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Amines

JEE Main 2025 (Online) 3rd April Evening ShiftMCQHard📈 Topic in 18% papers+4−1
Ques. 1JEE 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]

Answer

(3) [figure]

Solution

(3): 4-nitroaniline; -NH2 strongly activates and directs Br to both its ortho positions; then -NH2 is replaced by Br through the diazonium salt (Sandmeyer reaction) and -NO2 is reduced to -NH2:

[figure]

The product is 3,4,5-tribromoaniline.

Other sequences:

(1) Bromination of nitrobenzene gives m-bromonitrobenzene 3-bromoaniline after reduction.

(2) Bromobenzene does not react with NH3 under ordinary conditions (C–Br bond of haloarene has partial double bond character).

(4) Aniline with bromine water gives 2,4,6-tribromoaniline.

Answer: (3)

Keep practising — Amines

Ques. 2JEE 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
Ques. 3JEE Main 2025, 2 April (Evening Shift)

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
Ques. 4JEE Main 2025, 29 January (Morning Shift)

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)

Ques. 5JEE Main 2025, 29 January (Evening Shift)

Which one of the following reaction sequences will give an azo dye?

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
Ques. 6JEE Main 2025, 28 January (Morning Shift)

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]

Ques. 7JEE Main 2025, 28 January (Evening Shift)

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
Ques. 8JEE Main 2025, 24 January (Evening Shift)

For reaction [figure] The correct order of set of reagents for the above conversion is :

  1. Ac2O, Br2, H2O( ), NaOH
  2. H2SO4, Ac2O, Br2, H2O( ), NaOH
  3. Br2 | FeBr3, H2O( ), NaOH
  4. Ac2O, H2SO4, Br2, NaOH
Ques. 9JEE Main 2025, 23 January (Morning Shift)

Which among the following react with Hinsberg's reagent? Choose the correct answer from the options given below :

  1. A, C and E only
  2. A, B and E only
  3. C and D only
  4. B and D only
Ques. 10JEE Main 2025, 23 January (Morning Shift)

Consider the following sequence of reactions to produce major product (A) [figure] Molar mass of product (A) is g mol-1. (Given molar mass in g mol-1 of C : 12, H : 1, O : 16, Br : 80, N : 14, P : 31)

Ques. 11JEE Main 2025, 23 January (Evening Shift)

Consider the following sequence of reactions. [figure] Total number of sp3 hybridised carbon atoms in the major product C formed is .

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