Alcohols can be prepared in many ways. They are prepared from alkenes by acid-catalysed hydration and by hydroboration–oxidation. Alcohols are also prepared from aldehydes and ketones by reduction. The reduction of carboxylic acids and esters also gives alcohols. Alcohols are produced by the reaction of Grignard reagents with aldehydes and ketones. Phenol, also known as carbolic acid, was first isolated in the early nineteenth century from coal tar. Nowadays, phenol is commercially produced synthetically. In the laboratory, phenols are prepared from benzene derivatives in many ways.
Which of the following, on fusion with NaOH at 623 K and 320 atm followed by acidification, gives phenol?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Chlorobenzene
Phenol is \( \ce{C6H5OH} \): an –OH group on a benzene ring. It is made from a haloarene by replacing the halogen with –OH.
The C–Cl bond of chlorobenzene is hard to break, so drastic conditions are needed: fusion with NaOH at 623 K and 320 atm gives sodium phenoxide, and acid then sets free phenol.
\( \displaystyle \ce{C6H5Cl + 2NaOH} \xrightarrow[\text{320 atm}]{\text{623 K}} \ce{C6H5ONa + NaCl + H2O} \)
\( \displaystyle \ce{C6H5ONa + HCl -> C6H5OH + NaCl} \)
Why the other options are incorrect
Option (1) Chloropentane — Incorrect. It is an alkyl halide with no benzene ring; replacing its chlorine by –OH gives pentanol, an alcohol, not phenol.
Option (2) Chlorohexane — Incorrect. Six carbons in an open chain are not an aromatic ring; it gives hexanol, not phenol.
Option (4) Bromohexane — Incorrect. This is also an alkyl halide; it gives hexanol, not phenol.
NCERT (2026-27 edition):
7.4.2 Preparation of Phenols – “1. From haloarenes
Chlorobenzene is fused with NaOH at 623K and 320 atmospheric pressure. Phenol is obtained by acidification of sodium phenoxide so produced (Unit 6, Class XII).”