JEE Main 2026 — Jan 22, Shift 2 — Chemistry solution

JEE Main2026Jan 22, Shift 2Chemistry

The dibromo compound [P] (molecular formula : $C_{9} H_{10} B r_{2}$ ) when heated with excess sodamide followed by treatment with dilute HCl gives [Q]. On warming [Q] with mercuric sulphate and dilute sulphuric acid yields [R] which gives positive Iodoform test but negative Tollen's test. The compound [P] is :

Solution

The geminal dibromide [P] on $N a N H_{2}$ followed by acidic workup gives the terminal alkyne [Q]: 4-methyl-phenylacetylene ( $p$ -tolyl-C $\equiv$ CH).

$H g S O_{4} / d i l . H_{2} S O_{4}$ on the terminal alkyne (Markovnikov hydration via enol) gives the methyl ketone [R]: $p$ -tolyl-CO-CH $_{3}$ (4-methylacetophenone).

[R] gives a positive iodoform test (it contains a $- C O - C H_{3}$ group) and a negative Tollens test (it is a ketone, not an aldehyde) — consistent with the data.

Hence [P] is the gem-dibromide $p$ -tolyl-CBr $_{2}$ -CH $_{3}$ .

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