Model Test Questions XII Chemistry Chapter # 6 Alkyl Halides and Amines
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🧪📘 Model Test Questions — XII Chemistry
⚗️ Alkyl Halides and Amines
Chapter # 6 • Test # 8 ✨
🔬 Halides • 💊 Amines • 🧠 Mechanisms • 🧪 Reactions
🧪 Short Questions of Alkyl Halides
Q1. How are alkyl halides prepared by the reaction of alcohol with?
(i) HX (ii) PX₃ (iii) SOCl₂. Give the equations.
Q2. How would you prepare alkyl halide from alkenes and alkanes?
Q3. Draw all the possible isomers of an alkyl halide with composition C₅H₁₁Cl (8 isomers).
Q4. Why the alkyl part of Grignard’s reagent is nucleophilic in nature? OR Give reason of high reactivity of Grignard’s reagent.
Q5. How can you define a nucleophile? Write the names of four nucleophiles with their typical reagents.
Q6. Convert the following:
(a) Methyl magnesium bromide into acetone
(b) Ethyl chloride into ethyl amine
(c) Ethyl chloride into ethyl alcohol
(d) Ethyl amine into imine
Q7. How will you obtain the following?
(i) Ethane from methyl magnesium chloride [Alkylation]
(ii) Ethanoic acid from methyl magnesium chloride [Carbonation]
(iii) tertiarybutyl alcohol from ketone [Nucleophilic carbonyl addition of GR]
(iv) Ethyl alcohol from methyl magnesium iodide [Nucleophilic carbonyl addition of GR]
(v) secondary alcohol from Grignard’s reagent [Nucleophilic carbonyl addition of GR]
Q8. How can you justify the fact that alkyl halides are water insoluble?
Q9. Explain the following by giving scientific reasons:
▶ Why β-elimination reactions are not possible in methyl halides? (due to lack of β-H)
▶ Why Sɴ₂ reaction is not favourable in tertiary alkyl halides? (due to steric hindrance)
▶ Why tertiary carbocation is more stable than secondary and primary carbocations?
▶ E₂ reaction is of 2ᵑᵈ order while that of E₁ is of 1st order. (E₂ is bimolecular and E₁ unimolecular)
▶ 2° alkyl halides give Sɴ₁ mechanism in presence of polar solvent. (polar solvents help in ionization)
▶ Alkyl group behaves as a nucleophile in Grignard’s reagent. (EP Mg makes R nucleophilic)
▶ Why the mechanism of Sɴ₂ reaction completes in one step?
Q10. Why alkyl halide undergoes nucleophilic substitution reaction? Which reagent is required to convert a methyl iodide into:
(i) Methanol (ii) Methyl cyanide (iii) Dimethyl ether (iv) Thiol
💊 Short Questions of Amines
Q1. Why are secondary and tertiary amines more alkaline than primary amines?
Q2. How is primary amine converted into secondary and tertiary amines, give the equations.
Q3. Give an account on the basicity of amines.
Q4. How are amines prepared from nitriles, give the equations.
📘 Descriptive Questions
Q1. How can you define nucleophilic substitution reactions? Describe the mechanisms of Sɴ₁ and Sɴ₂ reactions. Write down 5 differences between Sɴ₁ and Sɴ₂.
OR
Outline the step-wise reaction mechanism of the following:
(i) Sɴ₂ reaction between Bromomethane and NaOH
(ii) Sɴ₁ reaction between 2-chloro-2-methylpropane and NaCN
Q2. What is β-elimination? Discuss the mechanisms of E₁ and E₂ reactions.
Q3. What are organometallic compounds? What are the key properties of organometallic compounds? How is Grignard’s reagent prepared? Write down the reactions of Grignard’s reagent with water, carbon dioxide, ester and amines.
OR
How do you convert Grignard’s reagent into three types of alcohols? Give only general equations.
Q4. What are Alkyl Halides? How are they classified? Define primary, secondary and tertiary alkyl halides. Give their general structures and examples.
Q5. Draw the orbital structure of methyl iodide and explain the type of hybridization in it.
Q6. Give a comparative study between nucleophilic substitution reactions and elimination reactions of alkyl halides.
Q7. Complete and balance the following reactions:
R–MgX + CO₂ → R–CO₂MgX → R–CO₂H + MgX₂
R–OH + SOCl₂ → R–OCl + SO₂ + HCl
H₃C–MgCl + CO₂ → H₃C–CO₂MgX → H₃C–CO₂H + MgX₂
H₅C₂–Br + AgNO₂ → H₅C₂–NO₂ + AgBr
R–OH + PCl₅ → R–Cl + POCl₃ + HCl
H₃C–I + NaCN → H₃C–CN + NaI
R–MgX + H₃C–NH₂ → R–H + Mg(H₃C–NH)X
R–MgX + H–OH → R–H + Mg(OH)X
🧪 Multiple Choice Questions on Alkyl Halides and Amines 📘
Q1. Which of the following composition justifies the secondary alkyl halide?
(a) R₃CX
(b) R₂CHX ✅
(c) RCH₂X
(d) CH₃X
✔ Correct: (b) R₂CHX → carbon attached to two alkyl groups = secondary halide.
Q2. Which alkyl halide cannot produce an alkene with alcoholic KOH?
(a) Methyl bromide ✅
(b) Ethyl bromide
(c) Propyl bromide
(d) Butyl bromide
✔ Correct: (a) Methyl bromide → no β-hydrogen, so β-elimination not possible.
Q3. Ethyl magnesium bromide with CO₂ yields:
(a) Methanoic acid
(b) Ethanoic acid ✅
(c) Propanoic acid
(d) Butanoic acid
✔ Correct: (b) Ethanoic acid → carbonation of ethyl magnesium bromide gives CH₃COOH.
Q4. Grignard’s reagent with ester produces:
(a) Aldehyde
(b) Carboxylic acid
(c) Ketone ✅
(d) Ether
✔ Correct: (c) Ketone → Grignard reagent reacts with ester to form ketones.
Q5. Amine act as bases because:
(a) They accept OH⁻
(b) They donate OH⁻
(c) They donate H⁺
(d) They accept H⁺ ✅
✔ Correct: (d) Amines accept protons via lone pair on nitrogen → basic character.
Q6. The structure of primary amine is:
(a) Planar trigonal
(b) Regular tetrahedral
(c) Tetrahedral pyramidal ✅
(d) Linear
✔ Correct: (c) Lone pair on N makes geometry tetrahedral pyramidal.
Q7. Alkyl amine reacts with nitrous acid + HCl to yield:
(a) Diazonium salt ✅
(b) Aldehyde
(c) Ketone
(d) Alcohol
✔ Correct: (a) Primary amines form diazonium salts with nitrous acid.
Q8. Sɴ₂ reaction occurs most easily if the substrate molecule is:
(a) Methyl iodide ✅
(b) Ethyl iodide
(c) 2-iodopropane
(d) 2-iodobutane
✔ Correct: (a) Methyl iodide → least steric hindrance, fastest SN2 reaction.
Q9. Suitable reagent required for the synthesis of propane from methyl magnesium iodide is:
(a) H₂O
(b) NH₃
(c) CH₃OH
(d) CH₃NH₂
❌ All given options are incorrect.
✔ Correct option: C₂H₅Cl → reacts with CH₃MgI to give propane (via alkylation).
Q10. 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: (a) Rate-determining step is carbocation formation → depends only on substrate concentration.
📘 MCQs on Alkyl Halides and Amines (Past Papers)
Q1. Which of the following would not yield alkane on reaction with Grignard’s reagent?
(a) Ammonia
(b) Primary amine
(c) Secondary amine
(d) Tertiary amine ✅
✔ Correct: (d) Tertiary amine → lacks active hydrogen, cannot form alkane with Grignard’s reagent.
Q2. Grignard’s reagent gives primary alcohol on reaction with:
(a) Methanal ✅
(b) Ethanal
(c) Acetone
(d) CO₂
✔ Correct: (a) Methanal → addition of Grignard reagent gives primary alcohol.
Q3. Grignard’s reagent adds to the C=O bond of carbonyl compounds to form:
(a) Magnesium halide alkoxides ✅
(b) Secondary alcohol
(c) Primary alcohol
(d) Tertiary alcohol
✔ Correct: (a) First product is magnesium halide alkoxide, which on hydrolysis gives alcohol.
Q4. Which one of the following is NOT used for identification of alkyl halide in the laboratory?
(a) Ammonical silver nitrate
(b) Silver oxide
(c) Silver nitrite
(d) Silver sulphate ✅
✔ Correct: (d) Silver sulphate → not used for alkyl halide identification.
Q5. Which of the following will produce alcohol by the action of alkyl halide on?
(a) Alcoholic caustic potash
(b) Aqueous ammonia
(c) Aqueous caustic potash ✅
(d) Water
✔ Correct: (c) Aqueous caustic potash → nucleophilic substitution gives alcohol.
Q6. Sɴ₂ reactions complete in ______ steps.
(a) 1 ✅
(b) 2
(c) 3
(d) 4
✔ Correct: Sɴ₂ is a single-step mechanism with simultaneous bond breaking and bond forming.
Q7. Sɴ₁ reactions complete in ______ steps.
(a) 1
(b) 2 ✅
(c) 3
(d) 4
✔ Correct: Sɴ₁ occurs in two steps → carbocation formation, then nucleophilic attack.
Q8. Which of the following undergo Sɴ₁ reactions?
(a) (CH₃)₂CH–Cl
(b) CH₃Cl
(c) (CH₃)₃C–Cl ✅
(d) All of them
✔ Correct: (c) Tertiary halides undergo Sɴ₁ easily due to stable carbocation formation.
Q9. The composition of sodium-lead alloy is:
(a) Na₄Pb✅
(b) Na₂Pb
(c) Na–Pb
(d) NaPb₂
✔ Correct: (a) Na₄Pb is the known sodium-lead alloy composition.
Q10. A primary alkyl halide would prefer to undergo _____________.
(a) Sɴ₁ reaction
(b) Sɴ₂ reaction ✅
(c) α–Elimination
(d) Racemization
✔ Correct: (b) Primary halides undergo Sɴ₂ due to less steric hindrance and unstable carbocation.
Q11. Which of the following alkyl halides will undergo SN1 reaction most readily?
(a) (CH₃)₃C—F
(b) (CH₃)₃C—Cl
(c) (CH₃)₃C—Br
(d) (CH₃)₃C—I ✅
✔ Correct: (d) Tertiary iodide → weakest C–I bond, stable carbocation → fastest SN1 reaction.
Q12. Tertiary alkyl halides are practically inert to substitution by SN2 mechanism because of:
(a) Insolubility
(b) Instability
(c) Inductive effect
(d) Steric hindrance ✅
✔ Correct: (d) Steric hindrance → bulky groups block backside attack required in SN2.
Q13. For which mechanisms, the first step involved is the same:
(a) E₁ and E2
(b) E₁ and Sɴ₁ ✅
(c) E₂ and Sɴ₁
(d) E₂ and Sɴ₂
✔ Correct: (b) Both E₁ and Sɴ₁ start with carbocation formation as the first step.
Q14. The alcohol-free and moisture-free organic solvent used in Grignard’s reagent preparation is called:
(a) Dry acid
(b) Dry ether
(c) Dry anhydrous ether ✅
(d) Both b and c
✔ Correct: (c) Dry anhydrous ether → prevents decomposition of Grignard reagent by moisture.
Q15. Which of the following reactions are favoured by polar aprotic solvent?
(a) Both Sɴ₁ and Sɴ₂
(b) Sɴ₂ reactions ✅
(c) Sɴ₁ reactions
(d) None of them
✔ Correct: (b) Polar aprotic solvents stabilize cations but not anions → enhance Sɴ₂ nucleophilic attack.
Q16. Which one of the following do not form Grignard’s reagent?
(a) CH₃Cl
(b) CH₃Br
(c) CH₃I
(d) CH₃F ✅
✔ Correct: (d) CH₃F → C–F bond is too strong, does not react with Mg to form Grignard reagent.
Q17. Which C–X bond has the highest bond energy per mole?
(a) C–F ✅
(b) C–Br
(c) C–Cl
(d) C–I
✔ Correct: (a) C–F bond is the strongest due to high electronegativity and small bond length.
Q18. Neo-pentyl bromide refers to follow which mechanism during substitution reactions?
(a) Not show Sɴ reactions ✅
(b) Sɴ₁
(c) Sɴ₂
(d) Both a and c
✔ Correct: (c) Neo-pentyl bromide → being primary, undergoes Sɴ₂ easily.
Q19. If a nucleophile is the attacking reagent, which one would be the most reactive?
(a) R–F
(b) R–Cl
(c) R–Br
(d) R–I ✅
✔ Correct: (d) R–I → weakest C–I bond, best leaving group → most reactive in nucleophilic substitution.
Q20. CH₃–Cl can show which of the following reaction with easiness?
(a) Sɴ₂ ✅
(b) Sɴ₁
(c) E₂
(d) Both a and b
✔ Correct: (a) Primary halide (CH₃Cl) → undergoes Sɴ₂ easily due to minimal steric hindrance.
Q21. Neutral nucleophiles among the following is:
(a) CN⁻
(b) NH₃ ✅
(c) Cl⁻
(d) C₂H₅O⁻
✔ Correct: (b) Ammonia (NH₃) → neutral molecule with lone pair, acts as nucleophile without charge.
Q22. 100% inversion of configuration takes place during:
(a) Sɴ₁
(b) Sɴ₂ ✅
(c) E₁
(d) E₂
✔ Correct: (b) Sɴ₂ → backside attack causes complete inversion of configuration (Walden inversion).
Q23. During Sɴ₂ mechanism carbon atom changes its state of hybridization as:
(a) sp → sp²
(b) sp² → sp³
(c) sp³ → sp
(d) sp³ → sp² ✅
✔ Correct: (d) Sɴ₂ transition state → carbon shifts from sp³ to partial sp² hybridization during attack.
📘 Smart Answers of XII Chemistry Model Test Questions
Short Questions — Chapter #6: Alkyl Halides and Amines
❓ Q1. How are alkyl halides prepared by the reaction of alcohol with (i) HX, (ii) PX₃, (iii) SOCl₂? Give the equations.
✅ Answer:
🧪 Preparation of Alkyl Halide by the Action of Halogen Acids (HX) on Alcohols
• In presence of anhydrous ZnCl₂ as catalyst.
• General Reaction: R–OH + HX → (ZnCl₂) → R–X + H₂O
📊 Order of Reactivity of Halogen Acids: HI > HBr > HCl
📈 Order of Reactivity of Alcohols: Tertiary > Secondary > Primary
🧪 Lucas Reagent: Mixture of conc. HCl + ZnCl₂ → used to test alcohol reactivity.
🎯 Purpose: Distinguish between primary, secondary, and tertiary alcohols based on reaction rate.
👀📊 Lucas Test Observations:
➡️ Tertiary Alcohol: 💥 Immediate cloudy ppt of alkyl halide
➡️ Secondary Alcohol: ⏱️ Turbidity in 5–10 minutes
➡️ Primary Alcohol: 🔥 No visible reaction unless heated
🧪 Equations:
• CH₃OH + HCl → (ZnCl₂, heat) → CH₃Cl + H₂O
• CH₃CH₂OH + HCl → (ZnCl₂, heat) → CH₃CH₂Cl + H₂O
• CH₃CHOHCH₃ + HCl → (ZnCl₂, 5–10 min) → CH₃CHClCH₃ + H₂O
• (CH₃)₃COH + HCl → (no catalyst, immediate) → (CH₃)₃CCl + H₂O
🎯📋 Conclusion: Lucas Test helps identify the type of alcohol by observing how quickly it forms an alkyl halide.
🧪 Preparation of Alkyl Halide by Halogenated Agents (PX₃, PX₅, SOCl₂)
🔬 Concept: Alcohols (R–OH) can be converted into alkyl halides (R–X) using reagents like SOCl₂, PCl₃, PCl₅.
⚗️ Reagents & Conditions:
➡️ SOCl₂ (Thionyl chloride): ✅ Preferred method — products (SO₂ & HCl) are gases. Pyridine absorbs HCl.
➡️ PX₃ (Phosphorus trihalides): 🔁 Reacts with 3 alcohol molecules → alkyl halide + H₃PO₃.
➡️ PX₅ (Phosphorus pentahalides): 🔥 Reacts with 1 alcohol molecule → alkyl halide + HX + POX₃.
🧪 Chemical Equations:
• R–OH + SOCl₂ → R–Cl + SO₂ + HCl (in presence of pyridine)
• 3R–OH + PX₃ → 3R–X + H₃PO₃
• R–OH + PX₅ → R–X + HX + POX₃
📌 Conclusion:
These methods efficiently convert alcohols into alkyl halides. ✅ Thionyl chloride is preferred due to gaseous by-products and cleaner separation. Pyridine acts as a base to neutralize HCl and drive the reaction forward.
🧪 Preparation of Alkyl Halide by the Action of Halogen Acids (HX) on Alcohols
• In presence of anhydrous ZnCl₂ as catalyst.
• General Reaction: R–OH + HX → (ZnCl₂) → R–X + H₂O
📊 Order of Reactivity of Halogen Acids: HI > HBr > HCl
📈 Order of Reactivity of Alcohols: Tertiary > Secondary > Primary
🧪 Lucas Reagent: Mixture of conc. HCl + ZnCl₂ → used to test alcohol reactivity.
🎯 Purpose: Distinguish between primary, secondary, and tertiary alcohols based on reaction rate.
👀📊 Lucas Test Observations:
➡️ Tertiary Alcohol: 💥 Immediate cloudy ppt of alkyl halide
➡️ Secondary Alcohol: ⏱️ Turbidity in 5–10 minutes
➡️ Primary Alcohol: 🔥 No visible reaction unless heated
🧪 Equations:
• CH₃OH + HCl → (ZnCl₂, heat) → CH₃Cl + H₂O
• CH₃CH₂OH + HCl → (ZnCl₂, heat) → CH₃CH₂Cl + H₂O
• CH₃CHOHCH₃ + HCl → (ZnCl₂, 5–10 min) → CH₃CHClCH₃ + H₂O
• (CH₃)₃COH + HCl → (no catalyst, immediate) → (CH₃)₃CCl + H₂O
🎯📋 Conclusion: Lucas Test helps identify the type of alcohol by observing how quickly it forms an alkyl halide.
🧪 Preparation of Alkyl Halide by Halogenated Agents (PX₃, PX₅, SOCl₂)
🔬 Concept: Alcohols (R–OH) can be converted into alkyl halides (R–X) using reagents like SOCl₂, PCl₃, PCl₅.
⚗️ Reagents & Conditions:
➡️ SOCl₂ (Thionyl chloride): ✅ Preferred method — products (SO₂ & HCl) are gases. Pyridine absorbs HCl.
➡️ PX₃ (Phosphorus trihalides): 🔁 Reacts with 3 alcohol molecules → alkyl halide + H₃PO₃.
➡️ PX₅ (Phosphorus pentahalides): 🔥 Reacts with 1 alcohol molecule → alkyl halide + HX + POX₃.
🧪 Chemical Equations:
• R–OH + SOCl₂ → R–Cl + SO₂ + HCl (in presence of pyridine)
• 3R–OH + PX₃ → 3R–X + H₃PO₃
• R–OH + PX₅ → R–X + HX + POX₃
📌 Conclusion:
These methods efficiently convert alcohols into alkyl halides. ✅ Thionyl chloride is preferred due to gaseous by-products and cleaner separation. Pyridine acts as a base to neutralize HCl and drive the reaction forward.
Q2. 🌈 How would you prepare alkyl halide from alkenes and alkanes? 🌈
✨ Preparation of Alkyl halide By the Photochemical Halogenation of Alkane ☀️
⚡ Alkanes react with halogens in the presence of sunlight (hν) or high temperature to form alkyl halides.
⚡ This is a substitution reaction, where hydrogen atoms are replaced by halogen atoms.
📌 General Reaction: Alkane + Cl₂ —hν→ Alkyl halides + HCl
🧪 Example Reactions:
➡️ CH₄ + Cl₂ —hν→ CH₃Cl + CH₂Cl₂ + CHCl₃ + CCl₄ + HCl
➡️ CH₃–CH₃ + Cl₂ —hν→ H₃C–CH₂Cl + H₃C–CHCl₂ + H₃C–CCl₃ + ClCH₂–CH₂Cl + ClCH₂–CHCl₂ + ClCH₂–CCl₃ + Cl₂CH–CHCl₂ + Cl₂CH–CCl₃ + Cl₃C–CCl₃ + HCl
⚠️ Note:
This method gives a mixture of mono‑, di‑, and poly‑halogenated products, making separation difficult. Hence, it is not a very good method for preparing pure alkyl halides.
✨ Preparation of Alkyl halide By the Hydrohalogenation of Alkene 🔬
⚡ Alkenes react with hydrogen halides (HX) to form alkyl halides by addition reaction.
📌 General Reaction: >C=C< + HX → >CH–XC<
🧪 Examples:
🔹 Symmetrical Alkene: H₂C=CH₂ (Ethene) + HI → H₃C–CH₂–I (Iodoethane)
🔹 Unsymmetrical Alkene (Markovnikov’s Rule): H₃C–HC=CH₂ (Propene) + HBr → H₃C–CH(Br)–CH₃ (2‑bromopropane)
📊 Reactivity Order of Hydrogen Halides: HI > HBr > HCl
⚡ Alkanes react with halogens in the presence of sunlight (hν) or high temperature to form alkyl halides.
⚡ This is a substitution reaction, where hydrogen atoms are replaced by halogen atoms.
📌 General Reaction: Alkane + Cl₂ —hν→ Alkyl halides + HCl
🧪 Example Reactions:
➡️ CH₄ + Cl₂ —hν→ CH₃Cl + CH₂Cl₂ + CHCl₃ + CCl₄ + HCl
➡️ CH₃–CH₃ + Cl₂ —hν→ H₃C–CH₂Cl + H₃C–CHCl₂ + H₃C–CCl₃ + ClCH₂–CH₂Cl + ClCH₂–CHCl₂ + ClCH₂–CCl₃ + Cl₂CH–CHCl₂ + Cl₂CH–CCl₃ + Cl₃C–CCl₃ + HCl
⚠️ Note:
This method gives a mixture of mono‑, di‑, and poly‑halogenated products, making separation difficult. Hence, it is not a very good method for preparing pure alkyl halides.
✨ Preparation of Alkyl halide By the Hydrohalogenation of Alkene 🔬
⚡ Alkenes react with hydrogen halides (HX) to form alkyl halides by addition reaction.
📌 General Reaction: >C=C< + HX → >CH–XC<
🧪 Examples:
🔹 Symmetrical Alkene: H₂C=CH₂ (Ethene) + HI → H₃C–CH₂–I (Iodoethane)
🔹 Unsymmetrical Alkene (Markovnikov’s Rule): H₃C–HC=CH₂ (Propene) + HBr → H₃C–CH(Br)–CH₃ (2‑bromopropane)
📊 Reactivity Order of Hydrogen Halides: HI > HBr > HCl
Q3. Draw all the possible isomers of an alkyl halide with composition C₅H₁₁Cl (8 isomers).
🧪 n‑Pentane skeleton (positional isomers)
1️⃣ 1‑Chloropentane — CH₃–CH₂–CH₂–CH₂–CH₂Cl
2️⃣ 2‑Chloropentane — CH₃–CH₂–CH₂–CH(Cl)–CH₃
3️⃣ 3‑Chloropentane — CH₃–CH₂–CH(Cl)–CH₂–CH₃
🧩 2‑Methylbutane skeleton (positional isomers)
4️⃣ 1‑Chloro‑2‑methylbutane — CH₃–CH(CH₃)–CH₂–CH₂Cl
5️⃣ 2‑Chloro‑2‑methylbutane — CH₃–C(Cl)(CH₃)–CH₂–CH₃
6️⃣ 3‑Chloro‑2‑methylbutane — CH₃–CH(CH₃)–CH(Cl)–CH₃
7️⃣ 1‑Chloro‑3‑methylbutane — CH₃–CH₂–CH(CH₃)–CH₂Cl
🧱 2,2‑Dimethylpropane skeleton (neopentyl)
8️⃣ 1‑Chloro‑2,2‑dimethylpropane (neopentyl chloride) — (CH₃)₃C–CH₂Cl
🌟 Note: These eight are the distinct constitutional isomers for C₅H₁₁Cl (primary, secondary, and tertiary centers across three carbon skeletons).![]()
1️⃣ 1‑Chloropentane — CH₃–CH₂–CH₂–CH₂–CH₂Cl
2️⃣ 2‑Chloropentane — CH₃–CH₂–CH₂–CH(Cl)–CH₃
3️⃣ 3‑Chloropentane — CH₃–CH₂–CH(Cl)–CH₂–CH₃
🧩 2‑Methylbutane skeleton (positional isomers)
4️⃣ 1‑Chloro‑2‑methylbutane — CH₃–CH(CH₃)–CH₂–CH₂Cl
5️⃣ 2‑Chloro‑2‑methylbutane — CH₃–C(Cl)(CH₃)–CH₂–CH₃
6️⃣ 3‑Chloro‑2‑methylbutane — CH₃–CH(CH₃)–CH(Cl)–CH₃
7️⃣ 1‑Chloro‑3‑methylbutane — CH₃–CH₂–CH(CH₃)–CH₂Cl
🧱 2,2‑Dimethylpropane skeleton (neopentyl)
8️⃣ 1‑Chloro‑2,2‑dimethylpropane (neopentyl chloride) — (CH₃)₃C–CH₂Cl
🌟 Note: These eight are the distinct constitutional isomers for C₅H₁₁Cl (primary, secondary, and tertiary centers across three carbon skeletons).
Q4. ✨ Why the alkyl part of Grignard’s reagent is nucleophilic in nature? ✨ OR ✨ Give reason of high reactivity of Grignard’s reagent. ✨
🧪 Reason for High Reactivity of Grignard’s Reagent (RMgX) / Nucleophilic Nature of Grignard’s Reagent:
🔗 The C–Mg bond in Grignard’s reagent is highly polar.
⚡ Carbon is more electronegative than magnesium, so:
➡️ Carbon (R–) gets δ⁻ (partial negative charge)
➡️ Magnesium (MgX) gets δ⁺ (partial positive charge)
🧠 Thus, the alkyl carbon behaves like a carbanion (R⁻).
🎯 This electron‑rich carbon acts as a nucleophile and attacks electrophilic carbon atoms, forming C–C bonds.
🔥 The unusual negative character on carbon makes Grignard’s reagents highly reactive.
🔄 Therefore, they mainly undergo nucleophilic substitution (SN) and nucleophilic addition (AN) reactions, which are mostly exothermic.
✅ Conclusion:
The alkyl part of Grignard’s reagent is nucleophilic due to the polar C–Mg bond, which gives the carbon a negative charge, making it highly reactive toward electrophiles.![]()
🔗 The C–Mg bond in Grignard’s reagent is highly polar.
⚡ Carbon is more electronegative than magnesium, so:
➡️ Carbon (R–) gets δ⁻ (partial negative charge)
➡️ Magnesium (MgX) gets δ⁺ (partial positive charge)
🧠 Thus, the alkyl carbon behaves like a carbanion (R⁻).
🎯 This electron‑rich carbon acts as a nucleophile and attacks electrophilic carbon atoms, forming C–C bonds.
🔥 The unusual negative character on carbon makes Grignard’s reagents highly reactive.
🔄 Therefore, they mainly undergo nucleophilic substitution (SN) and nucleophilic addition (AN) reactions, which are mostly exothermic.
✅ Conclusion:
The alkyl part of Grignard’s reagent is nucleophilic due to the polar C–Mg bond, which gives the carbon a negative charge, making it highly reactive toward electrophiles.
✨ Short Answer: ✨
➡️ In Grignard’s reagent (RMgX), the C–Mg bond is highly polar.
➡️ Carbon is more electronegative, so it gets a partial negative charge (δ⁻), while Mg gets δ⁺.
➡️ The alkyl carbon behaves like a carbanion (nucleophile).
➡️ This makes Grignard’s reagent highly reactive and able to attack electrophilic centres to form C–C bonds.
➡️ Carbon is more electronegative, so it gets a partial negative charge (δ⁻), while Mg gets δ⁺.
➡️ The alkyl carbon behaves like a carbanion (nucleophile).
➡️ This makes Grignard’s reagent highly reactive and able to attack electrophilic centres to form C–C bonds.
Q5. ✨ How can you define a nucleophile? Write the names of four nucleophiles with their typical reagents. ✨
✅ Answer
🔬 Definition:
A nucleophile is a chemical species that donates a pair of electrons to an electrophilic atom to form a covalent (coordinate) bond. OR Nucleophiles are electron‑rich species that attack electrophilic centres by donating an electron pair.
🧪 Examples of Nucleophiles with Typical Reagents:
🟢 OH⁻ (Hydroxide ion) → NaOH or KOH
🟢 SH⁻ (Sulphide ion) → NaSH or KSH
🟢 OR⁻ (Alkoxide ion) → NaOR
🟢 NH₂⁻ (Amide ion) → NaNH₂
🟢 RCOO⁻ (Carboxylate ion) → RCOONa
🟢 CN⁻ (Cyanide ion) → NaC≡N or NaCN
🟢 I⁻ (Iodide ion) → KI
🟢 N₃⁻ (Azide ion) → NaN₃
🟢 SR⁻ (Thiolate ion) → NaSR
🔬 Definition:
A nucleophile is a chemical species that donates a pair of electrons to an electrophilic atom to form a covalent (coordinate) bond. OR Nucleophiles are electron‑rich species that attack electrophilic centres by donating an electron pair.
🧪 Examples of Nucleophiles with Typical Reagents:
🟢 OH⁻ (Hydroxide ion) → NaOH or KOH
🟢 SH⁻ (Sulphide ion) → NaSH or KSH
🟢 OR⁻ (Alkoxide ion) → NaOR
🟢 NH₂⁻ (Amide ion) → NaNH₂
🟢 RCOO⁻ (Carboxylate ion) → RCOONa
🟢 CN⁻ (Cyanide ion) → NaC≡N or NaCN
🟢 I⁻ (Iodide ion) → KI
🟢 N₃⁻ (Azide ion) → NaN₃
🟢 SR⁻ (Thiolate ion) → NaSR
Q6. ✨ Convert the following: ✨
➡️ (a) Methyl magnesium bromide → Acetone 🔹
🧪 ⚡ Nucleophilic addition to ester yielding halo magnesium dialkoxide as adduct followed by hydrolysis
H₃Cᵟ⁻Mgᵟ⁺Br + CH₃Cᵟ⁺OOᵟ⁻C₂H₅ → (CH₃)₂C(O⁻Mg⁺Br)OC₂H₅ — H₃O⁺ → CH₃–CO–CH₃ (Acetone) + Mg(OC₂H₅)Br
➡️ (b) Ethyl chloride → Ethyl amine 🔹
🧪 ⚡ Nucleophilic substitution (SN) by –NH₂ group of NH₃ (ammonia)
CH₃CH₂ᵟ⁺Clᵟ⁻ + Hᵟ⁺–ᵟ⁻NH₂ — SN₂ → HCl + C₂H₅NH₂ (Ethyl amine; 1° amine)
➡️ (c) Ethyl chloride → Ethyl alcohol 🔹
🧪 ⚡ Nucleophilic substitution (SN) by –OH group of aqueous KOH
CH₃CH₂ᵟ⁺Clᵟ⁻ + K⁺OH⁻ (aq) → KCl + CH₃CH₂OH (Ethyl alcohol; 1° alcohol)
➡️ (d) Ethyl amine → Imine (Schiff base) 🔹
⚡ Nucleophilic addition–elimination reaction of primary amine with aldehyde/ketone
R–NH₂ + R′–CHO — condensation (elimination) → R′–CH=NR + H₂O
R–NH₂ + R′–CO–R″ — condensation (elimination) → R′–C(R″)=NR + H₂O
![]()
🧪 ⚡ Nucleophilic addition to ester yielding halo magnesium dialkoxide as adduct followed by hydrolysis
H₃Cᵟ⁻Mgᵟ⁺Br + CH₃Cᵟ⁺OOᵟ⁻C₂H₅ → (CH₃)₂C(O⁻Mg⁺Br)OC₂H₅ — H₃O⁺ → CH₃–CO–CH₃ (Acetone) + Mg(OC₂H₅)Br
➡️ (b) Ethyl chloride → Ethyl amine 🔹
🧪 ⚡ Nucleophilic substitution (SN) by –NH₂ group of NH₃ (ammonia)
CH₃CH₂ᵟ⁺Clᵟ⁻ + Hᵟ⁺–ᵟ⁻NH₂ — SN₂ → HCl + C₂H₅NH₂ (Ethyl amine; 1° amine)
➡️ (c) Ethyl chloride → Ethyl alcohol 🔹
🧪 ⚡ Nucleophilic substitution (SN) by –OH group of aqueous KOH
CH₃CH₂ᵟ⁺Clᵟ⁻ + K⁺OH⁻ (aq) → KCl + CH₃CH₂OH (Ethyl alcohol; 1° alcohol)
➡️ (d) Ethyl amine → Imine (Schiff base) 🔹
⚡ Nucleophilic addition–elimination reaction of primary amine with aldehyde/ketone
R–NH₂ + R′–CHO — condensation (elimination) → R′–CH=NR + H₂O
R–NH₂ + R′–CO–R″ — condensation (elimination) → R′–C(R″)=NR + H₂O
Q7. How will you obtain the following? 🧪✨
(i) Ethane from methyl magnesium chloride [Alkylation]
(ii) Ethanoic acid from methyl magnesium chloride [Carbonation]
(iii) tertiarybutyl alcohol from ketone [Nucleophilic carbonyl addition of GR]
(iv) Ethyl alcohol from methyl magnesium iodide [Nucleophilic carbonyl addition of GR]
(v) secondary alcohol from Grignard’s reagent [Nucleophilic carbonyl addition of GR]
(ii) Ethanoic acid from methyl magnesium chloride [Carbonation]
(iii) tertiarybutyl alcohol from ketone [Nucleophilic carbonyl addition of GR]
(iv) Ethyl alcohol from methyl magnesium iodide [Nucleophilic carbonyl addition of GR]
(v) secondary alcohol from Grignard’s reagent [Nucleophilic carbonyl addition of GR]
🧪 (i) Ethane from Methyl Magnesium Chloride ⚡🖇️
⚡ Alkylation/ methylation (Nucleophilic substitution (Sɴ) of halide of alkyl halide)
✨ Reagent: methyl chloride (CH₃Cl)
H₃Cᵟ⁻Mgᵟ⁺Cl + H₃Cᵟ⁺Clᵟ⁻ — Sɴ₂ → MgCl₂ + CH₃–CH₃ (Ethane)
🧪 (ii) Ethanoic Acid from Methyl Magnesium Chloride 💨💧
💨 Carbonation / nucleophilic addition of CO₂ followed by Acidification to yield Ethanoic acid
🧬 Halo magnesium carboxylate as an Intermediate adduct
H₃Cᵟ⁻Mgᵟ⁺Cl + Cᵟ⁺Oᵟ⁻₂ — Aɴ → CH₃COO⁻Mg⁺Cl — H₃O⁺ or H₂O/HX → Mg(OH)Cl + CH₃COOH (acetic acid)
🧪 (iii) Tertiary Butyl Alcohol from Ketone ➕💧
➕ Nucleophilic addition to ketone followed by 💧 Acid work-up yields tertiary butyl alcohol
🧬 Halo magnesium alkoxide as an Intermediate adduct
(CH₃)₂Cᵟ⁺=Oᵟ⁻ + H₃Cᵟ⁻Mgᵟ⁺Cl — Aɴ → (CH₃)₃C–Oᵟ⁻Mgᵟ⁺Cl — H₃O⁺ or H₂O/HX → Mg(OH)Cl + (CH₃)₃C–OH
🧪 (iv) Ethyl Alcohol from Methyl Magnesium Iodide ➕💧
➕ Nucleophilic addition to formaldehyde followed by Acidification to yield ethyl alcohol
H₃Cᵟ⁻Mgᵟ⁺Cl + Cᵟ⁺H₂Oᵟ⁻ — Aɴ → CH₃–CH₂–O⁻Mg⁺I — H₃O⁺ or H₂O/HX → Mg(OH)Cl + CH₃–CH₂–OH
🧪 (v) Secondary Alcohol from Grignard Reagent ➕💧
➕ Nucleophilic addition to aldehyde followed by Acidification (Acid work-up) yields secondary alcohol
🧬 Halo magnesium alkoxide as an Intermediate adduct
Rᵟ⁻–Mgᵟ⁺X + R'ᵟ⁺–CHOᵟ⁻ — Aɴ → R–CH(O⁻Mg⁺X)–R' — H₃O⁺ or H₂O/HX → Mg(OH)Cl + R–CH(OH)–R'![]()
⚡ Alkylation/ methylation (Nucleophilic substitution (Sɴ) of halide of alkyl halide)
✨ Reagent: methyl chloride (CH₃Cl)
H₃Cᵟ⁻Mgᵟ⁺Cl + H₃Cᵟ⁺Clᵟ⁻ — Sɴ₂ → MgCl₂ + CH₃–CH₃ (Ethane)
🧪 (ii) Ethanoic Acid from Methyl Magnesium Chloride 💨💧
💨 Carbonation / nucleophilic addition of CO₂ followed by Acidification to yield Ethanoic acid
🧬 Halo magnesium carboxylate as an Intermediate adduct
H₃Cᵟ⁻Mgᵟ⁺Cl + Cᵟ⁺Oᵟ⁻₂ — Aɴ → CH₃COO⁻Mg⁺Cl — H₃O⁺ or H₂O/HX → Mg(OH)Cl + CH₃COOH (acetic acid)
🧪 (iii) Tertiary Butyl Alcohol from Ketone ➕💧
➕ Nucleophilic addition to ketone followed by 💧 Acid work-up yields tertiary butyl alcohol
🧬 Halo magnesium alkoxide as an Intermediate adduct
(CH₃)₂Cᵟ⁺=Oᵟ⁻ + H₃Cᵟ⁻Mgᵟ⁺Cl — Aɴ → (CH₃)₃C–Oᵟ⁻Mgᵟ⁺Cl — H₃O⁺ or H₂O/HX → Mg(OH)Cl + (CH₃)₃C–OH
🧪 (iv) Ethyl Alcohol from Methyl Magnesium Iodide ➕💧
➕ Nucleophilic addition to formaldehyde followed by Acidification to yield ethyl alcohol
H₃Cᵟ⁻Mgᵟ⁺Cl + Cᵟ⁺H₂Oᵟ⁻ — Aɴ → CH₃–CH₂–O⁻Mg⁺I — H₃O⁺ or H₂O/HX → Mg(OH)Cl + CH₃–CH₂–OH
🧪 (v) Secondary Alcohol from Grignard Reagent ➕💧
➕ Nucleophilic addition to aldehyde followed by Acidification (Acid work-up) yields secondary alcohol
🧬 Halo magnesium alkoxide as an Intermediate adduct
Rᵟ⁻–Mgᵟ⁺X + R'ᵟ⁺–CHOᵟ⁻ — Aɴ → R–CH(O⁻Mg⁺X)–R' — H₃O⁺ or H₂O/HX → Mg(OH)Cl + R–CH(OH)–R'
Q8. How can you justify the fact that alkyl halides are water insoluble?
✨ Why Alkyl Halides Are Insoluble in Water 🌊🚫
Alkyl halides, in spite of their polar nature (dipole moment ≈ 2.05–2.15 D), are slightly soluble or almost insoluble in water due to:
➡️ Polar but not enough: Their polarity is insufficient to overcome strong water–water attractions.
➡️ No hydrogen bonding: Alkyl halides cannot form hydrogen bonds with water molecules.
➡️ Strong H‑bonding in water 💧: Water molecules are held together by strong hydrogen bonds, which alkyl halides cannot break.
➡️ Hydrophobic alkyl group 🛑: The non‑polar hydrocarbon part repels water.
➡️ Energy factor ⚡: The energy released on mixing alkyl halides with water is less than the energy required to break hydrogen bonds in water.
✅ Conclusion: Even though alkyl halides are polar, their inability to form hydrogen bonds and disrupt water’s hydrogen‑bond network makes them slightly soluble or insoluble in water.
Alkyl halides, in spite of their polar nature (dipole moment ≈ 2.05–2.15 D), are slightly soluble or almost insoluble in water due to:
➡️ Polar but not enough: Their polarity is insufficient to overcome strong water–water attractions.
➡️ No hydrogen bonding: Alkyl halides cannot form hydrogen bonds with water molecules.
➡️ Strong H‑bonding in water 💧: Water molecules are held together by strong hydrogen bonds, which alkyl halides cannot break.
➡️ Hydrophobic alkyl group 🛑: The non‑polar hydrocarbon part repels water.
➡️ Energy factor ⚡: The energy released on mixing alkyl halides with water is less than the energy required to break hydrogen bonds in water.
✅ Conclusion: Even though alkyl halides are polar, their inability to form hydrogen bonds and disrupt water’s hydrogen‑bond network makes them slightly soluble or insoluble in water.
Q9. Explain the following by giving scientific reasons:
🔥 Explanations with Scientific Reasons (Short & To the Point) ⚗️✨
➡️ Why β-elimination is not possible in methyl halides?
❌ Methyl halides do not have β-hydrogen, which is essential for elimination.
➡️ Why Sɴ₂ reaction is not favourable in tertiary alkyl halides?
🚧 Severe steric hindrance around the carbon blocks backside attack by nucleophile.
➡️ Why tertiary carbocation is more stable than secondary and primary carbocations?
➕ More alkyl groups donate electrons (+I effect & hyperconjugation), stabilizing positive charge.
➡️ Why E₂ reaction is 2nd order while E₁ is 1st order?
📊 E₂ depends on both substrate and base (one-step) → 2nd order
📉 E₁ depends only on substrate (carbocation formation is slow step) → 1st order
➡️ Why 2° alkyl halides give Sɴ₁ mechanism in presence of polar solvent?
💧 Polar solvents stabilize carbocation and halide ion, favoring Sɴ₁ reaction.
➡️ Why alkyl group behaves as a nucleophile in Grignard’s reagent?
⚡ Mg makes C–Mg bond polar (Cδ⁻) → alkyl group becomes nucleophilic.
➡️ Why Sɴ₂ reaction completes in one step?
🔄 Bond breaking and bond formation occur simultaneously via backside attack (no intermediate).
➡️ Why β-elimination is not possible in methyl halides?
❌ Methyl halides do not have β-hydrogen, which is essential for elimination.
➡️ Why Sɴ₂ reaction is not favourable in tertiary alkyl halides?
🚧 Severe steric hindrance around the carbon blocks backside attack by nucleophile.
➡️ Why tertiary carbocation is more stable than secondary and primary carbocations?
➕ More alkyl groups donate electrons (+I effect & hyperconjugation), stabilizing positive charge.
➡️ Why E₂ reaction is 2nd order while E₁ is 1st order?
📊 E₂ depends on both substrate and base (one-step) → 2nd order
📉 E₁ depends only on substrate (carbocation formation is slow step) → 1st order
➡️ Why 2° alkyl halides give Sɴ₁ mechanism in presence of polar solvent?
💧 Polar solvents stabilize carbocation and halide ion, favoring Sɴ₁ reaction.
➡️ Why alkyl group behaves as a nucleophile in Grignard’s reagent?
⚡ Mg makes C–Mg bond polar (Cδ⁻) → alkyl group becomes nucleophilic.
➡️ Why Sɴ₂ reaction completes in one step?
🔄 Bond breaking and bond formation occur simultaneously via backside attack (no intermediate).
Q10. Why alkyl halide undergoes nucleophilic substitution reaction? Which reagent is required to convert a methyl iodide into:
(i) Methanol (ii) Methyl cyanide (iii) Dimethyl ether (iv) Thiol
(i) Methanol (ii) Methyl cyanide (iii) Dimethyl ether (iv) Thiol
🔁 ✨ Why Alkyl Halides Undergo Nucleophilic Substitution Reaction? ⚗️🧪
➡️ Polar C–X bond: Halogen is more electronegative → Cᵟ⁺–Xᵟ⁻ bond forms.
➡️ Electrophilic α-carbon: The carbon atom attached to halogen (Cᵟ⁺) acts as an electrophilic centre, attracting nucleophiles.
➡️ Good leaving group: Halide ion (X⁻) leaves easily, stabilizing the reaction.
➡️ Nucleophilic attack: A nucleophile donates its electron pair to the α‑carbon, forming a new C–Nu bond while C–X bond breaks (heterolytic cleavage). 🔄
General Representation:
Nu⁻ + R–Cᵟ⁺H₂–Xᵟ⁻ — Sɴ → R–CH₂–Nu + X⁻
⭐ Nu⁻ → Attacking nucleophile
⭐ R–CH₂–X → Substrate (alkyl halide)
⭐ Cᵟ⁺ → Electrophilic centre
⭐ Xᵟ⁻ → Leaving group
⭐ R–CH₂–Nu → New carbon–nucleophile bond
⭐ X⁻ → Leaving nucleophile
⚗️🔁 Suitable Reagent for the Conversion of Methyl Iodide (CH₃I) into Required Product:
🔄 (i) Methanol (Alcohol) 🍺 → Nucleophile: OH⁻
CH₃I + KOH (aq) → KI + CH₃OH (methanol)
🔄 (ii) Methyl Cyanide (Nitrile) 🧬 → Nucleophile: CN⁻ (carbon end attacks)
CH₃I + KCN → KI + CH₃CN (methyl cyanide)
🔄 (iii) Dimethyl Ether 🌬️ → Nucleophile: OCH₃⁻
CH₃I + NaOCH₃ → KI + CH₃–O–CH₃ (dimethyl ether)
🔄 (iv) Triol (Methanethiol) 🧪 → Nucleophile: SH⁻
CH₃I + KSH → KI + CH₃SH (thiol)
✅ Key Takeaway:
The polarity of the C–X bond and the presence of a good leaving group make alkyl halides highly susceptible to nucleophilic substitution. Methyl iodide mainly undergoes Sɴ₂ reaction (one‑step, backside attack) 🎯
➡️ Polar C–X bond: Halogen is more electronegative → Cᵟ⁺–Xᵟ⁻ bond forms.
➡️ Electrophilic α-carbon: The carbon atom attached to halogen (Cᵟ⁺) acts as an electrophilic centre, attracting nucleophiles.
➡️ Good leaving group: Halide ion (X⁻) leaves easily, stabilizing the reaction.
➡️ Nucleophilic attack: A nucleophile donates its electron pair to the α‑carbon, forming a new C–Nu bond while C–X bond breaks (heterolytic cleavage). 🔄
General Representation:
Nu⁻ + R–Cᵟ⁺H₂–Xᵟ⁻ — Sɴ → R–CH₂–Nu + X⁻
⭐ Nu⁻ → Attacking nucleophile
⭐ R–CH₂–X → Substrate (alkyl halide)
⭐ Cᵟ⁺ → Electrophilic centre
⭐ Xᵟ⁻ → Leaving group
⭐ R–CH₂–Nu → New carbon–nucleophile bond
⭐ X⁻ → Leaving nucleophile
⚗️🔁 Suitable Reagent for the Conversion of Methyl Iodide (CH₃I) into Required Product:
🔄 (i) Methanol (Alcohol) 🍺 → Nucleophile: OH⁻
CH₃I + KOH (aq) → KI + CH₃OH (methanol)
🔄 (ii) Methyl Cyanide (Nitrile) 🧬 → Nucleophile: CN⁻ (carbon end attacks)
CH₃I + KCN → KI + CH₃CN (methyl cyanide)
🔄 (iii) Dimethyl Ether 🌬️ → Nucleophile: OCH₃⁻
CH₃I + NaOCH₃ → KI + CH₃–O–CH₃ (dimethyl ether)
🔄 (iv) Triol (Methanethiol) 🧪 → Nucleophile: SH⁻
CH₃I + KSH → KI + CH₃SH (thiol)
✅ Key Takeaway:
The polarity of the C–X bond and the presence of a good leaving group make alkyl halides highly susceptible to nucleophilic substitution. Methyl iodide mainly undergoes Sɴ₂ reaction (one‑step, backside attack) 🎯
📘 Smart Answers of XII Chemistry Model Test Questions
Smart Answers of Short Questions of Amines
Q1. Why are secondary and tertiary amines more alkaline than primary amines?
Reason of Greater Basicity of Secondary and Tertiary Amines than Primary Amines 💡
➡️ More alkyl groups in secondary and tertiary amines donate electron density (+I effect or positive inductive effect) to the nitrogen.
➡️ This increases the availability of the lone pair on nitrogen for protonation (to accept a proton) thereby increasing basicity.
➡️ Hence, secondary and tertiary amines are more basic (alkaline) than primary amines.
⚠️ Note: Very bulky tertiary amines may have steric hindrance reducing basicity slightly.
Order of Basicity (in aqueous medium): Secondary amines > Tertiary amines > Primary amines > Ammonia
✏️ Additional info:
In aqueous solution, secondary amines are the most basic due to the balance between inductive effect and solvation, while tertiary amines are slightly less basic because steric hindrance reduces solvation efficiency.
💡 Extra tip: Basicity of amines depends on the groups attached to nitrogen:
➡️ Electron-donating groups (e.g., –CH₃, –OCH₃) increase basicity by boosting electron density on nitrogen.
➡️ Electron-withdrawing groups (e.g., –NO₂, –CN, halogens) decrease basicity by pulling electron density away.
🏁 Secondary and tertiary amines have more alkyl groups, so they are more basic than primary amines.![]()
➡️ More alkyl groups in secondary and tertiary amines donate electron density (+I effect or positive inductive effect) to the nitrogen.
➡️ This increases the availability of the lone pair on nitrogen for protonation (to accept a proton) thereby increasing basicity.
➡️ Hence, secondary and tertiary amines are more basic (alkaline) than primary amines.
⚠️ Note: Very bulky tertiary amines may have steric hindrance reducing basicity slightly.
Order of Basicity (in aqueous medium): Secondary amines > Tertiary amines > Primary amines > Ammonia
✏️ Additional info:
In aqueous solution, secondary amines are the most basic due to the balance between inductive effect and solvation, while tertiary amines are slightly less basic because steric hindrance reduces solvation efficiency.
💡 Extra tip: Basicity of amines depends on the groups attached to nitrogen:
➡️ Electron-donating groups (e.g., –CH₃, –OCH₃) increase basicity by boosting electron density on nitrogen.
➡️ Electron-withdrawing groups (e.g., –NO₂, –CN, halogens) decrease basicity by pulling electron density away.
🏁 Secondary and tertiary amines have more alkyl groups, so they are more basic than primary amines.
Q2. How is primary amine converted into secondary and tertiary amines, give the equations.
🧪➡️ Conversion of Primary Amine into Secondary & Tertiary Amines by Alkylation using Alkyl halides ✨
⚗️ Method: Alkylation using alkyl halides 🧪
🔄 Reaction: Hydrogen atoms of the primary amine are replaced by alkyl groups.
🎯 Reaction Type: Sɴ₂ reaction (Each reaction adds one alkyl group, converting primary → secondary → tertiary amine)
✏️ Equations:
Primary → Secondary Amine: R–NH₂ + R'–X — Alkylation (Sɴ) → HX + R–NHR'
Secondary → Tertiary Amine: R–NHR' + R”–X — Alkylation (Sɴ) → HX + R–NR”R'
📝 Example:
CH₃–NH₂ + CH₃'–I — Alkylation (Sɴ) → HI + CH₃–NH–CH₃' or (CH₃)₂NH (dimethyl amine)
CH₃–NH–CH₃' + CH₃”–I — Alkylation (Sɴ) → HI + CH₃–N(CH₃”)CH₃' or (CH₃)₃N (trimethyl amine)
![]()
⚗️ Method: Alkylation using alkyl halides 🧪
🔄 Reaction: Hydrogen atoms of the primary amine are replaced by alkyl groups.
🎯 Reaction Type: Sɴ₂ reaction (Each reaction adds one alkyl group, converting primary → secondary → tertiary amine)
✏️ Equations:
Primary → Secondary Amine: R–NH₂ + R'–X — Alkylation (Sɴ) → HX + R–NHR'
Secondary → Tertiary Amine: R–NHR' + R”–X — Alkylation (Sɴ) → HX + R–NR”R'
📝 Example:
CH₃–NH₂ + CH₃'–I — Alkylation (Sɴ) → HI + CH₃–NH–CH₃' or (CH₃)₂NH (dimethyl amine)
CH₃–NH–CH₃' + CH₃”–I — Alkylation (Sɴ) → HI + CH₃–N(CH₃”)CH₃' or (CH₃)₃N (trimethyl amine)
Q3. Give an account on the basicity of amines.
🔥 Basicity of Amines ⚡🧪
🟥➤ Amines are basic due to the lone pair on nitrogen (Lewis base).
🟧➤ Lone pair can accept a proton (Bronsted-Lowry base) → gives alkaline solution 💧.
🟨➤ Effect of groups on nitrogen:
🟢 ➕ Electron-donating groups (–CH₃, –OCH₃, –NH₂) increase basicity ⚡
🔵 ➖ Electron-withdrawing groups (–NO₂, –CN, –X) decrease basicity ❌
🟪➤ Aliphatic amines are more basic than ammonia and aromatic amines 🧬⬆️
🟣 Aromatic amines: lone pair participates in resonance, less available for protonation 🔄
🟣 Aliphatic amines: lone pair fully available for donation ✅
💡 Quick summary line:
Lone pair availability + electron-donating groups ↑ basicity; resonance & electron-withdrawing groups ↓ basicity.![]()
🟥➤ Amines are basic due to the lone pair on nitrogen (Lewis base).
🟧➤ Lone pair can accept a proton (Bronsted-Lowry base) → gives alkaline solution 💧.
🟨➤ Effect of groups on nitrogen:
🟢 ➕ Electron-donating groups (–CH₃, –OCH₃, –NH₂) increase basicity ⚡
🔵 ➖ Electron-withdrawing groups (–NO₂, –CN, –X) decrease basicity ❌
🟪➤ Aliphatic amines are more basic than ammonia and aromatic amines 🧬⬆️
🟣 Aromatic amines: lone pair participates in resonance, less available for protonation 🔄
🟣 Aliphatic amines: lone pair fully available for donation ✅
💡 Quick summary line:
Lone pair availability + electron-donating groups ↑ basicity; resonance & electron-withdrawing groups ↓ basicity.
Q4. How are amines prepared from nitriles, give the equations.
🧪➡️ Preparation of Amines from Nitriles by its Reduction by Nascent H Using LiAlH₄ or by H₂ Using Rh-Al₂O₃, Pt or Raney Nickel 👤
🔹 Method: Chemical or Catalytic Reduction of nitriles using nascent H (LiAlH₄) or H₂ with catalysts (Rh–Al₂O₃, Pt, Raney Ni) ⚡
🔹 Product: Produces primary amines 👤
🔹 Key Note: The amine has one carbon more than the parent nitrile ➕1C
✨ Reaction:
R–C≡N + 2H₂ or 4[H] — Reduction (Rh–Al₂O₃, Pt, Raney Ni/🔺 or LiAlH₄) → R–CH₂–NH₂
✨ Examples:
CH₃–C≡N + 2H₂ or 4[H] — Reduction (Rh–Al₂O₃, Pt, Raney Ni/🔺 or LiAlH₄) → CH₃–CH₂–NH₂
⚡ Quick summary line:
💡 Nitrile + H → primary amine with one extra carbon.![]()
🔹 Method: Chemical or Catalytic Reduction of nitriles using nascent H (LiAlH₄) or H₂ with catalysts (Rh–Al₂O₃, Pt, Raney Ni) ⚡
🔹 Product: Produces primary amines 👤
🔹 Key Note: The amine has one carbon more than the parent nitrile ➕1C
✨ Reaction:
R–C≡N + 2H₂ or 4[H] — Reduction (Rh–Al₂O₃, Pt, Raney Ni/🔺 or LiAlH₄) → R–CH₂–NH₂
✨ Examples:
CH₃–C≡N + 2H₂ or 4[H] — Reduction (Rh–Al₂O₃, Pt, Raney Ni/🔺 or LiAlH₄) → CH₃–CH₂–NH₂
⚡ Quick summary line:
💡 Nitrile + H → primary amine with one extra carbon.
📘 Smart Answers of XII Chemistry Model Test Questions
Smart Answers of Descriptive Questions
Q1. How can you define nucleophilic substitution reactions? Describe the mechanisms of Sɴ₁ and Sɴ₂ reactions. Write down 5 differences between Sɴ₁ and Sɴ₂.
Definition of Nucleophilic Substitution Reactions or Sɴ reactions ⚡🧪
It is the substitution reaction in which a strong nucleophile (Nu⁻) replaces a weaker nucleophile as leaving group (X⁻) from an electrophilic carbon of alkyl halides (substrate).
S = Substitution, N = Nucleophilic.
🔢 Types of Sɴ Reactions:
➡️ Sɴ₂ or Bimolecular Nucleophilic Substitution Reactions (Bimolecular, one-step) 🔄
➡️ Sɴ₁ or Unimolecular Nucleophilic Substitution Reactions (Unimolecular, two-step) 🔄
General Representation:
Nu⁻ + R–Cᵟ⁺H₂–Xᵟ⁻ — Sɴ → R–CH₂–Nu + X⁻
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⭐ Nu⁻ → Attacking nucleophile
⭐ R–CH₂–X → Substrate (alkyl halide)
⭐ Cᵟ⁺ → Electrophilic centre
⭐ Xᵟ⁻ → Leaving group
⭐ R–CH₂–Nu → New carbon–nucleophile bond
⭐ X⁻ → Leaving nucleophile
🧪 Some Common Strong Nucleophiles along with their Typical Reagents and Class of Product:
🟢 OH⁻ → NaOH/KOH → 🍺 Alcohol
🟢 SH⁻ → NaSH/KSH → 🧴 Thiol
🟢 OR⁻ → NaOR/KOR → 🌬️ Ether
🟢 NH₂⁻ → NaNH₂ → 🧑🔬 Primary Amine
🟢 CN⁻ → NaCN → ⚛️ Nitrile
🟢 RCOO⁻ → RCOONa → 🌸 Ester
🟢 SR⁻ → NaSR/KSR → 🧪 Thioether
🟢 I⁻ → KI → 🔑 Alkyl Iodide
🟢 N₃⁻ → NaN₃ → ⚡ Azide
🟢 –C≡CR → NaC≡CR → 🔥 Higher Alkyne
🟢 –NHR → RNHR → 🧑🔬 Secondary Amine
🟢 –NR₂ → RNR₂ → 🧑🔬 Tertiary Amine
🟢 NR₃ → NR₃ → 💼 Quaternary Ammonium Ion
➡️ Sɴ₂ (Bimolecular, one-step) 🔄
📚 Definition: One-step nucleophilic substitution involving backside attack; simultaneous bond formation & C–X bond breaking.
⚙️ Mechanism: Transition state with both Nu⁻ and substrate.
⏱️ Rate: Depends on [alkyl halide] & [Nu⁻] → 2nd order.
🔬 Stereochemistry: Inversion (Walden inversion).
🌟 Favoured by: Primary alkyl halides, polar aprotic solvents (DMSO, DMF).
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➡️ Sɴ₁ (Unimolecular, two-step) 🔄
📚 Definition: Two-step reaction involving carbocation formation after leaving group departs, followed by nucleophile attack.
⚙️ Mechanism:
➡️ Slow ionization: C–X → carbocation (R⁺) + X⁻ (rate-determining step)
➡️ Fast nucleophilic attack: R⁺ + Nu⁻ → R–Nu
⏱️ Rate: Depends only on [alkyl halide] → 1st order.
🔬 Stereochemistry: Racemic mixture (inversion + retention).
🌟 Favoured by: Tertiary alkyl halides, polar protic solvents (H₂O, alcohols).
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Quick Mnemonics:
🌀 Sɴ₂: “Backside attack, One step, Bimolecular” 🔙
🧩 Sɴ₁: “Carbocation formed, Two steps, Unimolecular” 🧩
It is the substitution reaction in which a strong nucleophile (Nu⁻) replaces a weaker nucleophile as leaving group (X⁻) from an electrophilic carbon of alkyl halides (substrate).
S = Substitution, N = Nucleophilic.
🔢 Types of Sɴ Reactions:
➡️ Sɴ₂ or Bimolecular Nucleophilic Substitution Reactions (Bimolecular, one-step) 🔄
➡️ Sɴ₁ or Unimolecular Nucleophilic Substitution Reactions (Unimolecular, two-step) 🔄
General Representation:
Nu⁻ + R–Cᵟ⁺H₂–Xᵟ⁻ — Sɴ → R–CH₂–Nu + X⁻
⭐ R–CH₂–X → Substrate (alkyl halide)
⭐ Cᵟ⁺ → Electrophilic centre
⭐ Xᵟ⁻ → Leaving group
⭐ R–CH₂–Nu → New carbon–nucleophile bond
⭐ X⁻ → Leaving nucleophile
🧪 Some Common Strong Nucleophiles along with their Typical Reagents and Class of Product:
🟢 OH⁻ → NaOH/KOH → 🍺 Alcohol
🟢 SH⁻ → NaSH/KSH → 🧴 Thiol
🟢 OR⁻ → NaOR/KOR → 🌬️ Ether
🟢 NH₂⁻ → NaNH₂ → 🧑🔬 Primary Amine
🟢 CN⁻ → NaCN → ⚛️ Nitrile
🟢 RCOO⁻ → RCOONa → 🌸 Ester
🟢 SR⁻ → NaSR/KSR → 🧪 Thioether
🟢 I⁻ → KI → 🔑 Alkyl Iodide
🟢 N₃⁻ → NaN₃ → ⚡ Azide
🟢 –C≡CR → NaC≡CR → 🔥 Higher Alkyne
🟢 –NHR → RNHR → 🧑🔬 Secondary Amine
🟢 –NR₂ → RNR₂ → 🧑🔬 Tertiary Amine
🟢 NR₃ → NR₃ → 💼 Quaternary Ammonium Ion
➡️ Sɴ₂ (Bimolecular, one-step) 🔄
📚 Definition: One-step nucleophilic substitution involving backside attack; simultaneous bond formation & C–X bond breaking.
⚙️ Mechanism: Transition state with both Nu⁻ and substrate.
⏱️ Rate: Depends on [alkyl halide] & [Nu⁻] → 2nd order.
🔬 Stereochemistry: Inversion (Walden inversion).
🌟 Favoured by: Primary alkyl halides, polar aprotic solvents (DMSO, DMF).
📚 Definition: Two-step reaction involving carbocation formation after leaving group departs, followed by nucleophile attack.
⚙️ Mechanism:
➡️ Slow ionization: C–X → carbocation (R⁺) + X⁻ (rate-determining step)
➡️ Fast nucleophilic attack: R⁺ + Nu⁻ → R–Nu
⏱️ Rate: Depends only on [alkyl halide] → 1st order.
🔬 Stereochemistry: Racemic mixture (inversion + retention).
🌟 Favoured by: Tertiary alkyl halides, polar protic solvents (H₂O, alcohols).
🤜 Sɴ₂ Reactions
🔹 Bimolecular (molecularity = 2)
🔹 Overall order = 2 (2nd order)
🔹 One step, concerted
🔹 Given by primary alkyl halides
🔹 Favoured in polar aprotic solvents
🔹 Strong base required
🔹 Only inversion product formed
🔹 Reactivity: CH₃X > 1° > 2° >> 3°
🔹 No ionization step
🔹 Rate-determining step irreversible
🔹 Transition state only
🔹 No rearrangement
🔹 Backside attack only
🔹 Rate = k[Substrate][Nu⁻]
🔹 Example:
RCH₂-X + + Nu⁻— Slow → Nu-----RCH₂-----X ⁻— Fast → Nu-CH₂R+ X⁻
🔹 Bimolecular (molecularity = 2)
🔹 Overall order = 2 (2nd order)
🔹 One step, concerted
🔹 Given by primary alkyl halides
🔹 Favoured in polar aprotic solvents
🔹 Strong base required
🔹 Only inversion product formed
🔹 Reactivity: CH₃X > 1° > 2° >> 3°
🔹 No ionization step
🔹 Rate-determining step irreversible
🔹 Transition state only
🔹 No rearrangement
🔹 Backside attack only
🔹 Rate = k[Substrate][Nu⁻]
🔹 Example:
RCH₂-X + + Nu⁻— Slow → Nu-----RCH₂-----X ⁻— Fast → Nu-CH₂R+ X⁻
🤛 Sɴ₁ Reactions
🔹 Unimolecular (molecularity = 1)
🔹 Overall order = 1 (1st order)
🔹 Two-step, via carbocation
🔹 Given by tertiary alkyl halides
🔹 Favoured in polar protic solvents
🔹 Weak base sufficient
🔹 Inversion + retention (racemic)
🔹 Reactivity: 3° > 2° >> 1° >> CH₃X
🔹 Ionization occurs
🔹 Rate-determining step reversible
🔹 Carbocation intermediate formed
🔹 Rearrangement possible
🔹 Nucleophile attack from either side
🔹 Rate = k[Substrate]
🔹 Example:
R₃C-X → R₃C⁺ + X⁻ (slow)
R₃C⁺ + Nu⁻ → R₃C-Nu + Nu- CR₃ (fast)
🔹 Unimolecular (molecularity = 1)
🔹 Overall order = 1 (1st order)
🔹 Two-step, via carbocation
🔹 Given by tertiary alkyl halides
🔹 Favoured in polar protic solvents
🔹 Weak base sufficient
🔹 Inversion + retention (racemic)
🔹 Reactivity: 3° > 2° >> 1° >> CH₃X
🔹 Ionization occurs
🔹 Rate-determining step reversible
🔹 Carbocation intermediate formed
🔹 Rearrangement possible
🔹 Nucleophile attack from either side
🔹 Rate = k[Substrate]
🔹 Example:
R₃C-X → R₃C⁺ + X⁻ (slow)
R₃C⁺ + Nu⁻ → R₃C-Nu + Nu- CR₃ (fast)
Quick Mnemonics:
🌀 Sɴ₂: “Backside attack, One step, Bimolecular” 🔙
🧩 Sɴ₁: “Carbocation formed, Two steps, Unimolecular” 🧩
Q1. Outline the step-wise reaction mechanism of the following:
(i) Sɴ₂ reaction between Bromomethane and NaOH
(ii) Sɴ₁ reaction between 2-chloro-2-methylpropane and NaCN
(i) Sɴ₂ reaction between Bromomethane and NaOH
(ii) Sɴ₁ reaction between 2-chloro-2-methylpropane and NaCN
🧪 (i) Sɴ₂ reaction between Bromomethane and NaOH ⚗️
✨ Overall equation: CH₃–Br + OH⁻ → CH₃–OH + Br⁻
🔄 Backside Nucleophilic attack: OH⁻ approaches the carbon opposite the C–Br bond (nucleophile targets δ⁺ carbon).
🌀 Single transition state: Pentavalent-like TS where C is partially bonded to OH and Br simultaneously [C···OH and C···Br].
⚡ Bond-making/bond-breaking concerted: C–OH bond forms as C–Br bond breaks in one step (no intermediate).
🔬 Inversion (Walden): Product configuration inverts at the reaction center (for chiral centers).
✅ Products: Methanol (CH₃–OH) and bromide ion (Br⁻).
🧾 By product (Counterion pairing): Na⁺ pairs with Br⁻ to form NaBr (spectator).
🧭 👉 Key feature: One step mechanism with inversion of configuration (Walden inversion).
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✨ Overall equation: CH₃–Br + OH⁻ → CH₃–OH + Br⁻
🔄 Backside Nucleophilic attack: OH⁻ approaches the carbon opposite the C–Br bond (nucleophile targets δ⁺ carbon).
🌀 Single transition state: Pentavalent-like TS where C is partially bonded to OH and Br simultaneously [C···OH and C···Br].
⚡ Bond-making/bond-breaking concerted: C–OH bond forms as C–Br bond breaks in one step (no intermediate).
🔬 Inversion (Walden): Product configuration inverts at the reaction center (for chiral centers).
✅ Products: Methanol (CH₃–OH) and bromide ion (Br⁻).
🧾 By product (Counterion pairing): Na⁺ pairs with Br⁻ to form NaBr (spectator).
🧭 👉 Key feature: One step mechanism with inversion of configuration (Walden inversion).
🧪 (ii) Sɴ₁ reaction between 2-chloro-2-methylpropane and NaCN ⚗️
✨ Overall equation: (CH₃)₃C–Cl + NaCN → (CH₃)₃C–CN + NaCl
🐢 Ionization (slow, rate-determining): The C–Cl bond heterolytically cleaves to form the tert-butyl carbocation (CH₃)₃C⁺ and Cl⁻.
🧩 Carbocation intermediate: A planar, stabilized tertiary carbocation forms (no stereochemical preference).
⚡ Nucleophilic attack (fast): CN⁻ attacks the carbocation to give tert-butyl cyanide (t Bu–CN/(CH₃)₃C–CN).
✅ Products: tert-butyl cyanide (CH₃)₃C–CN) and chloride ion (Cl⁻).
🧾 By product (Counterion pairing): Na⁺ pairs with Cl⁻ to form NaCl (spectator).
🧭 👉 Key feature: Rate depends only on substrate concentration; polar protic solvents favor carbocation formation.
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✨ Overall equation: (CH₃)₃C–Cl + NaCN → (CH₃)₃C–CN + NaCl
🐢 Ionization (slow, rate-determining): The C–Cl bond heterolytically cleaves to form the tert-butyl carbocation (CH₃)₃C⁺ and Cl⁻.
🧩 Carbocation intermediate: A planar, stabilized tertiary carbocation forms (no stereochemical preference).
⚡ Nucleophilic attack (fast): CN⁻ attacks the carbocation to give tert-butyl cyanide (t Bu–CN/(CH₃)₃C–CN).
✅ Products: tert-butyl cyanide (CH₃)₃C–CN) and chloride ion (Cl⁻).
🧾 By product (Counterion pairing): Na⁺ pairs with Cl⁻ to form NaCl (spectator).
🧭 👉 Key feature: Rate depends only on substrate concentration; polar protic solvents favor carbocation formation.
Q2. What is β-elimination? Discuss the mechanisms of E1 and E2 reactions.
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 (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. 🔥
🧪 Examples:
• Dehydrohalogenation of alkyl halide with a base.
• Acid catalysed dehydration of alcohols with acid.
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🔹 Types of Elimination Reaction (By Hughes and Ingold; 1941):
⚡📊 Bimolecular β-elimination reaction (E₂ reaction).
🔥⏳ Unimolecular β-elimination reaction (E₁ reaction).
🌀 Mechanism of E₂ Reactions (One-step, Bimolecular, concerted, via a transition state) 🔥
⚡ One step, bimolecular mechanism — both the base and the leaving group interact in the rate-determining step. 🧪
🧴 Strong base (e.g., KOH, NaOH) removes β-H on the β-carbon (adjacent to the halogenated carbon) while halide (X⁻) leaves simultaneously.
⏳ Occurs via a transition state (unstable, high-energy).
🧮 The rate depends on the concentration of both substrate and base → bimolecular.
🏁 Net result: HX eliminated → alkene formed.
✨ Example: Dehydrohalogenation of alkyl halides.
🍃 Key Example: Conversion of 2-bromo-2-methylpropane into 2-methylpropene by KOH occurs via an E₂ mechanism.
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🌀 Mechanism of E₁ Reactions (Two-steps, Unimolecular, non-concerted, carbocation pathway) 💥
⚡ Two-step process: The formation of the carbocation is the rate-determining step.
🧴 Weak base is sufficient (e.g., water, alcohol).
Step 1: 🧩 Slow ionization → halide leaves, forming carbocation intermediate at the α-carbon (rate-determining).
Step 2: 🧴 Base removes β-H from the adjacent carbon, forming double bond (alkene) in this fast step.
⚖️ Rate depends only on the concentration of the substrate → unimolecular.
🍻 Example: In the case of 3-bromo-3-methylbutane with H₂SO₄, the first step is the departure of the bromine atom forming a tertiary carbocation, followed by deprotonation to form 2-methylbutene.
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✨ Quick Memory Aid:
⚡ E₂ = One-step, fast, strong base
🌀 E₁ = Two-step, carbocation intermediate, weak base
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 (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. 🔥
🧪 Examples:
• Dehydrohalogenation of alkyl halide with a base.
• Acid catalysed dehydration of alcohols with acid.
⚡📊 Bimolecular β-elimination reaction (E₂ reaction).
🔥⏳ Unimolecular β-elimination reaction (E₁ reaction).
🌀 Mechanism of E₂ Reactions (One-step, Bimolecular, concerted, via a transition state) 🔥
⚡ One step, bimolecular mechanism — both the base and the leaving group interact in the rate-determining step. 🧪
🧴 Strong base (e.g., KOH, NaOH) removes β-H on the β-carbon (adjacent to the halogenated carbon) while halide (X⁻) leaves simultaneously.
⏳ Occurs via a transition state (unstable, high-energy).
🧮 The rate depends on the concentration of both substrate and base → bimolecular.
🏁 Net result: HX eliminated → alkene formed.
✨ Example: Dehydrohalogenation of alkyl halides.
🍃 Key Example: Conversion of 2-bromo-2-methylpropane into 2-methylpropene by KOH occurs via an E₂ mechanism.
⚡ Two-step process: The formation of the carbocation is the rate-determining step.
🧴 Weak base is sufficient (e.g., water, alcohol).
Step 1: 🧩 Slow ionization → halide leaves, forming carbocation intermediate at the α-carbon (rate-determining).
Step 2: 🧴 Base removes β-H from the adjacent carbon, forming double bond (alkene) in this fast step.
⚖️ Rate depends only on the concentration of the substrate → unimolecular.
🍻 Example: In the case of 3-bromo-3-methylbutane with H₂SO₄, the first step is the departure of the bromine atom forming a tertiary carbocation, followed by deprotonation to form 2-methylbutene.
⚡ E₂ = One-step, fast, strong base
🌀 E₁ = Two-step, carbocation intermediate, weak base
Q3. What are organometallic compounds? What are the key properties of organometallic compounds? How is Grignard’s reagent prepared? Write down the reactions of Grignard’s reagent with water, carbon dioxide, ester and amines.
📖 Definition of Organometallic Compounds
Organic compounds containing at least one C–Metal bond.
🔹 Examples of Common Organometallic Compounds:
⚗️ Grignard’s reagent (R–Mg–X) → organic synthesis
🐝 Dimethyl zinc → insecticide
⛽ Tetraethyl lead → anti-knock (knock inhibitor) in petrol
💊 Methylcobalamine → Vitamin B12 derivative
❤️ Haemoglobin & 🌱 Chlorophyll → biological importance (in fact not Organometallic Compounds)
⚡ Key Properties of Organometallic Compounds:
🧊 State: liquid/solid (esp. aromatic groups)
💧 Solubility: insoluble in water, soluble in ether
🔗 Bond: C–M bond is highly covalent
🔥 Reactivity: very reactive, stored in organic solvents
📉 Electronegativity: metals < 2
➖ Reducers: act as reducing agents
💥 Spontaneous combustion: Li/Na organometallics volatile
☠️ Toxicity: volatile compounds toxic to humans
📘 Definition of Grignard’s Reagent:
Grignard’s reagent or alkyl magnesium halide is a well-known organometallic compound that contains a carbon-magnesium (C–Mg) bond formulated as R–Mg–X [Where; R = alkyl or aryl group, X = Cl, Br or I].
⚗️ Preparation of Grignard’s Reagent:
General Equation: R–X + Mg — dry ether (C₂H₅OC₂H₅) → R–Mg–X (alkyl magnesium halide)
R = alkyl/aryl group, X = Cl, Br, I
Example: CH₃CH₂–X + Mg — dry ether (C₂H₅OC₂H₅) → CH₃CH₂–Mg–X (Ethyl magnesium halide)
🧴 Requires dry ether as solvent.
🔑 Factors Controlling Ease of Formation of Grignard’s Reagent:
📏 Size of alkyl group → larger = harder to form.
🧪 Nature of halogen → ease order: I > Br > Cl (larger halogen size → weaker C–X bond easier to break).
🔬 Reactions of Grignard’s Reagent:
💧 With Water (Hydrolysis):
R–Mg–X + H₂O — H⁺/ Sɴ → R–H + Mg(OH)X
Example: CH₃–Mg–Br + H₂O — H⁺/ Sɴ → CH₄ + Mg(OH)Br
🧴 With Primary Amine (Aminolysis):
R–Mg–X + R′–NH₂ — Sɴ → R–H + R′–NH–MgX
Example: CH₃–Mg–X + C′H₃–NH₂ — Sɴ → CH₄ + C′H₃–NH–MgX
🌱 With Carbon Dioxide (Carboxylation/Carbonation):
R–Mg–X + CO₂ — Aɴ → R–COOMgX — H₃O⁺ → R–COOH + Mg(OH)X
Example: CH₃–Mg–Br + CO₂ — Aɴ → CH₃–COOMgBr — H₃O⁺ → CH₃–COOH + Mg(OH)Br
🔥 With Ester (Nucleophilic Addition):
R–Mg–X + R′–COOR′′ — Aɴ → RR′–C(OR′′)OMgX — rearrangement → R–COR′ + Mg(OR′′)X
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Organic compounds containing at least one C–Metal bond.
🔹 Examples of Common Organometallic Compounds:
⚗️ Grignard’s reagent (R–Mg–X) → organic synthesis
🐝 Dimethyl zinc → insecticide
⛽ Tetraethyl lead → anti-knock (knock inhibitor) in petrol
💊 Methylcobalamine → Vitamin B12 derivative
❤️ Haemoglobin & 🌱 Chlorophyll → biological importance (in fact not Organometallic Compounds)
⚡ Key Properties of Organometallic Compounds:
🧊 State: liquid/solid (esp. aromatic groups)
💧 Solubility: insoluble in water, soluble in ether
🔗 Bond: C–M bond is highly covalent
🔥 Reactivity: very reactive, stored in organic solvents
📉 Electronegativity: metals < 2
➖ Reducers: act as reducing agents
💥 Spontaneous combustion: Li/Na organometallics volatile
☠️ Toxicity: volatile compounds toxic to humans
📘 Definition of Grignard’s Reagent:
Grignard’s reagent or alkyl magnesium halide is a well-known organometallic compound that contains a carbon-magnesium (C–Mg) bond formulated as R–Mg–X [Where; R = alkyl or aryl group, X = Cl, Br or I].
⚗️ Preparation of Grignard’s Reagent:
General Equation: R–X + Mg — dry ether (C₂H₅OC₂H₅) → R–Mg–X (alkyl magnesium halide)
R = alkyl/aryl group, X = Cl, Br, I
Example: CH₃CH₂–X + Mg — dry ether (C₂H₅OC₂H₅) → CH₃CH₂–Mg–X (Ethyl magnesium halide)
🧴 Requires dry ether as solvent.
🔑 Factors Controlling Ease of Formation of Grignard’s Reagent:
📏 Size of alkyl group → larger = harder to form.
🧪 Nature of halogen → ease order: I > Br > Cl (larger halogen size → weaker C–X bond easier to break).
🔬 Reactions of Grignard’s Reagent:
💧 With Water (Hydrolysis):
R–Mg–X + H₂O — H⁺/ Sɴ → R–H + Mg(OH)X
Example: CH₃–Mg–Br + H₂O — H⁺/ Sɴ → CH₄ + Mg(OH)Br
🧴 With Primary Amine (Aminolysis):
R–Mg–X + R′–NH₂ — Sɴ → R–H + R′–NH–MgX
Example: CH₃–Mg–X + C′H₃–NH₂ — Sɴ → CH₄ + C′H₃–NH–MgX
🌱 With Carbon Dioxide (Carboxylation/Carbonation):
R–Mg–X + CO₂ — Aɴ → R–COOMgX — H₃O⁺ → R–COOH + Mg(OH)X
Example: CH₃–Mg–Br + CO₂ — Aɴ → CH₃–COOMgBr — H₃O⁺ → CH₃–COOH + Mg(OH)Br
🔥 With Ester (Nucleophilic Addition):
R–Mg–X + R′–COOR′′ — Aɴ → RR′–C(OR′′)OMgX — rearrangement → R–COR′ + Mg(OR′′)X
OR
How do you convert Grignard’s reagent into three types of alcohols? Give only general equations.
How do you convert Grignard’s reagent into three types of alcohols? Give only general equations.
🔹 Key Idea
Grignard’s reagent (R–Mg–X) reacts with carbonyl compounds (formaldehyde, aldehydes, ketones) via nucleophilic addition → forms alkoxide salts → hydrolysis gives alcohols.
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🧪 Types of Alcohols Formed:
➡️ Grignard’s reagent with formaldehyde (H–CHO) → Primary Alcohols (1°)
⚡ R–Mg–X + H–CHO — Ether (Aɴ) → R–CH₂–OMgX — H₃O⁺ (Hydrolysis) → R–CH₂–OH
➡️ Grignard’s reagent with higher aldehydes (R′–CHO) → Secondary Alcohols (2°)
⚡ R–Mg–X + R′–CHO — Ether (Aɴ) → R′–CH(OMgX)–R — H₃O⁺ (Hydrolysis) → R′–CH(OH)–R
➡️ Grignard’s reagent with ketones (R′–CO–R′′) → Tertiary Alcohols (3°)
⚡ R–Mg–X + R′–CO–R′′ — Ether (Aɴ) → R′–C(OMgX)(R)(R′′) — H₃O⁺ (Hydrolysis) → R′–C(OH)(R)(R′′) or R₃COH
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✨ Quick Recap:
🍶 Formaldehyde → Primary alcohol
🍸 Aldehyde → Secondary alcohol
🍹 Ketone → Tertiary alcohol
Grignard’s reagent (R–Mg–X) reacts with carbonyl compounds (formaldehyde, aldehydes, ketones) via nucleophilic addition → forms alkoxide salts → hydrolysis gives alcohols.
➡️ Grignard’s reagent with formaldehyde (H–CHO) → Primary Alcohols (1°)
⚡ R–Mg–X + H–CHO — Ether (Aɴ) → R–CH₂–OMgX — H₃O⁺ (Hydrolysis) → R–CH₂–OH
➡️ Grignard’s reagent with higher aldehydes (R′–CHO) → Secondary Alcohols (2°)
⚡ R–Mg–X + R′–CHO — Ether (Aɴ) → R′–CH(OMgX)–R — H₃O⁺ (Hydrolysis) → R′–CH(OH)–R
➡️ Grignard’s reagent with ketones (R′–CO–R′′) → Tertiary Alcohols (3°)
⚡ R–Mg–X + R′–CO–R′′ — Ether (Aɴ) → R′–C(OMgX)(R)(R′′) — H₃O⁺ (Hydrolysis) → R′–C(OH)(R)(R′′) or R₃COH
🍶 Formaldehyde → Primary alcohol
🍸 Aldehyde → Secondary alcohol
🍹 Ketone → Tertiary alcohol
Q4. What are Alkyl Halides? How are they classified? Define primary, secondary and tertiary alkyl halides. Give their general structures and examples.
📖 Definition
➡️ Alkyl halides (haloalkanes) = monohalogen derivatives of alkanes.
➡️ Formed by replacing one H atom in an aliphatic hydrocarbon with a halogen (Cl, Br, I).
❌ Fluorine usually excluded (fluorocarbons behave differently).
📌 Hybridization: sp³
📌 General formula: CₙH₂ₙ₊₁–X
📌 Type Formula: R–X (R stands for alkyl groups and X represents Cl, Br or I).
📌 Examples: CH₃Cl, C₂H₅Br, C₃H₇I etc.
🗂️ Classification of Alkyl Halides 🔹
Alkyl halides are classified as primary, secondary and tertiary depending upon whether the halide (–X) group is attached to primary, secondary or tertiary carbon atom respectively.
❶ Primary (1°) Alkyl Halides:
📝 Definition: Halogen-bearing carbon (α-carbon) attached to one other carbon.
🧱 General structure: R–CH₂–X
💡 Example: CH₃–CH₂–Cl (ethyl chloride)
❷ Secondary (2°) Alkyl Halides:
📝 Halogen-bearing carbon attached to two other carbons.
🧱 General structure: R–CHX–R'
💡 Example: CH₃–CHCl–CH₃ (isopropyl chloride)
❸ Tertiary (3°) Alkyl Halides:
📝 Halogen-bearing carbon attached to three other carbons.
🧱 General structure: R₃C–X
💡 Example: (CH₃)₃C–Cl (tert-butyl chloride)
✨ Quick Recap:
🟢 Primary (1°) alkyl halides → one carbon neighbor (–X attached to 1°C) → (R–CH₂–X) e.g. ethyl chloride
🟡 Secondary (2°) alkyl halides → two carbon neighbors (–X attached to 2°C) → (R–CHX–R') e.g. isopropyl chloride
🔴 Tertiary (3°) alkyl halides → three carbon neighbors (–X attached to 3°C) → (R₃C–X) e.g. tert-butyl chloride
➡️ Alkyl halides (haloalkanes) = monohalogen derivatives of alkanes.
➡️ Formed by replacing one H atom in an aliphatic hydrocarbon with a halogen (Cl, Br, I).
❌ Fluorine usually excluded (fluorocarbons behave differently).
📌 Hybridization: sp³
📌 General formula: CₙH₂ₙ₊₁–X
📌 Type Formula: R–X (R stands for alkyl groups and X represents Cl, Br or I).
📌 Examples: CH₃Cl, C₂H₅Br, C₃H₇I etc.
🗂️ Classification of Alkyl Halides 🔹
Alkyl halides are classified as primary, secondary and tertiary depending upon whether the halide (–X) group is attached to primary, secondary or tertiary carbon atom respectively.
❶ Primary (1°) Alkyl Halides:
📝 Definition: Halogen-bearing carbon (α-carbon) attached to one other carbon.
🧱 General structure: R–CH₂–X
💡 Example: CH₃–CH₂–Cl (ethyl chloride)
❷ Secondary (2°) Alkyl Halides:
📝 Halogen-bearing carbon attached to two other carbons.
🧱 General structure: R–CHX–R'
💡 Example: CH₃–CHCl–CH₃ (isopropyl chloride)
❸ Tertiary (3°) Alkyl Halides:
📝 Halogen-bearing carbon attached to three other carbons.
🧱 General structure: R₃C–X
💡 Example: (CH₃)₃C–Cl (tert-butyl chloride)
✨ Quick Recap:
🟢 Primary (1°) alkyl halides → one carbon neighbor (–X attached to 1°C) → (R–CH₂–X) e.g. ethyl chloride
🟡 Secondary (2°) alkyl halides → two carbon neighbors (–X attached to 2°C) → (R–CHX–R') e.g. isopropyl chloride
🔴 Tertiary (3°) alkyl halides → three carbon neighbors (–X attached to 3°C) → (R₃C–X) e.g. tert-butyl chloride
Q5. Draw the orbital structure of methyl iodide and explain the type of hybridization in it.
🌟 Orbital Structure of Methyl Iodide
🔹 Hybridization:
Carbon in alkyl halides is sp³-hybridized.
Each carbon atom forms 4 sigma bonds → tetrahedral geometry.
📐 Bond angle ≈ 109°.
⚗️ Orbital Overlap in CH₃I:
• Carbon uses 4 sp³ hybrid orbitals.
• Three orbitals overlap with 1s orbitals of H atoms → C–H sigma bonds.
• One orbital overlaps with p-orbital of iodine → C–I sigma bond.
📐 Geometry:
• Tetrahedral shape due to sp³ hybridization.
• Symmetrical distribution of bonds around carbon.
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✨ Quick Recap:
🌀 Carbon = sp³ hybridized
📐 Bond angle ≈ 109°
⚗️ 3 C–H bonds + 1 C–I bond
🔺 Overall geometry = tetrahedral
🔹 Hybridization:
Carbon in alkyl halides is sp³-hybridized.
Each carbon atom forms 4 sigma bonds → tetrahedral geometry.
📐 Bond angle ≈ 109°.
⚗️ Orbital Overlap in CH₃I:
• Carbon uses 4 sp³ hybrid orbitals.
• Three orbitals overlap with 1s orbitals of H atoms → C–H sigma bonds.
• One orbital overlaps with p-orbital of iodine → C–I sigma bond.
📐 Geometry:
• Tetrahedral shape due to sp³ hybridization.
• Symmetrical distribution of bonds around carbon.
🌀 Carbon = sp³ hybridized
📐 Bond angle ≈ 109°
⚗️ 3 C–H bonds + 1 C–I bond
🔺 Overall geometry = tetrahedral
Q6. Give a comparative study between nucleophilic substitution reactions and elimination reactions of alkyl halides.
🔄 Comparative Study: Substitution vs Elimination in Alkyl Halides (Competition between E₂ and SN₂ Reactions)
📖 Key Idea:
🌀 Nucleophilic Substitution (SN₂/SN₁): nucleophile attacks α-carbon (carbon bonded to halogen).
⚡ Elimination (E₂/E₁): base removes β-hydrogen → double bond forms.
⚖️ Both can compete since nucleophile can also act as a base.
✨ Formation of Two Products:
When substrate R–X is treated with an electron pair donor species (e.g., base or nucleophilic reagents), there will always be two products.
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✨⚡ Reaction Preference:
💥 Strong base (OH⁻, RO⁻): β-hydrogen removal → favors E₂ (elimination), SN₂ side reaction.
🌀 Weak base but strong nucleophile (I⁻, Br⁻, CH₃COO⁻): α-carbon attack → favors SN₂ (substitution), E₂ side reaction.
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✨🔹 Factors Affecting Pathway:
🔎 Solvent Effect 🧴
➡️ Polar solvents → favor substitution (SN₂/SN₁).
➡️ Non-polar solvents → favor elimination (E₂/E₁).
🔎 Base Effect ⚗️
➡️ Strong base → promotes elimination.
➡️ Weak base → promotes substitution.
🔎 Temperature Effect 🌡️
➡️ High temperature → favors elimination.
➡️ Low temperature → favors substitution.
🔎 Substrate Effect 🧬
➡️ Primary alkyl halide → favors substitution (SN₂).
➡️ Tertiary alkyl halide → favors elimination (E₂).
➡️ Secondary alkyl halide → can undergo both.
🏁 Conclusion:
⚡ E₂ favored by: high temp 🌡️, bulky substrate (3°) 🧬, strong base ⚡, non-polar solvent 🧴.
🌀 SN₂ favored by: low temp ❄️, less crowded substrate (1°) 🧬, weak base but strong nucleophile 🌀, polar solvent 💧.
✨ Quick Recap:
🌀 SN₂ → α-carbon attack
⚡ E₂ → β-hydrogen removal
🔥 Strong base + high temp → elimination
💧 Weak base + polar solvent → substitution
📖 Key Idea:
🌀 Nucleophilic Substitution (SN₂/SN₁): nucleophile attacks α-carbon (carbon bonded to halogen).
⚡ Elimination (E₂/E₁): base removes β-hydrogen → double bond forms.
⚖️ Both can compete since nucleophile can also act as a base.
✨ Formation of Two Products:
When substrate R–X is treated with an electron pair donor species (e.g., base or nucleophilic reagents), there will always be two products.
💥 Strong base (OH⁻, RO⁻): β-hydrogen removal → favors E₂ (elimination), SN₂ side reaction.
🌀 Weak base but strong nucleophile (I⁻, Br⁻, CH₃COO⁻): α-carbon attack → favors SN₂ (substitution), E₂ side reaction.
🔎 Solvent Effect 🧴
➡️ Polar solvents → favor substitution (SN₂/SN₁).
➡️ Non-polar solvents → favor elimination (E₂/E₁).
🔎 Base Effect ⚗️
➡️ Strong base → promotes elimination.
➡️ Weak base → promotes substitution.
🔎 Temperature Effect 🌡️
➡️ High temperature → favors elimination.
➡️ Low temperature → favors substitution.
🔎 Substrate Effect 🧬
➡️ Primary alkyl halide → favors substitution (SN₂).
➡️ Tertiary alkyl halide → favors elimination (E₂).
➡️ Secondary alkyl halide → can undergo both.
🏁 Conclusion:
⚡ E₂ favored by: high temp 🌡️, bulky substrate (3°) 🧬, strong base ⚡, non-polar solvent 🧴.
🌀 SN₂ favored by: low temp ❄️, less crowded substrate (1°) 🧬, weak base but strong nucleophile 🌀, polar solvent 💧.
✨ Quick Recap:
🌀 SN₂ → α-carbon attack
⚡ E₂ → β-hydrogen removal
🔥 Strong base + high temp → elimination
💧 Weak base + polar solvent → substitution
Q7. Complete and balance the following reactions:
⚗️ Completed & Balanced Reactions
1️⃣ Reaction with CO₂ (Carboxylation)
R–MgX + CO₂ → R–COOMgX → H₃O⁺ + R–COOH + MgX₂ ➡️ Forms carboxylic acid 🧪
2️⃣ Alcohol with Thionyl Chloride (SOCl₂)
R–OH + SOCl₂ → R–Cl + SO₂ + HCl ➡️ Converts alcohol → alkyl chloride 🌬️
3️⃣ Methyl Magnesium Chloride with CO₂
CH₃–MgCl + CO₂ → CH₃–COOMgCl → H₃O⁺ + CH₃–COOH + MgCl₂ ➡️ Produces acetic acid 🍶
4️⃣ Ethyl Bromide with Silver Nitrite (AgNO₂)
C₂H₅–Br + AgNO₂ → C₂H₅–NO₂ + AgBr ➡️ Forms nitroethane 💥
5️⃣ Alcohol with Phosphorus Pentachloride (PCl₅)
R–OH + PCl₅ → R–Cl + POCl₃ + HCl ➡️ Converts alcohol → alkyl chloride ⚡
6️⃣ Methyl Iodide with Sodium Cyanide (NaCN)
CH₃–I + NaCN → CH₃–CN + NaI ➡️ Forms methyl cyanide (acetonitrile) 🧴
✨ Quick Recap:
🧪 Grignard + CO₂ → Carboxylic acid
🌬️ Alcohol + SOCl₂ → Alkyl chloride
🍶 CH₃MgCl + CO₂ → Acetic acid
💥 C₂H₅Br + AgNO₂ → Nitroethane
⚡ Alcohol + PCl₅ → Alkyl chloride
🧴 CH₃I + NaCN → Acetonitrile
1️⃣ Reaction with CO₂ (Carboxylation)
R–MgX + CO₂ → R–COOMgX → H₃O⁺ + R–COOH + MgX₂ ➡️ Forms carboxylic acid 🧪
2️⃣ Alcohol with Thionyl Chloride (SOCl₂)
R–OH + SOCl₂ → R–Cl + SO₂ + HCl ➡️ Converts alcohol → alkyl chloride 🌬️
3️⃣ Methyl Magnesium Chloride with CO₂
CH₃–MgCl + CO₂ → CH₃–COOMgCl → H₃O⁺ + CH₃–COOH + MgCl₂ ➡️ Produces acetic acid 🍶
4️⃣ Ethyl Bromide with Silver Nitrite (AgNO₂)
C₂H₅–Br + AgNO₂ → C₂H₅–NO₂ + AgBr ➡️ Forms nitroethane 💥
5️⃣ Alcohol with Phosphorus Pentachloride (PCl₅)
R–OH + PCl₅ → R–Cl + POCl₃ + HCl ➡️ Converts alcohol → alkyl chloride ⚡
6️⃣ Methyl Iodide with Sodium Cyanide (NaCN)
CH₃–I + NaCN → CH₃–CN + NaI ➡️ Forms methyl cyanide (acetonitrile) 🧴
✨ Quick Recap:
🧪 Grignard + CO₂ → Carboxylic acid
🌬️ Alcohol + SOCl₂ → Alkyl chloride
🍶 CH₃MgCl + CO₂ → Acetic acid
💥 C₂H₅Br + AgNO₂ → Nitroethane
⚡ Alcohol + PCl₅ → Alkyl chloride
🧴 CH₃I + NaCN → Acetonitrile
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Tags
Alkyl Halides Amines reactions
Chapter 6 Alkyl Halides & Amines
Chemistry MCQs & short-answer questions
Karachi Board XII Chemistry
Second year
solved questions