Question

The major product expected from the following reaction is :

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

The CH2OH group can form a five-membered lactone with one of the neighbouring carbonyl groups under acid; the carboxylic acid is far more reactive than the amide.

Answer 1

The Correct Option is C

The substrate is a benzene ring carrying CO2H, CH2OH and CONH2 on three adjacent carbons (CH2OH in the middle) and an OH ortho to the amide.
With dry HCl(g) in \(\mathrm{CCl_4}\) (acid catalysis, no water) the benzylic CH2OH attacks a neighbouring carbonyl to close a five-membered lactone (phthalide type) with loss of a small molecule. The carboxylic acid is a much better acyl donor than the amide, which is the least reactive acid derivative, so intramolecular esterification with CO2H takes place: \(\mathrm{-CH_2OH + HO_2C- \rightarrow}\) lactone \(\mathrm{+ H_2O}\).
The amide \(\mathrm{CONH_2}\) and phenolic OH remain unchanged (HCl does not replace an aryl OH by Cl, which rules out (2) and (4)). The lactone formed from the CO2H group, with CONH2 and OH retained, is structure (3).
Answer: (3) option (3) as drawn