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

JEE Main2026Jan 22, Shift 2Chemistry

Consider the following reaction :

The product Y formed is :

A
2-methylhex-2-yne
B
5-methylhex-2-yne
C
2-methylhex-3-yne
D
Isopropylbut-1-yne

Solution

Step 1: The geminal dibromide on treatment with two equivalents of $N a N H_{2}$ undergoes double E2 elimination to give a terminal alkyne (the more stable / less substituted product), $C H \equiv C - C H_{2} - C H (C H_{3}) - C H_{3}$ — 4-methylpent-1-yne.

Step 2: $N a N H_{2}$ deprotonates the terminal $\equiv C - H$ to form the acetylide, which on alkylation with $C H_{3} B r$ gives an internal alkyne:

$C H_{3} - C \equiv C - C H_{2} - C H (C H_{3}) - C H_{3}$ , i.e., 2-methylhex-3-yne.

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