JEE Main 2020 · 9 January · evening · Single correct · +4 / −1

Consider the following reactions,
[P]{→((i) NaNO₂/HCl, 0-5°C (ii) β-naphthol/NaOH) Colored Solid; →(Br₂/H₂O) C₇H₆NBr₃
The compound [P] is :

  1. (a)
    Option (a)
  2. (b)
    Option (b)
  3. (c)
    Option (c)
  4. (d)
    Option (d)
Answer and solution

Answer (b)

Solution

Chemistry ma’am · 3 steps

Read both clues: the azo dye needs a primary aryl amine, and three Br need three free ortho/para spots.

  1. Dye test. NaNO₂/HCl at 0 to 5 °C turns only a primary aromatic amine into a diazonium salt. That salt couples with β-naphthol in NaOH to give a coloured azo dye. A secondary amine only forms an N-nitrosamine, so it gives no dye.
  2. Bromine test. Toluidine is C₇H₉N, and C₇H₆NBr₃ means three ring H are replaced by Br. Bromine water attacks every free position ortho and para to -NH₂, so all three of those positions (2, 4 and 6) must be free.
    OptionPrimary amine?Free positions ortho/para to NH₂Bromine water product
    (a) o-toluidineyes4, 6 (2 is blocked by CH₃)C₇H₇NBr₂
    (b) m-toluidineyes2, 4, 6C₇H₆NBr₃
    (c) N-methyl-m-toluidineno, secondary2, 4, 6C₈H₈NBr₃, wrong formula
    (d) p-toluidineyes2, 6 (4 is blocked by CH₃)C₇H₇NBr₂
  3. So. Only m-toluidine passes both tests. It gives 2,4,6-tribromo-3-methylaniline, C₇H₆NBr₃, and its CH₃ also directs to these same three positions. The answer is (b).

The idea to keep. With bromine water, an aniline loses one ring H for every free ortho and para position. Count those free spots to match the formula, and remember that only primary aryl amines give azo dyes.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Amines & Diazonium Salts

Asked about 170 times in JEE Main since 2020. This question is from 2020.