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Hydrocarbons
Preparation of Alkynes
1 question from 1 paper, each with its answer.
- question
- 1
- paper
- 1
- of Hydrocarbons
- 1.33%
Concepts tested
- Double dehydrohalogenation of gem-dihalide
All Questions
2002 AIPMT 1 question
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When $CH_3CH_2CHCl_2$ is treated with $NaNH_2$, the product formed is : -$CH_3CH_2CHCl_2$ (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, $CH_3-C\equiv CH$.