Propyne is treated successively with (i) NaNH2; (ii) CH3CH2CH2Br; (iii) H2, Lindlar's catalyst. The final product is

JEE Advanced levelSingle correct optionDifficulty 3 of 3hc089

Propyne is treated successively with (i) $NaNH X_{2}$ ; (ii) $CH X_{3} CH X_{2} CH X_{2} Br$ ; (iii) $H X_{2}$ , Lindlar's catalyst. The final product is

Steps (i) and (ii) alkylate the acetylide:

CH X_{3} - C \equiv C X^{-}

CH X_{3} CH X_{2} CH X_{2} Br

gives

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

, hex-2-yne. Lindlar hydrogenation is syn, so it gives (Z)-hex-2-ene. Na in liquid ammonia would have given the E isomer.