Question

Consider the following sequence of reactions given below: $ \text{Br-C}_6\text{H}_5 \xrightarrow{\text{alc. KOH}} \text{C}_6\text{H}_4 \xrightarrow{\text{NaNH}_2} \text{C}_6\text{H}_3 \xrightarrow{Hg^{2+}} \text{C}_6\text{H}_2 \xrightarrow{Zn-Hg} P $ The product P is:

• A.
• B.
• C.
• D.

Hint:

In organic chemistry, elimination reactions, especially with alcoholic KOH, often lead to alkyne formation. Reductions such as the Clemmensen reduction typically convert carbonyl compounds to hydrocarbons.

Answer 1

The Correct Option is A

Step 1: Reaction with alc.
KOH: The reaction of \(\text{C}_6\text{H}_5\text{Br}\) with alcoholic KOH leads to the elimination of HBr, forming phenylacetylene.
Step 2: Reaction with NaNH$_2$:
The reaction with NaNH$_2$ results in the formation of an alkyne by dehydrohalogenation.
Step 3: Reaction with Hg$^{2+$}:
The reaction with mercuric ions (Hg$^{2+}$) leads to the formation of a ketone via a hydration reaction, creating phenyl ketone.
Step 4: Reduction with Zn-Hg: Finally, reduction with Zn-Hg (Clemmensen reduction) reduces the ketone to benzene.