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In the following sequence of reactions $CH_3-Br \xrightarrow{KCN} A \xrightarrow{H_3O^+} B \xrightarrow[ether]{LiAlH_4} C$, the end product (C) is
A
Ethyl alcohol
B
Acetone
C
Methane
D
Acetaldehyde
Detailed Solution
$CH_3Br \xrightarrow{KCN} CH_3CN$ (A, methyl cyanide) by nucleophilic substitution
$CH_3CN \xrightarrow{H_3O^+} CH_3COOH$ (B, acetic acid) by hydrolysis
$CH_3COOH \xrightarrow[ether]{LiAlH_4} CH_3CH_2OH$ (C) by reduction
So C is ethyl alcohol.
$CH_3CN \xrightarrow{H_3O^+} CH_3COOH$ (B, acetic acid) by hydrolysis
$CH_3COOH \xrightarrow[ether]{LiAlH_4} CH_3CH_2OH$ (C) by reduction
So C is ethyl alcohol.