Question

At low temperature, phenol on reaction with \( \mathrm{Br_2} \) in \( \mathrm{CS_2} \) gives

• A. 2, 4, 6-Tribromophenol
• B. a mixture of ortho-and para-bromophenol
• C. ortho-bromophenol only
• D. para-bromophenol only

Hint:

Non-polar CS2 and low temperature stop at monobromination. The OH group directs to ortho and para, so a mixture forms, with para as the major product.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The OH group of phenol is strongly activating and directs the incoming group to the ortho and para positions.
The product depends on the solvent and the temperature.

Step 2: Case 1, bromine water:
In a polar solvent like water, phenol ionises partly to the phenoxide ion, which is even more reactive.
All three activated positions are brominated and 2,4,6-tribromophenol forms as a white precipitate.

Step 3: Case 2, \( \mathrm{Br_2} \) in \( \mathrm{CS_2} \) at low temperature:
\( \mathrm{CS_2} \) is a non-polar solvent, so ionisation of phenol is suppressed.
The ring is less reactive and the reaction stops at one bromine atom.
\[ \mathrm{C_6H_5OH + Br_2 \xrightarrow{CS_2,\ low\ temp.} o\text{-}BrC_6H_4OH + p\text{-}BrC_6H_4OH + HBr} \]
The para isomer is the major product because it has less steric hindrance, but the ortho isomer is also formed.

Step 4: Check the options:
(A) is the product with bromine water, not with \( \mathrm{CS_2} \). So (A) is wrong.
(C) and (D) each name only one isomer. The NCERT text says a mixture forms, so both are wrong.
(B) names the mixture, so (B) is right.

Final Answer:
Phenol gives a mixture of o-bromophenol and p-bromophenol, option (B).
\[ \boxed{\text{mixture of ortho- and para-bromophenol}} \]