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$CH_3CH_2Cl \xrightarrow{NaCN} X \xrightarrow{Ni/H_2} Y \xrightarrow{\text{acetic anhydride}} Z$. Z in the above reaction sequence is : -
A
$CH_3CH_2CH_2NHCOCH_3$
B
$CH_3CH_2CH_2NH_2$
C
$CH_3CH_2CH_2CONHCH_3$
D
$CH_3CH_2CH_2CONHCOCH_3$
Detailed Solution
Step 1: Nucleophilic substitution by cyanide gives propanenitrile (X): $CH_3CH_2Cl + NaCN \rightarrow CH_3CH_2CN + NaCl$
Step 2: Catalytic hydrogenation reduces the nitrile to a primary amine (Y): $CH_3CH_2C\equiv N + 2H_2 \xrightarrow{Ni} CH_3CH_2CH_2NH_2$ (n-propylamine)
Step 3: Acetic anhydride acetylates the amine to an amide (Z): $CH_3CH_2CH_2NH_2 + (CH_3CO)_2O \rightarrow CH_3CH_2CH_2NHCOCH_3 + CH_3COOH$
So Z is N-propylacetamide, $CH_3CH_2CH_2NHCOCH_3$.
Step 2: Catalytic hydrogenation reduces the nitrile to a primary amine (Y): $CH_3CH_2C\equiv N + 2H_2 \xrightarrow{Ni} CH_3CH_2CH_2NH_2$ (n-propylamine)
Step 3: Acetic anhydride acetylates the amine to an amide (Z): $CH_3CH_2CH_2NH_2 + (CH_3CO)_2O \rightarrow CH_3CH_2CH_2NHCOCH_3 + CH_3COOH$
So Z is N-propylacetamide, $CH_3CH_2CH_2NHCOCH_3$.