Question

Identify the product B in the following conversion.
\[ \text{Chlorobenzene} + \mathrm{H_2O} \xrightarrow[\text{Pressure}]{\mathrm{Cu},\,673\,K} A \xrightarrow[\;373\,K\;]{\text{conc. } \mathrm{H_2SO_4}} B \]

• A. 4–Hydroxybenzene sulphonic acid
• B. Benzene sulphonic acid
• C. 3–Hydroxybenzene sulphonic acid
• D. 2–Hydroxybenzene sulphonic acid

Hint:

Phenol directs electrophilic substitution mainly to ortho and para positions.

Answer 1

The Correct Option is A

Step 1: First conversion.
Chlorobenzene reacts with water in presence of copper catalyst at high temperature and pressure to form phenol.

Step 2: Second conversion.
Phenol undergoes sulphonation with concentrated sulphuric acid at 373 K.

Step 3: Orientation of substitution.
The –OH group is an activating ortho–para directing group, and at higher temperature, para substitution predominates.

Step 4: Conclusion.
Thus, the product formed is 4–hydroxybenzene sulphonic acid.