Consider the following statements about the electrophilic substitution reactions of phenol.
(A) When phenol is treated with bromine in a solvent of low polarity such as CS2 at low temperature, monobromophenols are formed.
(B) In the bromination of phenol, a Lewis acid such as FeBr3 is needed to polarise the bromine molecule.
(C) With dilute nitric acid at 298 K, phenol yields a mixture of ortho and meta nitrophenols.
(D) The ortho and para nitrophenols formed by the nitration of phenol can be separated by steam distillation.
Which of the statements given above are correct? Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) and (D) only
The –OH group activates the benzene ring of phenol towards electrophilic substitution and directs the incoming group to the ortho and para positions, which become electron rich by resonance.
(A) When phenol is treated with bromine in a solvent of low polarity such as CS2 at low temperature, monobromophenols are formed. — True. Under these mild conditions only one bromine atom enters the ring:
\( \displaystyle \ce{C6H5OH + Br2} \xrightarrow[\text{low temperature}]{\ce{CS2}} \text{o- and p-}\ce{BrC6H4OH + HBr} \)
(B) In the bromination of phenol, a Lewis acid such as FeBr3 is needed to polarise the bromine molecule. — False. Benzene needs FeBr3, but with phenol the bromine molecule is polarised even without a Lewis acid, because the –OH group activates the ring so strongly.
(C) With dilute nitric acid at 298 K, phenol yields a mixture of ortho and meta nitrophenols. — False. The –OH group is ortho, para directing, so the mixture is of ortho and para nitrophenols:
\( \displaystyle \ce{C6H5OH + HNO3} \xrightarrow[\text{298 K}]{\text{dil.}} \text{o- and p-}\ce{O2NC6H4OH + H2O} \)
(D) The ortho and para nitrophenols formed by the nitration of phenol can be separated by steam distillation. — True. o-Nitrophenol is steam volatile because of intramolecular hydrogen bonding; p-nitrophenol is less volatile, as intermolecular hydrogen bonds hold its molecules together.
NCERT (2026-27 edition):
7.4.4 Chemical Reactions – “(i) Nitration: With dilute nitric acid at low temperature (298 K), phenol yields a mixture of ortho and para nitrophenols.”
“The ortho and para isomers can be separated by steam distillation. o-Nitrophenol is steam volatile due to intramolecular hydrogen bonding while p-nitrophenol is less volatile due to intermolecular hydrogen bonding which causes the association of molecules.”
“(a) When the reaction is carried out in solvents of low polarity such as CHCl3 or CS2 and at low temperature, monobromophenols are formed.”
“The usual halogenation of benzene takes place in the presence of a Lewis acid, such as FeBr3 (Unit 6, Class XII), which polarises the halogen molecule. In case of phenol, the polarisation of bromine molecule takes place even in the absence of Lewis acid. It is due to the highly activating effect of –OH group attached to the benzene ring.”