Two isomers (A) and (B) with Molar mass 184 g/mol and elemental composition C, 52.2%; H, 49% and Br 42.9% gave benzoic acid and p-bromobenzoic acid, respectively on oxidation with KMnO$$_4$$. Isomer 'A' is optically active and gives a pale yellow precipitate when warmed with alcoholic AgNO$$_3$$. Isomer 'A' and 'B' are, respectively:

Given:

$$\%C = 52.2,\quad \%H = 4.9,\quad \%Br = 42.9$$

For 100 g compound: 

$$C = 52.2\text{ g}$$

$$H = 4.9\text{ g}$$

$$Br = 42.9\text{ g}$$

Moles:

$$C=\frac{52.2}{12}=4.35$$

$$H=\frac{4.9}{1}=4.9$$

$$Br=\frac{42.9}{80}=0.536$$

Dividing by the smallest value:

$$C:H:Br=8:9:1$$

Hence molecular formula is:

$$C_8H_9Br$$

Molar mass:

$$8(12)+9(1)+80=185\approx184$$

Thus the molecular formula is: $${C_8H_9Br}$$

Oxidation with $$KMnO_4$$:

Compound $$A$$ gives: $$\text{Benzoic acid}$$

Therefore bromine is present in the side chain.

Compound $$B$$ gives:

$$p\text{-Bromobenzoic acid}$$

Therefore bromine is present on the benzene ring at para position.

For isomer $$A$$:

It is optically active and gives a pale yellow precipitate with alcoholic $$AgNO_3$$.

This indicates the presence of a reactive benzylic bromide.

The structure:

$$C_6H_5CH(Br)CH_3$$ contains a chiral carbon atom.

Hence it is optically active.

Therefore: $$A={C_6H_5CH(Br)CH_3}$$ (1-bromo-1-phenylethane)

For isomer $$B$$:

To give $$p$$-bromobenzoic acid on oxidation, bromine must be on the ring and the side chain must be oxidizable.

Thus:

$$B={p\text{-Br}C_6H_4CH_2CH_3}$$ $$(p$$-bromoethylbenzene)

Therefore,

$$A = C_6H_5CH(Br)CH_3,$$ $$B = p\text{-Br}C_6H_4CH_2CH_3$$