CUET UG Chemistry Booster Test - 3 Preparation of Amines -

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CUET UG Chemistry Booster Test - 3 Preparation of Amines

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Chemistry Booster Test - 3 Preparation of Amines

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QUESTION 1 OF 20

Critically evaluate the following regarding nitro reduction to amines:
(A) Nitroalkanes can be effectively reduced to alkanamines.
(B) Nitroarenes are catalytically reduced to arylamines.
(C) The solid catalyst (Ni, Pd, Pt) must be finely divided.
(D) Hydrogen gas is passed through the system.

Answer Review

1 Critically evaluate the following regarding nitro reduction to amines:
(A) Nitroalkanes can be effectively reduced to alkanamines.
(B) Nitroarenes are catalytically reduced to arylamines.
(C) The solid catalyst (Ni, Pd, Pt) must be finely divided.
(D) Hydrogen gas is passed through the system.

A A, B, C, D

B B, C, D

C A, C, D

D A, B

💡 Short Explanation

�� Nitroalkanes reduce to alkanamines. �� Nitroarenes reduce to arylamines. �� H₂ with finely divided Ni/Pd/Pt is used.

✅ Correct Answer Explanation

Nitro compounds are reduced to amines by catalytic hydrogenation using hydrogen gas in the presence of finely divided nickel, palladium or platinum. → Nitroalkanes give alkanamines. → Nitroarenes give arylamines. → Finely divided catalysts provide large surface area. → Hydrogen gas acts as the reducing agent. Therefore, all statements A, B, C and D are correct.

❓ Why Other Options Are Incorrect

�� Option B → Omits Statement A, which is correct.
�� Option C → Omits Statement B, which is correct.
�� Option D → Omits Statements C and D, which are correct.

🎯 Strategy

Used

Option Grouping

Application:

�� Check each statement with catalytic hydrogenation of nitro compounds.

Final Logic:

�� All four statements correctly describe nitro reduction.

🧠 Memory Trick

NO₂ + H₂/Ni = NH₂

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Preparation of Amines by Reduction of Nitro Compounds

2 What is the IUPAC name of the target product formed by the vigorous reduction of 1-nitropropane with metals in acidic medium?

A Propan-1-amine

B Propan-2-amine

C N-Methylpropanamine

D Butan-1-amine

💡 Short Explanation

�� 1-Nitropropane has three carbons. �� Reduction converts –NO₂ to –NH₂. �� The amino group remains at carbon-1.

✅ Correct Answer Explanation

1-Nitropropane has the structure CH₃CH₂CH₂NO₂. On reduction with metals in acidic medium, the nitro group is converted into an amino group. CH₃CH₂CH₂NO₂ → CH₃CH₂CH₂NH₂ The product is Propan-1-amine.

❓ Why Other Options Are Incorrect

�� Option B → Propan-2-amine would have NH₂ on the second carbon.
�� Option C → N-Methylpropanamine is a secondary amine.
�� Option D → Butan-1-amine has four carbon atoms.

🎯 Strategy

Used

Substitution

Application:

�� Replace –NO₂ with –NH₂ without changing carbon skeleton.

Final Logic:

�� 1-Nitropropane reduces to Propan-1-amine.

🧠 Memory Trick

NO₂ becomes NH₂

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Reduction of Nitroalkanes

3 Why are transition metal catalysts like nickel, palladium, or platinum required to be 'finely divided' when deployed in the hydrogenation of nitro compounds?

A To deliberately decrease the reactive surface area.

B To massively increase the surface area for superior adsorption of hydrogen gas.

C To chemically lower the standard temperature of the reaction.

D To alter the physical phase of the catalyst.

💡 Short Explanation

�� Finely divided catalysts have greater surface area. �� More hydrogen gets adsorbed. �� This increases catalytic efficiency.

✅ Correct Answer Explanation

In catalytic hydrogenation, hydrogen gas is adsorbed on the surface of the metal catalyst. Finely divided nickel, palladium or platinum provides a large surface area, allowing better adsorption of hydrogen and more effective reduction of nitro compounds to amines.

❓ Why Other Options Are Incorrect

�� Option A → Finely divided catalyst increases, not decreases, surface area.
�� Option C → The main purpose is surface adsorption, not directly lowering standard temperature.
�� Option D → The catalyst remains solid; its phase is not changed.

🎯 Strategy

Used

Elimination

Application:

�� Identify the role of surface area in catalysis.

Final Logic:

�� Greater surface area improves hydrogen adsorption.

🧠 Memory Trick

Fine Catalyst = More Surface

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Catalytic Hydrogenation of Nitro Compounds

4 In the presence of finely divided palladium or platinum, what chemical species acts as the direct reducing agent for transforming nitro compounds into amines?

A Carbon dioxide

B Hydrogen gas

C Iron scrap

D Hydrochloric acid

💡 Short Explanation

�� H₂ supplies hydrogen for reduction. �� Pd/Pt acts only as catalyst. �� Nitro group converts to amino group.

✅ Correct Answer Explanation

In catalytic hydrogenation, palladium or platinum provides the surface for reaction, but the actual reducing species is hydrogen gas. Hydrogen converts the nitro group (–NO₂) into the amino group (–NH₂).

❓ Why Other Options Are Incorrect

�� Option A → Carbon dioxide is not a reducing agent here.
�� Option C → Iron scrap is used in metal-acid reduction, not Pd/Pt catalytic hydrogenation.
�� Option D → Hydrochloric acid provides acidic medium in Fe/HCl reduction, but is not the direct reducing gas.

🎯 Strategy

Used

Elimination

Application:

�� Separate catalyst from reducing agent.

Final Logic:

�� Pd/Pt catalyzes; H₂ reduces.

🧠 Memory Trick

Catalyst Helps, H₂ Reduces

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Catalytic Reduction of Nitro Compounds

List 1 (Operational Component)	List 2 (Primary Role/Nature)
1. Fe	a. Hydrolyses to continuously release acid
2. HCl	b. Starting reactant targeted to be reduced
3. FeCl₂	c. Source providing the acidic medium
4. RNO₂	d. Metal reducing agent

A 1-d, 2-c, 3-a, 4-b

B 1-a, 2-b, 3-c, 4-d

C 1-d, 2-a, 3-c, 4-b

D 1-c, 2-d, 3-a, 4-b

💡 Short Explanation

�� Fe acts as metal reducing agent. �� HCl gives acidic medium. �� FeCl₂ hydrolyses and releases HCl.

✅ Correct Answer Explanation

In Fe/HCl reduction of nitro compounds: → Fe acts as the metal reducing agent. → HCl provides the acidic medium. → FeCl₂ formed during the reaction undergoes hydrolysis and releases HCl. → RNO₂ is the starting nitro compound targeted for reduction. Correct matching: 1-d, 2-c, 3-a, 4-b

❓ Why Other Options Are Incorrect

�� Option B → Fe is incorrectly matched with acid release instead of reducing action.
�� Option C → HCl and FeCl₂ roles are interchanged.
�� Option D → Fe and HCl roles are interchanged.

🎯 Strategy

Used

Option Grouping

Application:

�� Match each component with its chemical role in Fe/HCl reduction.

Final Logic:

�� Fe reduces, HCl acidifies, FeCl₂ regenerates acid, RNO₂ is reduced.

🧠 Memory Trick

Fe Reduces, FeCl₂ Releases HCl

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Reduction of Nitro Compounds using Fe/HCl

6 Consider the critical role of hydrolysis in the Fe/HCl reduction framework:
(A) It rapidly consumes all available hydrochloric acid.
(B) It cyclically releases hydrochloric acid during the reaction.
(C) It suppresses the reaction from proceeding to completion.
(D) It restricts the method to small-scale laboratory use only.

A A, C

B B

C B, D

D A, D

💡 Short Explanation

�� FeCl₂ undergoes hydrolysis. �� Hydrolysis releases HCl. �� This reduces the amount of acid required.

✅ Correct Answer Explanation

In Fe/HCl reduction, FeCl₂ formed during the reaction undergoes hydrolysis and releases hydrochloric acid. This regenerated HCl keeps the acidic medium active, so only a small amount of HCl is initially required. Only Statement B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Statements A and C are incorrect. Hydrolysis does not consume all HCl or stop the reaction.
�� Option C → Statement D is incorrect; the method is preferred and practical.
�� Option D → Both statements are incorrect.

🎯 Strategy

Used

Elimination

Application:

�� Remove statements opposite to NCERT explanation of FeCl₂ hydrolysis.

Final Logic:

�� Hydrolysis releases HCl during the reaction.

🧠 Memory Trick

Hydrolysis Helps HCl Return

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Fe/HCl Reduction and Acid Regeneration

📄 Passage

The carbon - halogen bond in alkyl or benzyl halides can be easily cleaved by a nucleophile. Hence, an alkyl or benzyl halide on reaction with an ethanolic solution of ammonia undergoes nucleophilic substitution reaction in which the halogen atom is replaced by an amino (–NH₂) group. This process of cleavage of the C–X bond by ammonia molecule is known as ammonolysis. The primary amine thus obtained behaves as a nucleophile and can further react with alkyl halide.

What foundational type of organic mechanism dictates the reaction when an alkyl halide is exposed to an ethanolic solution of ammonia?

A Electrophilic substitution

B Nucleophilic substitution

C Electrophilic addition

D Nucleophilic elimination

💡 Short Explanation

�� Ammonia acts as a nucleophile. �� Halogen is replaced by –NH₂. �� The reaction is nucleophilic substitution.

✅ Correct Answer Explanation

The passage states that alkyl or benzyl halides react with ethanolic ammonia through nucleophilic substitution. In this reaction, the halogen atom is replaced by the amino group (–NH₂). Therefore, the mechanism is nucleophilic substitution.

❓ Why Other Options Are Incorrect

�� Option A → Ammonia is a nucleophile, not an electrophile.
�� Option C → No addition across a multiple bond occurs.
�� Option D → The reaction involves substitution, not elimination.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Use the exact mechanism stated in the passage.

Final Logic:

�� Alkyl halide + ethanolic NH₃ = nucleophilic substitution.

🧠 Memory Trick

NH₃ attacks, X leaves

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Ammonolysis of Alkyl Halides

📄 Passage

According to the passage, why is the primary amine capable of reacting further with the alkyl halide?

A Because it rapidly transforms into an electrophile.

B Because it retains an unshared electron pair and behaves as a nucleophile.

C Because it acts as a Lewis acid in the mixture.

D Because it functions as a specialized catalyst.

💡 Short Explanation

�� Primary amine contains lone pair on nitrogen. �� It behaves as a nucleophile. �� Therefore, it can react further with alkyl halide.

✅ Correct Answer Explanation

Amines have an unshared pair of electrons on nitrogen. This lone pair allows the primary amine to behave as a nucleophile. Hence, after formation, it can further attack alkyl halides and produce higher amines.

❓ Why Other Options Are Incorrect

�� Option A → Primary amine does not become an electrophile.
�� Option C → Amines generally act as Lewis bases, not Lewis acids.
�� Option D → It reacts as a nucleophile, not as a catalyst.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Connect "reacts further" with nucleophilic lone-pair behavior.

Final Logic:

�� Lone pair on nitrogen makes primary amine nucleophilic.

🧠 Memory Trick

Amine N has Lone Pair

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Ammonolysis and Higher Amine Formation

9 If the ammonolysis of alkyl halides is experimentally conducted at 373 K in an open vessel rather than a sealed tube, what is the principal cause of reaction failure?

A The alkyl halide immediately polymerizes.

B Ammonia gas escapes, denying the system the necessary nucleophile.

C The temperature uncontrollably rises, inducing combustion.

D An exclusive quaternary salt solidifies instantly.

💡 Short Explanation

�� Ammonia is volatile. �� Sealed tube prevents NH₃ escape. �� NH₃ is required as nucleophile.

✅ Correct Answer Explanation

Ammonolysis is carried out in a sealed tube at 373 K because ammonia is volatile. In an open vessel, ammonia gas may escape, reducing the concentration of the nucleophile required for substitution. This lowers or prevents formation of amines.

❓ Why Other Options Are Incorrect

�� Option A → Alkyl halide polymerization is not the principal issue.
�� Option C → Combustion is not the normal reaction problem here.
�� Option D → Quaternary salt formation is not immediate or exclusive.

🎯 Strategy

Used

Elimination

Application:

�� Identify why sealed conditions are required.

Final Logic:

�� Sealed tube retains volatile ammonia.

🧠 Memory Trick

Seal NH₃ Inside

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Reaction Conditions for Ammonolysis

10 The temperature condition specified is 373 K. This absolute unit 'K' implies a thermodynamic temperature scale in which the standard freezing point of water sits approximately at:

A 273 K

B 0 K

C 100 K

D 373 K

💡 Short Explanation

�� Kelvin is absolute temperature scale. �� Freezing point of water is 0°C. �� 0°C corresponds to 273 K.

✅ Correct Answer Explanation

The Kelvin scale is the SI unit scale for thermodynamic temperature. On this scale, the freezing point of water is approximately 273 K, while the boiling point is approximately 373 K.

❓ Why Other Options Are Incorrect

�� Option B → 0 K is absolute zero.
�� Option C → 100 K is far below water's freezing point.
�� Option D → 373 K is boiling point of water, not freezing point.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

�� Convert known Celsius reference points to Kelvin.

Final Logic:

�� 0°C = 273 K.

🧠 Memory Trick

Water freezes at 273 K, boils at 373 K

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Reaction Conditions and Temperature Units

11 The analytical treatment of the ultimate quaternary ammonium salt (formed in ammonolysis) with a strong base to liberate the free primary/secondary amine is defined by which chemical action?

A Catalytic Oxidation

B Acid-base neutralization/displacement

C Step-growth Polymerization

D Aldol Condensation

💡 Short Explanation

�� Quaternary ammonium salt is ionic. �� Strong base helps liberate free amine. �� This is an acid-base displacement process.

✅ Correct Answer Explanation

In ammonolysis, amines are often obtained initially as ammonium salts. Treatment with a strong base liberates the free amine from its salt. This involves acid-base neutralization or displacement, where the base removes the acidic proton from the ammonium species and releases the free amine.

❓ Why Other Options Are Incorrect

�� Option A → No oxidation occurs in this step.
�� Option C → Polymer formation is not involved.
�� Option D → Aldol condensation involves carbonyl compounds, not ammonium salts.

🎯 Strategy

Used

Elimination

Application:

�� Identify the chemical role of strong base in releasing amine from ammonium salt.

Final Logic:

�� Strong base liberates free amine through acid-base displacement.

🧠 Memory Trick

Base frees amine

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Ammonolysis and Liberation of Free Amines

12 Arrange the following compounds sequentially based on the order they form during the exhaustive ammonolysis of an alkyl halide (assuming 1:1 reaction steps):
(A) Quaternary ammonium salt
(B) Primary amine
(C) Tertiary amine
(D) Secondary amine

A (B), (D), (C), (A)

B (A), (B), (C), (D)

C (B), (C), (D), (A)

D (A), (D), (C), (B)

💡 Short Explanation

�� First alkylation gives primary amine. �� Further alkylation gives secondary and tertiary amines. �� Final product is quaternary ammonium salt.

✅ Correct Answer Explanation

During exhaustive ammonolysis, ammonia first forms a primary amine. The primary amine can further react with alkyl halide to form a secondary amine. The secondary amine then forms a tertiary amine, and further alkylation gives a quaternary ammonium salt. Sequence: Primary amine → Secondary amine → Tertiary amine → Quaternary ammonium salt Therefore: (B), (D), (C), (A)

❓ Why Other Options Are Incorrect

�� Option B → Starts with quaternary ammonium salt, which is the final product.
�� Option C → Places tertiary amine before secondary amine.
�� Option D → Gives reverse-like incorrect sequence.

🎯 Strategy

Used

Ordering

Application:

�� Follow stepwise alkylation of nitrogen.

Final Logic:

�� 1° → 2° → 3° → 4° ammonium salt.

🧠 Memory Trick

Primary, Secondary, Tertiary, Quaternary

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Successive Alkylation in Ammonolysis

13 Deduce the IUPAC name of the organic product obtained when ethanenitrile is exhaustively reduced using LiAlH₄.

A Methanamine

B Ethanamine

C Propanamine

D Butanamine

💡 Short Explanation

�� Ethanenitrile has two carbon atoms. �� Reduction converts –C≡N into –CH₂NH₂. �� Product is ethanamine.

✅ Correct Answer Explanation

Ethanenitrile, CH₃CN, contains two carbon atoms. On reduction with LiAlH₄, nitriles are converted into primary amines. CH₃CN → CH₃CH₂NH₂ The product is Ethanamine.

❓ Why Other Options Are Incorrect

�� Option A → Methanamine has one carbon atom.
�� Option C → Propanamine has three carbon atoms.
�� Option D → Butanamine has four carbon atoms.

🎯 Strategy

Used

Substitution

Application:

�� Convert nitrile group into –CH₂NH₂ and retain total carbon count.

Final Logic:

�� Ethanenitrile reduces to Ethanamine.

🧠 Memory Trick

Nitrile carbon becomes CH₂NH₂

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Reduction of Nitriles

14 In the strategic "ascent of amine series" using nitrile reduction, the final primary amine contains how many more carbon atoms compared to the initial amine starting material?

A Zero

B One

C Two

D Three

💡 Short Explanation

�� Nitrile carbon becomes part of amine. �� Carbon chain increases by one. �� Hence, it is ascent of series.

✅ Correct Answer Explanation

In ascent of amine series, the nitrile route adds one carbon atom to the final amine chain because the carbon of the –CN group becomes the carbon of –CH₂NH₂ after reduction. Therefore, the final amine contains one more carbon atom than the starting material.

❓ Why Other Options Are Incorrect

�� Option A → No increase would not be ascent.
�� Option C → Nitrile reduction adds only one carbon.
�� Option D → Three-carbon increase does not occur.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

�� Count the carbon added through the –CN group.

Final Logic:

�� –CN contributes one carbon atom.

🧠 Memory Trick

CN adds 1 carbon

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Ascent of Amine Series

15 Which reagent specifically functions to transform the carbonyl moiety of an amide directly into a methylene group (–CH₂–), consequently forming an amine?

A Aqueous Br₂/NaOH

B Dilute H₂O/H⁺

C Lithium aluminium hydride (LiAlH₄)

D Ethanolic NH₃

💡 Short Explanation

�� LiAlH₄ is a strong reducing agent. �� It reduces amide carbonyl to –CH₂–. �� Product formed is an amine.

✅ Correct Answer Explanation

Lithium aluminium hydride (LiAlH₄) reduces amides to amines by converting the carbonyl group (C=O) into a methylene group (–CH₂–). General conversion: RCONH₂ → RCH₂NH₂ Thus, LiAlH₄ is the correct reagent.

❓ Why Other Options Are Incorrect

�� Option A → Br₂/NaOH causes Hofmann bromamide degradation with one carbon loss.
�� Option B → Hydrolysis does not reduce amides to amines.
�� Option D → Ethanolic NH₃ is used in ammonolysis of alkyl halides.

🎯 Strategy

Used

Elimination

Application:

�� Identify the reagent that performs strong reduction of amides.

Final Logic:

�� LiAlH₄ reduces C=O to –CH₂–.

🧠 Memory Trick

LiAlH₄ reduces C=O

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Reduction of Amides

16 Assess the following statements on the chemical reduction of amides:
(A) The reaction efficiently utilizes LiAlH₄.
(B) It forms amines containing the exact same number of carbon atoms as the parent amide.
(C) It forms primary amines with one carbon less than the parent amide.
(D) It yields secondary amines exclusively.

A A, B

B A, C

C B, D

D A, D

💡 Short Explanation

�� LiAlH₄ reduces amides to amines. �� Carbon number is retained in this reduction. �� One-carbon loss occurs in Hofmann degradation, not LiAlH₄ reduction.

✅ Correct Answer Explanation

Statement A is correct because LiAlH₄ is the standard reagent used for reducing amides to amines. → Statement B is correct because amide reduction converts the carbonyl group into –CH₂– without removing the carbonyl carbon. Example: RCONH₂ → RCH₂NH₂ Thus, the amine has the same number of carbon atoms as the parent amide. → Statement C is incorrect because one-carbon loss is seen in Hoffmann bromamide degradation, not LiAlH₄ reduction. → Statement D is incorrect because primary amides give primary amines; the reaction does not yield secondary amines exclusively.

❓ Why Other Options Are Incorrect

�� Option B → Includes Statement C, which is incorrect.
�� Option C → Includes Statement D and omits Statement A.
�� Option D → Includes Statement D, which is incorrect.

🎯 Strategy

Used

Option Grouping

Application:

�� Separate LiAlH₄ reduction from Hoffmann bromamide degradation.

Final Logic:

�� LiAlH₄ reduces amides without carbon loss.

🧠 Memory Trick

LiAlH₄ keeps carbon count

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Reduction of Amides

17 Examine the molecular steps within Gabriel synthesis:
(A) Ethanolic KOH abstracts a proton to form a potassium salt.
(B) N-Alkylphthalimide emerges as the key intermediate upon alkyl halide addition.
(C) It yields pure secondary and tertiary amines.
(D) Final alkaline hydrolysis cleaves the intermediate to form a primary amine.

A A, B, C

B A, B, D

C B, C, D

D A, C, D

💡 Short Explanation

�� Ethanolic KOH forms potassium phthalimide. �� Alkyl halide gives N-alkylphthalimide. �� Hydrolysis gives primary amine.

✅ Correct Answer Explanation

Statement A is correct because ethanolic KOH removes the acidic hydrogen of phthalimide, forming potassium phthalimide. → Statement B is correct because potassium phthalimide reacts with alkyl halide to form N-alkylphthalimide. → Statement C is incorrect because Gabriel synthesis is used to prepare primary amines, not secondary and tertiary amines. → Statement D is correct because alkaline hydrolysis of N-alkylphthalimide gives a primary amine.

❓ Why Other Options Are Incorrect

�� Option A → Includes Statement C, which is incorrect.
�� Option C → Includes Statement C, which is incorrect.
�� Option D → Includes Statement C and omits Statement B.

🎯 Strategy

Used

Option Grouping

Application:

�� Follow the Gabriel synthesis sequence step by step.

Final Logic:

�� KOH → potassium salt → N-alkylphthalimide → primary amine.

🧠 Memory Trick

Gabriel gives 1° amine only

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Gabriel Phthalimide Synthesis

18 The fundamental chemical reason Gabriel synthesis is restricted solely to primary aliphatic amines and fails for anilines is that:

A Aryl halides undergo facile electrophilic substitution instead.

B The phthalimide anion is sterically too bulky to approach aryl halides.

C Aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.

D Phthalimide instantly decomposes aromatic ring systems.

💡 Short Explanation

�� Gabriel synthesis needs nucleophilic substitution. �� Alkyl halides undergo this reaction. �� Aryl halides do not react suitably.

✅ Correct Answer Explanation

Gabriel synthesis involves nucleophilic substitution between potassium phthalimide and an alkyl halide. Aryl halides do not undergo nucleophilic substitution with the phthalimide anion under these conditions because the aryl C–X bond has partial double-bond character and is difficult to cleave. Therefore, anilines cannot be prepared by Gabriel synthesis.

❓ Why Other Options Are Incorrect

�� Option A → Electrophilic substitution on aryl halides is not the reason Gabriel synthesis fails.
�� Option B → Steric bulk is not the main NCERT reason.
�� Option D → Phthalimide does not instantly decompose aromatic systems.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Identify the key reaction step Gabriel synthesis requires.

Final Logic:

�� No nucleophilic substitution by aryl halides means no aniline formation.

🧠 Memory Trick

Gabriel: Alkyl works, Aryl fails

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Limitation of Gabriel Synthesis

19 Mechanistically, during the Hoffmann bromamide degradation reaction, the migrating alkyl or aryl group transfers from which atom to which atom?

A From the nitrogen atom to the carbonyl carbon

B From the adjacent oxygen to the nitrogen atom

C From the carbonyl carbon directly to the nitrogen atom

D From the alpha-carbon to the beta-carbon

💡 Short Explanation

�� Hofmann degradation involves rearrangement. �� Alkyl/aryl group migrates. �� Migration occurs from carbonyl carbon to nitrogen.

✅ Correct Answer Explanation

In Hoffmann bromamide degradation, the alkyl or aryl group attached to the carbonyl carbon migrates to the nitrogen atom during rearrangement. This rearrangement leads to the formation of an isocyanate intermediate, which finally hydrolyses to a primary amine with one carbon less.

❓ Why Other Options Are Incorrect

�� Option A → Direction is reversed.
�� Option B → Oxygen is not the migrating center.
�� Option D → No alpha-to-beta carbon migration occurs.

🎯 Strategy

Used

Elimination

Application:

�� Focus on the known rearrangement step of Hoffmann degradation.

Final Logic:

�� R group migrates from carbonyl carbon to nitrogen.

🧠 Memory Trick

Carbonyl to Nitrogen Migration

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Hoffmann Bromamide Degradation Mechanism

20 If a chemist targets the preparation of pure Propanamine utilizing Hoffmann bromamide degradation, which exact amide precursor must they synthesize first?

A Propanamide

B Butanamide

C Ethanamide

D Pentanamide

💡 Short Explanation

�� Hoffmann degradation removes one carbon. �� Propanamine has three carbons. �� Starting amide must have four carbons.

✅ Correct Answer Explanation

Hoffmann bromamide degradation converts an amide into a primary amine with one carbon atom less than the starting amide. Target product: Propanamine = 3 carbons Required starting amide: Butanamide = 4 carbons Reaction: Butanamide → Propanamine Therefore, butanamide is the correct precursor.

❓ Why Other Options Are Incorrect

�� Option A → Propanamide would give Ethanamine.
�� Option C → Ethanamide would give Methanamine.
�� Option D → Pentanamide would give Butanamine.

🎯 Strategy

Used

Substitution

Application:

�� Add one carbon to the target amine to identify the amide precursor.

Final Logic:

�� Target 3C amine requires 4C amide.

🧠 Memory Trick

Hofmann: Amide Carbon −1

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Hoffmann Bromamide Degradation

CUET UG Chemistry Booster Test – 3 Preparation of Amines

CUET UG Chemistry Booster Test - 3 Preparation of Amines

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