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Amines

JEE Main 2026 (Online) 21st January Morning ShiftMCQEasy📈 Topic in 24% papers+4−1
Ques. 1JEE Main 2026, 21 January (Morning Shift)

An organic compound (P) on treatment with aqueous ammonia under hot condition forms compound (Q) which on heating with Br2 and KOH forms compound (R) having molecular formula C6H7N. Names of P, Q and R respectively are.

  1. Toluic acid, methylbenzamide, 2-methylaniline
  2. Benzoic acid, benzamide, aniline
  3. Benzoic acid, 4-methylbenzamide, 4-methylaniline
  4. Phenylethanoic acid, phenylethanamide, benzamine

Answer

(2) Benzoic acid, benzamide, aniline

Solution

R = C6H7N = C6H5NH2 (aniline).

Hoffmann bromamide degradation: RCONH2 + Br2 + 4KOH RNH2 + K2CO3 + 2KBr + 2H2O

The amine has one carbon atom less than the amide Q = C6H5CONH2 (benzamide).

An acid on heating with ammonia gives the ammonium salt, which loses water to give the amide:

C6H5COOH + NH3 C6H5COO-NH4+ C6H5CONH2 + H2O

P = benzoic acid, Q = benzamide, R = aniline.

(Option (3) would give 4-methylaniline, C7H9N; option (4) would give benzylamine, C7H9N.)

Answer: (2)

Keep practising — Amines

Ques. 2JEE Main 2026, 21 January (Morning Shift)

A hydrocarbon 'P' (C4H8) on reaction with HCl gives an optically active compound 'Q' (C4H9Cl) which on reaction with one mole of ammonia gives compound 'R' ( C4H11N). 'R' on diazotization followed by hydrolysis gives 'S'. Identify P, Q, R and S.

  1. P=CH3-CH=CH-CH3, Q=CH3-CH2-CH2-CH2-Cl, R=CH3-CH2-CH2- NH2 |CH2, S=CH3-CH2- OH |CH2 |CH2
  2. P=CH3-CH=CH-CH3, Q=CH3-CH2- Cl |CH-CH3, R=CH3-CH2- NH2 |CH-CH3, S=CH3-CH2- OH |CH-CH3
  3. P = [figure] , Q = [figure] , R = [figure] , S = [figure]
  4. P=CH3-CH2-CH=CH2, Q=CH3-CH2-CH2- Cl |CH2, R=CH3-CH2-CH2- CH3 |NH, S=CH3-CH2- OH | CH3 |CH
Ques. 3JEE Main 2026, 21 January (Morning Shift)

Consider the following reaction sequence [figure] The percentage of nitrogen in product 'T' formed is %. (Nearest integer) (Given molar mass in g mol-1 H : 1, C : 12, N : 14, O : 16)

Ques. 4JEE Main 2026, 21 January (Evening Shift)

[figure] Consider the above sequence of reactions. The number of bromine atom(s) in the final product (P) will be :

  1. 6
  2. 5
  3. 1
  4. 3
Ques. 5JEE Main 2025, 7 April (Morning Shift)

Which of the following amine(s) show(s) positive carbylamine test? Choose the correct answer from the options given below:

  1. C Only
  2. B, C and D Only
  3. A and E Only
  4. A and C Only
Ques. 6JEE Main 2025, 7 April (Evening Shift)

The descending order of basicity of the following amines is : Choose the correct answer from the options given below :

  1. E > A > D > C > B
  2. E > D > A > B > C
  3. E > D > B > A > C
  4. B > E > D > A > C
Ques. 7JEE Main 2025, 4 April (Morning Shift)

The major product (A) formed in the following reaction sequence is [figure]

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
Ques. 8JEE Main 2025, 3 April (Morning Shift)

Identify [A], [B] and [C], respectively in the following reaction sequence : [figure]

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
Ques. 9JEE Main 2025, 3 April (Morning Shift)

In the following reactions, which one is NOT correct?

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
Ques. 10JEE Main 2025, 3 April (Evening Shift)

The sequence from the following that would result in giving predominantly 3,4,5-Tribromoaniline is :

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
Ques. 11JEE Main 2025, 2 April (Morning Shift)

The correct order of basic nature in aqueous solution for the bases NH3, H2N-NH2, CH3CH2NH2, (CH3CH2)2NH and (CH3CH2)3N is :

  1. NH3 < H2N-NH2 < (CH3CH2)3N < CH3CH2NH2 < (CH3CH2)2NH
  2. H2N-NH2 < NH3 < (CH3CH2)3N < CH3CH2NH2 < (CH3CH2)2NH
  3. H2N-NH2 < NH3 < CH3CH2NH2 < (CH3CH2)3N < (CH3CH2)2NH
  4. NH3 < H2N-NH2 < CH3CH2NH2 < (CH3CH2)2NH < (CH3CH2)3N

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