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$Phenol \overset{NaOH}{\to} A \to_{2) H^{+}}^{1) {CO}_{2}} B \to_{H^{+}}^{{({CH}_{3} CO)}_{2} O} C$ .
Incorrect statement among the following is

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Preparation of ‘B’ from phenol is called Kolbe’s reaction

‘B’ is steam volatile

‘C’ has a free Phenolic– OH group of ‘B’

‘C’ has a free – OH group of COOH

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detailed solution

Correct option is C

‘B’ is salicylic acid. Salicylic acid forms when phenol is reacted with NaOH/CO₂ follows by acidic treatment. This reaction also known as Kolbe's reaction. It is a steam volatile compound.

‘C’ is Aspirin (acetylsalicylic acid) and it has a free – OH group of COOH.

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Phenol→NaOHA→2)H+1)CO2B→H+CH3CO2OC .Incorrect statement among the following is