Question
In the reaction,
In the reaction,
\[\begin{align}
& H-C\equiv CH\xrightarrow[\text{(ii)}\,C{{H}_{3}}C{{H}_{2}}Br]{\text{(i)}\,\frac{NaN{{H}_{2}}}{liq.N{{H}_{3}}}} \\
& X\xrightarrow[\text{(ii)}\,C{{H}_{3}}C{{H}_{2}}Br]{\text{(i)}\,\frac{NaN{{H}_{2}}}{liq.N{{H}_{3}}}}Y \\
\end{align}\]
$$X$$ and $$Y$$ are
A.
$$X = 2 - butyne;Y = 3 - hexyne$$
B.
$$X = 2 - butyne;Y = 2 - hexyne$$
C.
$$X = 1 - butyne;Y = 2 - hexyne$$
D.
$$X = 1 - butyne;Y = 3 - hexyne$$
Answer :
$$X = 1 - butyne;Y = 3 - hexyne$$
Solution :
Since, $${\frac{{NaN{H_2}}}{{liq.N{H_3}}}}$$ behaves as a base, so it abstracts proton from acetylene to form acetylide anion followed by alkylation to give compound $$(X)$$
i.e. $$1-butyne.$$ $$(X)$$ further reacts with $${\frac{{NaN{H_2}}}{{liq.N{H_3}}}}$$ followed by alkylation with ethyl bromide yields $$3-hexyne$$ $$(Y).$$
Since, $${\frac{{NaN{H_2}}}{{liq.N{H_3}}}}$$ behaves as a base, so it abstracts proton from acetylene to form acetylide anion followed by alkylation to give compound $$(X)$$
i.e. $$1-butyne.$$ $$(X)$$ further reacts with $${\frac{{NaN{H_2}}}{{liq.N{H_3}}}}$$ followed by alkylation with ethyl bromide yields $$3-hexyne$$ $$(Y).$$