JEE Main 2022 · 29 Jun · Shift 1 · Chemistry Q76

← JEE Main 2022 · 29 Jun · Shift 1

ChemistrySingle correct+4 / -1Question 76 of 90

Consider the above reactions, the product A and product B respectively are

Aniline with excess Br2 in H2O gives A; aniline treated with (i) (CH3CO)2O, (ii) Br2 in CH3COOH, (iii) HCl gives B.

4-Bromoaniline and 4-bromoaniline

2,4,6-Tribromoaniline and 2,4,6-tribromoaniline

2,4,6-Tribromoaniline and 4-bromoaniline

4-Bromoaniline and 2,4,6-tribromoaniline

Answer and worked solution

Official answer: C

A. Free $\mathrm{NH_2}$ activates fully: 2,4,6-tribromoaniline. B. Acetylation moderates it; bromination goes para and HCl removes the acetyl: 4-bromoaniline.

(C)

Topics:Amines Aromatic hydrocarbons and electrophilic substitution

← Previous question Next question →

JEE Main 2022 · 29 Jun · Shift 1, Chemistry question 76