The final product in the following reaction sequence is
p - chloroaniline `underset(0-5^@C)overset(NaNO_2,HCl)rarroverset(KCN ?)rarroverset(LiAlH_4 ?)rarr`

Options

A. p - chlorophenol
B. p - chlorobenzamide
C. p - chlorobenzylamine
D. p - chlorobenzyl alcohol

Answer

To solve the question regarding the final product of the reaction sequence involving p-chloroaniline, we will break down the steps involved in the reactions.

### Step-by-Step Solution:

1. **Starting Material**: The reaction begins with p-chloroaniline, which has the structure:
\[ \text{Cl-C}_6\text{H}_4\text{NH}_2 \]

2. **Diazotization Reaction**: The first step involves treating p-chloroaniline with sodium nitrite (NaNO2) and hydrochloric acid (HCl) at a low temperature (0-5°C). This reaction leads to the formation of a diazonium salt:
\[ \text{Cl-C}_6\text{H}_4\text{N}_2^+\text{Cl}^- \]
Here, the amino group (-NH2) is converted into a diazonium group (-N2+), while the chlorine atom remains unchanged.

3. **Nucleophilic Substitution with KCN**: The diazonium salt is then treated with potassium cyanide (KCN). In this step, the diazonium group is replaced by a cyanide group (-CN) through nucleophilic substitution:
\[ \text{Cl-C}_6\text{H}_4\text{CN} \]
The product formed is p-chlorobenzonitrile.

4. **Reduction with LiAlH4**: The final step involves reducing the nitrile group (-CN) using lithium aluminum hydride (LiAlH4), which is a strong reducing agent. This reduction converts the nitrile group into a primary amine:
\[ \text{Cl-C}_6\text{H}_4\text{CH}_2\text{NH}_2 \]
The final product is p-chlorobenzylamine.

### Final Product:
The final product of the reaction sequence is:
\[ \text{p-chlorobenzylamine} \]