Use the reaction with alcoholic $AgNO_3$

A white precipitate with alcoholic $AgNO_3$ indicates that bromine is present in a form that can ionize easily, i.e. typically a benzylic bromide or an alkyl bromide, not an aryl bromide.

So in $A\,(C_8H_9Br)$, the Br is most likely on a side chain attached to benzene.

Degree of unsaturation / likely skeleton

For $C_8H_9Br$, the likely aromatic structure is a benzene ring ($C_6$) plus a two-carbon side chain containing Br.

Possible benzylic bromides include:

• $C_6H_5CH_2CH_2Br$
• $C_6H_5CH(Br)CH_3$
• bromoxylene type structures, etc.

Use oxidation data

Oxidation of $A$ gives an acid $B$ of formula $C_8H_6O_4$.

This molecular formula corresponds to benzene dicarboxylic acid, $C_6H_4(COOH)_2$.

The statement that $B$ easily forms anhydride on heating identifies it specifically as phthalic acid (ortho-benzenedicarboxylic acid), because only the ortho isomer readily forms a cyclic anhydride.

Hence, $B = o\text{-}C_6H_4(COOH)_2$

What must $A$ be to oxidize to phthalic acid?

Strong oxidation converts any benzylic carbon having at least one H into $-COOH$.

To get two adjacent carboxyl groups, the starting compound must have two adjacent side chains on benzene, each oxidizable to $-COOH$.

Since the molecular formula of $A$ is $C_8H_9Br$, the ring must carry:

• one methyl group $(-CH_3)$
• one bromomethyl group $(-CH_2Br)$

because total carbon count is 8: $C_6H_4 + CH_3 + CH_2Br = C_8H_9Br$

On oxidation: $o\text{-}CH_3C_6H_4CH_2Br \xrightarrow{[O]} o\text{-}HOOC\,C_6H_4\,COOH$

This gives phthalic acid.

Check the $AgNO_3$ test

In $o\text{-}CH_3C_6H_4CH_2Br$, bromine is in the benzyl bromide position ($-CH_2Br$), which readily reacts with alcoholic $AgNO_3$ to give $AgBr$ white precipitate.

So this fits both clues.

Identify $A$

Therefore, compound $A$ is: $o\text{-}CH_3C_6H_4CH_2Br$ i.e. o-methyl benzyl bromide / 2-methylbenzyl bromide / o-bromomethyl toluene.

Comparison with stored answer

The derived answer corresponds to the option representing $o\text{-}CH_3C_6H_4CH_2Br$, which is given as Option D.

Hence, the derived answer agrees with the stored correct answer.