Question

When $1$ -alkyne is treated with $Na + Liq. NH_3$ 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)

$R-C\equiv C.H \xrightarrow{Na/liq \,NH_3} R -C \equiv \overset{-}{C} .Na^+$
$\xrightarrow{ClCH_3}R-C\equiv C -CH_3$
Higher alkyne having one carbon more than $1$-alkyne.