JEE Chemistry Amines MCQs Set 17

Multiple Choice Questions (MCQs) for JEE Chemistry: Amines

Access targeted multiple-choice questions for Amines designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Practice Amines MCQs for JEE Chemistry

Access the complete set of multiple-choice questions for Amines below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: Among the following, which does not act as an intermediate in the Hofmann rearrangement?
(a) RNCO
(b) RCON:
(c) RCON(H)Br
(d) RNC
Show Answer & Explanation

Answer: (d) RNC

Explanation:
1. The steps are N-bromoamide, bromamide anion, acyl nitrene (or concerted shift), isocyanate, then hydrolysis.
2. An isocyanide RNC does not form.

Teacher's Note:
1. Always link acid strength to the stability of the carboxylate ion.
2. Electron-withdrawing groups increase acidity and electron-donating groups decrease it.

Question 2: Aniline reacts with which of these to form a Schiff base?
(a) acetic acid
(b) benzaldehyde
(c) acetone
(d) NH3
Show Answer & Explanation

Answer: (b) benzaldehyde

Explanation:
1. A primary aromatic amine condenses with an aldehyde to give an imine, C6H5N=CHC6H5, from benzaldehyde.

Teacher's Note:
1. Note the name reaction and write its equation once in your notebook.
2. Such one-step recall questions are common in the JEE and NEET style papers.

Question 3: Which of the following does not reduce Tollens' reagent?
(a) CH3CHO
(b) HCOOH
(c) C6H5NHOH
(d) C6H5NO2
Show Answer & Explanation

Answer: (d) C6H5NO2

Explanation:
1. Nitrobenzene is at a high oxidation level and cannot reduce Ag+.
2. The other three are reducing agents.

Teacher's Note:
1. Count carbons before and after the reaction to avoid mistakes in chain-length changes.
2. Check whether the reagent is a reducing, oxidising or dehydrating agent before choosing the product.

Question 4: Which one of the following is the most basic? (A) N-propylaniline, (B) N-ethylbenzylamine, (C) N-phenylpropanamide
(a) A
(b) B
(c) C
(d) all are equally basic
Show Answer & Explanation

Answer: (b) B

Explanation:
1. In B the nitrogen is attached only to alkyl-type carbons and the lone pair is free.
2. In A the lone pair is delocalised into the ring and in C into the carbonyl group.

Teacher's Note:
1. Eliminate options by checking functional groups on both sides of the reaction.
2. Only one option will have the correct functional group and carbon skeleton.

Question 5: Which one of the following methods is neither meant for the synthesis nor for the separation of amines?
(a) Hinsberg method
(b) Hofmann method
(c) Wurtz reaction
(d) Curtius reaction
Show Answer & Explanation

Answer: (c) Wurtz reaction

Explanation:
1. The Wurtz reaction couples alkyl halides to alkanes and does not involve amines.

Teacher's Note:
1. Revise the physical properties such as hydrogen bonding and boiling points together with the chemistry.
2. Compare similar compounds in pairs when you are asked to rank them.

Question 6: Aniline is converted in a set of reactions: aniline → (NaNO2/HCl) A → (CuCN) B → (H2/Ni) C → (HNO2) D. The product D is
(a) C6H5CH2NH2
(b) C6H5NHCH2CH3
(c) C6H5NHOH
(d) C6H5CH2OH
Show Answer & Explanation

Answer: (d) C6H5CH2OH

Explanation:
1. A = benzenediazonium chloride; B = benzonitrile; C = benzylamine.
2. Nitrous acid converts the primary aliphatic-type amine C6H5CH2NH2 to the alcohol C6H5CH2OH.

Teacher's Note:
1. Learn the reactivity order of acid derivatives: acyl chloride, anhydride, ester, amide.
2. Most questions of this type are solved by recalling the standard reagent and product pair.

Question 7: Electrolytic reduction of nitrobenzene in a weakly acidic medium gives
(a) aniline
(b) nitrosobenzene
(c) N-phenylhydroxylamine
(d) p-hydroxyaniline
Show Answer & Explanation

Answer: (c) N-phenylhydroxylamine

Explanation:
1. In weakly acidic or neutral medium the reduction stops at the hydroxylamine stage.
2. Strongly acidic medium gives p-aminophenol by rearrangement.

Teacher's Note:
1. Always link acid strength to the stability of the carboxylate ion.
2. Electron-withdrawing groups increase acidity and electron-donating groups decrease it.

Question 8: Among C3H7NH2, NH3, CH3NH2, C2H5NH2 and C6H5NH2, the least basic compound is
(a) C3H7NH2
(b) NH3
(c) CH3NH2
(d) C6H5NH2
Show Answer & Explanation

Answer: (d) C6H5NH2

Explanation:
1. Delocalisation of the lone pair into the ring makes aniline the weakest base (pKb 9.4).

Teacher's Note:
1. Note the name reaction and write its equation once in your notebook.
2. Such one-step recall questions are common in the JEE and NEET style papers.

Question 9: The reduction of which of the following would yield a secondary amine?
(a) alkyl nitrite
(b) carbylamine (isocyanide)
(c) primary amine
(d) secondary nitro compound
Show Answer & Explanation

Answer: (b) carbylamine (isocyanide)

Explanation:
1. R-N≡C + 4[H] → R-NH-CH3.

Teacher's Note:
1. Count carbons before and after the reaction to avoid mistakes in chain-length changes.
2. Check whether the reagent is a reducing, oxidising or dehydrating agent before choosing the product.

Question 10: An azo dye is made by the coupling of phenol and
(a) diazonium chloride
(b) o-nitroaniline
(c) benzoic acid
(d) chlorobenzene
Show Answer & Explanation

Answer: (a) diazonium chloride

Explanation:
1. Phenol couples with the diazonium ion (electrophile) at the para position in weakly alkaline medium.

Teacher's Note:
1. Eliminate options by checking functional groups on both sides of the reaction.
2. Only one option will have the correct functional group and carbon skeleton.

Question 11: C6H5NH2 → (NaNO2/HCl) X → (Cu2(CN)2) Y → (H2O/H+) Z. Z is
(a) C6H5-NH-CH3
(b) C6H5-COOH
(c) C6H5-CH2-NH2
(d) C6H5-CH2-COOH
Show Answer & Explanation

Answer: (b) C6H5-COOH

Explanation:
1. X = diazonium chloride; Y = benzonitrile.
2. Acid hydrolysis of the nitrile gives benzoic acid.

Teacher's Note:
1. Revise the physical properties such as hydrogen bonding and boiling points together with the chemistry.
2. Compare similar compounds in pairs when you are asked to rank them.

Question 12: Acetamide is treated with bromine and caustic soda. The product is
(a) acetic acid
(b) bromoacetic acid
(c) methylamine
(d) ethylamine
Show Answer & Explanation

Answer: (c) methylamine

Explanation:
1. CH3CONH2 + Br2 + 4NaOH → CH3NH2 + Na2CO3 + 2NaBr + 2H2O.

Teacher's Note:
1. Learn the reactivity order of acid derivatives: acyl chloride, anhydride, ester, amide.
2. Most questions of this type are solved by recalling the standard reagent and product pair.

Question 13: Cyclohexanone, on reaction with dimethylamine in the presence of a catalytic amount of acid with continuous removal of water, forms a compound generally known as
(a) a Schiff's base
(b) an enamine
(c) an imine
(d) an amine
Show Answer & Explanation

Answer: (b) an enamine

Explanation:
1. A secondary amine condenses with a ketone that has alpha hydrogens to give an enamine, since the carbinolamine cannot form a neutral imine.

Teacher's Note:
1. Always link acid strength to the stability of the carboxylate ion.
2. Electron-withdrawing groups increase acidity and electron-donating groups decrease it.

Question 14: R-NH-CHO →(POCl3, pyridine) product. The product is
(a) R-N=C=O
(b) R-N+≡C-
(c) R-C≡N
(d) none of these
Show Answer & Explanation

Answer: (b) R-N+≡C-

Explanation:
1. Dehydration of an N-substituted formamide gives the isocyanide.

Teacher's Note:
1. Note the name reaction and write its equation once in your notebook.
2. Such one-step recall questions are common in the JEE and NEET style papers.

Question 15: Which of the following is a secondary pollutant?
(a) CO
(b) NO
(c) PAN (peroxyacetyl nitrate)
(d) SO2
Show Answer & Explanation

Answer: (c) PAN (peroxyacetyl nitrate)

Explanation:
1. Secondary pollutants are formed in the atmosphere from primary ones, and PAN forms from hydrocarbons, NO2 and oxygen in sunlight.

Teacher's Note:
1. Count carbons before and after the reaction to avoid mistakes in chain-length changes.
2. Check whether the reagent is a reducing, oxidising or dehydrating agent before choosing the product.

Question 16: Nitration of aniline also gives m-nitroaniline in strongly acidic medium because
(a) in an electrophilic substitution reaction the amino group is meta-directive
(b) in spite of substituents, the nitro group always goes to the m-position
(c) in strong acidic medium, nitration of aniline is a nucleophilic substitution reaction
(d) in strong acidic medium, aniline is present as anilinium ion
Show Answer & Explanation

Answer: (d) in strong acidic medium, aniline is present as anilinium ion

Explanation:
1. Protonation gives -NH3+, a strongly deactivating, meta-directing group.

Teacher's Note:
1. Eliminate options by checking functional groups on both sides of the reaction.
2. Only one option will have the correct functional group and carbon skeleton.

Question 17: Identify the product obtained from 3,4,5-tribromoaniline by (i) diazotisation and (ii) H3PO2.
(a) 3,4,5-tribromobenzene
(b) 1,2,3-tribromobenzene
(c) 2,4,6-tribromobenzene
(d) 3,4,5-tribromonitrobenzene
Show Answer & Explanation

Answer: (b) 1,2,3-tribromobenzene

Explanation:
1. H3PO2 replaces N2+ by H, so only the three Br atoms remain, on adjacent carbons.
2. That is 1,2,3-tribromobenzene.

Teacher's Note:
1. Revise the physical properties such as hydrogen bonding and boiling points together with the chemistry.
2. Compare similar compounds in pairs when you are asked to rank them.

Question 18: The correct order of basicity of the amines (I) C6H5NH2, (II) CH3NH2, (III) (CH3)2NH, (IV) (CH3)3N in increasing order in water is
(a) I < IV < II < III
(b) IV < I < III < II
(c) I < II < III < IV
(d) II < III < IV < I
Show Answer & Explanation

Answer: (a) I < IV < II < III

Explanation:
1. Aniline is the weakest.
2. Among the methylamines the aqueous order is (CH3)3N < CH3NH2 < (CH3)2NH.

Teacher's Note:
1. Learn the reactivity order of acid derivatives: acyl chloride, anhydride, ester, amide.
2. Most questions of this type are solved by recalling the standard reagent and product pair.

Question 19: When acetamide is treated with bromine and caustic soda, we get
(a) acetic acid
(b) bromoacetic acid
(c) methylamine
(d) ethane
Show Answer & Explanation

Answer: (c) methylamine

Explanation:
1. Hofmann degradation removes the carbonyl carbon.

Teacher's Note:
1. Always link acid strength to the stability of the carboxylate ion.
2. Electron-withdrawing groups increase acidity and electron-donating groups decrease it.

Question 20: In organic compounds, nitrogen is tested in Lassaigne's test as
(a) NaNH2
(b) NaCN
(c) NaNO2
(d) NaNO3
Show Answer & Explanation

Answer: (b) NaCN

Explanation:
1. Sodium fusion converts N and C into sodium cyanide, which is then converted to Prussian blue.

Teacher's Note:
1. Note the name reaction and write its equation once in your notebook.
2. Such one-step recall questions are common in the JEE and NEET style papers.

Question 21: Liebermann's nitroso reaction is used for testing
(a) primary amines
(b) secondary amines
(c) tertiary amines
(d) all of the above
Show Answer & Explanation

Answer: (b) secondary amines

Explanation:
1. Secondary amines form N-nitroso compounds with nitrous acid, giving a characteristic colour change in the Liebermann test.

Teacher's Note:
1. Count carbons before and after the reaction to avoid mistakes in chain-length changes.
2. Check whether the reagent is a reducing, oxidising or dehydrating agent before choosing the product.

Question 22: The nauseating smell in the carbylamine test for primary amines is due to the formation of
(a) isocyanide
(b) chloroform
(c) cyanide
(d) DDT
Show Answer & Explanation

Answer: (a) isocyanide

Explanation:
1. RNC has a very foul smell.

Teacher's Note:
1. Eliminate options by checking functional groups on both sides of the reaction.
2. Only one option will have the correct functional group and carbon skeleton.

Question 23: When a primary amine is heated with CS2 in the presence of excess mercuric chloride, an isothiocyanate is formed. This reaction is called
(a) Hofmann bromide reaction
(b) Hofmann mustard oil reaction
(c) carbylamine reaction
(d) Perkin reaction
Show Answer & Explanation

Answer: (b) Hofmann mustard oil reaction

Explanation:
1. The isothiocyanates have mustard-oil smell, so the reaction is called the Hofmann mustard oil reaction.

Teacher's Note:
1. Revise the physical properties such as hydrogen bonding and boiling points together with the chemistry.
2. Compare similar compounds in pairs when you are asked to rank them.

Question 24: Diazo-coupling is useful to prepare some
(a) dyes
(b) proteins
(c) pesticides
(d) vitamins
Show Answer & Explanation

Answer: (a) dyes

Explanation:
1. The coupling of diazonium salts with phenols or amines gives azo dyes.

Teacher's Note:
1. Learn the reactivity order of acid derivatives: acyl chloride, anhydride, ester, amide.
2. Most questions of this type are solved by recalling the standard reagent and product pair.

Question 25: The carbylamine test is used in the detection of
(a) aliphatic secondary amines
(b) aromatic primary amines
(c) aliphatic primary amines
(d) both aliphatic and aromatic primary amines
Show Answer & Explanation

Answer: (d) both aliphatic and aromatic primary amines

Explanation:
1. Any compound with a free -NH2 on an aliphatic or aromatic carbon gives the test.

Teacher's Note:
1. Always link acid strength to the stability of the carboxylate ion.
2. Electron-withdrawing groups increase acidity and electron-donating groups decrease it.

Question 26: Which of the following substances does not give the iodoform test?
(a) C6H5CN
(b) RNH2
(c) CH3OH
(d) all of these
Show Answer & Explanation

Answer: (d) all of these

Explanation:
1. None has the CH3CO- or CH3CH(OH)- unit required.

Teacher's Note:
1. Note the name reaction and write its equation once in your notebook.
2. Such one-step recall questions are common in the JEE and NEET style papers.

Question 27: Which one of the following, when heated with KOH and a primary amine, gives the carbylamine test?
(a) CHCl3
(b) CH3Cl
(c) CH3OH
(d) CH3CN
Show Answer & Explanation

Answer: (a) CHCl3

Explanation:
1. Chloroform loses HCl to form dichlorocarbene, which reacts with the amine.

Teacher's Note:
1. Count carbons before and after the reaction to avoid mistakes in chain-length changes.
2. Check whether the reagent is a reducing, oxidising or dehydrating agent before choosing the product.

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Practice MCQs for JEE Chemistry Amines

Chapter MCQs with Answers for JEE Chemistry

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