The major product obtained in the following reaction is:

We need to determine the major product formed when the given amino alcohol intermediate reacts with acetic anhydride ($$\text{(CH}_3\text{CO)}_2\text{O}$$) in the presence of one equivalent of pyridine at room temperature.

Reaction Mechanism and Chemoselectivity:

1. Nucleophilicity Comparison:

   The substrate contains two distinct nucleophilic functional groups: a secondary alcohol group ($$\text{--OH}$$) and a primary amine group ($$\text{--NH}_2$$). Because nitrogen is significantly less electronegative than oxygen, its lone pair is held less tightly and is much more available for bonding. Consequently, the amino nitrogen is a far stronger nucleophile than the hydroxyl oxygen:

   $$\text{Nucleophilicity: --NH}_2 > \text{--OH}$$



2. Selective N-Acetylation:

   Under mild conditions (room temperature) and limited reagent quantities (1 equivalent of pyridine), the more nucleophilic amine group selectively attacks one of the electrophilic carbonyl centers of the acetic anhydride molecule. This undergoes a nucleophilic acyl substitution, displacing an acetate leaving group:

   $$\text{R--NH}_2 + \text{(CH}_3\text{CO)}_2\text{O} \xrightarrow{\text{Pyridine}} \text{R--NHCOCH}_3 + \text{CH}_3\text{COOH}$$

   The secondary alcohol group ($$\text{--OH}$$) remains completely unreacted and intact throughout this selective transformation.

Conclusion:

The chemoselective reaction smoothly yields the N-acetylated amide derivative while leaving the secondary hydroxyl group untouched.

Answer: Option B