JEE Main 2026 · 22 January · evening · Single correct · +4 / −1
Consider the following reaction :
The product Y formed is :
- (a) 2-methylhex-3-yne
- (b) 5-methylhex-2-yne
- (c) Isopropylbut-1-yne
- (d) 2-methylhex-2-yne
Answer and solution
Answer (a) 2-methylhex-3-yne
Solution
Chemistry ma’am · 5 steps
Two eliminations build the triple bond, then the acetylide ion picks up the isopropyl group.
- Make X. The starting compound is CH₃CH₂CH₂CHBr₂, a gem-dibromide with both Br on C-1. Two equivalents of NaNH₂ remove two molecules of HBr, so X is but-1-yne, CH₃CH₂C≡CH.
- Acetylide ion. The terminal C-H of but-1-yne is acidic. NaNH₂ removes this proton and gives the butynide ion, CH₃CH₂C≡C⁻.
- Alkylation. The acetylide ion attacks (CH₃)₂CHBr and displaces Br⁻. The isopropyl group bonds directly to the alkyne carbon, so the triple bond does not move.
CH₃CH₂CH₂CHBr₂ →(2NaNH₂) CH₃CH₂C≡CH →(NaNH₂) CH₃CH₂C≡C⁻ →((CH₃)₂CHBr) CH₃CH₂C≡C-CH(CH₃)₂
- Name Y. Number from the end that gives the triple bond the lower number: (CH₃)₂CH-C≡C-CH₂CH₃ is a six-carbon chain with the triple bond at C-3 and a methyl group at C-2.
- So. Y is 2-methylhex-3-yne, option (a).
The idea to keep. A gem or vic dihalide with 2 NaNH₂ gives an alkyne. A terminal alkyne with NaNH₂ and then R–X lengthens the chain, and R joins exactly at the alkyne carbon.
Answer from the NTA final answer key. Solution by PrepWiser.