Identify acid $B$ from its formula and clue

We are told that oxidation gives acid $B$ with molecular formula $C_8H_6O_4$.

Also, the anhydride of $B$ is used in the preparation of phenolphthalein. That anhydride is phthalic anhydride, so acid $B$ must be phthalic acid:

$B = \text{phthalic acid} = C_6H_4(COOH)_2$

Infer the side chain in compound $A$

Oxidation with alkaline $KMnO_4$ converts any alkyl side chain on benzene having at least one benzylic hydrogen into $-COOH$.

Since oxidation of $A$ gives phthalic acid, the starting compound must be a benzene derivative with two oxidizable side-chain carbons in ortho positions, or one side chain that on oxidation gives two adjacent carboxyl groups.

Use the molecular formula of $A$

$A = C_9H_{10}O$

Degree of unsaturation:

$\text{DU} = \frac{2(9)+2-10}{2} = \frac{10}{2} = 5$

A benzene ring accounts for 4, and one more unsaturation likely comes from the aromatic ring count overall with one oxygen-containing substituent but no extra double bond outside aromatic ring required if it is an alcohol/ether attached to benzene.

Use the positive iodoform test

A positive iodoform test is given by compounds containing either:

• $CH_3CO-$, or
• $CH_3CH(OH)-$

Since $A$ has only one oxygen and must oxidize to phthalic acid, a likely structure is a benzene ring bearing:

• one methyl group, and
• one $CH(OH)CH_3$ group

These together would give formula:

$C_6H_4 + CH_3 + CH(OH)CH_3 = C_9H_{10}O$

Check carefully:

Ring with two substituents contributes $C_6H_4$.

Add $CH_3$ and $CH(OH)CH_3$:

$C_6H_4 + C_1H_3 + C_2H_5O = C_9H_{12}O$

This does not match.

Try an aromatic ketone that gives iodoform test

A methyl ketone gives positive iodoform test. Consider:

$o\text{-methyl acetophenone} = C_6H_4(CH_3)(COCH_3)$

Formula check:

$C_6H_4 + CH_3 + COCH_3 = C_9H_{10}O$

This matches perfectly.

Check oxidation of $o$-methyl acetophenone

Structure: benzene with adjacent substituents $-CH_3$ and $-COCH_3$.

On strong oxidation with alkaline $KMnO_4$:

• the benzylic $-CH_3$ group oxidizes to $-COOH$
• the side chain of aryl methyl ketone $-COCH_3$ also oxidizes to $-COOH$

Thus:

$o\text{-methyl acetophenone} \xrightarrow[\text{KOH}]{KMnO_4} o\text{-benzene dicarboxylic acid}$

i.e.

$\text{phthalic acid}$

Check iodoform test again

The group $-COCH_3$ is a methyl ketone, so it gives a positive iodoform test.

Conclusion

Therefore, compound $A$ is:

$\boxed{o\text{-methyl acetophenone} \; (2\text{-methylacetophenone})}$

This corresponds to Option D.