$\mathrm{Phenol}\stackrel{\mathrm{NaOH}}{\to }\mathrm{A}\underset{2){\mathrm{H}}^{+}}{\overset{1){\mathrm{CO}}_2}{\to }}\mathrm{B}\underset{{\mathrm{H}}^{+}}{\overset{{\left({\mathrm{CH}}_3\mathrm{CO}\right)}_2\mathrm{O}}{\to }}\mathrm{C}$ .

Incorrect statement among the following is

‘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.