In the following reaction sequence,

the major product B is

We need to determine the major product B formed in the given reaction sequence starting from p-toluidine.

Step-by-Step Reaction Mechanism:

1. Step 1: Acetylation (Formation of Compound A)

   When p-toluidine (4-methylaniline) reacts with acetic anhydride ($$\text{CH}_3\text{CO-O-COCH}_3$$), the amino group ($$-\text{NH}_2$$) undergoes acetylation. This converts the strongly activating amino group into a moderately activating acetamido group ($$-\text{NHCOCH}_3$$):

   $$\text{p-Toluidine} + (\text{CH}_3\text{CO})_2\text{O} \longrightarrow \text{p-Methylacetanilide (Compound A)} + \text{CH}_3\text{COOH}$$

   This protection step softens the reactivity of the nitrogen lone pair, preventing multiple halogenations.




2. Step 2: Electrophilic Monobromination (Formation of Product B)

   Compound A is then treated with bromine in acetic acid ($$\text{Br}_2 / \text{AcOH}$$):

   • Both the $$-\text{NHCOCH}_3$$ group and the $$-\text{CH}_3$$ group are ortho/para-directing.
   • The para-position relative to both groups is completely blocked.
   • Since the acetamido group ($$-\text{NHCOCH}_3$$) is a significantly stronger activating group than the alkyl group ($$-\text{CH}_3$$) due to resonance ($$+R$$ vs $$+I/\text{hyperconjugation}$$), the incoming electrophile ($$\text{Br}^\oplus$$) is directed to the position ortho to the $$-\text{NHCOCH}_3$$ group.