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Welcome to Learn Chemistry by Dr. Inam Jazbi! 🎓 I’m Dr. Inam Jazbi, and this blog is your ultimate guide to Class XII Chemistry Chapter 6: Alkyl Halides and Amines. Here, you’ll find detailed model test questions, solved examples, and step-by-step explanations designed to make learning easy, clear, and effective. Whether you are preparing for Karachi Board exams, practice tests, or final term assessments, this guide will help you master important reactions, mechanisms, and problem-solving techniques to boost your grades.
.jpg "XII Chemistry Model Test Questions Chapter 6: Alkyl Halides & Amines")
1. Chapter Overview:
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Introduction to Alkyl Halides and Amines
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Importance in Organic Chemistry
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Common Reactions & Applications
2. Key Concepts to Remember:
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Nomenclature of Alkyl Halides and Amines
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Types: Primary, Secondary, Tertiary
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Physical and Chemical Properties
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Important Reactions (Substitution, Elimination, Addition)
3. Model Test Questions (MCQs & Short Questions):
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Multiple Choice Questions with Answers
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Short Answer Questions with Step-by-Step Solutions
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Numerical Problems (if applicable)
4. Mechanism-Based Questions:
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SN1 and SN2 Reactions
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Hofmann and Gabriel Synthesis
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Reactions of Amines (E.g., Diazotization, Hoffmann Bromamide)
5. Tips for Exam Preparation:
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Focus on reaction mechanisms and exceptions
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Solve previous year papers
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Revise nomenclature rules and reaction types
6. Practice Section / Downloadable Content:
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Free downloadable PDF of Chapter 6 model test questions
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Quick revision sheet for Alkyl Halides & Amines
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“💡 Bookmark this page for quick revision before exams!”
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“Struggling with Alkyl Halides and Amines? Let’s simplify it with solved model test questions!”
Model Test Questions XII Chemistry Chapter # 6 Alkyl Halides and Amines
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🧪 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 SN1 and SN2 reactions. Write down 5 differences between SN1 and SN2.
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?
✔ Correct: (b) R₂CHX → carbon attached to two alkyl groups = secondary halide.
Q2. Which alkyl halide cannot produce an alkene with alcoholic KOH?
✔ Correct: (a) Methyl bromide → no β-hydrogen, so β-elimination not possible.
Q3. Ethyl magnesium bromide with CO₂ yields:
✔ Correct: (b) Ethanoic acid → carbonation of ethyl magnesium bromide gives CH₃COOH.
Q4. Grignard’s reagent with ester produces:
✔ Correct: (c) Ketone → Grignard reagent reacts with ester to form ketones.
Q5. Amine act as bases because:
✔ Correct: (d) Amines accept protons via lone pair on nitrogen → basic character.
Q6. The structure of primary amine is:
✔ Correct: (c) Lone pair on N makes geometry tetrahedral pyramidal.
Q7. Alkyl amine reacts with nitrous acid + HCl to yield:
✔ Correct: (a) Primary amines form diazonium salts with nitrous acid.
Q8. Sɴ₂ reaction occurs most easily if the substrate molecule is:
✔ Correct: (a) Methyl iodide → least steric hindrance, fastest SN2 reaction.
Q9. Suitable reagent required for the synthesis of propane from methyl magnesium iodide is:
❌ All given options are incorrect.
✔ Correct option: C₂H₅Cl → reacts with CH₃MgI to give propane (via alkylation).
Q10. The rate of Sɴ₁ mechanism depends upon:
✔ 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?
✔ Correct: (d) Tertiary amine → lacks active hydrogen, cannot form alkane with Grignard’s reagent.
Q2. Grignard’s reagent gives primary alcohol on reaction with:
✔ 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:
✔ 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?
✔ 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?
✔ Correct: (c) Aqueous caustic potash → nucleophilic substitution gives alcohol.
Q6. Sɴ₂ reactions complete in ______ steps.
✔ Correct: Sɴ₂ is a single-step mechanism with simultaneous bond breaking and bond forming.
Q7. Sɴ₁ reactions complete in ______ steps.
✔ Correct: Sɴ₁ occurs in two steps → carbocation formation, then nucleophilic attack.
Q8. Which of the following undergo Sɴ₁ reactions?
✔ Correct: (c) Tertiary halides undergo Sɴ₁ easily due to stable carbocation formation.
Q9. The composition of sodium-lead alloy is:
✔ Correct: (a) Na₄Pb is the known sodium-lead alloy composition.
Q10. A primary alkyl halide would prefer to undergo _____________.
✔ 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?
✔ 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:
✔ Correct: (d) Steric hindrance → bulky groups block backside attack required in SN2.
Q13. For which mechanisms, the first step involved is the same:
✔ 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:
✔ Correct: (c) Dry anhydrous ether → prevents decomposition of Grignard reagent by moisture.
Q15. Which of the following reactions are favoured by polar aprotic solvent?
✔ 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?
✔ 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?
✔ 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?
✔ 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?
✔ 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?
✔ Correct: (a) Primary halide (CH₃Cl) → undergoes Sɴ₂ easily due to minimal steric hindrance.
Q21. Neutral nucleophiles among the following is:
✔ Correct: (b) Ammonia (NH₃) → neutral molecule with lone pair, acts as nucleophile without charge.
Q22. 100% inversion of configuration takes place during:
✔ Correct: (b) Sɴ₂ → backside attack causes complete inversion of configuration (Walden inversion).
Q23. During Sɴ₂ mechanism carbon atom changes its state of hybridization as:
✔ 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
🧪 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.
⚡ 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
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).
🔗 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.
➡️ 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.
🔬 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
🧪 ⚡ 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
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