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When phenol is treated with $CHCl_3$ and NaOH, the product formed is :-
A
Benzaldehyde
B
Salicylaldehyde
C
Salicylic acid
D
Benzoic acid
Detailed Solution
This is the Reimer–Tiemann reaction. Chloroform and NaOH form dichlorocarbene ($:CCl_2$), an electrophile.
It attacks the phenoxide ion mainly at the ortho position. The $-CHCl_2$ group formed is hydrolysed by the alkali to $-CHO$.
$C_6H_5OH + CHCl_3 + 3NaOH \rightarrow o-HOC_6H_4CHO + 3NaCl + 2H_2O$ (after acidification)
The product is o-hydroxybenzaldehyde, i.e. salicylaldehyde.
(With $CCl_4$ instead of $CHCl_3$, salicylic acid is formed.)
It attacks the phenoxide ion mainly at the ortho position. The $-CHCl_2$ group formed is hydrolysed by the alkali to $-CHO$.
$C_6H_5OH + CHCl_3 + 3NaOH \rightarrow o-HOC_6H_4CHO + 3NaCl + 2H_2O$ (after acidification)
The product is o-hydroxybenzaldehyde, i.e. salicylaldehyde.
(With $CCl_4$ instead of $CHCl_3$, salicylic acid is formed.)