JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Haloalkanes and Haloarenes

Double dehydrohalogenation of $1,1$-dibromobutane by two equivalents of sodamide gives but-$1$-yne, whose terminal hydrogen is then removed by the excess base to form the acetylide. This acetylide displaces bromide from isopropyl bromide in an ${\mathrm{S}}_{\mathrm{N}}2$ step, attaching an isopropyl group to the terminal carbon to give $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{CH}-\mathrm{C}\equiv \mathrm{C}-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$. Numbering the six-carbon chain places the methyl at C-2 and the triple bond between C-3 and C-4, so the product Y is 2-methylhex-3-yne.