Given below is a reaction sequence: $\text{CH}_3\text{CH}_2\text{Cl} \xrightarrow{\text{NaCN}} X \xrightarrow{\text{H}_2/\text{Ni}} Y \xrightarrow{\text{Acetic Anhydride}} Z$

The product '$Z$' in the above reaction is:

The given reaction sequence involves the following three steps:

1. Nucleophilic substitution: Ethyl chloride ($\text{CH}_3\text{CH}_2\text{Cl}$) reacts with sodium cyanide ($\text{NaCN}$) undergoing a substitution reaction. Because cyanide is an ambident nucleophile and $\text{NaCN}$ is predominantly ionic, the attack takes place mainly through the carbon atom, resulting in the formation of propanenitrile ($X$) . $\text{CH}_3\text{CH}_2\text{Cl} + \text{NaCN} \rightarrow \text{CH}_3\text{CH}_2\text{CN} \text{ (X)} + \text{NaCl}$

2. Reduction: Propanenitrile ($X$) undergoes catalytic hydrogenation in the presence of $\text{H}_2/\text{Ni}$ to form a primary amine, propan-1-amine ($Y$). $\text{CH}_3\text{CH}_2\text{CN} + 2\text{H}_2 \xrightarrow{\text{Ni}} \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2 \text{ (Y)}$

3. Acylation: The primary amine, propan-1-amine ($Y$), undergoes acylation upon treatment with acetic anhydride to form an N-substituted amide, N-propylacetamide ($Z$). $\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2 + (\text{CH}_3\text{CO})_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{NHCOCH}_3 \text{ (Z)} + \text{CH}_3\text{COOH}$

Thus, the final product '$Z$' is $\text{CH}_3\text{CH}_2\text{CH}_2\text{NHCOCH}_3$.