Phenol is treated with $NaOH$ , then with $CO X_{2}$ at $400 K$ and 4-7 atm, and the mixture is acidified to give A. A is warmed with ethanoic anhydride and a few drops of concentrated $H X_{2} SO X_{4}$ to give B. B is

A2-acetoxybenzoic acid
Bphenyl ethanoate
C2-hydroxybenzoic acid
Dmethyl 2-hydroxybenzoate

Answer: A

This is the Kolbe-Schmitt reaction: sodium phenoxide and

CO X_{2}

give 2-hydroxybenzoic acid (A) after acidification. Acetylation of the phenolic OH by ethanoic anhydride gives 2-acetoxybenzoic acid (aspirin, B). The carboxylic acid group stays free.

Practice Structural Identification and Practical Organic Chemistry More on Structure determination from reaction sequences

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Phenol is treated with NaOH, then with COX2​ at 400 K and 4-7 atm, and the mixture is acidified to give A. A is warmed with ethanoic anhydride and a few drops of concentrated HX2​SOX4​ to give B. B is

Answer: A

More questions on Structure determination from reaction sequences

Phenol is treated with $NaOH$ , then with $CO X_{2}$ at $400 K$ and 4-7 atm, and the mixture is acidified to give A. A is warmed with ethanoic anhydride and a few drops of concentrated $H X_{2} SO X_{4}$ to give B. B is