Question

Write chemical equations for obtaining the following from phenol:
(i) 4-Nitrophenol
(ii) Picric acid
(iii) 4-Bromophenol
(iv) Salicylaldehyde
(v) Benzoquinone

Hint:

Phenol is highly reactive towards electrophilic substitution due to the activating effect of –OH group. By controlling the conditions (dilute vs. concentrated reagents, catalysts, solvents), selective products such as mono-, tri-nitro, halo, or formyl-substituted phenols can be obtained.

Answer 1

(i) Preparation of 4-Nitrophenol:
Phenol undergoes nitration with dilute HNO$_3$ at room temperature.
\[ C_6H_5OH + HNO_3 \;\longrightarrow\; o\text{-nitrophenol} + p\text{-nitrophenol} \] Para product (4-nitrophenol) is the major one due to steric effects.
(ii) Preparation of Picric acid (2,4,6-trinitrophenol):
On treatment with concentrated HNO$_3$ in presence of concentrated H$_2$SO$_4$, phenol gives picric acid.
\[ C_6H_5OH \;\xrightarrow[\;conc.\;H_2SO_4\;]{conc.\;HNO_3}\; 2,4,6\text{-trinitrophenol (picric acid)} \] (iii) Preparation of 4-Bromophenol:
Phenol reacts with bromine water to give 2,4,6-tribromophenol. But to obtain monobromo derivative, phenol is treated with bromine in CS$_2$ at low temperature.
\[ C_6H_5OH + Br_2 \;\xrightarrow[\;CS_2\;]{273K}\; 4\text{-bromophenol} \] (iv) Preparation of Salicylaldehyde:
This is obtained by the Reimer–Tiemann reaction. Phenol is treated with chloroform (CHCl$_3$) and alkali (KOH/NaOH).
\[ C_6H_5OH + CHCl_3 + 3NaOH \;\longrightarrow\; o\text{-HO–C}_6H_4–CHO + 3NaCl + 2H_2O \] (Main product is salicylaldehyde at ortho-position.)
(v) Preparation of Benzoquinone:
Phenol is oxidised by strong oxidising agents like Na$_2$Cr$_2$O$_7$/H$_2$SO$_4$.
\[ C_6H_5OH \;\xrightarrow[\;oxidation\;]{}\; p\text{-benzoquinone} \] \[ \boxed{\text{Thus, phenol can yield nitrophenols, bromophenols, aldehydes, and quinones via suitable reactions.}} \]