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Amines · Question 31

JEE Main 2025 (Online) 7th April Evening ShiftMCQMedium📈 Topic in 10% papers+4−1
1

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

Answer

(3) E > D > B > A > C

Solution

Aliphatic amines are more basic than aromatic amines because in aryl amines the lone pair of N is delocalised over the ring.

(CH3)2NH (E) > CH3NH2 (D) (+I effect of two methyl groups; in water, pKb 3.27 and 3.38)

Among the anilines:

B: –OMe at the para position donates electrons by +R effect more basic than aniline (pKb 8.70)

A: aniline (pKb 9.38)

C: -NO2 at the para position withdraws electrons by –R and –I effects least basic (pKb 13.0)

E > D > B > A > C

Answer: (3)

JEE Main 2025 (Online) 7th April Evening Shift

Keep practising — Amines

2

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
3

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
4

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
5

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
6

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
7

When a concentrated solution of sulphanilic acid and 1-naphthylamine is treated with nitrous acid (273 K) and acidified with acetic acid, the mass (g) of 0.1 mole of product formed is: (Given molar mass in g mol-1 H : 1, C : 12, N : 14, O : 16, S : 32)

  1. 330
  2. 33
  3. 343
  4. 66
JEE Main 2025 (Online) 2nd April Evening Shift
8

Given below are some nitrogen containing compounds : [figure] Each of them is treated with HCl separately. 1.0 g of the most basic compound will consume mg of HCl. (Given molar mass in g mol-1 C : 12, H : 1, O : 16, Cl : 35.5)

JEE Main 2025 (Online) 29th January Morning Shift
9

Consider the following sequence of reactions: [figure] 11.25 mg of chlorobenzene will produce 10-1 mg of product B. (Consider the reactions result in complete conversion.) [Given molar mass of C, H, O, N and Cl as 12, 1, 16, 14 and 35.5 g mol-1 respectively]

JEE Main 2025 (Online) 28th January Morning Shift
10

Identify correct statements : Choose the correct answer from the options given below :

  1. (B) and (C) only
  2. (D) and (E) only
  3. (A) and (B) only
  4. (B) and (D) only
JEE Main 2025 (Online) 28th January Evening Shift

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