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Amines · Question 26

JEE Main 2026 (Online) 21st January Morning ShiftMCQEasy📈 Topic in 26% papers+4−1
1

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)

JEE Main 2026 (Online) 21st January Morning Shift

Keep practising — Amines

2

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
JEE Main 2026 (Online) 21st January Morning Shift
3

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)

JEE Main 2026 (Online) 21st January Morning Shift
4

[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
JEE Main 2026 (Online) 21st January Evening Shift
5

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
JEE Main 2025 (Online) 7th April Morning Shift
6

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
JEE Main 2025 (Online) 7th April Evening Shift
7

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

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

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

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

In the following reactions, which one is NOT correct?

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
JEE Main 2025 (Online) 3rd April Morning Shift
10

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

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
JEE Main 2025 (Online) 3rd April Evening Shift
11

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
JEE Main 2025 (Online) 2nd April Morning Shift

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