Ethyne is treated successively with (i) one equivalent of NaNH2, (ii) 1-bromopropane, (iii) NaNH2 (again), (iv) iodomethane, and (v) Na in liquid ammonia. The final product is

JEE Advanced levelSingle correct optionDifficulty 3 of 3hc123

Ethyne is treated successively with (i) one equivalent of $NaNH X_{2}$ , (ii) 1-bromopropane, (iii) $NaNH X_{2}$ (again), (iv) iodomethane, and (v) Na in liquid ammonia. The final product is

Steps (i)-(ii) give

HC \equiv C - CH X_{2} CH X_{2} CH X_{3}

(pent-1-yne). Steps (iii)-(iv) methylate the terminal carbon to give

CH X_{3} - C \equiv C - CH X_{2} CH X_{2} CH X_{3}

(hex-2-yne). Na/liq.

NH X_{3}

gives anti reduction, so the product is the trans alkene, (E)-hex-2-ene.