The conversions which are represented correctly are
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (B), (C) and (D) only
The –OH group makes the benzene ring of phenol very reactive, and the ring itself can be oxidised or freed of the –OH group. Each conversion is judged by its reagent and the product it really gives.
(A) Phenol with chloroform and NaOH giving salicylic acid — Incorrect. This is the Reimer-Tiemann reaction, which introduces a –CHO group at the ortho position and gives salicylaldehyde, not salicylic acid. Salicylic acid is obtained from sodium phenoxide and carbon dioxide (Kolbe’s reaction).
(B) Phenol with \( \ce{Na2Cr2O7} \), \( \ce{H2SO4} \) and heat giving benzoquinone — Correct. Oxidation of phenol with chromic acid gives the conjugated diketone benzoquinone:
\( \displaystyle \ce{C6H5OH} \xrightarrow[\Delta]{\ce{Na2Cr2O7},\ \ce{H2SO4}} \ce{O=C6H4=O} \)
(C) Phenol with chloroform and NaOH giving salicylaldehyde — Correct. In the Reimer-Tiemann reaction a substituted benzal chloride is formed at the ortho position and is hydrolysed by the alkali:
\( \displaystyle \ce{C6H5OH} \xrightarrow[\text{(ii) }\ce{H+}]{\text{(i) }\ce{CHCl3},\ \text{aq. NaOH}} o\text{-}\ce{HOC6H4CHO} \)
(D) Phenol heated with zinc dust giving benzene — Correct. Zinc dust removes the oxygen and is itself converted to zinc oxide:
\( \displaystyle \ce{C6H5OH + Zn} \xrightarrow{\Delta} \ce{C6H6 + ZnO} \)
NCERT (2026-27 edition):
Reimer-Tiemann Reaction
