Question

When $1$ -alkyne is treated with $Na+Liq.N{H}_3$ and product is reacted with methyl chloride, the end product of the reaction will be

• A. Lower alkyne having two carbons less than $1$-alkyne
• B. Lower alkyne having one carbon less than $1$-alkyne
• C. Higher alkyne having one carbon more than $1$ -alkyne
• D. Higher alkyne having two carbons more than $1$ -alkyne

Answer 1

Answer: (C)

$R-C\equiv C.H\stackrel{Na/liqN{H}_3}{\to }R-C\equiv \stackrel{-}{C}.N{a}^{+}$
$\stackrel{ClC{H}_3}{\to }R-C\equiv C-C{H}_3$
Higher alkyne having one carbon more than $1$-alkyne.