MCQ
When $CH_3CH_2CH_2CHCl_2$ is treated with $2\, gram$ equivalent $NaNH_2$, the product formed is
- ✓$C{H_3}C{H_2}C \equiv CH$
- B$C{H_3}C{H_2}-CH = C{H_2}$
- C
- D
✓
Answer
Correct option: A.
$C{H_3}C{H_2}C \equiv CH$
a
$CH_3-CH_2-CH_2-CHCl_2$ $\xrightarrow{{NaN{H_2}}}C{H_3} - C{H_2} - C \equiv CH$
$CH_3-CH_2-CH_2-CHCl_2$ $\xrightarrow{{NaN{H_2}}}C{H_3} - C{H_2} - C \equiv CH$
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