOₓ, HC → CO₂, N₂, H₂O ✅

OR
How was ozone layer formed? Explain the causes of depletion of ozone layer.

✍️ Answer

🌍 Formation of Ozone Layer ☀️

☀️ UV rays from the Sun split oxygen molecules into individual oxygen atoms (O₂ → O + O).
🔗 Free oxygen atoms (O) combine with O₂ to form ozone (O₂ + O → O₃).
🛡️ Ozone accumulates in the stratosphere, creating a protective layer by absorbing harmful UV radiation 🌞

⚠️ Causes of Ozone Layer Depletion

🧪 Chlorofluorocarbons (CFCs): Found in old refrigerators, ACs, aerosols – release chlorine atoms that destroy ozone.
🌬️ Halons & Other Ozone-Depleting Substances (ODS): Used in fire extinguishers, release bromine that destroy ozone.
🚗 Vehicle emissions: Nitrogen Oxides (NOₓ) from vehicles and power plants contribute to ozone breakdown.
🌡️ Human activities → excessive fossil fuel burning increases pollutants.
☁️ Natural causes: Volcanic eruptions can also release gases that reduce ozone locally.

✨ Bonus Info

🔥 What are Halons (bromine containing fire extinguishing gases) and Ozone-Depleting Substances (ODS)?
Halons are man-made compounds containing carbon, fluorine, and bromine. They were widely used in fire extinguishers because they are very effective at stopping combustion. In the stratosphere, they release bromine atoms that rapidly destroy ozone, contributing to ozone layer depletion.

Ozone-Depleting Substances (ODS) are chemicals that destroy ozone (O₃) in the stratosphere, causing ozone layer depletion 🌍⚠️.
Common ODS:
CFCs (Chlorofluorocarbons) – from refrigerators, ACs, aerosols 🧊💨
Halons – used in fire extinguishers 🔥
Carbon Tetrachloride (CCl₄) – industrial solvent 🏭
Methyl Chloroform (CH₃CCl₃) – cleaning agent 🧴

OR
Define greenhouse gases. How greenhouse gases cause global warming?

✍️ Answer

Definition of Greenhouse Gases (GHGs) 🌍

Greenhouse gases (GHGs) are atmospheric heat-trapping gases (like CO₂, CH₄, N₂O, O₃, and water vapor) that absorb and re-emit infrared radiation (heat), keeping the Earth warm. They act like a “blanket” around Earth, trapping heat in the atmosphere.

Examples of GHGs

CO₂ (Carbon dioxide), CH₄ (Methane), N₂O (Nitrous oxide), H₂O vapor, O₃ (Ozone) 🌿

🌍 Contribution of GHGs to Global Warming

Greenhouse gases cause global warming by absorbing infrared radiation and re-radiating it back to Earth, raising surface temperatures, but too much leads to overheating! 🛏️🌍

☀️ How They Cause Global Warming 🌡️

🌞 Sunlight enters Earth’s atmosphere. ☀️
🌍 Earth’s surface absorbs energy and re-emits it as infrared (heat). 🌍🔥
🛑 Greenhouse gases trap part of this heat, preventing it from escaping into space. 🌫️
📈 Increased GHGs from human activities (burning fossil fuels, deforestation, industry) intensify this effect, raising global temperatures, leading to global warming and climate change. 🌡️

(ix) What is industrial smog and how is it formed?

✍️ Answer

Origin of Term Smog

The term Smog is the combination of smoke and fog.

🏭 Definition of Industrial Smog

A type of air (sulfur based) pollution caused mainly in industrial cities by burning large amounts of coal in industries. It is a mixture of smoke + fog, containing sulfur dioxide, particulate matter, and water droplets, making the air thick and harmful.

💡 Other Names

Classical smog or London smog (because it was common in London in the 19th–20th century)

🏭 Harmful by-products of Industrial Activities

Factories release SO₂ when burning coal and fossil fuels (oil) 🔥
Emit solid particles: smoke, soot, particulates like metal oxides, salts, soil dust 🪨
💧 In humid, cold conditions, SO₂ dissolves in water droplets → forms sulfuric acid (H₂SO₄) mist droplets 💧

⚙️ How is Industrial Smog Formed?

🌫️ Mixing of SO₂ (acidic droplets or fog) + Smoke (soot) + aerosols + volatile organic compounds (VOCs) → dense brown-yellow industrial smog that reduces visibility and harms health.

⚠️ Effects

Human health: Respiratory problems 😷
Plants: Reduced growth 🌱
Environment: Major contributor to air pollution 🌍

OR
Explain four fundamental methods for the testing of waste water.

✍️ Answer

🧪 Four Fundamental Methods for Wastewater Testing

👃 Physical Test

Checks odour, colour, taste and turbidity of water.
Indicates visible pollution (impurities) and basic water quality.

⚗️ Chemical Test

Measures pH, dissolved oxygen (DO), biochemical oxygen demand (BOD), chemical oxygen demand (COD).
Detects toxic chemicals, salts, and harmful compounds.

🦠 Microbiological Test

Detects harmful microorganisms like bacteria and other microbes in wastewater.
Identifies pathogens and assesses ecological impact.

🌿 Organic Test

Checks for pesticides and volatile organic solvents like petrol, benzene, toluene.

OR
What information about the structure of a molecule we can get from mass spectroscopy? Give the applications of mass spectroscopy.

✍️ Answer

🔬 Information from Mass Spectroscopy

➡️Mass spectroscopy provides data about the molecular weight and fragmentation pattern of a molecule.
➡️It helps identify the molecular formula and structural features.
➡️The fragmentation peaks reveal how the molecule breaks down, giving clues about bonding and arrangement.
➡️Isotopic peaks show presence of isotopes like Cl, Br.
➡️It can distinguish between similar compounds by their unique mass spectra.

📌 Applications of Mass Spectroscopy

➡️Used in organic chemistry for structural determination.
➡️Helps in drug design and pharmaceutical analysis.
➡️Applied in environmental testing for pollutants.
➡️Used in forensic science for chemical identification.
➡️Important in biochemistry for protein and peptide analysis.
➡️Widely used in industry for quality control and material testing.

(x) Give the scope of pharmaceutical industries in Pakistan. Write down names of five drugs with their uses.

✍️ Answer

🏭 Scope of Pharmaceutical Industries in Pakistan

Definition: Pharmaceutical industries are companies that manufacture drugs for medical use. 💊

Importance / Scope

➡️ Contribute significantly to the healthcare sector 🏥
➡️ Manufacture medicines for disease control ⚕️
➡️ Invest in research & development (R&D) to discover new drugs, therapies, and treatments 🔬

💊 Classification of Drugs Based on Pharmacological Effect

📌 Aspirin – Analgesic/pain reliever; ➡️ reduces fever, inflammation, and acts as an antithrombotic🔹
📌 Penicillin–Antibiotic;➡️ fights bacterial infections by suppressing growth or killing microorganisms🦠
📌 Paracetamol – Antipyretic;➡️ lowers raised body temperature to reduce fever 🌡️
📌 Fluconazole – Antifungal; ➡️ kills fungi causing skin infections 🦠🧴
📌 Ibuprofen – Anti-inflammatory;➡️ reduces swelling, inflammation, and relieves pain 💪
📌 Diphenhydramine – Anti-allergic / Antihistamine; ➡️ relieves allergies, sneezing, and itching 🤧
📌 Chloroquine – Anti-protozoal / Anti-malarial; ➡️ treats mosquito-borne diseases like malaria 🦟

OR
What is meant by acid rain? Describe its causes and adverse effects on human life style and health. What measures can be taken to prevent acid rain?

✍️ Answer

🌧️ Definition of Acid Rain

Term introduced by: Robert Angus Smith (1872).
Acid rain refers to rainfall or precipitation (rain, snow, fog, dust) with a pH lower than 5.5, caused by the presence of sulfuric acid (H₂SO₄), nitric acid (HNO₃), and carbonic acid (H₂CO₃) in the atmosphere.
It forms when pollutant (SO₂ and NOₓ) gases mix with water droplets in the air.
Normal rain has pH 6–6.5, so acid rain is more acidic 🌡️⚠️

⚠️ Causes of Acid Rain

🏭 Industrial Emissions → Burning coal, oil, and fuel releases SO₂ and NOₓ
🚗 Vehicle Exhausts → Cars, trucks, and airplanes emit NOₓ
🌋 Natural Causes → Volcanic eruptions release SO₂
🌡️ Human activities → excessive fossil fuel use intensifies acid deposition

⚙️ Mechanism of Acid Rain

🏭 Pollutants released: SO₂, NOₓ, CO₂ from industries & vehicles
💧 React with water: Form acids → H₂SO₄, HNO₃, H₂CO₃
🌧️ Falls as rain: Acidic rain with pH < 5.5

💔 Adverse Effects on Human Life & Health 🚑

🌊 Impact on Aquatic Ecosystem: Harms fish & plants 🐟🌱
🌾 Damage to crops & forests: Lowers soil fertility
🏚️ Corrosion of buildings & infrastructure
💧 Pollutes underground water 🚱
👩‍⚕️ Respiratory problems: asthma, bronchitis, lung irritation 😷
🌍 Overall → environmental pollution & ecosystem imbalance

🛡️ Measures to Prevent Acid Rain

⚡ Use cleaner fuels (solar, wind, hydro, natural gas) 🔋🌞
🚗 Promote public transport & electric vehicles 🚊
🏭 Install scrubbers & filters in industries 🏭
🌳 Plant more trees 🌳
📉 Reduce fossil fuel consumption globally
📜 Strict environmental regulations ✅

OR
Describe the preparation and two properties of nylon, terylene and PVC.

✍️ Answer

🧪 Important Synthetic Polymers

Nylon 6,6 🧵

Type: Synthetic polyamide fiber (forms amide linkages −NH−CO−)
Properties: High strength, lightweight, excellent mechanical properties
Uses: Ropes, tents, parachutes, fishing nets, tires ⛺🪢
Nature of Polymer: Condensation polymer
Preparation: Condensation polymerization of hexamethylenediamine + adipic acid with elimination of water 💧
Equation: n H₂N-(CH₂)₆-NH₂ + n HOOC-(CH₂)₄-COOH ⟶ [−NH−(CH₂)₆−NH−CO−(CH₂)₄−CO−]ₙ + 2nH₂O
Chemical name: Poly(hexamethylene adipamide)

Terylene / Dacron (Polyester) 🌸

Type: Synthetic polyester fiber (forms ester linkages −COO−)
Uses: Textiles, packaging, beverage containers 🥤👕
Nature of Polymer: Condensation polymer
Preparation: Condensation polymerization of terephthalic acid + ethylene glycol
Equation: n HO-CH₂CH₂-OH + n HOOC-C₆H₄-COOH ⟶ [−O−CH₂CH₂−O−CO−C₆H₄−CO−]ₙ + 2nH₂O
Chemical name: Polyethylene Terephthalate (PET)

Polyvinyl Chloride (PVC) 🟦

Type: Thermoplastic polymer
Properties: Lightweight, durable, good electric insulation, low cost
Uses: Pipes, bottles, medical tubes, wire insulation ⚡🏥
Nature of Polymer: Addition polymer
Preparation: By addition polymerization of vinyl chloride at 60–70°C using H₂O₂ as initiator
Equation: nCH₂=CHCl —H₂O₂, 60–70°C → [−CH₂−CHCl−]ₙ

(xi) Write down complete balanced action of following reactions:

✍️ Answer

➡️ Magnesium is heated with nitrogen gas

Mg + N₂ → Mg₃N₂

➡️ Potassium is put into ethyl alcohol

2K + 2C₂H₅OH → 2C₂H₅OK + H₂

➡️ Chlorine reacts with nitrogen

N₂(g) + 3Cl₂(g) → 2NCl₃(g)

➡️ Fluorine reacts with oxygen

2F₂ + O₂ → 2OF₂

➡️ Aluminium reacts with water

2Al + 6H₂O → 2Al(OH)₃ + 3H₂

➡️ Silicon reacts with steam

Si + 2H₂O → SiO₂ + 2H₂

➡️ Bleaching powder is dissolved in water

CaOCl₂ + H₂O → Ca(OH)₂ + Cl₂

➡️ Phosphorus reacts vigorously with water

2P₄ + 12H₂O → 3H₃PO₄ + 5PH₃↑

➡️ Reaction of chromium with steam

2Cr + 3H₂O → Cr₂O₃ + 3H₂

➡️ Reaction of dichromate with ferrous salt

6Fe²⁺ + Cr₂O₇²⁻ + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O

➡️ Reaction of manganese with air

3Mn + 2O₂ → Mn₃O₄

➡️ Sulphur reacts at high temperature with water

S + 2H₂O → SO₂ + 2H₂

➡️ Silicon is heated with nitrogen at high temperatures

3Si + 2N₂ → Si₃N₄

➡️ Phosphorus reacts with nitrogen at high temperature

6P + 2N₂ → 2P₃N₅

➡️ Nitrogen reacts with oxygen in the presence of catalyst

2N₂ + O₂ → 2N₂O

➡️ A piece of aluminium is dropped into concentrated sulphuric acid

2Al + 6H₂SO₄ → Al₂(SO₄)₃ + 3SO₂ + 6H₂O

➡️ Ferric chloride is mixed in an aqueous solution of caustic soda

FeCl₃ + 3NaOH → Fe(OH)₃ + 3NaCl

➡️ Chlorine gas is passed through an aqueous solution of caustic soda

Cl₂ + 2NaOH → NaCl + NaClO + H₂O

➡️ Reaction of conc. nitric acid with copper

Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O

➡️ Reaction of conc. sulphuric acid with copper

Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O

➡️ Reaction of permanganate with oxalic acid

2KMnO₄ + 5H₂C₂O₄ + 3H₂SO₄ → 2MnSO₄ + 10CO₂ + 8H₂O + K₂SO₄

➡️ Reaction of manganese with dilute sulphuric acid

Mn + H₂SO₄ → MnSO₄ + H₂

(xii) Complete and balance the following chemical equations:

✍️ Answer

➡️ 2Li₍ₛ₎ + H₂(g) → 2LiH
➡️ 2Na₍ₛ₎ + Cl₂(g) → 2NaCl(s)
➡️ 6Na₍ₛ₎ + N₂(g) → 2Na₃N₍ₛ₎
➡️ 3Ca₍ₛ₎ + N₂(g) → Ca₃N₂₍ₛ₎
➡️ 6Li₍ₛ₎ + N₂(g) → 2Li₃N₍ₛ₎
➡️ 4Li₍ₛ₎ + O₂(g) → 2Li₂O₍ₛ₎
➡️ 4Na₍ₛ₎ + O₂(g) → 2Na₂O₍ₛ₎
➡️ 2Na₍ₛ₎ + O₂(g) —Excess O₂→ Na₂O₂₍ₛ₎
➡️ K₍ₛ₎ + O₂(g) → KO₂₍ₛ₎
➡️ Rb₍ₛ₎ + O₂(g) → RbO₂₍ₛ₎
➡️ 2Be₍ₛ₎ + O₂(g) → 2BeO₍ₛ₎
➡️ Sr₍ₛ₎ + O₂(g) → SrO₂₍ₛ₎
➡️ Si₍ₛ₎ + 2H₂O₍ₗ₎ → SiO₂₍ₛ₎ + 2H₂↑
➡️ 3Si₍ₛ₎ + 2N₂(g) → Si₃N₄₍ₛ₎
➡️ Si₍ₛ₎ + 2H₂(g) → SiH₄(g)
➡️ Si₍ₛ₎ + 2H₂O₍ₗ₎ → SiO₂(aq) + 2H₂↑
➡️ 4P(g) + 5O₂(g) —Excess O₂→ 2P₂O₅(g)
➡️ 4P(g) + 3O₂(g) —Limited O₂→ 2P₂O₃(g)
➡️ 2N₂(g) + O₂(g) —Catalyst→ 2N₂O(g)
➡️ 2Na + 2C₂H₅OH(aq) → 2C₂H₅ONa(aq) + H₂(g)↑
➡️ 6OH⁻ + 3Cl₂ —Excess O₂→ 5Cl⁻ + ClO₃⁻ + 3H₂O
➡️ 6NaOH + 3Cl₂ —Excess O₂→ 5NaCl + NaClO₃ + 3H₂O
➡️ CaOCl₂ + 2H₂O₍ₗ₎ → Ca(OH)₂ + 2HOCl
➡️ 2K₍ₛ₎ + 2HCl(aq) + H₂O₍ₗ₎ → 2KCl(aq) + H₂(g)↑
➡️ 4B₍ₛ₎ + 3O₂(g) → 2B₂O₃₍ₛ₎
➡️ N₂(g) + O₂(g) → 2NO(g)
➡️ N₂(g) + 3Cl₂(g) → 2NCl₃(g)
➡️ 2P₄₍ₛ₎ + 12H₂O₍ₗ₎ → 3H₃PO₄(aq) + 5PH₃↑
➡️ 6P₍ₛ₎ + 5N₂(g) → 2P₃N₅₍ₛ₎
➡️ 2F₂(g) + O₂(g) → 2OF₂(g)
➡️ 2Al₍ₛ₎ + 6H₂O₍ₗ₎ → 2Al(OH)₃₍ₛ₎ + 3H₂↑
➡️ 2Al₍ₛ₎ + N₂(g) → 2AlN₍ₛ₎
➡️ Zn₍ₛ₎ + 2NaOH(g) → Na₂ZnO₍aq₎ + H₂↑
➡️ 2Na/Hg₍ₗ₎ + 2H₂O₍ₗ₎ → 2NaOH₍ₐq₎ + H₂↑ + 2Hg₍ₗ₎

(xiii) Give reasons of any 4 of the following:
➡️ Ionization energy decreases from top to bottom in s-block elements.
➡️ H₂O and NH₃ act as ligands but H₃O⁺ and NH₄⁺ do not.
➡️ Fluorine is the strongest oxidizing agent and lithium is the strongest reducing agent.
➡️ Alkali metals are good conductor of electricity.
➡️ Melting and boiling points of zinc are exceptionally low.
➡️ Multidentate ligands are known as chelating agents.
➡️ The transition metals complexes are coloured.
➡️ Mn²⁺ show maximum paramagnetic character amongst the bivalent ions of first transition series.
➡️ Melting point of d-block elements increase up to middle of the series and then decrease.
➡️ Configuration of Cr is 3d⁵ 4s¹ instead of 3d⁴ 4s² & Configuration of Cu is 3d¹⁰ 4s¹ instead of 3d⁹ 4s².
➡️ Why Cu²⁺ ions is blue but Zn²⁺ is colourless.
➡️ Acidity of hydrogen halide increases from HF to HI.
➡️ Gallium has smaller atomic radii than aluminum despite being below the aluminum in group IIIA.
➡️ Electronegativity decreases regularly from top to bottom in s-block elements.
➡️ Boiling point of halogens increases down the group in the periodic table.
➡️ Transition elements have ability to form alloys.
➡️ Why are the binding energy of Mn²⁺ and Fe³⁺ ions the highest and that of zinc is least in 3d series?
➡️ Why do transition elements show variable oxidation states?
➡️ Paramagnetic behaviour is the strongest for Fe³⁺ and Mn²⁺.
➡️ The maximum oxidation state increases up to middle of series & then decreases afterward.
➡️ How the given reaction is avoided during the preparation of NaOH? (6OH⁻ + Cl₂ ➔ Cl⁻ + ClO₃⁻ + 3H₂O).
➡️ Ligands are generally called Lewis bases.

✍️ Answer

➡️ Ionization energy decreases down s block ⬇️

💡 Atomic size increases → weaker nuclear pull → easier electron removal ⚡

➡️ H₂O & NH₃ act as ligands, H₃O⁺ & NH₄⁺ do not

💡 Lone pairs available in H₂O/NH₃; in H₃O⁺/NH₄⁺ lone pairs are used in bonding (NH₄⁺ has no lone pair).

➡️ Fluorine strongest oxidizer, Lithium strongest reducer

💡 F is highly electronegative → gains ē easily → strongest oxidizing agent ⚡; Li loses ē easily → strongest reducing agent 🔥

➡️ Alkali metals conduct electricity ⚡

One free valence electron → mobile ē → easy current flow 🔋

➡️ Zn has low melting/boiling points 🧊

Weak metallic bonding due to filled 3d¹⁰ shell → less delocalized electrons → low mp & bp 🔥

➡️ Multidentate ligands = Chelating agents 🧲

They form stable ring-like complexes by binding at multiple donor sites 🧩

➡️ Transition metal complexes are coloured 🌈

d-orbitals split in ligand field → absorb visible light → d–d transitions → colour 🎨

➡️ Mn²⁺ shows max paramagnetism 🧲

Has 5 unpaired d electrons → maximum magnetic moment

➡️ Melting points rise mid-series then fall in d block ⬆️⬇️

More unpaired d-electrons up to middle → strong bonding; weaker after due to filled orbitals ⚖️

➡️ Cr & Cu have special configurations 🧬

Extra stability from half-filled (3d⁵) and fully filled (3d¹⁰) orbitals 🔑

➡️ Cu²⁺ is blue, Zn²⁺ is colourless 🎨

Cu²⁺ has partially filled d-orbitals → d–d transitions → coloured; Zn²⁺ has d¹⁰ → no transitions

➡️ Acidity of HX increases HF → HI ⚡

H–X bond strength decreases down group ⬇️ → easier H⁺ release 💧

➡️ Gallium smaller than Al 🟢

d-electron contraction → ineffective shielding → smaller atomic size 📉

➡️ Electronegativity decreases down s block ⬇️

Larger size → outer electron farther → weaker nuclear attraction ⬇️

➡️ Boiling point of halogens increases down group 🧊

Larger size → stronger van der Waals forces → higher boiling point 🌡️

➡️ Transition metals form alloys ⚙️

Similar atomic sizes → atoms mix easily in lattices

➡️ Binding energy Mn²⁺ & Fe³⁺ highest, Zn least 🔋

Stable half-filled d-orbitals → stronger lattice energy; Zn²⁺ has d¹⁰ → least

➡️ Transition metals show variable oxidation states 🔄

Close energy of (n–1)d and ns orbitals → multiple oxidation states possible

➡️ Paramagnetism strongest in Fe³⁺ & Mn²⁺ 🧲

Maximum unpaired d electrons → strong magnetism

➡️ Max oxidation state rises then falls in series 🔼⬇️

More unpaired electrons mid-series; fewer later → decline 📈

➡️ Avoid reaction in NaOH prep (6OH⁻ + Cl₂ …) ⚠️

Controlled conditions prevent disproportionation of Cl₂ 🛡️

➡️ Ligands = Lewis bases 🧪

💡 They donate electron pairs to metal ions → coordinate bond formation

✏️ Smart Answers of Short-Answer Questions of ⚗️ Organic Chemistry Section 🧪 ✏️

(xiv) Define any FOUR of the following:
➡️ Catenation
➡️ Reforming
➡️ Homologous series
➡️ Functional group
➡️ Chiral carbon
➡️ Enantiomers
➡️ Knocking
➡️ Octane number
➡️ Polymerization (polymers)
➡️ Isomerism (isomers)
➡️ Metamerism
➡️ Knock-inhibitor
➡️ Carbonization
➡️ Heterocyclics
➡️ Saponification
➡️ Glycosidic linkage
➡️ Peptide bond
➡️ Zwitterion
➡️ Electrophile
➡️ Nucleophile
➡️ Hückel rule
➡️ Aromaticity

✍️ Answer

1. Catenation 🔗

Ability of carbon to bond with itself forming chains, branched chains, rings, or combined chain-ring.

2. Reforming 🔄

Conversion of straight-chain alkanes into branched or aromatic hydrocarbons to increase octane number and prevent knocking, thereby improving fuel quality.

3. Homologous Series 📚

Group of organic compounds with the same functional group and similar chemical properties, arranged by increasing molecular mass; successive members differ by –CH₂– (14 amu).

4. Functional Group ⚛️

Specific atom/group or multiple bond in an organic molecule responsible for its chemical behavior and characteristic properties.

5. Polymerization / Polymers 🧵

Reaction where small molecules (monomers) join to form large macromolecules (polymers) with repeating units; may be linear, branched, or 3D.

6. Isomerism 🔁

Existence of two or more compounds possessing the same molecular formula but different properties. Such compounds are isomers.

7. Metamerism ⚖️

Type of structural isomerism where a polyvalent functional group (like –O–, –NH–) is attached to different alkyl or aryl groups on each side; such compounds are metamers.

8. Knocking 💥

Sharp metallic sound in engines caused by premature fuel ignition (abnormal combustion) before spark.

9. Octane Number ⛽

Measure of fuel’s ability to resist knocking; expressed as % of isooctane in an isooctane-heptane mixture with same knocking tendency.

10. Knock-Inhibitor 🛑

Substances like tetraethyl lead (TEL) added to fuel that reduce engine knocking by increasing octane number.

11. Carbonization 🔥

Thermal decomposition of carbon-rich material (like coal) in absence of air, producing solids, liquids, and gases.

12. Heterocyclics 🔄⭕

Cyclic compounds containing one or more atoms (hetero atoms) other than carbon (N, O, S) in the ring. Examples: Pyrrole, Furan, Thiophene, Pyridine.

13. Chiral Carbon 🖐️

Carbon atom attached to four different atoms or groups, leading to handedness.

14. Enantiomers 🪞

Pair of molecules that are mirror images of each other but cannot be superimposed.

15. Saponification 🧼

Hydrolysis of fats or oils with alkali to form soap and glycerol.

16. Glycosidic Linkage 🍬

Covalent bond that joins two monosaccharides in carbohydrates.

17. Peptide Bond 🧬

Amide bond (–CO–NH–) formed between two amino acids.

18. Zwitterion ⚡

Molecule carrying both positive and negative charges but overall electrically neutral (e.g., amino acids in water).

19. Electrophile ➕

Electron-deficient species that accepts an electron pair.

20. Nucleophile ➖

Electron-rich species that donates an electron pair.

21. Hückel Rule 🔢

A planar ring molecule is aromatic if it has (4n + 2) π electrons (n = 0,1,2…).

22. Aromaticity 🌸

Property of cyclic, planar, conjugated molecules with delocalized π electrons that follow Hückel rule and show extra stability.

(xiv, OR) Write down Two differences between the following:

✍️ Answer

⚛️ (a) Reducing vs Non-Reducing Sugars 🍬

Reducing Sugars 🧪	Non-Reducing Sugars 🚫
Have free aldehyde/ketone group 🔓	No free aldehyde/ketone group 🔒
Reduce Fehling’s/Tollen’s reagent ✅	Do not reduce these reagents ❌
Can form glycosidic bonds easily	Already bonded, no free group
Act as reducing agents	Cannot act as reducing agents
Examples: glucose, fructose, maltose 🍯	Example: sucrose 🍭

⚛️ (b) Aliphatic vs Aromatic Compounds 🧴🌸

Aliphatic Compounds 🧴	Aromatic Compounds 🌸
Open-chain (straight or branched) 🔗	Contain benzene rings ⭕
No bond between first & last carbon 🚫	Closed ring structure 🔄
May be saturated/unsaturated 🔁	Always unsaturated with π system 🔁
Burn with non-sooty flame 🔥	Burn with sooty flame 🖤
Lower C:H ratio ⚖️	Higher C:H ratio ⚖️
No delocalized π electrons ❌	Have delocalized π electrons ✅
No characteristic smell 👃	Pleasant aromatic smell 🌺
Generally more reactive ⚡	Comparatively stable 😌
No fixed bond pattern 🔀	Alternate single & double bonds 🔄
Less stable ❗	Extra stable due to resonance ⭐
Undergo addition & substitution 🔁	Mainly substitution 🔄
Examples: alkanes, alkenes 🧪	Examples: benzene, naphthalene 🧫

⚛️ (c) Saturated 🟢 vs Unsaturated 🔵 Hydrocarbons

Saturated Hydrocarbons 🟢	Unsaturated Hydrocarbons 🔵
All valencies satisfied by single bonds 🔗	Contain double/triple bonds ⚡
Only C–C single bonds 🧪	At least one C=C or C≡C 🔥
No π bonds ❌	Have π bonds ✅
Carbon atoms sp³ hybridized 🧬	Some carbons sp²/sp hybridized 🧬
More stable ⭐	Less stable ❗
Less reactive 😌	More reactive ⚡
Substitution reactions 🔄	Addition reactions ➕
Higher hydrogen content 🧪	Lower hydrogen content ⚖️
No polymerization 🚫	Readily polymerize 🧵
Burn with blue flame 🔵🔥	Burn with yellow sooty flame 🟡🔥
Examples: alkanes, cycloalkanes 🧴	Examples: alkenes, alkynes 🌿
Eg: Methane (CH₄), Ethane (C₂H₆)	Eg: Ethene (C₂H₄), Ethyne (C₂H₂)

⚛️ (d) Total vs Partial Synthesis 🏭

Total Synthesis 🏗️	Partial Synthesis 🛠️
Made entirely in lab from simple raw materials ⚗️	Prepared by modifying a natural substance 🌱
Complex molecule from cheap starting materials 🧪	Complex molecule from natural compound 🌿
No natural starting material 🚫🌿	Natural product used ✅🌿
Used for complex molecules 🧬	Easier and cheaper 💰
Starts from simple molecules 🔬	Starts from complex natural molecule 🧬
Time-consuming ⏳	Relatively faster ⏱️
Useful in new drug discovery 💊✨	Useful in modifying existing drugs 💊🔄
Many steps 📚	Fewer steps ⚡
Generally expensive 💰	Cheaper 💵
Used when natural source unavailable 🚫🌍	Used when total synthesis is too difficult 😓
Careful planning of pathway 🧠🧪	Chemical modification of existing compound 🔧
Examples: Total synthesis of paclitaxel (Taxol) 💊, lab synthesis of urea 🧪	Examples: Semi-synthetic penicillin 💉, ibuprofen from cumene 💊

(xiv, OR) Define Bucky Ball. Explain its structure and mention its some properties and uses.

✍️ Answer

✍️ Definition of Bucky Ball 🏀

Bucky Ball (Fullerene or Buckminsterfullerene, C₆₀) is an allotrope of carbon made up of 60 carbon atoms arranged in a hollow spherical cage structure called fullerene (C₆₀).
It was discovered in 1985 by Richard Smalley and Harry Kroto.

🏗️ Structure

➡️ Molecular formula: C₆₀ (60 carbon atoms) 🧪
➡️ Shape: Resembles a geodesic dome (soccer ball) designed by Richard Buckminster Fuller 🏟️
➡️ Other Name: Fullerene ⚛️
➡️ Composition: 12 pentagons 🔷 + 20 hexagons 🔶
➡️ Arrangement: Each pentagon surrounded by hexagons 🔄
➡️ Hybridization: Each carbon bonded to three neighbors (sp² hybridized) 🧬
➡️ Final Structure: Hollow, cage-like spherical structure 🎈

✨ Properties of Bucky Ball (C₆₀)

➡️ Hollow spherical structure → diameter ≈ 1 nm 📏
➡️ Large surface area → high reactivity 🧪
➡️ High symmetry → exceptional stability ⭐
➡️ High melting point (~2800°C) → thermally stable 🔥
➡️ Lightweight structure → useful in nanomaterials ⚖️
➡️ Delocalized π electrons → unique electronic properties ⚡🧬
➡️ Strong C–C bonds → high tensile strength 💪
➡️ sp² hybridized carbon network → good conductivity ⚡
➡️ Soluble in organic solvents (benzene, toluene) 🧪🟣
➡️ Acts as strong electron acceptor → useful in electronics ⚡🔋

🛠️ Uses of Bucky Ball (C₆₀)

➡️ Nanotechnology → building blocks for advanced materials 🧫
➡️ Medicine → drug delivery, antiviral agents, MRI contrast 💊
➡️ Electronics → organic solar cells, superconductors ⚡
➡️ Lubricants → reduce friction at nanoscale 🛢️
➡️ Research → model for carbon structures & quantum effects 🔬

(xv) We often use the term iso and neo in the common system of naming of alkanes. Explain with examples.

✍️ Answer

🌿 Branched Alkanes (Iso and Neo) & Common Prefixes

In the common system of nomenclature, the prefixes iso and neo are used to describe specific branching patterns in alkanes.
Certain branched alkanes are named using prefixes to distinguish them from straight-chain (linear) alkanes.
Common prefixes include: iso-, neo-, sec-, tert-.
Examples: n-pentane, isopentane, neopentane 🧪

📛 Iso-Alkanes / 2-Methyl Alkanes 🔹

Definition: Iso means the molecule has a branch at the second carbon atom of the chain.
These are alkanes where one methyl group (–CH₃) is attached to the penultimate carbon (second-last carbon).
Prefix: iso- (added before parent chain).
Chain ending: –CH(CH₃)₂ group.
Key Point: Used for 2-methyl alkanes.

Examples:
Iso-butane (C₄H₁₀) → CH₃–CH(CH₃)–CH₃ → 2-methylpropane 🧪
Iso-pentane (C₅H₁₂) → CH₃–CH(CH₃)–CH₂–CH₃ → 2-methylbutane 🌿

🌸 Neo-Alkanes / 2,2-Dimethyl Alkanes 🔹

Definition: Neo means the molecule has two branches at the second carbon atom of the chain.
These are alkanes where two methyl groups (–CH₃) are attached to the penultimate carbon.
Prefix: neo- (added before parent chain).
Chain ending: –C(CH₃)₃ group.
Key Point: Penultimate carbon is trisubstituted (3 methyl groups).

Examples:
Neo-pentane (C₅H₁₂) → C(CH₃)₄ → 2,2-dimethylpropane 🧪
Neo-hexane (C₆H₁₄) → C(CH₃)₃–CH₂–CH₃ → 2,2-dimethylbutane 🌿

(xv, OR) If an organic compound contains both double and triple bond in the main carbon chain, what rules you follow to write its IUPAC names. Explain by giving an example.

✍️ Answer

🧬 Definition of Alkenyne

A carbon chain containing both double (C=C) and triple (C≡C) bonds is called an alkenyne.
Name is derived by replacing –ane of the parent alkane with –en–yne ⚛️
❌ Avoid the term “alkeneyne” used in some books.

🧪 IUPAC Naming Rules for Alkenynes ⚡

When a compound contains both double and triple bonds:

📖 Summarized Rules 📌
✨ If double & triple bonds equidistant → double bond gets priority 🔢
✨ If bonds at different positions → start numbering nearest to first multiple bond 📏

📖 Detailed Rules 📌
➡️ Longest chain rule → must include both bonds 🧬
➡️ Lowest locant rule → assign lowest possible numbers 🔢
➡️ Priority → double bond (–ene) gets preference ⚡
➡️ Suffix order → name ends with –en–yne (double bond first)
➡️ Indicate positions of both bonds with locants
➡️ Substituents listed alphabetically 📝

📌 Examples

✨ CH₂=CH–C≡CH → But-1-en-3-yne ✅
✨ CH₂=CH–C≡C–CH₃ → Pent-1-en-3-yne ✅
✨ CH₃–CH=CH–C≡CH → Pent-2-en-4-yne ✅

(xvi) How is coal produced under the earth crust? Write the name of four types of coal and mention the %age of carbon content in them. Explain destructive distillation of coal and various products obtained from it.

✍️ Answer

🪨 Coal

Coal is the most important solid fossil fuel and it is a plant-derived black coloured mineral found beneath the earth’s crust.

🔨 Formation of Coal 🌑

Coal is formed under the Earth’s crust anaerobic decay of plant remains buried in swamps and sediments over millions of years 🌱⏳.
Heat and pressure and bacterial and chemical processes gradually dehydrate and carbonize the plant material, forming coal. This gradual process is called coalification 🔥🌍.

🪨 Coal Types

➡️ Peat 🌿 → Carbon content: 45–60%; earliest stage, soft, light, high moisture, burns with smoke 🔥
➡️ Lignite (Brown Coal) 🟤 → Carbon content: 60–70%; brownish black, soft
➡️ Bituminous Coal ⚫ → Carbon content: 75–85%; black colored soft coal
➡️ Anthracite ⚫✨ → Carbon content: 90–95%; dark black coloured hard and driest coal, burns without smoke

🔥 Carbonization / Destructive Distillation of Coal

Definition: Thermal decomposition of coal in the absence of air to produce solids, liquids, and gases ⚗️.
Process: Coal is heated (400–900°C) in a closed vessel (iron retort) → breaks down into different components.

🛠️ Four Principal Products of Destructive Distillation 🔥

➡️ Coke 🔥 → Grayish black hard, nearly pure carbon (98–99%), used as a fuel and in metallurgy (reducing agent).
➡️ Coal Gas 💨 → Mixture of H₂ (50–53%), CH₄ (26–35%), CO (7–8%), N₂ (11%) and heavier gases (3%); used as fuel and for lighting.
➡️ Coal Tar 🏺 → Thick dark liquid, hidden treasure aromatics (~215 compounds); used in dyes, chemicals, medicines, road tar.
➡️ Ammoniacal Liquor 🧪 → Contains ammonium compounds and liquid ammonia (usually obtained from bituminous coal), used to prepare fertilizers.

(xvi) OR
Define homologous series and write its three general properties.

✍️ Answer

📚 Definition of Homologous Series

A homologous series is a group of organic compounds that have the same functional group and similar chemical properties, where successive members differ by –CH₂– (methylene group) or 14 amu in molecular mass 🔗.
Individual members are called homologues and the phenomenon is called homology.

Examples:
➡️ Hydrocarbons: Alkanes, Alkenes, Alkynes 🧪
➡️ Halogen derivatives: Alkyl halides 🌿
➡️ Alcohols, Aldehydes, Ketones, Carboxylic acids, Esters, Amines, etc.

✨ General Properties of Homologous Series

⚛️ Identical structure & composition – All members have similar structure. Successive members differ by –CH₂– (14 amu).

⚛️ Similar chemical properties – All members of a series react in similar ways due to the same functional group.
e.g., All alkanes are largely unreactive but undergo combustion and halogenation.

⚛️ Same general formula 🧪 – Members of each series follow a common molecular formula 🧮
e.g.: Alkanes → CₙH₂ₙ₊₂, Alkenes → CₙH₂ₙ, Alkynes → CₙH₂ₙ₋₂, Alcohols → CₙH₂ₙ₊₁OH.

⚛️ Common or general method of preparation 🛠️ – Members of a series can often be synthesized using similar chemical reactions.
e.g., Alkanes via reduction of alkyl halides.

⚛️ Gradation in physical properties 🌡️ – Boiling point, melting point, density change gradually with molecular mass.
e.g., In alkanes, m.p., b.p., and density increase as chain length increases.

(xvii) Draw the orbital structure of ethane or ethene and ethyne and explain how ethyne is distinguished from ethene by a simple chemical test.

✍️ Answer

⚛️ Orbital structure of ethane or ethene and ethyne

(xvii)OR
What is free radical? Give stepwise mechanism for the chlorination of methane.

✍️ Answer

⚛️ Distinguishing Ethyne from Ethene 🔬

⚗️ Test: Reaction with Ammoniacal Silver Nitrate / Ammoniacal Silver Nitrate Test
🧪 Reagent: Ammoniacal silver nitrate solution (AgNO₃ + NH₃, prepared by adding ammonia to silver nitrate)

✅ Observation
➡️ Ethyne (C₂H₂) → Forms a white precipitate
➡️ Ethene (C₂H₄) → No precipitate

🧾 Reason
➡️ Ethyne is a terminal alkyne and has an acidic hydrogen atom (–C≡C–H).
➡️ It reacts with ammoniacal silver nitrate where silver replaces acidic hydrogen atom of ethyne to form white precipitate of silver acetylide.
➡️ Ethene (an alkene) does not have acidic hydrogen, so it does not react and no precipitate formed.

⚖️ Chemical Equation
C₂H₂ + 2[Ag(NH₃)₂]OH → Ag₂C₂↓ + 4NH₃ + 2H₂O

⚛️ Free Radical

A free radical is a neutral atom or group of atoms that contains an unpaired electron.
Because of the unpaired electron, free radicals are highly reactive and usually formed by homolytic bond cleavage (breaking of a covalent bond so that each atom takes one electron).
📌 Example: ➡️ Cl₂ → 2Cl· (each chlorine atom has one unpaired electron)

🔥 Mechanism of Halogenation (Free Radical Substitution)

✦ Halogenation of methane (reaction of methane with chlorine or bromine) in presence of UV light / high temperature is a free radical substitution chain reaction.
✦ It proceeds through homolytic fission and occurs in three main steps:

➡️ Chain Initiation Step (Photochemical/Thermal Homolysis of Cl₂ into chlorine free radicals, Cl•)
✦ Chlorine molecule breaks under UV light or heat (endothermic step).
✦ Homolytic cleavage of Cl–Cl bond forms two chlorine free radicals.
➡️ Cl₂ —hv/Δ→ Cl• + Cl• ------ ΔH = +242 kJ/mol

➡️ Chain Propagation Step (Radical reacts and regenerates another radical)
✦ Chlorine radical reacts with methane to form methyl radical.
✦ Methyl radical reacts with another chlorine molecule.
✦ Product is formed and new chlorine radical is regenerated.
➡️ Cl• + CH₄ → HCl + CH₃•
➡️ CH₃• + Cl₂ → CH₃Cl + Cl• (The regenerated Cl• continues the chain)

➡️ Chain Termination Step (Radicals combine and stop the reaction)
✦ Two free radicals combine to form stable molecules in dark terminating the chain reaction.
➡️ Cl• + Cl• → Cl–Cl
➡️ CH₃• + Cl• → CH₃–Cl
➡️ CH₃• + CH₃• → CH₃–CH₃

(xviii) Why benzene show stability towards addition reaction? Why benzene gives electrophilic substitution reaction? Write the chemical equations for the Friedel-Craft reaction and Sulphonation. Also write stepwise the mechanism of nitration or acylation of benzene.

✍️ Answer

✨ Reason of Stability towards Addition

Benzene is highly stable due to delocalized 6π electrons (aromaticity) with high resonance stabilization energy; addition would destroy this aromatic stability (break π-electron cloud) and requires energy to proceed. Therefore, benzene resists addition reactions.

⚡ Reason of Electrophilic Substitution

Benzene is electron-rich acting as a nucleophile and attracts electrophiles (E⁺), and substitution restores aromaticity after reaction. Thus, benzene prefers electrophilic substitution rather than addition.

📌 1. Friedel–Crafts Alkylation Reaction

General Reaction: C₆H₆ + RCl —AlCl₃→ C₆H₅R + HCl (Catalyst: Anhydrous AlCl₃)
Example: C₆H₆ + CH₃Cl —AlCl₃→ C₆H₅CH₃ + HCl
Product: Alkylbenzene like toleune etc.

📌 Friedel–Crafts Acylation Reaction

General Reaction: C₆H₆ + RCOCl —AlCl₃→ C₆H₅COR + HCl (Catalyst: Anhydrous AlCl₃)
Example: C₆H₆ + CH₃COCl —AlCl₃→ C₆H₅COCH₃ + HCl
Product: Acylbenzene like acetophenone etc.

📌 2. Sulphonation of Benzene

Reaction: C₆H₆ + H₂SO₄(conc.) → C₆H₅SO₃H + H₂O
Product: Benzene sulphonic acid
Electrophile: SO₃

🔥 Stepwise Mechanism of Nitration of Benzene

Step 1: Generation of Electrophile (Nitronium ion)

Reagents: Conc. HNO₃ + Conc. H₂SO₄
Product: Nitrobenzene
Electrophile formed: NO₂⁺ (nitronium ion)

HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O

Step 2: Formation of Sigma Complex (Arenium ion)

➡️ C₆H₆ + NO₂⁺ → [C₆H₆NO₂]⁺ (Sigma complex / Carbocation intermediate)

Step 3: Deprotonation and Restoration of Aromaticity

➡️ [C₆H₆NO₂]⁺ + HSO₄⁻ → H₂SO₄ + C₆H₅NO₂ (Nitrobenzene)

🔥 Stepwise Mechanism of Friedel–Crafts Acylation (Alternative Question)

Step 1: Generation of Electrophile (Acylium Ion)

➡️ RCOCl + AlCl₃ → AlCl₄⁻ + RCO⁺ (Electrophile: Acylium ion)

Step 2: Formation of Sigma Complex (Arenium ion)

➡️ C₆H₆ + RCO⁺ → C₆H₆⁺COR (Sigma complex / Carbocation intermediate)

Step 3: Deprotonation and Restoration of Aromaticity

➡️ C₆H₆⁺COR + AlCl₄⁻ → C₆H₅COR + HCl + AlCl₃

(xix, OR) Identify each of the following with one laboratory test:
➡️ Alcohol ➡️ Phenol ➡️ Alkene ➡️ Aldehyde

✍️ Answer

🧪 Alcohol (With Sodium metal) 🔥

👀 Observation: 👉 Brisk effervescence of H₂ gas → Alcohol present
Reaction: 2ROH + 2Na → 2RONa + H₂↑ (Hydrogen gas evolved)

🌿 Phenol (With Neutral FeCl₃ solution) 💜

👀 Observation: 👉 Violet/purple colour appears → Phenol present
Reaction: 3 C₆H₅OH + FeCl₃ → 3HCl + Fe(O–C₆H₅)₃ (Violet complex formed)

⚡ Alkene (With Bromine Water) 🟠

👀 Observation: 👉 Orange colour decolourises → Alkene present
Reaction: CH₂=CH₂ + Br₂ → CH₂Br–CH₂Br (Bromine decolourised)

🍎 Aldehyde (Tollens’ Test using Tollens’ reagent i.e. ammoniacal silver nitrate) 🪞

👀 Observation: 👉 Silver mirror formed → Aldehyde present ✔️
Reaction: RCHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → RCOO⁻ + 4NH₃ + 2H₂O + 2Ag↓ (Silver mirror formed)

(xix) What is meant by stereo isomerism, chiral carbon and plane polarized light? Define optical and cis and trans isomers with examples. Explain optical isomerism briefly.

OR
What is meant by isomerism? Explain four different types of structural isomers and two types of stereoisomerism and give one example of each. Draw all possible isomers of pentyl alcohol (or pentyl chloride), pentene, butyne & compound with formula C₃H₆O and C₂H₄O₂.

✍️ Answer

🧬 Stereo Isomerism

Stereo isomerism is a type of isomerism in which compounds have the same molecular formula and structural formula but differ in the spatial arrangement of atoms.

🔹 Chiral Carbon

✦ A chiral carbon is a carbon atom attached to four different atoms or groups.
✦ Example: 2-butanol (CH₃–CHOH–CH₂–CH₃) — the second carbon is chiral.

🌈 Plane Polarized Light

Plane polarized light is light in which vibrations occur in only one plane.

👁️ Optical Isomers

Optical isomers are stereoisomers that are non-superimposable mirror images of each other and rotate plane polarized light.
Example: Lactic acid (CH₃–CHOH–COOH)
✦ One form rotates light to right (+) → Dextrorotatory
✦ Other rotates to left (–) → Levorotatory

🔁 Cis–Trans Isomers (Geometrical Isomers)

These arise due to restricted rotation around a double bond.
Example: But-2-ene
✦ Cis-but-2-ene: Similar groups on same side
✦ Trans-but-2-ene: Similar groups on opposite sides

✨ Optical Isomerism (Brief Explanation)

✦ Optical isomerism occurs due to the presence of a chiral carbon atom.
✦ The two mirror-image forms (enantiomers):
✦ Have identical physical properties (except optical rotation)
✦ Rotate plane polarized light in opposite directions
✦ Cannot be superimposed on each other
✔️ Hence, optical isomerism is a type of stereoisomerism caused by molecular asymmetry.

🧬 Isomerism

Isomerism is the phenomenon where compounds have the same molecular formula but different structures or arrangements of atoms, leading to different properties.

🧩 Structural Isomerism

➡️ Chain isomerism: Different carbon chain. ✦ Example: Butane and 2-methylpropane (C₄H₁₀)
➡️ Position isomerism: Functional group at different positions. ✦ Example: 1-butanol and 2-butanol (C₄H₉OH)
➡️ Functional group isomerism: Different functional groups. ✦ Example: Ethanol (C₂H₅OH) and Dimethyl ether (CH₃OCH₃)
➡️ Metamerism: Same functional group, different alkyl groups. ✦ Example: Diethyl ether (C₂H₅–O–C₂H₅) and Methyl propyl ether (CH₃–O–C₃H₇)

🔄 Stereoisomerism

➡️ Geometrical (cis-trans) isomerism: Restricted rotation around double bond. ✦ Example: Cis-but-2-ene and Trans-but-2-ene
➡️ Optical isomerism: Non-superimposable mirror images due to chiral carbon. ✦ Example: Lactic acid (CH₃–CHOH–COOH)

📌 Examples of Isomers

✦ Pentyl alcohol (C₅H₁₂O): 1-pentanol, 2-pentanol, 3-pentanol; chain isomers: n-pentanol, isopentanol, neopentanol
✦ Pentyl chloride (C₅H₁₁Cl): 1-chloropentane, 2-chloropentane, 3-chloropentane
✦ Pentene (C₅H₁₀): 1-pentene, 2-pentene; cis-2-pentene, trans-2-pentene
✦ Butyne (C₄H₆): 1-butyne, 2-butyne
✦ C₃H₆O: Propanal (CH₃–CH₂–CHO), Acetone (CH₃–CO–CH₃)
✦ C₂H₄O₂: Acetic acid (CH₃–COOH), Methyl formate (H–COO–CH₃)

(xx, OR) Write note on classification of organic compounds and Natural sources of Organic Compounds

✍️ Answer

🧪 Classification of Organic Compounds ✨

1️⃣ Hydrocarbons
Hydrocarbons are organic compounds containing only carbon and hydrogen.
Called the “mother” of organic compounds, as introducing functional groups creates countless derivatives.

2️⃣ Types of Hydrocarbons Based on molecular structure (carbon skeleton) and nature of bonds:
➡️ Aliphatic / Acyclic / Open Chain hydrocarbons
➡️ Cyclic or Closed Chain hydrocarbons / Ring Compounds

2.1 Open Chain / Aliphatic / Acyclic Hydrocarbons

📖 Definition: Carbon atoms arranged in straight or branched open chains with single, double, or triple bonds.
🔓 Structure: No link between first and last carbon atoms.
🔥 Flame Characteristic: Non-sooty (Complete Combustion)
🛢️ Common Name: Aliphatic hydrocarbons (Greek origin: oil/fat-like).
🔹 Subtypes:
➡️ Saturated hydrocarbons (Alkanes, Cycloalkanes): All C–C bonds single.
➡️ Unsaturated hydrocarbons (Alkenes, Alkynes): Contain double/triple bonds.
✦ Alkenes: CₙH₂ₙ
✦ Alkynes: CₙH₂ₙ₋₂

2.2 Cyclic / Ring Hydrocarbons

📖 Definition: Carbon atoms form closed chains (rings); may include heteroatoms.
🔒 Structure: First and last carbon atoms linked to form ring.
🔹 Subtypes:
➡️ Alicyclic (Non-Benzenoid): Saturated/unsaturated rings without benzene.
✦ Cycloalkanes: CₙH₂ₙ
✦ Cycloalkenes: CₙH₂ₙ₋₂
➡️ Aromatic (Benzenoid): Rings containing benzene (C₆H₆) or derivatives.

Features of Aromatic Compounds:
➡️ Planar hexagonal rings with alternating single/double bonds.
➡️ Follow Hückel’s rule: 4n + 2 π electrons → extra stability.
➡️ Modern term: Arenes (benzene derivatives).
➡️ Initially called “aromatic” due to smell, now based on stability.

🧪 Natural Sources of Organic Compounds ✨

1️⃣ Natural Products from Plants and Animals 🌿🐾
Organic compounds isolated from plants, animals, microbes are called natural products.
Produced by metabolic processes, extracted via distillation, filtration, purification.
Used in medicines, cosmetics, nutrition.
Examples: Glucose, cellulose, insulin, cholesterol, caffeine, nicotine, menthol, peppermint oil.

🟥 Sources:
Animals: Proteins, fats, vitamins, hormones, urea, uric acid.
Plants: Proteins, oils, vitamins, sugars, starch, cellulose, dyes, drugs, fibers (cotton).

2️⃣ Fossil Fuels / Fossil Remains ⛏️
Derived from ancient plant/animal remains in rocks/sediments.
Formation: Organisms die → buried → compressed → transformed.
Non-renewable energy sources.

🟥 Types of Fossil Fuels
➡️ Coal 🏭
➡️ Natural Gas 🔥
➡️ Petroleum / Crude Oil 🛢️

(a) Coal 🏭

Solid fuel from plants.
Formation: Anaerobic decay of buried plant matter under heat/pressure.
Largest reserves in Pakistan: Thar Desert, Sindh (~175 billion tons lignite).
Coal Types:
➡️ Peat: 45–60% C, soft, smoky flame.
➡️ Lignite: 60–70% C, soft brownish coal.
➡️ Bituminous: 70–85% C, black coal, electricity/steel, produces coke & tar.
➡️ Anthracite: 90–95% C, hard, smokeless, furnaces/power/domestic fuel.

(b) Natural Gas 🔥

Formation: Decomposition of marine microorganisms.
Composition: Mostly methane (>85%), plus ethane, propane, butane.
Uses: Cleaner energy, less pollution.
Occurrence in Pakistan: Discovered 1952 at Sui, Baluchistan; also in Sindh.

(c) Petroleum / Crude Oil / Rock Oil 🛢️

Definition: “Petra” = rock, “oleum” = oil → thick black liquid.
Composition: Mixture of hydrocarbons + small N, O, S compounds.
Formation: Anaerobic decay of buried marine animals.
Uses: Fuel (transport, power), chemical feedstock.
Fractional Distillation Products: Gasoline, kerosene, diesel, naphtha, paraffin wax, lubricating oils.

(xx) Define organo-metallic compound with examples. What is Grignard’s reagent? Write the equation of its (methyl magnesium iodide) like reaction with:
➡️ Water ➡️ Dry ice ➡️ Formalin ➡️ Methyl amine ➡️ Carbonyl compounds

✍️ Answer

🧪 Organo-Metallic Compound

Definition: An organo-metallic compound is a compound in which a metal is directly bonded to a carbon atom of an organic group.
Examples: Methyl lithium: CH₃Li, Grignard reagent: CH₃MgI, Phenyl sodium: C₆H₅Na

⚡ Grignard’s Reagent

Definition: Grignard reagent is an organomagnesium compound of the general formula R–Mg–X, where R = alkyl/aryl, X = halogen.
Preparation: Prepared by reacting alkyl or aryl halide with magnesium in dry ether:
CH₃I + Mg —dry ether→ CH₃MgI

🧬 Reactions of Methyl Magnesium Iodide (CH₃MgI)

🧪 With Water

CH₃MgI + H₂O → CH₄ + Mg(OH)I
👉 Product: Methane

🧪 With Dry Ice (CO₂)

CH₃MgI + CO₂ → CH₃COOMgI —H⁺/H₂O→ CH₃COOH + Mg(OH)I
👉 Product: Acetic acid (after acid workup)

🧪 With Formalin (HCHO)

CH₃MgI + HCHO → CH₃CH₂OMgI —H⁺/H₂O→ CH₃CH₂OH + Mg(OH)I
👉 Product: Ethanol (after hydrolysis)

🧪 With Methyl Amine (CH₃NH₂)

CH₃MgI + CH₃NH₂ → CH₄ + CH₃N⁻MgI⁺
👉 Product: Methane

🧪 With Carbonyl Compounds

🌟 Aldehyde (R–CHO):
CH₃MgI + R–CHO → R–CH(OMgI)–CH₃ —H⁺/H₂O→ R–CH(OH)–CH₃
👉 Product: Secondary alcohol

🌟 Ketone (R–CO–R'):
CH₃MgI + R–CO–R' → R–C(OMgI)–CH₃–R' —H⁺/H₂O→ R–C(OH)–CH₃–R'
👉 Product: Tertiary alcohol

(xxi, OR) Write the equation for the reaction of acetaldehyde with the following:
Chromic acid, lithium aluminium hydride, Zinc-mercury amalgam, hydroxylamine, ammonia, hydrogen cyanide.

✍️ Answer

🧪 Reactions of Acetaldehyde (CH₃CHO)

1️⃣ With Chromic Acid (Oxidation)

➡️ CH₃CHO + [O] → CH₃COOH
👉 Product: Acetic acid

2️⃣ With Lithium Aluminium Hydride (Reduction)

➡️ CH₃CHO + 2[H] → CH₃CH₂OH
👉 Product: Ethanol

3️⃣ With Zinc-Mercury Amalgam (Clemmensen Reduction)

➡️ CH₃CHO + 2[H] —Zn-Hg/HCl→ CH₃CH₃
👉 Product: Ethane

4️⃣ With Hydroxylamine (Formation of Oxime)

➡️ CH₃CHO + NH₂OH → CH₃CH=NOH + H₂O
👉 Product: Acetaldehyde oxime

5️⃣ With Ammonia (Formation of Imine / Schiff Base)

➡️ CH₃CHO + NH₃ → CH₃CH=NH + H₂O
👉 Product: Acetaldehyde imine

6️⃣ With Hydrogen Cyanide (Formation of Cyanohydrin)

➡️ CH₃CHO + HCN → CH₃CH(OH)CN
👉 Product: Acetaldehyde cyanohydrin

(xxi) Why are alkyl amines basic in nature? How a primary alkyl amine is converted into secondary & tertiary amine?

✍️ Answer

🟥 Why Are Alkyl Amines Basic in Nature?

1. Lone Pair on Nitrogen ⚡
Nitrogen has a lone pair that can accept H⁺.
Acts as a Brønsted–Lowry base.
RNH₂ + H⁺ → RNH₃⁺

2. +I (Electron-Donating) Effect of Alkyl Groups ➕
Alkyl groups (–R) push electron density toward nitrogen through the inductive effect.
This increases the availability of the lone pair for protonation, making alkyl amines more basic than ammonia (NH₃).

3. Stability of Conjugate Acid 🛡️
The more stable the alkyl ammonium ion formed, the stronger the base.
Basicity order (aqueous):
Secondary amine > Primary amine > Tertiary amine > Ammonia (due to solvation and steric effects)

🟥 Conversion of Primary Alkyl Amine into Secondary and Tertiary Amine

A primary amine (RNH₂) can be converted into secondary (R₂NH) and tertiary amines (R₃N) by alkylation (reaction with alkyl halides).
➡️ Reaction Type: Nucleophilic substitution (SN) of primary amine with alkyl halide in alcoholic medium.
➡️ Note: Each step replaces H atoms of amines with alkyl groups.
➡️ Result: Mixture of primary, secondary, and tertiary amines.

🟥 Stepwise Alkylation

Primary → Secondary Amine:
RNH₂ + R–X → R₂NH

Secondary → Tertiary Amine:
R₂NH + R–X → R₃N

🟥 Further Reaction (Optional Information)

Tertiary Amine → Quaternary Ammonium Salt (Exhaustive Alkylation)
R₃N + R′X → R₄N⁺X⁻

(xxii, OR) Define primary, secondary and tertiary amines. Why are amines basic in nature? Explain why secondary amines are more basic than primary amines. How can we prepare ethyl amine from the following compounds:
* Ethyl iodide * Methyl cyanide * Ethanamide

✍️ Answer

🧪 Classification of Amines (Definition of Primary, Secondary and Tertiary Amines)

➡️ Primary (1°) amines: One hydrogen of ammonia is replaced by an alkyl/aryl group.
✦ General formula: R–NH₂ (e.g., CH₃NH₂, methylamine).

➡️ Secondary (2°) amines: Two hydrogens of ammonia are replaced by alkyl/aryl groups.
✦ General formula: R₂NH (e.g., (CH₃)₂NH, dimethylamine).

➡️ Tertiary (3°) amines: All three hydrogens of ammonia are replaced by alkyl/aryl groups.
✦ General formula: R₃N (e.g., (CH₃)₃N, trimethylamine).

⚗️ Why Amines Are Basic?

(1) Lone Pair on Nitrogen ⚡
Nitrogen has one lone pair of electrons which can accept a proton (H⁺), forming substituted ammonium ions (RNH₃⁺, R₂NH₂⁺, etc.).
Hence, amines act as Brønsted–Lowry bases (as well as Lewis bases).
RNH₂ + H⁺ → RNH₃⁺

(2) +I Effect of Alkyl Groups ➕
Alkyl groups donate electron density toward nitrogen (positive inductive effect), increasing electron density on nitrogen and making the lone pair more available for protonation.
Thus, alkyl amines are more basic than NH₃.

⚗️ Why Are Secondary Amines More Basic Than Primary Amines?

✦ Basicity Order (aqueous): Secondary amine > Primary amine > Tertiary amine > NH₃
✦ Reason:
Secondary amines are more basic than primary amines because two alkyl groups exert a stronger +I effect, increasing electron density on nitrogen and making the lone pair more available for protonation.
Their conjugate acid (R₂NH₂⁺) is also more stabilized in aqueous solution, increasing basic strength.

⚗️ Preparation of Ethylamine (C₂H₅NH₂)

(A) From Ethyl Iodide (Ammonolysis of Alkyl Halide)
By heating with alcoholic ammonia:
C₂H₅I + NH₃ → C₂H₅NH₂ + HI

(B) From Methyl Cyanide / Acetonitrile (CH₃CN)
Reduction of nitrile using H₂/Ni or LiAlH₄:
CH₃CN + 2H₂ → C₂H₅NH₂

(C) From Ethanamide (CH₃CONH₂)
Reduction of amide using LiAlH₄:
CH₃CONH₂ + 4[H] → CH₃CH₂NH₂ + H₂O

(xxii) Name four derivatives of carboxylic acids with their class formulae and write the equations of their preparation.

✍️ Answer

🧪 Derivatives of Carboxylic Acids & Their Preparation ⚗️

🟥 Acid Chloride (Acyl Chloride) 🔥

✦ Class formula: RCOCl
✦ Preparation: RCOOH + SOCl₂ → RCOCl + SO₂ + HCl

🟥 Acid Anhydride 🌡️

✦ Class formula: (RCO)₂O
✦ Preparation (Dehydration): 2RCOOH —(P₂O₅)→ (RCO)₂O + H₂O

🟥 Ester 🍃

✦ Class formula: RCOOR′
✦ Preparation (Esterification): RCOOH + R′OH —(conc. H₂SO₄)→ RCOOR′ + H₂O

🟥 Amide 📘

✦ Class formula: RCONH₂
✦ Preparation: RCOOH + NH₃ → RCOONH₄ —(Δ)→ RCONH₂ + H₂O

(xxiii, OR) Discuss the acidic nature of carboxylic acid. How is carboxylic acid prepared by:
➡️ Hydrolysis of alkyl nitrile (Methyl Nitrile)
➡️ Oxidation of primary alcohols (Ethanal)
➡️ Oxidation of ketone (Acetone)
➡️ Carbonation of Grignard’s reagent (Methyl magnesium chloride)

✍️ Answer

🧪 Acidic Nature of Carboxylic Acids ⚗️

Carboxylic acids (RCOOH) are acidic because they can donate a proton (H⁺):
RCOOH ⇌ RCOO⁻ + H⁺

🔎 Reason for Acidity

1️⃣ Resonance Stabilization:
The carboxylate ion (RCOO⁻) is stabilized by resonance; the negative charge is delocalized over two oxygen atoms.

2️⃣ –I Effect of Carbonyl Group:
The C=O group withdraws electron density (–I Effect), weakening the O–H bond and facilitating proton release.
Hence, carboxylic acids are more acidic than alcohols.

⚗️ Preparation of Carboxylic Acids

➡️ 1️⃣ By Acidic Hydrolysis of Alkyl Nitrile (Methyl Nitrile) 🧪

Methyl nitrile (acetonitrile, CH₃CN) on hydrolysis gives acetic acid.
CH₃CN + 2H₂O + HCl → CH₃COOH + NH₄Cl

➡️ 2️⃣ By Oxidation of Primary Alcohol (Ethanal route) 🔥

Primary alcohol first forms aldehyde, then acid (Ethanol → Ethanal → Acetic acid).
CH₃CH₂OH + [O] → CH₃CHO
CH₃CHO + [O] → CH₃COOH

➡️ 3️⃣ By Oxidation of Ketone (Acetone) ⚡

On strong oxidation, ketones undergo cleavage (Acetone gives acetic acid).
CH₃COCH₃ + [O] → CH₃COOH + CO₂ + H₂O

➡️ 4️⃣ By Carbonation of Grignard Reagent (Methyl Magnesium Chloride) ❄️

Grignard reagent reacts with CO₂ followed by hydrolysis.
CH₃MgCl + CO₂ → CH₃COOMgCl
CH₃COOMgCl + HCl → CH₃COOH + MgCl₂

(xxiii) Define nucleophilic substitution reaction. How can we prepare following compounds using CH₃Br?
➡️ CH₃SH ➡️ CH₃OCH₃ ➡️ CH₃COOCH₃ ➡️ CH₃CN

✍️ Answer

🧪 Nucleophilic Substitution Reaction ⚡

Definition: A nucleophilic substitution reaction is a chemical reaction in which a nucleophile (electron-rich species) replaces a leaving group (usually a halogen) in a molecule.
General Reaction: R–X + Nu⁻ → R–Nu + X⁻ (X = halogen, Nu⁻ = nucleophile)

⚙️ Preparation of Compounds from CH₃Br

1️⃣ Methyl Mercaptan (CH₃SH) 🧤

Reaction: CH₃Br + NaSH → CH₃SH + NaBr

2️⃣ Dimethyl Ether (CH₃OCH₃) 💨

Reaction:
CH₃Br + NaOH → CH₃OH + NaBr
CH₃OH + CH₃Br → CH₃OCH₃ + HBr
(Or direct: CH₃Br + NaOCH₃ → CH₃OCH₃ + NaBr)

3️⃣ Methyl Acetate (CH₃COOCH₃) 🍃

Reaction: CH₃Br + CH₃COONa → CH₃COOCH₃ + NaBr

4️⃣ Methyl Cyanide (CH₃CN) ⚡

Reaction: CH₃Br + KCN → CH₃CN + KBr

(xxiv) Give the scope of pharmaceutical industries in Pakistan. Write down names of five drugs with their uses.

✍️ Answer

💊 Scope of Pharmaceutical Industry in Pakistan 🇵🇰

The pharmaceutical industry in Pakistan has a growing scope due to:
➡️ Increasing population and healthcare needs 👨‍👩‍👧‍👦
➡️ Rising chronic diseases like diabetes, hypertension 💉
➡️ Government incentives for local drug manufacturing 🏭
➡️ Export potential to neighboring countries 🌍
➡️ Research & development in generic and new drugs 🔬

It provides employment opportunities, strengthens the economy, and improves public health.

💊 Classification of Drugs Based on Pharmacological Effect & Their Uses 🏥

📌 1. Aspirin – Analgesic / pain reliever; ➡️ reduces fever, inflammation, and acts as an antithrombotic ❤️
📌 2. Penicillin – Antibiotic; ➡️ fights bacterial infections by suppressing growth or killing microorganisms 🦠
📌 3. Paracetamol (Acetaminophen) – Antipyretic; ➡️ pain reliever & fever reducer 🌡️
📌 4. Fluconazole – Antifungal; ➡️ kills fungi causing skin infections 🦠🧴
📌 5. Ibuprofen – Anti-inflammatory; ➡️ reduces swelling, inflammation, and relieves pain 💪
📌 6. Diphenhydramine – Anti-allergic / Antihistamine; ➡️ reduces histamine levels, relieving allergies, sneezing, and itching 🤧
📌 7. Chloroquine – Anti-protozoal / Anti-malarial; ➡️ treats mosquito-borne diseases like malaria 🦟

(xxv, OR) What is antihistamine drug? Give the symptoms in which it is used.

✍️ Answer

💊 What is an Antihistamine Drug? 🤧

Definition: An antihistamine is a drug that blocks the action of histamine (a chemical released during allergic reactions) by binding to histamine receptors in the body.
Function: It reduces allergy symptoms like swelling, itching, and mucus production.

🩺 Symptoms / Conditions Where Antihistamines Are Used

➡️ Sneezing 🤧
➡️ Itching ✋
➡️ Runny nose 🌬️
➡️ Other allergic reactions ⚡

(xxv) Consider the following structures and answer the following questions:
(A) CH₂=CH₂ (B) C₆H₆
(a) Draw the hybrid structure of A
(b) Write equation for the conversion of B into acetophenone.
(c) Write the equation for the conversion of A into acetic acid.
(d) Write the equations for the conversion of B into phenol and TNT.

✍️ Answer

(a) Hybrid Structure of A (Ethene) 🧪

(b) Conversion of Benzene (B) into Acetophenone 🧪

Reaction (Friedel–Crafts Acylation):
C₆H₆ + CH₃COCl —(AlCl₃ catalyst)→ C₆H₅COCH₃ + HCl
Benzene → Acetophenone (C₆H₅COCH₃)

(c) Conversion of Ethene (A) into Acetic Acid ⚡

Step 1: Hydration to Ethanol:
CH₂=CH₂ + H₂O —(H₂SO₄ catalyst/Δ)→ CH₃CH₂OH
Step 2: Oxidation to Acetic Acid:
CH₃CH₂OH + [O] —(K₂Cr₂O₇)→ CH₃COOH
Overall: CH₂=CH₂ → CH₃CH₂OH → CH₃COOH

(d) Conversion of Benzene (B) into Phenol & TNT 🔹

1️⃣ Phenol (C₆H₅OH):
C₆H₆ + Cl₂ —(FeCl₃)→ C₆H₅Cl (Chlorination)
C₆H₅Cl + NaOH —(aq, Δ)→ C₆H₅ONa —(HCl)→ C₆H₅OH

2️⃣ TNT (2,4,6-Trinitrotoluene, C₆H₂(NO₂)₃CH₃):
C₆H₆ + CH₃Cl —(AlCl₃)→ C₆H₅CH₃ (Toluene, via Friedel–Crafts alkylation)
C₆H₅CH₃ + 3HNO₃ —(H₂SO₄)→ C₆H₂(NO₂)₃CH₃ (TNT via successive nitration)

(xxv, OR) The structure of two organic compounds ‘A’ and ‘B’ are shown below:
(A) CH₂=CH₂ (B) HC≡CH
(a) Draw and explain the orbital structure of A and specify hybridization and bond angle.
(b) Write the equations, when A and B react with ozone (ozonolysis).

✍️ Answer

🧪 Organic Compounds: CH₂=CH₂ (A) & HC≡CH (B) ⚡

Given:
(A) CH₂=CH₂ → Ethene
(B) HC≡CH → Ethyne

(a) Orbital Structure of A (Ethene, CH₂=CH₂) 🔹

Hybridization: Each carbon in ethene is sp² hybridized.
Bond Angle: Approximately 120°, due to trigonal planar geometry around each carbon.

(b) Ozonolysis Reactions (Reaction with O₃) 🌪️

1️⃣ Ethene (CH₂=CH₂):
CH₂=CH₂ + O₃ —[Zn/H₂O]→ 2 HCHO (Formaldehyde)
Ozone cleaves the double bond to form aldehydes or ketones.

2️⃣ Ethyne (HC≡CH):
HC≡CH + O₃ —[H₂O₂]→ OHC–CHO (Glyoxal)
Ozone cleaves the triple bond to form glyoxal.

(xxvi) What are synthetic polymers? Write down the names of two synthetic and two natural polymers? Write down the preparation of two condensation polymers and one addition polymer with equations.

✍️ Answer

🧪 Synthetic Polymers ✨

Definition: Synthetic polymers are man-made macromolecules formed by joining many small monomer units through chemical reactions.

🌿 Examples of Polymers

1️⃣ Natural Polymers:
➡️ Cellulose – structural material in plants 🌱
➡️ Proteins (e.g., silk, wool) – made of amino acids 🐑

2️⃣ Synthetic Polymers:
➡️ Nylon-6,6 – synthetic fiber 🧵
➡️ Polystyrene (PS) – plastic used in packaging 📦

⚙️ Preparation of Polymers

1️⃣ Condensation Polymer: Nylon-6,6 🧵

Formed by hexamethylenediamine + adipic acid
Equation:
n H₂N–(CH₂)₆–NH₂ + n HOOC–(CH₂)₄–COOH → –[NH–(CH₂)₆–NH–CO–(CH₂)₄–CO]–ₙ + 2n H₂O
Monomers condense and eliminate water to form the polymer.

2️⃣ Condensation Polymer: Terylene (Polyester) 🏺

Formed by ethylene glycol + terephthalic acid
Equation:
n HO–CH₂–CH₂–OH + n HOOC–C₆H₄–COOH → –[–O–CH₂–CH₂–O–CO–C₆H₄–CO–]–ₙ + 2n H₂O

3️⃣ Addition Polymer: Polyethylene (PE) ♻️

Formed by polymerization of ethene. Double bonds open and join without elimination of small molecules.
Equation:
n CH₂=CH₂ → –[CH₂–CH₂]–ₙ

(xxvii, OR) Distinguish two types of polymers based on mode of polymerization and action of heat. Which polymer is obtained on polymerization of following monomers. Write complete reaction equation:
➡️ Vinyl chloride ➡️ Adipic acid and hexamethylene diamine ➡️ Ethylene glycol and Terephthalic acid

✍️ Answer

🧪 Types of Polymers Based on Polymerization 🔥

Addition Polymers (➕):
➡️ Formed by opening double bonds of monomers.
➡️ No by-product is formed.
➡️ Usually thermoplastic (softens on heating).

Condensation Polymers (💧):
➡️ Formed by condensation of two monomers.
➡️ Small molecules (H₂O, HCl) are eliminated.
➡️ Can be thermoplastic or thermosetting.

📌 Types of Polymers Based on Heat 🌡️

➡️ Thermoplastics ♻️: Soften on heating and regain shape on cooling.
➡️ Thermosetting Plastics 🔥: Undergo irreversible hardening on heating, become rigid.

⚙️ Polymers from Given Monomers

1️⃣ Vinyl Chloride → Polyvinyl Chloride (PVC)

Reaction: n CH₂=CHCl → –[CH₂–CHCl]–ₙ
Type: Addition polymer ♻️

2️⃣ Adipic Acid + Hexamethylene Diamine → Nylon-6,6

Reaction:
n H₂N–(CH₂)₆–NH₂ + n HOOC–(CH₂)₄–COOH → –[NH–(CH₂)₆–NH–CO–(CH₂)₄–CO]–ₙ + 2n H₂O
Type: Condensation polymer 🧵

3️⃣ Ethylene Glycol + Terephthalic Acid → Terylene (Polyester)

Reaction:
n HO–CH₂–CH₂–OH + n HOOC–C₆H₄–COOH → –[–O–CH₂–CH₂–O–CO–C₆H₄–CO–]–ₙ + 2n H₂O
Type: Condensation polymer 🏺

(xxvii) Give the equation and write the name of final product in the following process. (write only equation)

✍️ Answer

🧪 Organic Reaction Equations (xxvii) ⚡

1️⃣ Benzene diazonium chloride + water (Δ)
C₆H₅N₂⁺Cl⁻ + H₂O → C₆H₅OH + N₂ + HCl
Final Product: Phenol

2️⃣ Propyl alcohol + Thionyl chloride (SOCl₂)
CH₃CH₂CH₂OH + SOCl₂ → CH₃CH₂CH₂Cl + SO₂ + HCl
Final Product: Propyl chloride

3️⃣ Phenol + Conc. H₂SO₄
Low temp (~20°C): C₆H₅OH + H₂SO₄ → o-C₆H₄OH–SO₃H + H₂O (o-Phenol sulfonic acid)
High temp (~100°C): C₆H₅OH + H₂SO₄ → p-C₆H₄OH–SO₃H + H₂O (p-Phenol sulfonic acid)

4️⃣ Acetylene or Propyne + H₂O / H₂SO₄, HgSO₄ (75°C)
HC≡CH + H₂O → CH₂=CHOH → CH₃CHO (Acetaldehyde)
HC≡C–CH₃ + H₂O → CH₂=CH(OH)CH₃ → CH₃–CO–CH₃ (Acetone)

5️⃣ 1,2-Dibromoethane + alcoholic KOH
BrCH₂–CH₂Br + 2KOH → HC≡CH + 2KBr + 2H₂O
Final Product: Acetylene

6️⃣ Acetic acid + LiAlH₄
CH₃COOH + 4[H] → CH₃CH₂OH + H₂O
Final Product: Ethanol

7️⃣ Sodium ethanoate + soda lime
CH₃COONa + NaOH (Δ) → CH₄ + Na₂CO₃
Final Product: Methane

8️⃣ Acetone + Acidified K₂Cr₂O₇
CH₃COCH₃ + [O] → CH₃COOH + CO₂ + H₂O

9️⃣ Ethyne + HBr
HC≡CH + HBr → CH₂=CHBr (Bromoethene) + excess HBr → CH₃–CHBr₂
Final Product:1,2-dibromoethane

🔟 Propene + HBr
CH₃–CH=CH₂ + HBr → CH₃–CHBr–CH₃
Final Product: 2-Bromopropane

1️⃣1️⃣ Ethanol + Excess H₂SO₄ (Δ)
2 CH₃CH₂OH → CH₃CH₂OCH₂CH₃ + H₂O
Final Product: Diethyl ether

1️⃣2️⃣ Ethene or 2-Butene + O₃ (Ozonolysis)
CH₂=CH₂ + O₃ → Ozonide → 2 HCHO
CH₃CH=CHCH₃ + O₃ → Ozonide → 2 CH₃CHO
Final Product: Aldehydes

1️⃣3️⃣ Ethene + Peracetic acid (100°C)
CH₂=CH₂ + CH₃COOOH → (CH₂–O–CH₂) + CH₃COOH
Final Product: Ethylene oxide

(xxviii, OR) How can we prepare following compounds (any four):
(a) Ethanal from ethyne (b) Phenyl hydrazone from formaldehyde (c) Ethanol from organo-metallic compound (d) Oxime from acetaldehyde (e) Tert-butyl alcohol from Grignard’s reagent (f) Ethene from ethanol (g) Bromohydrin from ethene (h) Ethyne from ethene

✍️ Answer

🧪 Preparation of Compounds ⚡

(a) Ethanal from Ethyne

HC≡CH + H₂O —[H₂SO₄/HgSO₄]→ CH₂=CHOH → CH₃CHO
Method: Hydration of ethyne
Final Product: Ethanal

(b) Phenyl Hydrazone from Formaldehyde

HCHO + C₆H₅NHNH₂ → H₂C=NNHC₆H₅ + H₂O
Method: Reaction with phenylhydrazine
Final Product: Formaldehyde phenylhydrazone

(c) Ethanol from Organo-Metallic Compound

CH₃MgBr + H₂O → CH₃CH₂OH + Mg(OH)Br
Method: Hydrolysis of Grignard reagent
Final Product: Ethanol

(d) Oxime from Acetaldehyde

CH₃CHO + NH₂OH → CH₃CH=NOH + H₂O
Method: Reaction with hydroxylamine
Final Product: Acetaldehyde oxime

(e) Tert-Butyl Alcohol from Grignard’s Reagent

(CH₃)₃C–MgBr + H₂O → (CH₃)₃COH + MgBrOH
Method: Hydrolysis of tertiary Grignard
Final Product: Tert-butyl alcohol

(f) Ethene from Ethanol

CH₃CH₂OH —[conc. H₂SO₄/Δ]→ CH₂=CH₂ + H₂O
Method: Dehydration of alcohol
Final Product: Ethene

(g) Bromohydrin from Ethene

CH₂=CH₂ + Br₂ + H₂O → CH₂(OH)–CH₂Br
Method: Halohydrin formation
Final Product: Ethylene bromohydrin

(h) Ethyne from Ethene

CH₂=CH₂ + Br₂ → BrCH₂–CH₂Br
BrCH₂–CH₂Br + 2KOH(alc) → HC≡CH + 2KBr + 2H₂O
Method: Halogenation followed by dehydrohalogenation
Final Product: Ethyne

(xxviii, OR) Write only the equations for the following reactions:
Williamson’s synthesis, Dow’s process, Esterification, Saponification, Clemmensen Reduction.

✍️ Answer

🧪 Important Organic Reactions – Equations ⚡

1️⃣ Williamson’s Synthesis (Ether Formation)

R–X + R′–ONa → R–O–R′ + NaX
Example: CH₃Br + NaOCH₃ → CH₃OCH₃ + NaBr

2️⃣ Dow’s Process (Chlorination of Phenol)

C₆H₅OH + NaOH → C₆H₅ONa + H₂O
C₆H₅ONa + Cl₂ → C₆H₅Cl + NaCl + H₂O

3️⃣ Esterification (Carboxylic Acid + Alcohol)

RCOOH + R′OH —[conc. H₂SO₄/Δ]→ RCOOR′ + H₂O
Example: CH₃COOH + CH₃OH —[conc. H₂SO₄/Δ]→ CH₃COOCH₃ + H₂O

4️⃣ Saponification (Hydrolysis of Ester)

RCOOR′ + NaOH → RCOONa + R′OH
Example: CH₃COOCH₃ + NaOH → CH₃COONa + CH₃OH

5️⃣ Clemmensen Reduction (Ketone/Aldehyde → Alkane)

R–CO–R′ + HCl —[Zn(Hg)]→ R–CH₂–R′
Example: CH₃COCH₃ + HCl —[Zn(Hg)]→ CH₃CH₂CH₃

(xxix, OR) What is Lucas reagent and Lucas test? How is this test used to distinguish three types of alcohols?

✍️ Answer

🧪 Lucas Reagent & Lucas Test 🍶

1️⃣ Lucas Reagent:

A mixture of concentrated HCl and anhydrous ZnCl₂.
Function: Acts as a Lewis acid to convert alcohols into alkyl chlorides.

2️⃣ Lucas Test:

A qualitative test to distinguish primary (1°), secondary (2°), and tertiary (3°) alcohols based on rate of reaction with Lucas reagent.
General Reaction: Alcohol + Lucas Reagent → Alkyl chloride + H₂O

✅ Principle:

Tertiary alcohols react fastest due to carbocation stability (3° > 2° > 1°).
Formation of insoluble alkyl chloride produces turbidity.

⚡ Distinguishing Alcohols Using Lucas Test 🍶

Tertiary (3°) Alcohols:
R₃COH + HCl/ZnCl₂ → R₃CCl + H₂O
Observation: Instant turbidity (<1 min)

Secondary (2°) Alcohols:
R₂CHOH + HCl/ZnCl₂ → R₂CHCl + H₂O
Observation: Turbidity in 5–10 min

Primary (1°) Alcohols:
RCH₂OH + HCl/ZnCl₂ → RCH₂Cl + H₂O
Observation: Turbidity appears after hours

(xxx, OR) Write down two methods of preparation of ethers. How is oxonium salt of ether formed?

✍️ Answer

🧪 Methods of Preparation of Ethers ⚡

1️⃣ Williamson’s Synthesis (Alkoxide + Alkyl Halide)

Reaction: R–X + R′–ONa → R–O–R′ + NaX
Example: CH₃Br + NaOCH₃ → CH₃OCH₃ + NaBr

2️⃣ Dehydration of Alcohols (Acid-Catalyzed)

Reaction: 2 R–OH —[conc. H₂SO₄/Δ]→ R–O–R + H₂O
Example: 2 CH₃CH₂OH —[conc. H₂SO₄/Δ]→ CH₃CH₂OCH₂CH₃ + H₂O

⚡ Formation of Oxonium Salt of Ether 🧪

Ether reacts with strong acids (H⁺ donors) like HBr or HI.
Reaction: R–O–R + H⁺ → [R–O⁺–R]X⁻
Example: CH₃OCH₃ + HBr → [CH₃–O⁺–CH₃]Br⁻

Explanation: The oxygen of ether donates its lone pair to H⁺, forming a positively charged oxonium ion.

(xxix) Explain the following with scientific reasons:

(a) Ethanol is liquid but ethyl chloride is a gas at room temperature.
(b) Boiling point of alcohol is higher than ether and carbonyl compounds.
(c) Formaldehyde is highly soluble in water as compared to other aldehydes.
(d) Oxidation of aldehydes is faster than ketones.
(e) The boiling point of carboxylic acids are high than alcohol.
(f) The structure of carboxylic acid is trigonal planar.
(g) Tertiary alcohols cannot be oxidized.

✍️ Answer

🧪 Scientific Explanations ⚡

➡️ (a) Ethanol is liquid but ethyl chloride is gas
👉 Ethanol forms hydrogen bonds (–OH group) → strong intermolecular forces.
👉 Ethyl chloride (CH₃CH₂Cl) has only van der Waals forces → weak → gas.

➡️ (b) Boiling point of alcohol > ether & carbonyl compounds
👉 Alcohols have –OH group → hydrogen bonding, increasing boiling point.
👉 Ethers & carbonyls lack hydrogen bonding (or weaker) → lower boiling points.

➡️ (c) Formaldehyde highly soluble in water
👉 Small size + polar –CHO group forms hydrogen bonds with water easily.
👉 Larger aldehydes have bulky alkyl groups → less solubility.

➡️ (d) Oxidation of aldehydes faster than ketones
👉 Aldehydes have –H attached to carbonyl carbon, making them easily oxidizable.
👉 Ketones lack this –H → resistant to mild oxidation.

➡️ (e) Boiling point of carboxylic acids > alcohols
👉 Carboxylic acids form strong dimers via double hydrogen bonding → higher intermolecular forces.
👉 Alcohols form only single hydrogen bonds.

➡️ (f) Structure of carboxylic acid is trigonal planar
👉 Carbonyl carbon is sp² hybridized → bond angles ~120° → planar structure.

➡️ (g) Tertiary alcohols cannot be oxidized
👉 No alpha-hydrogen atom attached to the –OH-bearing carbon (alpha C) → oxidation to carbonyl is not possible.

(xxx, OR) Differentiate between atomic absorption and emission spectroscopy. What is the purpose of UV-visible spectroscopy? What are its applications in chemistry and biology?

✍️ Answer

⚡ Difference between Atomic Absorption and Emission Spectroscopy 🧪

Feature	Atomic Absorption Spectroscopy (AAS) 🔹	Atomic Emission Spectroscopy (AES) 🔸
Principle	Measures absorption of light by atoms	Measures light emitted by excited atoms
Energy Change	Atom absorbs energy → electron excitation	Atom emits energy → electron returns to lower state
Light Source	Hollow cathode lamp	Flame or plasma excites atoms
Use	Quantitative determination of metals	Quantitative & qualitative metal analysis

🎯 Purpose of UV-Visible Spectroscopy

Measures absorption of UV or visible light by molecules.
Determines electronic transitions in π → π* or n → π* orbitals.

🧬 Applications in Chemistry and Biology

Chemistry:
➡️ Determine concentration of solutions (Beer–Lambert law).
➡️ Study reaction kinetics and molecular structure.

Biology:
➡️ Measure protein and nucleic acid concentrations.
➡️ Study enzyme kinetics and ligand binding.

✏️ Smart Answers of ✍️ Detailed-Answer Questions of Guess Paper |Section-C ✏️
⚗️ Inorganic–General Chemistry Section 🧪

(Q3) Draw a flow diagram of contact process and describe various steps involved in the industrial manufacture of oil of vitriol. Write the reactions of concentrated Sulphuric acid with Al and sucrose.

✍️ Answer

🏭 Industrial Manufacture of Sulphuric Acid (Oil of Vitriol) – Contact Process

Developed by: Knietsch, Germany, early 19th century
Industrial use began: 1912
Principle: Catalytic oxidation of sulphur dioxide (SO₂) using vanadium pentoxide (V₂O₅) catalyst.

🏭 Plant and Process Overview

Three Main Parts / Steps:
➡️ Sulphur/Pyrite Burner – Formation of SO₂
➡️ Purification Unit – Removal of impurities from SO₂
➡️ Contact Tower & Absorption – Catalytic oxidation to SO₃ and absorption to form H₂SO₄

1️⃣ Formation of Sulphur Dioxide (SO₂)

S + O₂ —[400–450°C]→ SO₂ (impure)

2️⃣ Purification of SO₂

Removes impurities: dust, unburnt sulphur, NO, NO₂, As₂O₃, etc.
Achieved using: dust filters, scrubbers, drying towers, arsenic purifiers and tyndal box.
Purpose: Protect catalyst and ensure efficiency.

3️⃣ Catalytic Oxidation of SO₂ into SO₃

Purified SO₂ + air passed through contact tower.
Conditions: 400–500°C, 1–1.7 atm, catalyst V₂O₅ (promoted by K₂SO₄).
Yield: ~98% SO₃
Reaction: 2SO₂ + O₂ → 2SO₃ (∆H = –45 kcal/mol)

4️⃣ Absorption in Towers – Formation of Oleum & H₂SO₄

SO₃ absorbed in conc. H₂SO₄ (97–98%) → oleum (H₂S₂O₇).
Dilution with water → concentrated H₂SO₄ (98%).
Reactions:
2SO₃ + H₂SO₄ → H₂S₂O₇ (Oleum)
H₂S₂O₇ + H₂O → 2H₂SO₄

⚡ Reactions of Concentrated H₂SO₄

(a) With Aluminum (Al):
2Al + 6H₂SO₄ (conc) → Al₂(SO₄)₃ + 3SO₂ + 6H₂O
Observation: Redox reaction, SO₂ gas evolved.

(b) With Sucrose (C₁₂H₂₂O₁₁):
C₁₂H₂₂O₁₁ + H₂SO₄ (conc) → 12C + 11H₂O
Observation: Carbon “charcoal column” formed, dehydration reaction.

(Q3, OR) Describe various steps involve in the extraction of 99.99% pure copper from its chalcopyrite ore. (Draw diagram where necessary)

✍️ Answer

🏭 Extraction of Copper from Chalcopyrite (CuFeS₂) – Pyrometallurgy

Ore Used: Chalcopyrite (Copper Pyrite, CuFeS₂)
Method: Pyrometallurgy (Dry Process)

📌 Main Steps of Metallurgy

Extraction of Copper from sulphide ore involves the following steps:
1️⃣ Crushing and grinding of Ores into Powdered Form
2️⃣ Frothing and Concentration of Crushed Ore (removal of impurities)
3️⃣ Roasting (Formation of Cuprous sulphide and ferrous oxide)
4️⃣ Smelting (Formation of Molten Matte or course metal) 🥣
5️⃣ Bessemerisation (Formation of Blister Copper) ⚡
6️⃣ Electrolytic Refining of Blistered Copper 🔋

1️⃣ Crushing & Grinding

Ore is crushed and powdered to increase surface area.

2️⃣ Concentration (Ore Dressing / Beneficiation) 🧹

Purpose: Remove impurities (gangue: quartz, clay, limestone).
Method: Froth Flotation – powdered ore + pine oil + water, agitated with air blast.
Ore particles attach to oil → form froth → skimmed off (enriched ore).
Gangue → wetted by water → settles at bottom.

3️⃣ Roasting 🔥

Definition: Heating of concentrated ore in excess air at high temperature below melting point.
Furnace Used: Reverberatory (flat) furnace.
Objective: Convert sulphides to oxides and remove volatile impurities (SO₂ etc.).
Composition After Roasting: Cu₂S, FeO, FeS (unreacted).
Reaction: 2CuFeS₂ + 4O₂ → Cu₂S + 2FeO + 3SO₂

4️⃣ Smelting – Formation of Molten Matte 🥣

Definition: Reduction of ore oxides using coke to obtain molten metal.
Charge: Roasted ore + coke + silica.
Reaction: FeO + SiO₂ → FeSiO₃ (slag).
Product: Molten Matte (Cu₂S + FeS).

5️⃣ Bessemerisation ⚡

Furnace: Pear-shaped Bessemer converter lined with MgO.
Process:
Step 1: 2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
Step 2: 2Cu₂O + Cu₂S → 6Cu + SO₂
Product: Blister Copper (~99% pure).
Appearance: Bubbles on surface due to escaping SO₂.

6️⃣ Electrolytic Refining ⚡🔋

Impurities: Fe, As, Ag, Au, Zn, Pt, Ni, Pb.
Setup: Lead-lined tank, CuSO₄ + dilute H₂SO₄ electrolyte.
Anode: Blister copper.
Cathode: Pure copper plates.
Reactions:
Anode: Cu → Cu²⁺ + 2ē
Cathode: Cu²⁺ + 2ē → Cu (pure)
Result: Copper deposited at cathode is 99.99% pure.
Impurities: Noble metals (Ag, Au, Pt) settle as anode mud.

(Q4) What is water pollution? Discuss different types water pollution also explain any parameters of drinking water analysis.

✍️ Answer

💧 Definition of Water Pollution

Any undesirable change in water quality affecting life is known as water pollution.
Causes: Population growth, industrialization, urbanization, and human activities.

🏞️ Types of Water Pollution

1️⃣ Suspended Solids & Sediments 🌫️
Colloids: Small particles like dust, coal, microorganisms stay suspended.
Sediments: Larger particles like sand, clay settle at the bottom.
Effects: Cause turbidity and reduce sunlight for aquatic life.

2️⃣ Dissolved Solids 🧪
Inorganic: Minerals, salts, metal ions (Na⁺, K⁺, Ca²⁺, Mg²⁺), chlorides, carbonates, sulfates.
Organic: From animals, plants, microorganisms, sewage, and industrial waste.
Effects: Affect water quality and aquatic life.

🧾 Parameters of Drinking Water Analysis (WHO)

pH ⚖️: Acidity/alkalinity; 6.5–8.5 (WHO).
Turbidity 🌫️: Cloudiness from suspended particles; <5 NTU.
Total Dissolved Solids (TDS) 💎: All dissolved inorganic/organic solids; <600 mg/L.
Chlorine Residue 🧴: Remaining chlorine after disinfection; <0.2 mg/L.
BOD & COD 🔬: Biochemical & chemical oxygen demand; BOD <5 mg/L, COD <10 mg/L.

(Q4, OR) Define Troposphere and Stratosphere. Describe the chemistry involved due to the presence of oxides of carbon and nitrogen in the troposphere. (Write equations where necessary)

✍️ Answer

🌤️ Troposphere

Definition: The lowest layer of Earth's atmosphere, extending up to 8–15 km from the surface.
Characteristics:
➡️ Weather phenomena occur here
➡️ Temperature decreases with height
➡️ Contains most of the atmospheric mass

☀️ Stratosphere

Definition: The layer above the troposphere, extending from ~15 km to 50 km altitude.
Characteristics:
➡️ Contains ozone layer
➡️ Temperature increases with height due to absorption of UV radiation
➡️ Relatively stable and calm, little weather activity

🧪 Chemistry of Oxides of Nitrogen (NOₓ)

🟥 Common Oxides: Nitric oxide (NO), Nitrogen dioxide (NO₂).

🟥 Formation:
📌 Combustion of fuels (coal, petrol, natural gas) at high temperature:
N₂ + O₂ —(high T)→ 2NO
2NO + O₂ —(high T)→ 2NO₂

📌 Photochemical reactions in sunlight (photolysis):
NO₂ —(hv)→ NO + O
O₂ + O → O₃ (ozone formation)

🟥 Harmful Effects:
📌 Acid rain formation:
4NO₂ + 2H₂O + O₂ → 4HNO₃
📌 Ozone formation: NOₓ participates in photochemical smog, producing ground-level ozone (O₃), harmful to health and vegetation.

🧪 Chemistry of Oxides of Carbon (COₓ) 🌍

🟥 Common Oxides: Carbon monoxide (CO), Carbon dioxide (CO₂).

🟥 Formation:
📌 Incomplete combustion of fuels → CO.
📌 Complete combustion of fuels → CO₂.

🟥 Reactions:
CO + OH· → CO₂ + H· (removal of CO in troposphere)

🟥 Harmful Effects:
➡️ CO: Toxic, binds with hemoglobin reducing oxygen transport.
➡️ CO₂: Major greenhouse gas, contributes to global warming.

🧪 Chemistry of Oxides of Sulphur (SOₓ) 🌫️

🟥 Common Oxides: Sulphur dioxide (SO₂), Sulphur trioxide (SO₃).

🟥 Formation ⚡
📌 Natural volcanic eruptions & human activities like burning sulphur-rich coal in power plants 🌋🏭
📌 Photochemical reactions in sunlight:
2SO₂ + O₂ —(hv)→ 2SO₃ (photolytic oxidation)

🟥 Harmful Effects ☠️
➡️ Health Impact: Exposure leads to serious cardiac and respiratory diseases 💔🫁
➡️ Environmental Impact: Reduced crop yields due to acid rain 🌾🌧️

(Q5) Explain with the help of a diagram of Castner Kellner cell, how caustic soda is obtained by the electrolysis of aqueous sodium chloride. Write down merits and demerits of Castner Kellner process. Write down action of sodium hydroxide solution on zinc and aluminium metal.

✍️ Answer

⚡ Production of Caustic Soda – Castner-Kellner Process

🟥 Basic Principle: Electrolysis of aqueous sodium chloride (brine/25% W/W solution of NaCl) using moving mercury cathode to produce NaOH, with Cl₂ and H₂ as by-products.
Reaction: 2NaCl + 2H₂O —(Electric current)→ 2NaOH + Cl₂↑ + H₂↑

🏗️ Construction and Working of Castner-Kellner Cell ⚡

📌 Upper Chamber:
Contains NaCl solution.
Titanium anodes immersed to oxidize Cl⁻ → Cl₂ gas.
Small amount of OH⁻ from water also oxidize → O₂ gas also formed.

📌 Moving Mercury Cathode:
Mercury circulates continuously (pump/eccentric wheel).
Acts as intermediate cathode forming sodium amalgam (Na/Hg).

📌 Denuder (Graphite Chamber):
Sodium from amalgam reacts with water.
Produces NaOH and H₂ gas.

🟥 Electrolytic Reactions

📌 Ionization:
➡️ 2(NaCl(aq) ⇌ Na⁺ (aq) + Cl⁻ (aq))
➡️ 4(H₂O₍ₗ₎ ⇌ H⁺ (aq) + OH⁻ (aq))

📌 Cathode Reaction (Reduction):
➡️ Reduction of Na⁺: 2(Na⁺ (aq) + ē → Na₍ₗ₎)
➡️ Formation of Sodium Amalgam: 2(Na₍ₗ₎ + Hg₍ₗ₎ → Na/Hg₍ₗ₎) (Sodium amalgam)
➡️ Reaction of amalgam with Water: 2Na/Hg₍ₗ₎ + 2H₂O₍ₗ₎ → 2NaOH(aq) + H₂↑ + 2Hg(l) (Mercury is recycled)

📌 Anode Reaction (Oxidation):
➡️ Chloride ions oxidation: 2Cl⁻ (aq) → Cl₂↑ + 2ē
➡️ Hydroxide ions oxidation: 4OH⁻ (aq) → 2H₂O₍ₗ₎ + O₂↑ + 4ē (Partial Oxidation)

🟢 Advantages of Flowing Mercury Cathode in Castner-Kellner’s Process ✅

➡️ Rapid and efficient process
➡️ Produces high-purity NaOH
➡️ Separation of Cl₂ and NaOH avoids unwanted reaction

🔴 Disadvantages of Castner-Kellner’s Process ❌

➡️ Mercury vapors pollute water → bioaccumulation in marine life
➡️ Environmental hazard despite controls

🧪 Action of NaOH on Metals

Observation: Hydrogen gas released; soluble complex salts formed.
Zinc: Zn + 2NaOH → Na₂ZnO₂ + H₂↑
Aluminium: 2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂↑

(Q5, OR) Write the balanced chemical equations for the following chemical processes:

✍️ Answer

➡️ Bleaching powder is dissolved in water 🌊
CaOCl₂ + H₂O → Ca(OH)₂ + Cl₂

➡️ Fluorine reacts with oxygen 🌬️
2F₂ + O₂ → 2OF₂

➡️ Potassium is put into ethyl alcohol 🍶
2K + 2C₂H₅OH → 2C₂H₅OK + H₂

➡️ Carbon heated with nitrogen at high temperature 🔥
3C(s) + 2N₂(g) → C₃N₄(s)

➡️ Permanganate ion reacts with oxalic acid 🧪
2KMnO₄ + 5H₂C₂O₄ + 3H₂SO₄ → 2MnSO₄ + 10CO₂ + 8H₂O + K₂SO₄

➡️ Manganese reacts with air 🌬️
3Mn + 2O₂ → Mn₃O₄

➡️ Potassium dichromate dissolved in water at neutral pH 💧
Cr₂O₇²⁻ (red) + H₂O → 2CrO₄²⁻ (yellow) + 2H⁺

➡️ Phosphorous is put in water 💦
2P₄(s) + 12H₂O(l) → 3H₃PO₄(aq) + 5PH₃↑

➡️ Sodium burns in excess of air 🔥
2Na(s) + O₂(g) —Excess O₂→ Na₂O₂(s)

➡️ Silicon reacts with steam 💨
Si + 2H₂O → SiO₂ + 2H₂

➡️ Sulphur reacts at high temperature with water 🔥
S + 2H₂O → SO₂ + 2H₂

➡️ Phosphorus reacts with nitrogen at high temperature 🔥
6P + 5N₂ → 2P₃N₅

➡️ Chlorine reacts with nitrogen 🌫️
N₂(g) + 3Cl₂(g) → 2NCl₃(g)

➡️ Aluminium in concentrated sulphuric acid ⚡
2Al + 6H₂SO₄ → Al₂(SO₄)₃ + 3SO₂ + 6H₂O

➡️ Ferric chloride in caustic soda 🧪
FeCl₃ + 3NaOH → Fe(OH)₃ + 3NaCl

➡️ Chlorine gas in cold & hot NaOH 🌡️
Cold: Cl₂ + 2NaOH → NaCl + NaClO + H₂O
Hot: 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O

➡️ Cuprous oxide with cuprous sulphide 🔥
2Cu₂O + Cu₂S → 6Cu + SO₂

➡️ Chromium with steam at high temperature 🔥
2Cr + 3H₂O → Cr₂O₃ + 3H₂

➡️ Carbon and silicon heated at high temperature 🔥
3Si(s) + 2N₂(g) → Si₃N₄(s)

➡️ Bleaching powder with hydrochloric acid 🧴
CaOCl₂ + 2HCl → CaCl₂ + Cl₂↑ + H₂O

➡️ Copper with concentrated nitric acid ⚡
Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O

➡️ Chromium in dilute hydrochloric acid 🧪
2Cr + 6HCl → 2CrCl₃ + 3H₂

➡️ KMnO₄ and FeSO₄ in H₂SO₄ (ionic) 🔬
MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

➡️ KMnO₄ and oxalic acid in H₂SO₄ (ionic) 🔬
2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

➡️ K₂Cr₂O₇ and FeSO₄ in H₂SO₄ (ionic) 🔬
Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O

(Q6) (a) What is the main cause of Global warming? How does it effect on weather pattern?

✍️ Answer

🟥 Main Cause of Global Warming 🌍

The primary cause of global warming is the increase of greenhouse gases (GHGs) in the atmosphere, especially carbon dioxide (CO₂), methane (CH₄), and nitrous oxide (N₂O). These gases trap heat from the sun, causing the Earth’s temperature to rise.
Main Sources:
➡️ Burning of fossil fuels (coal, oil, gas) for energy
➡️ Deforestation
➡️ Industrial emissions
➡️ Agriculture and livestock activities

🟥 Effects on Weather Patterns ⛅

Global warming disrupts the normal climate, leading to:
➡️ Extreme temperatures – hotter summers, milder winters in some regions
➡️ Changes in rainfall – more intense storms, floods, or droughts
➡️ Rising sea levels – causing coastal flooding and erosion
➡️ Increased frequency of extreme events – hurricanes, cyclones, heatwaves

Summary: Global warming makes weather more unpredictable and extreme, affecting ecosystems and human life.

(Q6 b) Complete and balance the following chemical equations:

✍️ Answer

➡️ Chromium with water 💧
2Cr₍ₛ₎ + 3H₂O₍ₗ₎ → Cr₂O₃₍ₛ₎ + 3H₂↑

➡️ Lithium with hydrogen ⚡
2Li₍ₛ₎ + H₂₍g₎ → 2LiH₍ₛ₎

➡️ Sodium with nitrogen 🌬️
6Na₍ₛ₎ + N₂₍g₎ → 2Na₃N₍ₛ₎

➡️ Calcium with nitrogen 🌬️
3Ca₍ₛ₎ + N₂₍g₎ → Ca₃N₂₍ₛ₎

➡️ Lithium with nitrogen 🌬️
6Li₍ₛ₎ + N₂₍g₎ → 2Li₃N₍ₛ₎

➡️ Potassium with oxygen 🌬️
K₍ₛ₎ + O₂₍g₎ → KO₂₍ₛ₎

➡️ Chromium with hydrochloric acid 🧪
2Cr₍ₛ₎ + 6HCl(aq) → 2CrCl₃(aq) + 3H₂↑

➡️ Manganese with oxygen 🌬️
3Mn₍ₛ₎ + 2O₂ → Mn₃O₄₍ₛ₎

➡️ KMnO₄ with FeSO₄ in H₂SO₄ 🔬
MnO₄⁻₍aq₎ + 5Fe²⁺₍aq₎ + 8H⁺₍aq₎ → Mn²⁺₍aq₎ + 5Fe³⁺₍aq₎ + 4H₂O

➡️ KMnO₄ with oxalic acid in H₂SO₄ 🔬
2MnO₄⁻ + 5H₂C₂O₄ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

➡️ K₂Cr₂O₇ with oxalic acid in H₂SO₄ 🔬
2Cr₂O₇²⁻₍aq₎ + 3H₂C₂O₄₍aq₎ + 8H⁺₍aq₎ → 4Cr³⁺ + 6CO₂ + 7H₂O

✏️ Smart Answers of ✍️ Detailed-Answer Questions of Guess Paper |Section-C ✏️
⚗️ Organic Chemistry Section 🧪

(Q7) What is meant by nucleophile? Differentiate between Sɴ₁ and Sɴ₂ reactions. Explain the reaction mechanism of Sɴ₁ and Sɴ₂ reactions.

✍️ Answer

📌 What is a Nucleophile? 🧪

A nucleophile is a species that donates a pair of electrons to an electron-deficient carbon to form a new bond.
✅ Usually negative ions or molecules with lone pair
Examples: OH⁻, CN⁻, NH₃, Cl⁻

📌 Differentiate between Sɴ₁ and Sɴ₂ 🔄

🟢 Sɴ₁ Reaction
Molecularity → Unimolecular
Rate law → Rate = k[RX]
Mechanism → Two-step reaction
Intermediate → Carbocation formed
Substrate preference → 3° > 2° > 1°
Rearrangement → Possible
Stereochemistry → Racemization
Nucleophile strength → Weak nucleophile works
Solvent → Polar protic

🟢 Sɴ₂ Reaction
Molecularity → Bimolecular
Rate law → Rate = k[RX][Nu⁻]
Mechanism → One-step (concerted)
Intermediate → No intermediate (transition state only)
Substrate preference → 1° > 2° > 3°
Rearrangement → Not possible
Stereochemistry → Inversion (Walden inversion)
Nucleophile strength → Strong nucleophile required
Solvent → Polar aprotic

📌 🔄 Mechanism of Sɴ₁ ⚙️

✔ Two-step reaction
✔ Carbocation intermediate
⏱️ Rate: Depends only on [alkyl halide] → unimolecular.
🔬 Stereochemistry: Racemic mixture (inversion + retention).
🌟 Favoured by: Tertiary alkyl halides, polar protic solvents (H₂O, alcohols).

➡️ Step 1: Slow ionization of C–X → Leaving group (X⁻) departs → carbocation (rate-determining step/RDS)
R₃C–X ⇌ R₃C⁺ + X⁻ (slow)

➡️ Step 2: Fast nucleophilic attack on carbocation:
R₃C⁺ + Nu⁻ → R₃C–Nu + Nu–CR₃ (fast)

📌 🔄 Mechanism of Sɴ₂ ⚙️

✔ One-step reaction (concerted reaction)
✔ Backside attack
⏱️ Rate: Depends on [alkyl halide] & [Nu⁻] → bimolecular and 2nd order.
🔬 Stereochemistry: Inversion of configuration (Walden inversion).
🌟 Favoured by: Primary alkyl halides, polar aprotic solvents (DMSO, DMF).

RCH₂–X + + Nu⁻ —(Slow)→ Nu⁻-----RCH₂-----X⁻ [Transition State] —(Fast)→ Nu–CH₂R + X⁻

(Q7 OR) Write down names, type formula and characteristic groups of three types of monohaloalkane. Explain the mechanism of following reactions: (No description is required).

✍️ Answer

📌 Reactions and Mechanisms

🟥 (i) NaOH / KOH with 2-bromo-2-methylpropane

Substrate: (CH₃)₃CBr (3° halide)
Type of reaction: Sɴ₁ (tertiary halide)

Mechanism:
➡️ Generation of Nucleophile (Ionization of Reagent): NaOH ⇌ Na⁺ + OH⁻ (nucleophile)
➡️ Step 1: Slow ionization of C–X and Formation of carbocation (slow- RDS)
(CH₃)₃C–Br ⇌ (CH₃)₃C⁺ + Br⁻ (slow)
➡️ Step 2: Fast nucleophilic attack on carbocation:
(CH₃)₃C⁺ + OH⁻ → (CH₃)₃C–OH + HO–C(CH₃)₃ (fast)
➡️ Formation of Side Product: Na⁺ + Br⁻ → NaBr

🟥 (ii) NaSH / KSH with 2-chlorobutane (2° halide) in aprotic solvent

Substrate: CH₃CH(Cl)CH₂CH₃ (2° halide)
Type of reaction: Sɴ₂ in aprotic solvent

Mechanism:
➡️ Generation of Nucleophile: NaSH ⇌ Na⁺ + SH⁻ (nucleophile)
➡️ Formation of Inverted Substituted Product:
CH₃CH(Cl)CH₂CH₃ + HS⁻ —(Slow)→ HS⁻-----CH(CH₃)(CH₂CH₃)-----X⁻ [Transition State] —(Fast)→ HS–CH(CH₃)(CH₂CH₃) + Cl⁻
➡️ Formation of Side Product: Na⁺ + Cl⁻ → NaCl

🟥 (ii) NaSH / KSH with 2-chlorobutane (2° halide) in protic solvent

Substrate: CH₃CHClCH₂CH₃ (2° halide)
Type of reaction: Sɴ₁ in protic solvent

Mechanism (No description):
➡️ Generation of Nucleophile: NaSH ⇌ Na⁺ + SH⁻ (nucleophile)
➡️ Step 1: Slow ionization of C–X and Formation of carbocation (slow- RDS)
CH₃CH(Cl)CH₂CH₃ ⇌ CH₃CH⁺CH₂CH₃ + Cl⁻ (slow)
➡️ Step 2: Fast nucleophilic attack on carbocation:
CH₃CH⁺CH₂CH₃ + SH⁻ → (CH₂CH₃)(CH₃)CH–SH + HS–CH(CH₃)(CH₂CH₃) (fast)
➡️ Formation of Side Product: Na⁺ + Cl⁻ → NaCl

🟥 (iii) NaCN / KCN with 1-chlorobutane (1° halide)

Substrate: CH₃CH₂CH₂CH₂Cl (1° halide)
Type of reaction: Sɴ₂

Mechanism (No description):
➡️ Generation of Nucleophile: NaCN ⇌ Na⁺ + CN⁻ (nucleophile)
➡️ Formation of Inverted Substituted Product:
CH₃CH₂CH₂CH₂–Cl + CN⁻ —(Slow)→ CN⁻-----CH₃CH₂CH₂CH₂-----X⁻ [Transition State] —(Fast)→ CN–CH₃CH₂CH₂CH₂ + Cl⁻
➡️ Formation of Side Product: Na⁺ + Cl⁻ → NaCl

(Q8) Write down structural formulae of any 8 of the following organic molecules:

✍️ Answer

🧪 Structural Formulas of Organic Molecules ✨

➡️ TNT (2,4,6-Trinitrotoluene): C₆H₂(NO₂)₃CH₃ (Benzene ring with CH₃ at C1, NO₂ at C2, C4, C6)

➡️ Benzophenone: (C₆H₅)₂CO (Two phenyl rings attached to carbonyl; C=O)

➡️ Picric Acid (2,4,6-Trinitrophenol): C₆H₂(NO₂)₃OH (OH at C1, NO₂ at C2, C4, C6)

➡️ p-Cresol (4-Methylphenol): CH₃C₆H₄OH (OH at C1, CH₃ at C4)

➡️ Isobutyraldehyde: (CH₃)₂CHCHO or CH₃–CH(CH₃)–CHO

➡️ Oxalic Acid: HOOC–COOH or O=C(OH)–C(OH)=O

➡️ Ethylene Glycol: HO–CH₂–CH₂–OH

➡️ Neopentyl Alcohol (2,2-Dimethyl-1-propanol): (CH₃)₃C–CH₂OH

➡️ Di-isopropyl Ketone (4-Methylpentan-2-one): (CH₃)₂CH–CO–CH(CH₃)₂

➡️ Iso-valeric Acid (3-Methylbutanoic Acid): (CH₃)₂CH–CH₂–COOH

➡️ Resorcinol (1,3-Benzenediol): C₆H₄(OH)₂ (OH at C1 & C3)

➡️ 2-Ethoxyhexane: CH₃–CH(OCH₂CH₃)–CH₂–CH₂–CH₂–CH₃

➡️ Trifluoroacetic Acid: F₃C–COOH

➡️ Triphenylamine: (C₆H₅)₃N

➡️ Divinyl Acetylene: CH₂=CH–C≡C–CH=CH₂

➡️ Carbinol (Methanol): CH₃OH

➡️ Isobutyric Acid: (CH₃)₂CH–COOH

➡️ Glyoxal: OHC–CHO or H–C(=O)–C(=O)–H

➡️ Isobutanoyl Bromide: (CH₃)₂CH–COBr

➡️ Adipic Acid: HOOC–(CH₂)₄–COOH

➡️ Cyclopentane: C₅H₁₀ or ⬟ (pentagon ring)

➡️ Diphenyl Ether: C₆H₅–O–C₆H₅

➡️ 1,2,3-Benzentriol (Pyrogallol): C₆H₃(OH)₃ (OH at C1, C2, C3)

➡️ Isopropyl Butanoate: CH₃CH₂CH₂–COO–CH(CH₃)₂

➡️ 2-Methoxy-2-methylbutane: CH₃–C(CH₃)(OCH₃)–CH₂CH₃

➡️ α,β-Dimethylvaleric Acid: CH₃–CH₂–CH(CH₃)–CH(CH₃)–COOH

➡️ Isopropyl Propionate: CH₃CH₂COO–CH(CH₃)₂

➡️ Ethyl Neo-pentyl Ether: (CH₃)₃C–CH₂–O–CH₂CH₃

➡️ Neo-pentyl Iodide: (CH₃)₃C–CH₂–I

➡️ Benzamide: C₆H₅–CONH₂ or C₆H₅–C(=O)–NH₂

➡️ Terephthalic Acid (Benzene-1,4-dioic acid): HOOC–C₆H₄–COOH

(Q9) Define β-elimination reactions. Mention its two types and give difference between them and Outline the mechanism of reaction between sec-butyl chloride & alcoholic KOH in the presence of polar protic and polar aprotic solvents?

✍️ Answer

🟥 Definition of β-Elimination Reaction / 1,2-Elimination

A β-elimination reaction involves the removal of a β-hydrogen and an electronegative group (like X⁻ or OH⁻) from two adjacent carbon atoms in a substrate (e.g., alkyl halides). This leads to the formation of an unsaturated compound (i.e., an alkene) by attacking with a strong base like KOH. 💥
Key Points:
➡️ 1,2-elimination: Hydrogen and leaving group (like halide or OH) are removed from adjacent carbons.
🎯 Produces an alkene (double bond).
➡️ Common examples include dehydrohalogenation of alkyl halides and acid-catalyzed dehydration of alcohols. 🔥

🟥 Types of Elimination Reaction (By Hughes and Ingold; 1941)

⚡📊 Bimolecular β-elimination reaction (E₂ reaction).
🔥⏳ Unimolecular β-elimination reaction (E₁ reaction).

🟥 Key Differences between E₁ and E₂

📌 1. E₁ (Unimolecular Elimination)
➡️ Two-step mechanism
➡️ Formation of carbocation intermediate
➡️ First order reaction (rate depends only on substrate; Rate law = k[RX])
➡️ Favored by polar protic solvents (e.g., alcohol, water)
➡️ Weak base is sufficient

📌 2. E₂ (Bimolecular Elimination)
➡️ One-step (concerted) mechanism
➡️ No carbocation intermediate
➡️ Second order reaction (rate depends on substrate + base; Rate law = k[RX][Base])
➡️ Favored by polar aprotic solvents
➡️ Favored by strong base

🟥 Reaction of Sec-Butyl Chloride with Alcoholic KOH In Polar Protic Solvent

➡️ Compound: 2-Chlorobutane (Sec-butyl chloride)
➡️ Products: Mainly But-2-ene (Saytzeff’s rule)
👉 Mechanism: E₁ In Polar Protic Solvent (Alcoholic KOH, e.g., ethanol)

📌 Mechanism Steps:
➡️ Ionization (Slow step): 2-Chlorobutane → Secondary carbocation + Cl⁻
➡️ Deprotonation: Base (ROH/KOH) removes β-hydrogen → Formation of alkene

🟥 Reaction of Sec-Butyl Chloride with Alcoholic KOH In Polar Aprotic Solvent

➡️ Compound: 2-Chlorobutane (Sec-butyl chloride)
➡️ Products: Mainly But-2-ene (Saytzeff’s rule)
👉 Mechanism: E₂ In Polar Aprotic Solvent (e.g., DMSO, acetone) with Strong Base

📌 Mechanism Steps (Single step):
➡️ Strong base removes β-hydrogen
➡️ C=C bond forms simultaneously
➡️ Cl⁻ leaves at the same time

(Q9 OR) What are proteins? Classify various types of proteins on the basis of their function and structures. Also give biological significance and properties of proteins

✍️ Answer

🟥 📖 Definition of Proteins 🧬

Proteins (from Proteios meaning “first”) are naturally occurring complex nitrogenous organic macromolecules made up of long chains of α-amino acids linked by peptide bonds (–CONH–).
They are high molecular weight polypeptides (≈ 34,000 to 5,000,000 Da) that fold into specific three-dimensional structures to perform biological functions.
👉 Proteins are built from 22 different α-amino acids, and their sequence determines structure and function.

🟥 Classification of Proteins Based on their Functions 🧪

1️⃣ Catalytic Proteins (Enzymes)
➡️ Speed up biochemical reactions
➡️ Highly specific biological catalysts
➡️ Example: Lipase (digests fats), protease (digests proteints), amylase (digests starch)

2️⃣ Storage Proteins
➡️ Store nutrients or metal ions
➡️ Examples: Albumin, globulin, Casein, Ferritin

3️⃣ Transport Proteins
➡️ Carry substances in blood or across cellular membranes
➡️ Example: Hemoglobin (transports oxygen)

4️⃣ Hormonal / Regulatory Proteins
➡️ Regulate body processes and cell signaling
➡️ Example: Insulin

5️⃣ Protective Proteins
➡️ Defend the body
➡️ Example: Antibodies (γ-globulins)

6️⃣ Contractile Proteins
➡️ Help in movement and muscle contraction
➡️ Examples: Actin and Myosin

🟥 Classification of Proteins Based on their Structure 🧪

1️⃣ Primary Structure
➡️ Linear sequence of amino acids
➡️ Determines overall protein nature

2️⃣ Secondary Structure
➡️ Local folding of polypeptide chain
➡️ Forms: α-Helix and β-Pleated sheet
➡️ Stabilized by hydrogen bonds

3️⃣ Tertiary Structure
➡️ Three-dimensional folding of a single polypeptide chain
➡️ Stabilized by: Disulfide bonds, Ionic bonds (salt bridges), Hydrogen bonds, van der Waals forces
➡️ Example: Myoglobin

4️⃣ Quaternary Structure
➡️ Association of two or more polypeptide subunits
➡️ Example: Hemoglobin (4 subunits)

🟥 Biological Importance of Proteins 🌟

✔️ Energy Source (1 g = 4 kcal)
✔️ Build and repair body tissues (Structural proteins)
✔️ Oxygen transport / Respiratory Function (Hemoglobin, Myoglobin)
✔️ Enzymatic catalysis / Catalytic Function
✔️ Hormonal regulation (Insulin)
✔️ Muscle contraction (Actin & Myosin)
✔️ Immunity (Antibodies like globulins and γ-globulins)
✔️ Storage of nutrients (Ferritin)
✔️ Genetic regulation (Nucleoproteins)

🟥 Properties of Proteins ⚗️

🔸 Water soluble (due to polar groups)
🔸 Amphoteric (contain –COOH and –NH₂ groups)
🔸 High molecular weight
🔸 Specific 3D structure
🔸 Can be denatured by heat, extreme pH, or chemicals
🔸 Some are colored (e.g., hemoglobin)
🔸 Proteins exhibit flexibility due to rotation around peptide bonds and amino acid side chains.

(Q10) Explain ONE simple laboratory test to distinguish between the any 4 of the following pair of compounds:

✍️ Answer

🟥 (i) 🧴 Alkanes vs Alkenes

🔹 Test: Bromine Water Test
Procedure: Add bromine water to the sample.
✅ Observation:
➡️ Alkene → Decolorizes bromine water (reddish-brown → colorless)
➡️ Alkane → No change (in absence of sunlight)
🧾 Equation: CH₂=CH₂ + Br₂ → BrCH₂–CH₂Br
👉 Alkenes undergo addition reaction; alkanes do not.

🟥 (ii) 🧪 Propanal vs Propanone

🔹 Test: Tollens’ Test
✅ Observation:
➡️ Propanal (aldehyde) → Silver mirror formed
➡️ Propanone (ketone) → No reaction
🧾 Equation: R–CHO + 2[Ag(NH₃)₂]OH → R–COONH₄ + 2Ag↓ + 3NH₃ + H₂O
👉 Aldehydes reduce Tollens’ reagent; ketones do not.

🟥 (iii) 🍷 Alcohol vs Phenol

🔹 Test: Neutral Ferric Chloride (FeCl₃) Test
✅ Observation:
➡️ Phenol → Violet/purple/blue color
➡️ Alcohol → No color change
🧾 Equation: 6C₆H₅OH + FeCl₃ → H₃[Fe(C₆H₅O)₆] + 3HCl
👉 Phenol forms colored complex (ferric phenoxide) with FeCl₃.

🟥 (iv) 🛢 n-Hexane vs Benzene

🔹 Test: Soot or Ignition Test
✅ Observation:
➡️ Benzene → Burns with sooty (smoky) flame 🔥
➡️ n-Hexane → Burns with clean, non-sooty flame
👉 Aromatic compounds burn with sooty flame due to high carbon content.

🟥 (v) ⚗️ 1-Butyne vs 2-Butyne

🔹 Test: Ammoniacal Silver Nitrate Test
✅ Observation:
➡️ 1-Butyne (terminal alkyne) → White precipitate (silver acetylide)
➡️ 2-Butyne (internal alkyne) → No precipitate
🧾 Equation: RC≡CH + AgNO₃ —(NH₃)→ RC≡CAg↓ + HNO₃
👉 Only terminal alkynes give this test.

🟥 (vi) 🧬 Alkenes vs Alkynes

🔹 Test: Ammoniacal Silver Nitrate Test
✅ Observation:
➡️ Alkyne (terminal) → White precipitate
➡️ Alkene → No precipitate
🧾 Equation: HC≡CH + 2AgNO₃ —(NH₃)→ HC≡CAg↓ + 2HNO₃
👉 Alkynes form metal acetylides; alkenes do not.

(Q11) What is meant by orientation of benzene? Explain ortho-para and meta directing groups. Write the equation for the preparation of the following compounds from benzene.

✍️ Answer

🟥 🧬 Orientation of Benzene

Orient means “to determine the position.”
👉 The process by which the first substituent decides the position of the second incoming group in benzene is called Orientation.

Orientation of benzene refers to the position (ortho, meta, para) at which a new substituent enters the benzene ring during an electrophilic substitution reaction, depending on the nature of the already attached group.
👉 In simple words: The first substituent directs the position of the second substituent in monosubstituted benzene.

🧬 Positions in benzene:
In monosubstituted benzene (C₆H₅–G), three disubstituted products are possible:
➡️ Ortho (o-) → 1,2-position
➡️ Meta (m-) → 1,3-position
➡️ Para (p-) → 1,4-position
👉 The relative amount of these isomers depends on the nature of the first substituent (G).

🟥 🔹 Ortho–Para Directing Groups

These groups direct the incoming electrophile to ortho and para positions.
✅ Examples: –CH₃, –OH, –NH₂, –Cl, –Br
👉 Usually electron-donating groups (EDG)
👉 Increase electron density in ring
Example: Toluene on nitration gives o- and p-nitrotoluene.

🟥 🔹 Meta Directing Groups

These groups direct substitution to the meta position.
✅ Examples: –NO₂, –COOH, –CHO, –SO₃H
👉 Usually electron-withdrawing groups (EWG)
👉 Decrease electron density in ring
Example: Nitrobenzene on nitration gives m-dinitrobenzene.

🟥 Preparation Equations

➡️ TNT (2,4,6-Trinitrotoluene):
C₆H₆ + CH₃Cl—(AlCl₃, alkylation)→ C₆H₅CH₃ + 3HNO₃ —(H₂SO₄,successive nitration)→ C₆H₂(NO₂)₃CH₃

➡️ m-Nitrotoluene:
C₆H₆ + + HNO₃ —(H₂SO₄, nitration)→ C₆H₅NO₂ + CH₃Cl—(AlCl₃, nitration with –NO₂ m-directin)→ m-C₆H₄(NO₂)CH₃

➡️ m-Nitrobenzoic Acid:
C₆H₆ + CH₃Cl—(AlCl₃, alkylation)→ C₆H₅CH₃ + [O] —(KMnO₄, Oxidation)→ C₆H₅COOH —(HNO₃/H₂SO₄, nitration with –COOH m-directing)→ m-C₆H₄(NO₂)COOH

➡️ o- and p-Nitrotoluene:
C₆H₆ + CH₃Cl—(AlCl₃, alkylation)→ C₆H₅CH₃ + HNO₃ —(H₂SO₄, nitration with –CH₃ o,p-directin)→ o-C₆H₄(NO₂)CH₃ + p-C₆H₄(NO₂)CH₃

(Q11 OR) Complete and balance any five of the following reactions:

✍️ Answer

➡️ Dehydration of Propanol 🧪
CH₃CH₂CH₂OH —(Excess Conc. H₂SO₄/170°C)→ CH₃–CH=CH₂ + H₂O

➡️ Dehydrohalogenation of Ethyl Bromide with Alcoholic KOH ⚗️
C₂H₅Br + KOH —(Alcohol/Heat)→ CH₂=CH₂ + KBr + H₂O

➡️ Sɴ of 1-bromopropane with Aqueous KOH 🔬
CH₃CH₂CH₂Br + KOH —Aqueous→ CH₃CH₂CH₂OH + KBr

➡️ Dehydrohalogenation of Vicinal Dibromide ⚡
BrC₂H₄Br + 2KOH —(Alcohol/Heat)→ HC≡CH + 2KBr + 2H₂O

➡️ Electrophilic Addition of Propene 🌟
CH₃CH=CH₂ + HBr → CH₃CH₂CH₂Br

➡️ Substitution Reaction of Acetylene with CuCl 🔧
HC≡CH + CuCl ⟶ CuC≡CCu + HCl

➡️ Electrophilic Addition of Ethyne with HBr ⚗️
HC≡CH + 2HBr ⟶ CH₂–CHBr₂

➡️ Substitution Reaction of Alcohol with SOCl₂ 🧴
R–OH + SOCl₂ ⟶ R–Cl + SO₂ + HCl

➡️ Carbonation of GR with CO₂ into Acid 🌬️
CH₃MgCl + CO₂ ⟶ CH₃CO₂MgX —(H₂O)→ CH₃CO₂H + Mg(OH)Cl

➡️ Reaction of Ethyl Bromide with AgNO₂ 🔬
C₂H₅Br + AgNO₂ ⟶ C₂H₅NO₂ + AgBr↓

➡️ Sɴ Reaction of Iodomethane with NaCN ⚡
CH₃I + NaCN ⟶ CH₃CN + NaI

➡️ Ozonolysis of Ethene 🌫️
CH₂=CH₂ + O₃ ⟶ ozonide ⟶ 2CH₂O

➡️ Aɴ Reaction of Acetaldehyde with HCN 🧪
CH₃CHO + HCN ⟶ CH₃CH(CN)OH

➡️ Oxidation of Formaldehyde 🔥
HCHO + [O] ⟶ HCOOH

➡️ Reaction of Acetone with Phenylhydrazine 🧬
CH₃COCH₃ + NH₂NHC₆H₅ ⟶ (CH₃)₂C=NNHC₆H₅ + H₂O

➡️ Reaction of Formaldehyde with Hydroxylamine 🧪
HCHO + NH₂OH ⟶ CH₂=NOH + H₂O

➡️ Friedel-Crafts Acylation of Benzene with Acetyl Chloride ⚗️
C₆H₆ + CH₃COCl —(AlCl₃)→ C₆H₅COCH₃ + HCl

(Q13 OR) Write down the IUPAC names of any 8 of the following organic compounds:

✍️ Answer

➡️ Compound 1
CH₂=CH−HC=CH−C≡CH 👉 IUPAC Name: Hexa-1,3-diene-5-yne

➡️ Compound 2
CH₃−CH=CH−CH₂−COOH 👉 IUPAC Name: Pent-2-enoic acid

➡️ Compound 3
CH₃–CH₂–OC(CH₃)₂C₂H₅ 👉 IUPAC Name: 2-Ethoxy-2-methylbutane

➡️ Compound 4
(CH₃)₂CH−CO−CH(C₂H₅)₂ 👉 IUPAC Name: 2-methyl-4-ethylhexan-3-one

➡️ Compound 5
CH₂=CH−C≡C−CH=CH₂ 👉 IUPAC Name: Hexa-1,5-diene-3-yne

➡️ Compound 6
CH₂=CH−CH₂−C≡CH₂ 👉 IUPAC Name: 1-penten-4-yne

➡️ Compound 7
CH₂=CH−(CH₂)₄−COOH 👉 IUPAC Name: 6-hexenoic acid

➡️ Compound 8
(CH₃)₃C.CO−CH₂CH₂−CHO 👉 IUPAC Name: 5-methyl-4-oxohexanal

➡️ Compound 9
CH₃COCCl₃ 👉 IUPAC Name: 1,1,1-trichloro-2-propanone

➡️ Compound 10
F₃C−COOH 👉 IUPAC Name: trifluoroacetic acid

➡️ Compound 11
OHC–CHO 👉 IUPAC Name: 1,2-ethanedial (acet dialdehyde)

➡️ Compound 12
(CH₃)₃C–CO–CH₂CH₃ 👉 IUPAC Name: 4,4-Dimethylpentan-3-one

➡️ Compound 13
(C₆H₅)₃C–CHO 👉 IUPAC Name: triphenylethanal

➡️ Compound 14
CH₂=CH–CH(OH)–CH₂–COOH 👉 IUPAC Name: 3-Hydroxy-4-pentenoic acid

➡️ Compound 15
C₂H₅–O–CH₂–C(CH₃)₃ 👉 IUPAC Name: 1-Ethoxy-2,2-dimethylpropane

➡️ Compound 16
C₂H₅–CH₂–COO–CH(CH₃)₂ 👉 IUPAC Name: isopropyl butyrate

➡️ Compound 17
(CH₃)₂CH–CO–C(CH₃)₃ 👉 IUPAC Name: 2,2,4-trimethylpentan-3-one

➡️ Compound 18
CH₃–(CHOH)₂–CH₂OH 👉 IUPAC Name: 1,2,3-butanetriol

➡️ Compound 19
(CH₂)₂(OH)₂ 👉 IUPAC Name: 1,2-ethanediol (ethylene glycol)