JEE Main 2025 · Chemistry · Carbonyl Compounds, Carboxylic Acids and Amines · medium

Toluene is oxidised with alkaline KMnO4 followed by acidification to A. A reacts with SOCl2 to give B, B with excess NH3 gives C, and C with Br2/KOH…

Single correct option

Toluene is oxidised with alkaline

\mathrm{KMnO_4}

followed by acidification to A. A reacts with

\mathrm{SOCl_2}

to give B, B with excess

\mathrm{NH_3}

gives C, and C with

\mathrm{Br_2/KOH}

gives D. D is diazotised at

273

K and coupled with

\beta

-naphthol in alkaline medium to give E. E is

Show answer and solution

Answer: (C) 1-(phenylazo)naphthalen-2-ol

A = benzoic acid, B = benzoyl chloride, C = benzamide. Hofmann bromamide degradation removes one carbon:

\mathrm{C_6H_5CONH_2}\to\mathrm{C_6H_5NH_2}

(D = aniline). Diazotisation with

\mathrm{NaNO_2/HCl}

0

5^\circ

C gives benzenediazonium chloride, which couples at the

1

-position of

\beta

-naphthol in alkaline medium, giving the orange-red dye 1-(phenylazo)naphthalen-2-ol (E).

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