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Tardigrade
Question
Chemistry
The reagent used for the following conversion is
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Q. The reagent used for the following conversion is
Hydrocarbons
A
Alcoholic KOH
17%
B
Alcoholic KOH followed by $NaNH _2$
73%
C
Aqueous KOH followed by $NaNH _2$
3%
D
$Zn / CH _3 OH$
7%
Solution:
$BrCH _2 CH _2 Br \xrightarrow{\text { alc. } KOH } BrCH = CH _2 \xrightarrow{ NaNH _2} CH \equiv CH$
