In the following chemical reaction

$\mathrm{Acetylene} \underset{{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}}{\overset{{\mathrm{NaNH}}_2 / \mathrm{Liq}. {\mathrm{NH}}_3}{\to }} \mathrm{X} \underset{{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}}{\overset{{\mathrm{NaNH}}_2 / \mathrm{Liq}. {\mathrm{NH}}_3}{\to }} \mathrm{Y}$

X and Y are respectively?

• The reaction sequence involves reaction of alkyne with sodamide in liq. Ammonia

produces ethynyl sodium which was reacted with ethyl bromide produces 1-butyne.

•  1- butyne reacted with sodamide in liq. Ammonia produces but-1-yn-1-yl sodium which was reacted with ethyl bromide produces 3-Hexyne.

$\mathrm{Acetylene} \underset{ }{\overset{{\mathrm{NaNH}}_2 / \mathrm{Liq}. {\mathrm{NH}}_3}{\to }}\mathrm{CH}\equiv \mathrm{CNa}\stackrel{{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}}{\to } \mathrm{CH}\equiv {\mathrm{CCH}}_2{\mathrm{CH}}_3 \stackrel{{\mathrm{NaNH}}_2 / \mathrm{Liq}. {\mathrm{NH}}_3}{\to } \mathrm{NaC}\equiv {\mathrm{CCH}}_2{\mathrm{CH}}_3 \stackrel{{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}}{\to } {\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{C}\equiv {\mathrm{CCH}}_2\mathrm{C}{{\mathrm{H}}_{3 }}_{ }$

Therefore X and Y are 1-Butyne; 3-Hexyne

Hence the correct option is (A).