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Amines

JEE Main 2026 (Online) 24th January Morning ShiftMCQEasy📈 Topic in 5% papers+4−1
Ques. 1JEE Main 2026, 24 January (Morning Shift)

The correct stability order of the following diazonium salts is [figure]

  1. C > D > B > A
  2. A > C > D > B
  3. A > B > C > D
  4. C > A > D > B

Answer

(2) A > C > D > B

Solution

Arenediazonium ions are stabilised by delocalisation of the positive charge into the ring. An electron-donating group at the para position increases this delocalisation, an electron-withdrawing group decreases it.

A: –OCH3 (+R) donates electrons to the ring most stable.

C: no substituent.

D: –CN (–R, –I) withdraws electrons less stable than C.

B: –NO2 (strongest –R, –I) least stable.

A > C > D > B

Answer: (2)

Keep practising — Amines

Ques. 2JEE Main 2026, 24 January (Evening Shift)

Given below are two statements : In the light of the above statements, choose the correct answer from the options given below

  1. Both Statement I and Statement II are false
  2. Statement I is false but Statement II is true
  3. Statement I is true but Statement II is false
  4. Both Statement I and Statement II are true
Ques. 3JEE Main 2026, 23 January (Morning Shift)

Consider the following sequence of reactions. [figure] Assuming that the reaction proceeds to completion, then 137 mg of 4-nitrotoluene will produce mg of B. (Given molar mass in g mol-1 H : 1, C : 12, N : 14, O : 16, Br : 80)

  1. 208
  2. 228
  3. 146
  4. 301
Ques. 4JEE Main 2026, 23 January (Morning Shift)

Compound 'P' undergoes the following sequence of reactions : [figure] 'P' is :

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
Ques. 5JEE Main 2026, 23 January (Evening Shift)

Given below are two statements : In the light of the above statements, choose the correct answer from the options given below

  1. Both Statement I and Statement II are true
  2. Statement I is false but Statement II is true
  3. Statement I is true but Statement II is false
  4. Both Statement I and Statement II are false
Ques. 6JEE Main 2026, 23 January (Evening Shift)

A student has been given a compound 'x' of molecular formula C6H7N. 'x' is sparingly soluble in water. However, on addition of dilute mineral acid, 'x' becomes soluble in water. 'x' when treated with CHCl3 and KOH(alc), 'y' is produced. 'y' has a specific unpleasant smell. On treatment with benzenesulphonyl chloride, 'x' gives a compound 'z' which is soluble in alkali. The number of different "H" atoms present in 'z' is :

  1. 8
  2. 4
  3. 5
  4. 7
Ques. 7JEE Main 2026, 22 January (Morning Shift)

'A' is a neutral organic compound (M.F. : C8H9ON). On treatment with aqueous Br2/HO , 'A' forms a compound 'B' which is soluble in dilute acid. 'B' on treatment with aqueous NaNO2/HCl (0-5 C) produces a compound 'C' which on treatment with CuCN/NaCN produces 'D'. Hydrolysis of 'D' produces 'E' which is also obtainable from the hydrolysis of 'A'. 'E' on treatment with acidified KMnO4 produces 'F'. 'F' contains two different types of hydrogen atoms. The structure of 'A' is

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
Ques. 8JEE Main 2026, 22 January (Evening Shift)

[figure] The final product [B] is :

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
Ques. 9JEE Main 2026, 22 January (Evening Shift)

The mass of benzanilide obtained from the benzoylation reaction of 5.8 g of aniline, if yield of product is 82\%, is g (nearest integer). (Given molar mass in g mol-1 H : 1, C : 12, N : 14, O : 16)

Ques. 10JEE 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
Ques. 11JEE 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

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