Identify the product formed (A and E)

The starting compound is p-nitrotoluene.

On bromination with $$Br_2$$, the $$-CH_3$$ group acts as an ortho/para director and the $$-NO_2$$ group acts as a meta director. Both groups direct the incoming bromine to the same position, giving product A as 2-bromo-4-nitrotoluene.

Reduction with $$Sn/HCl$$ converts the nitro group into an amino group, forming product B (2-bromo-4-methylaniline).

Treatment with $$NaNO_2/HCl$$ at $$273-278\text{ K}$$ converts the amino group into the corresponding diazonium salt, giving product C.

Reaction with $$H_3PO_2/H_2O$$ replaces the diazonium group by hydrogen (deamination), yielding product D as o-bromotoluene.

Oxidation with $$KMnO_4/KOH$$ followed by acidic workup $$H_3O^+$$ converts the benzylic $$-CH_3$$ group into $$-COOH$$ without affecting the bromine substituent.

Thus, product E is o-bromobenzoic acid (2-bromobenzoic acid).

Therefore,

• Product A = 2-bromo-4-nitrotoluene

• Product E = 2-bromobenzoic acid

Hence, the correct option is $$\boxed{\text{Option B}}$$.