CUET UG Chemistry Booster Test - 2 Preparation of Amines -

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CUET UG Chemistry Booster Test - 2 Preparation of Amines

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Chemistry Booster Test - 2 Preparation of Amines

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QUESTION 1 OF 20

The chemical reaction where nitro compounds are converted to amines using H₂ gas over a finely divided Pd catalyst represents which type of organic conversion?

Answer Review

1 The chemical reaction where nitro compounds are converted to amines using H₂ gas over a finely divided Pd catalyst represents which type of organic conversion?

A Catalytic hydrogenation

B Electrophilic aromatic substitution

C Nucleophilic substitution

D Elimination

💡 Short Explanation

�� Hydrogen gas is used as the reducing agent. �� Palladium acts as a catalyst. �� Nitro group is converted into an amino group.

✅ Correct Answer Explanation

Nitro compounds are reduced to amines by passing hydrogen gas in the presence of finely divided catalysts such as Pd, Pt, or Ni. Example: R–NO₂ + 3H₂ → R–NH₂ + 2H₂O → Since hydrogen is added in the presence of a catalyst, the process is called catalytic hydrogenation.

❓ Why Other Options Are Incorrect

�� Option B → No electrophilic substitution occurs.
�� Option C → No nucleophile replaces a group.
�� Option D → No elimination of atoms/groups takes place.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Identify the reaction from the reagents H₂ and Pd.

Final Logic:

�� H₂ + catalyst = catalytic hydrogenation.

🧠 Memory Trick

H₂ + Pd = Hydrogenation

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

2 When nitro compounds are reduced using metals like Fe, the reaction environment must strictly be:

A Strongly basic

B Neutral

C Acidic

D Aqueous alkaline

💡 Short Explanation

�� Fe/HCl reduction is carried out in acidic medium. �� Acid supplies protons required for reduction. �� Amines are produced efficiently.

✅ Correct Answer Explanation

Nitro compounds can be reduced using metals such as iron in the presence of hydrochloric acid. Example: R–NO₂ + Fe/HCl → R–NH₂ → The reaction requires an acidic medium for successful reduction.

❓ Why Other Options Are Incorrect

�� Option A → Strongly basic medium is not used.
�� Option B → Neutral medium is ineffective.
�� Option D → Alkaline medium is not the standard condition.

🎯 Strategy

Used

Elimination

Application:

�� Recall Fe/HCl reduction conditions.

Final Logic:

�� Fe reduction proceeds in acidic medium.

🧠 Memory Trick

Fe + HCl = Acidic Reduction

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

3 During the catalytic hydrogenation of nitroethane utilizing a finely divided nickel catalyst, the product formed has the IUPAC name:

A Methanamine

B Ethanamine

C N-Methylethanamine

D Nitroethane

💡 Short Explanation

�� Nitro group converts into amino group. �� Carbon chain remains unchanged. �� Nitroethane becomes ethanamine.

✅ Correct Answer Explanation

Nitroethane (CH₃CH₂NO₂) undergoes reduction: CH₃CH₂NO₂ → CH₃CH₂NH₂ → The product CH₃CH₂NH₂ is called ethanamine.

❓ Why Other Options Are Incorrect

�� Option A → One-carbon amine.
�� Option C → Secondary amine.
�� Option D → Starting material itself.

🎯 Strategy

Used

Substitution

Application:

�� Replace –NO₂ with –NH₂.

Final Logic:

�� Nitroethane → Ethanamine.

🧠 Memory Trick

NO₂ → NH₂

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

4 Regarding catalysts in nitro reduction:
(A) Palladium effectively catalyzes the reduction.
(B) Platinum is also an effective catalyst.
(C) Catalysts must be finely divided for optimal function.
(D) Nitroalkanes can be similarly reduced to alkanamines.

A A, B, C, D

B A, B, D

C B, C, D

D A, C, D

💡 Short Explanation

�� Pd and Pt are effective catalysts. �� Finely divided catalysts provide large surface area. �� Nitroalkanes form alkanamines on reduction.

✅ Correct Answer Explanation

Palladium, platinum, and nickel are commonly used catalysts. → Catalysts are finely divided to increase surface area and hydrogen adsorption. → Nitroalkanes and nitroarenes are both reduced to corresponding amines. Therefore all statements are correct.

❓ Why Other Options Are Incorrect

�� Option B → Omits statement C.
�� Option C → Omits statement A.
�� Option D → Omits statement B.

🎯 Strategy

Used

Option Grouping

Application:

�� Verify each statement individually.

Final Logic:

�� All four statements are correct.

🧠 Memory Trick

Pd, Pt, Ni + H₂ = Amines

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

📄 Passage

Nitro compounds are reduced to amines by passing hydrogen gas in the presence of finely divided nickel, palladium or platinum and also by reduction with metals in acidic medium. Reduction with iron scrap and hydrochloric acid is preferred because FeCl₂ formed gets hydrolysed to release hydrochloric acid during the reaction. Thus, only a small amount of hydrochloric acid is required to initiate the reaction.

Based on the text, why is reduction with iron scrap and hydrochloric acid specifically preferred over other metal-acid combinations?

A Iron is the most reactive metal available.

B FeCl₂ is highly stable and does not participate further.

C FeCl₂ gets hydrolysed to continually release HCl during the reaction.

D It restricts the formation of secondary amines.

💡 Short Explanation

�� FeCl₂ is formed during reaction. �� FeCl₂ undergoes hydrolysis. �� HCl is regenerated continuously.

✅ Correct Answer Explanation

During Fe/HCl reduction, FeCl₂ is produced. → FeCl₂ undergoes hydrolysis and releases HCl back into the reaction mixture. → Thus only a small initial quantity of HCl is needed, making the process economical and efficient.

❓ Why Other Options Are Incorrect

�� Option A → Not the reason stated in NCERT.
�� Option B → FeCl₂ actively participates through hydrolysis.
�� Option D → Secondary amine formation is unrelated.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Extract the exact reason from the passage.

Final Logic:

�� Hydrolysis of FeCl₂ regenerates HCl.

🧠 Memory Trick

FeCl₂ → HCl Returns

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Fe/HCl Reduction Method

📄 Passage

The continuous release of HCl via hydrolysis during the Fe/HCl reduction implies that:

A No initial HCl is needed to start the reaction.

B Only a small amount of HCl is initially required to initiate the process.

C The reaction vessel must be constantly cooled to prevent excess acid.

D The reaction completely stops once FeCl₂ is formed.

💡 Short Explanation

�� FeCl₂ regenerates HCl. �� Acid is recycled. �� Small initial amount is sufficient.

✅ Correct Answer Explanation

Since FeCl₂ releases HCl through hydrolysis, hydrochloric acid is continuously regenerated. → Therefore, only a small amount of HCl is required initially to start the reaction.

❓ Why Other Options Are Incorrect

�� Option A → Initial HCl is still necessary.
�� Option C → Cooling is not mentioned.
�� Option D → FeCl₂ formation does not stop the reaction.

🎯 Strategy

Used

Elimination

Application:

�� Select the statement directly supported by the passage.

Final Logic:

�� Regeneration of HCl reduces acid requirement.

🧠 Memory Trick

Small HCl, Big Job

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Hydrolysis of FeCl₂

7 Arrange the following alkyl halides in decreasing order of their reactivity in ammonolysis:
(A) CH₃–Cl
(B) CH₃–I
(C) CH₃–Br

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

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

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

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

💡 Short Explanation

�� Better leaving group means higher reactivity. �� I⁻ leaves most easily. �� Cl⁻ leaves least easily.

✅ Correct Answer Explanation

Reactivity in ammonolysis depends on C–X bond cleavage. Leaving group ability: I⁻ > Br⁻ > Cl⁻ Therefore: CH₃I > CH₃Br > CH₃Cl

❓ Why Other Options Are Incorrect

�� Option A → Reverse order.
�� Option C → Places bromide above iodide.
�� Option D → Incorrect Cl–Br order.

🎯 Strategy

Used

Ordering

Application:

�� Rank halides by leaving group ability.

Final Logic:

�� I > Br > Cl.

🧠 Memory Trick

I Beats Br Beats Cl

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

8 Ammonolysis of alkyl halides is defined by:
(A) The cleavage of the C–X bond by an NH₃ molecule.
(B) The initial primary amine behaving as a nucleophile.
(C) The eventual formation of quaternary ammonium salts.
(D) Being a nucleophilic substitution reaction.

A A, C, D

B A, B, D

C A, B, C, D

D C, D

💡 Short Explanation

�� NH₃ cleaves C–X bond. �� Primary amine acts as nucleophile. �� Further alkylation can produce quaternary salts.

✅ Correct Answer Explanation

Ammonolysis is a nucleophilic substitution reaction where NH₃ replaces the halogen atom. → The primary amine formed can react further as a nucleophile, producing secondary amines, tertiary amines, and finally quaternary ammonium salts. Thus all statements are correct.

❓ Why Other Options Are Incorrect

�� Options A, B, D omit one or more correct statements.

🎯 Strategy

Used

Option Grouping

Application:

�� Verify each statement independently.

Final Logic:

�� All four statements are true.

🧠 Memory Trick

NH₃ → 1° → 2° → 3° → Quaternary

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Ammonolysis Mechanism

9 Why must the ammonolysis reaction be carried out in a sealed tube at 373 K?

A To prevent the escape of volatile ammonia gas at elevated temperatures.

B To cause the alkyl halide to solidify.

C To act as a solid-phase catalyst for the reaction.

D To selectively precipitate the secondary amine.

💡 Short Explanation

�� Ammonia is volatile. �� Heating causes gas loss. �� Sealed tube prevents escape.

✅ Correct Answer Explanation

Ammonolysis is performed at 373 K using ethanolic ammonia. → Since ammonia is volatile, heating in an open vessel would cause its escape. → A sealed tube maintains sufficient ammonia concentration for the reaction.

❓ Why Other Options Are Incorrect

�� Option B → No solidification occurs.
�� Option C → The tube is not a catalyst.
�� Option D → No selective precipitation occurs.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Connect ammonia volatility with reaction setup.

Final Logic:

�� Sealed tube prevents ammonia loss.

🧠 Memory Trick

Hot NH₃ Needs Closed Tube

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

10 The temperature designated for ammonolysis is 373 K. The unit 'K' corresponds to an absolute scale where 373 equates to which value in degrees Celsius (°C)?

A 100

B 273

C 373

D 0

💡 Short Explanation

�� K = °C + 273. �� 373 − 273 = 100. �� Therefore 373 K equals 100°C.

✅ Correct Answer Explanation

Conversion formula: °C = K − 273 → Therefore: 373 − 273 = 100°C Hence 373 K corresponds to 100°C.

❓ Why Other Options Are Incorrect

�� Option B → Kelvin offset value.
�� Option C → Same numerical value but wrong scale.
�� Option D → Corresponds to 273 K.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

�� Convert Kelvin to Celsius.

Final Logic:

�� 373 K = 100°C.

🧠 Memory Trick

K − 273 = °C

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

11 To obtain a primary amine as the major product rather than a mixture during ammonolysis, which condition must be met?

A Use a large excess of the alkyl halide.

B Use a large excess of ammonia.

C Conduct the reaction at exactly 500 K.

D Utilize a palladium catalyst.

💡 Short Explanation

�� Primary amines can further react with alkyl halides. �� Excess ammonia suppresses further alkylation. �� This increases the yield of primary amine.

✅ Correct Answer Explanation

During ammonolysis, the initially formed primary amine is more nucleophilic than ammonia and can further react with alkyl halides to produce secondary amines, tertiary amines, and quaternary ammonium salts. → To minimize these side reactions and obtain primary amine as the major product, a large excess of ammonia is used. → Excess NH₃ increases the probability of alkyl halide reacting with ammonia rather than with the amine product.

❓ Why Other Options Are Incorrect

�� Option A → Excess alkyl halide promotes formation of higher amines.
�� Option C → 500 K is not the required condition.
�� Option D → Palladium catalyst is used in hydrogenation, not ammonolysis.

🎯 Strategy

Used

Elimination

Application:

�� Identify the condition that suppresses further alkylation.

Final Logic:

�� Excess NH₃ favors formation of primary amine.

🧠 Memory Trick

More NH₃ = More 1° Amine

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

12 List 1 (Reacting Species) | List 2 (Resulting Product)

List 1	List 2
1. Primary amine + RX	a. Secondary amine
2. Secondary amine + RX	b. Tertiary amine
3. Tertiary amine + RX	c. Quaternary ammonium salt
4. Ammonium salt + Strong Base	d. Free amine

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

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

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

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

💡 Short Explanation

�� Primary amine undergoes alkylation to secondary amine. �� Secondary amine gives tertiary amine. �� Tertiary amine gives quaternary salt.

✅ Correct Answer Explanation

Successive alkylation during ammonolysis proceeds as: Primary amine + RX → Secondary amine Secondary amine + RX → Tertiary amine Tertiary amine + RX → Quaternary ammonium salt Ammonium salt + Strong Base → Free amine Thus: 1 → a 2 → b 3 → c 4 → d

❓ Why Other Options Are Incorrect

�� Option B → Incorrect product sequence.
�� Option C → Secondary amine does not directly produce quaternary salt.
�� Option D → Completely mismatched progression.

🎯 Strategy

Used

Option Grouping

Application:

�� Follow the sequential alkylation pathway.

Final Logic:

�� Each alkylation step increases substitution by one level.

🧠 Memory Trick

1° → 2° → 3° → 4° Salt

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Formation of Higher Amines

13 The chemical reduction of nitriles with LiAlH₄ is specifically favored when synthesizing primary amines designed for:

A Decreasing the carbon chain length (descent of series)

B Ascent of amine series

C Producing exclusive secondary amines

D Producing tertiary amines

💡 Short Explanation

�� Nitrile reduction forms primary amines. �� Carbon chain length increases by one carbon. �� Useful for ascent of amine series.

✅ Correct Answer Explanation

Nitriles are reduced by LiAlH₄ to give primary amines. Example: CH₃CN → CH₃CH₂NH₂ → The resulting amine contains one additional carbon compared with the starting amine from which the nitrile may have been prepared. → Therefore this method is widely used for the ascent of amine series.

❓ Why Other Options Are Incorrect

�� Option A → Chain length does not decrease.
�� Option C → Primary amines are formed.
�� Option D → Tertiary amines are not produced.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Recall the synthetic utility of nitrile reduction.

Final Logic:

�� Nitrile reduction increases carbon chain length by one.

🧠 Memory Trick

Nitrile Reduction = +1 Carbon

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

14 Which specific type of amine is successfully produced when a nitrile undergoes catalytic hydrogenation?

A Secondary amine

B Primary amine

C Quaternary salt

D Imide

💡 Short Explanation

�� Nitriles are reduced to amines. �� The –CN group converts into –CH₂NH₂. �� Product is always a primary amine.

✅ Correct Answer Explanation

Catalytic hydrogenation of nitriles converts the nitrile group into a primary amino group. General reaction: R–C≡N + 2H₂ → R–CH₂NH₂ → The product contains one amino group attached to a carbon chain and is therefore a primary amine.

❓ Why Other Options Are Incorrect

�� Option A → Secondary amines are not formed directly.
�� Option C → Quaternary salts are not formed.
�� Option D → Imides are unrelated products.

🎯 Strategy

Used

Elimination

Application:

�� Recall the direct product of nitrile reduction.

Final Logic:

�� Reduction of –CN gives –CH₂NH₂.

🧠 Memory Trick

CN → CH₂NH₂

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

15 What is the IUPAC name of the primary amine formed when Benzamide is completely reduced with LiAlH₄?

A Benzenamine

B Phenylmethanamine

C N-Methylaniline

D Benzylamine

💡 Short Explanation

�� Benzamide is reduced by LiAlH₄. �� Carbonyl group converts to –CH₂–. �� Product is benzylamine (IUPAC: Phenylmethanamine).

✅ Correct Answer Explanation

Benzamide (C₆H₅CONH₂) undergoes reduction with LiAlH₄. Reaction: C₆H₅CONH₂ → C₆H₅CH₂NH₂ → The product C₆H₅CH₂NH₂ is commonly called benzylamine. → Its correct IUPAC name is phenylmethanamine.

❓ Why Other Options Are Incorrect

�� Option A → Benzenamine is aniline (C₆H₅NH₂).
�� Option C → Contains an N-methyl substituent not present.
�� Option D → Benzylamine is the common name, not the IUPAC name.

🎯 Strategy

Used

Substitution

Application:

�� Replace amide carbonyl by methylene group.

Final Logic:

�� Benzamide reduction gives C₆H₅CH₂NH₂.

🧠 Memory Trick

Benzamide → Benzylamine

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Reduction of Amides with LiAlH₄

16 The conversion of an amide directly into an amine utilizing LiAlH₄ is best classified chemically as:

A Oxidation

B Reduction

C Hydrolysis

D Ammonolysis

💡 Short Explanation

�� LiAlH₄ is a powerful reducing agent. �� It converts the carbonyl group of an amide into a methylene group. �� The product formed is an amine.

✅ Correct Answer Explanation

Lithium aluminium hydride (LiAlH₄) reduces amides to amines by converting the carbonyl carbon (C=O) into a methylene group (–CH₂–). Example: CH₃CONH₂ ⟶ CH₃CH₂NH₂ → Since hydrogen is added and the oxidation state of carbon decreases, the process is classified as a reduction reaction.

❓ Why Other Options Are Incorrect

�� Option A → Oxidation involves increase in oxidation state, which does not occur.
�� Option C → Hydrolysis involves cleavage by water.
�� Option D → Ammonolysis involves reaction with ammonia.

🎯 Strategy

Used

Elimination

Application:

�� Identify the role of LiAlH₄ in organic reactions.

Final Logic:

�� LiAlH₄ converts amides into amines through reduction.

🧠 Memory Trick

LiAlH₄ = Powerful Reducer

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

17 In the Gabriel phthalimide synthesis process:
(A) Phthalimide is treated with ethanolic KOH.
(B) A potassium salt of phthalimide is generated.
(C) The salt is subsequently heated with an alkyl halide.
(D) Alkaline hydrolysis finally yields a primary amine.

A A, B, C

B B, C, D

C A, C, D

D A, B, C, D

💡 Short Explanation

�� Phthalimide forms potassium phthalimide. �� It reacts with alkyl halides. �� Hydrolysis gives primary amines.

✅ Correct Answer Explanation

Gabriel synthesis proceeds through the following steps: 1. Phthalimide + ethanolic KOH → Potassium phthalimide. 2. Potassium phthalimide + Alkyl halide → N-alkylphthalimide. 3. Alkaline hydrolysis of N-alkylphthalimide → Primary amine. → Therefore all four statements are correct.

❓ Why Other Options Are Incorrect

�� Option A → Omits the final hydrolysis step.
�� Option B → Omits the initial KOH treatment.
�� Option C → Omits potassium salt formation.

🎯 Strategy

Used

Option Grouping

Application:

�� Verify each step of Gabriel synthesis sequentially.

Final Logic:

�� All listed steps belong to Gabriel synthesis.

🧠 Memory Trick

KOH → Alkylation → Hydrolysis → 1° Amine

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

18 Which of the following compounds CANNOT be prepared using Gabriel phthalimide synthesis?

A Methanamine

B Ethanamine

C Propan-1-amine

D Aniline

💡 Short Explanation

�� Gabriel synthesis prepares primary aliphatic amines. �� Aryl halides do not undergo the required nucleophilic substitution. �� Therefore aniline cannot be prepared.

✅ Correct Answer Explanation

Gabriel phthalimide synthesis works through nucleophilic substitution of alkyl halides by the phthalimide anion. → Aryl halides generally do not undergo nucleophilic substitution under these conditions. → Since aniline (C₆H₅NH₂) is an aromatic primary amine, it cannot be synthesized by this method.

❓ Why Other Options Are Incorrect

�� Option A → Methanamine can be prepared.
�� Option B → Ethanamine can be prepared.
�� Option C → Propan-1-amine can be prepared.

🎯 Strategy

Used

Elimination

Application:

�� Identify which compound is aromatic rather than aliphatic.

Final Logic:

�� Gabriel synthesis fails for arylamines.

🧠 Memory Trick

Gabriel → Aliphatic Only

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

19 Identify the specific reaction type illustrated:
Amide + Br₂ + NaOH(aq) → Primary amine + Carbon loss

A Gabriel synthesis

B Hoffmann bromamide degradation

C Sandmeyer reaction

D Ammonolysis

💡 Short Explanation

�� Uses Br₂ and NaOH. �� Converts amide to primary amine. �� Product contains one carbon less.

✅ Correct Answer Explanation

Hoffmann bromamide degradation converts an amide into a primary amine using bromine and aqueous sodium hydroxide. General reaction: RCONH₂ + Br₂ + 4NaOH → RNH₂ + Na₂CO₃ + 2NaBr + 2H₂O → The amine formed contains one carbon atom less than the parent amide.

❓ Why Other Options Are Incorrect

�� Option A → Gabriel synthesis uses phthalimide.
�� Option C → Sandmeyer reaction involves diazonium salts.
�� Option D → Ammonolysis involves ammonia and alkyl halides.

🎯 Strategy

Used

Odd One Out

Application:

�� Recognize the characteristic reagent pair Br₂/NaOH.

Final Logic:

�� Br₂ + NaOH + amide indicates Hoffmann degradation.

🧠 Memory Trick

Br₂ + NaOH = Hoffmann

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

20 About the Hoffmann bromamide degradation:
(A) The amine formed contains one carbon more than the amide.
(B) An alkyl or aryl group explicitly migrates.
(C) Migration occurs from the carbonyl carbon to the nitrogen atom.
(D) The final amine contains one carbon less than the starting amide.

A B, D

B A, B

C B, C, D

D A, C

💡 Short Explanation

�� Rearrangement involves group migration. �� Product contains one less carbon atom. �� Carbonyl carbon is lost as carbonate/CO₂.

✅ Correct Answer Explanation

In Hoffmann bromamide degradation: • An alkyl/aryl group migrates during rearrangement. • The carbonyl carbon is removed from the carbon skeleton. • The resulting primary amine contains one carbon atom less than the original amide. Thus: ✓ (B) Correct ✓ (D) Correct ✗ (A) Incorrect ✗ (C) Incorrect

❓ Why Other Options Are Incorrect

�� Option B → Statement A is incorrect.
�� Option C → Statement C is incorrect. Migration is not described as carbonyl carbon moving to nitrogen.
�� Option D → Both statements are incorrect.

🎯 Strategy

Used

Elimination

Application:

�� Apply the key feature: one-carbon loss.

Final Logic:

�� Hoffmann degradation gives a primary amine with one less carbon and involves migration.

🧠 Memory Trick

Hoffmann = Minus One Carbon

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

CUET UG Chemistry Booster Test – 2 Preparation of Amines

CUET UG Chemistry Booster Test - 2 Preparation of Amines

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