Question

Identify the final product \([Z]\) formed when chlorobenzene undergoes the given series of reactions.

• A. a
• B. b
• C. c
• D. d

Hint:

In aromatic side-chain reactions, \(Cl_2/UV\) causes benzylic chlorination, while alcoholic \(KCN\) replaces benzylic halogen by \(-CN\), which on hydrolysis gives \(-COOH\).

Answer 1

The Correct Option is A

Step 1: Identify the first reaction.
Chlorobenzene reacts with \(CH_3Cl\) in the presence of anhydrous \(AlCl_3\).
This is a Friedel-Crafts alkylation reaction.
In this reaction, a methyl group \((-CH_3)\) is introduced on the benzene ring.
\[ C_6H_5Cl + CH_3Cl \xrightarrow{Anhydrous \, AlCl_3} CH_3-C_6H_4-Cl \]

Step 2: Directing effect of chlorine.
Chlorine is deactivating due to its \(-I\) effect, but it is ortho-para directing due to resonance donation.
Therefore, methylation of chlorobenzene gives mainly ortho and para products.
Among these, the para product is major due to less steric hindrance.
So, the major product \([W]\) is \(p\)-chlorotoluene.
\[ [W] = p\text{-chlorotoluene} \]

Step 3: Reaction of \(p\)-chlorotoluene with \(Cl_2\) in UV light.
The next reaction is done using \(Cl_2\) in the presence of UV light.
This condition causes free-radical chlorination at the benzylic position, not on the aromatic ring.
Thus, the methyl group \((-CH_3)\) is converted into chloromethyl group \((-CH_2Cl)\).
\[ p\text{-chlorotoluene} \xrightarrow{Cl_2/UV} p\text{-chlorobenzyl chloride} \] So,
\[ [X] = p\text{-chlorobenzyl chloride} \]

Step 4: Reaction with alcoholic \(KCN\).
The compound \([X]\), \(p\)-chlorobenzyl chloride, reacts with alcoholic \(KCN\).
Here, \(CN^-\) acts as a nucleophile and replaces the benzylic chlorine atom.
Thus, \(-CH_2Cl\) changes into \(-CH_2CN\).
\[ p\text{-chlorobenzyl chloride} \xrightarrow{Alcoholic \, KCN} p\text{-chlorobenzyl cyanide} \] So,
\[ [Y] = p\text{-chlorobenzyl cyanide} \]

Step 5: Hydrolysis of nitrile group.
The compound \([Y]\) is hydrolysed with excess dilute \(HCl\).
On acidic hydrolysis, the nitrile group \((-CN)\) is converted into carboxylic acid group \((-COOH)\).
Therefore, \(-CH_2CN\) becomes \(-CH_2COOH\).
\[ p\text{-chlorobenzyl cyanide} \xrightarrow{Hydrolysis \, with \, excess \, dil. \, HCl} p\text{-chlorophenylacetic acid} \]

Step 6: Identify the final product \([Z]\).
The final product contains chlorine on the benzene ring and a \(-CH_2COOH\) group para to chlorine.
Therefore, the final product is \(p\)-chlorophenylacetic acid.
This structure matches option (A).

Step 7: Final conclusion.
The reaction sequence is:
\[ Chlorobenzene \rightarrow p\text{-chlorotoluene} \rightarrow p\text{-chlorobenzyl chloride} \rightarrow p\text{-chlorobenzyl cyanide} \rightarrow p\text{-chlorophenylacetic acid} \] Hence, the correct answer is option (A).
Final Answer:
The final product \([Z]\) is:
\[ \boxed{p\text{-Chlorophenylacetic acid}} \]