2002 AIPMT HydrocarbonsPreparation of Alkynes Easy Question 70 of 75 in Hydrocarbons

When $CH_3CH_2CHCl_2$ is treated with $NaNH_2$, the product formed is : -

A $CH_3-CH=CH_2$
B $CH_3-C\equiv CH$
C -
D -

The answer is B.

Solution

$CH_3CH_2CHCl_2$ (1,1-dichloropropane) is a geminal dihalide.
Sodamide, $NaNH_2$, is a very strong base. It removes two molecules of HCl (two successive eliminations) to form a triple bond:
$CH_3CH_2CHCl_2 \xrightarrow{NaNH_2} CH_3CH=CHCl \xrightarrow{NaNH_2} CH_3C\equiv CH$
So the product is propyne, $CH_3-C\equiv CH$.

Concept Double dehydrohalogenation of gem-dihalide

Hydrocarbons in past papers

75 questions from this chapter have appeared across 26 exam years.