Compound (A), $$C_8H_9Cl$$, gives a white precipitate when warmed with alcoholic AgNO$$_3$$. Oxidation of (A) gives an acid (B), $$C_8H_6O_4$$. (B) easily forms anhydride on heating. Identify the compound (A).

We have compound (A) with the molecular formula $$C_8H_9Cl$$. A first clue is given by the test with alcoholic silver nitrate: a white precipitate of $$AgCl$$ appears when (A) is warmed with alcoholic $$AgNO_3$$. This reaction proceeds rapidly only if the chlorine atom is present in a benzylic or allylic alkyl chloride; an aryl chloride $$\left(-Cl\; \text{directly on the benzene ring}\right)$$ does not react under these conditions. Hence the $$-Cl$$ must be attached to a benzylic $$CH_2$$ group rather than to the ring itself.

Next, (A) is oxidised to give compound (B) whose formula is $$C_8H_6O_4$$. Let us examine this formula. A benzene ring contributes $$C_6$$, so only $$C_2$$ atoms remain outside the ring. The oxygen count (four oxygens) strongly suggests the presence of two carboxyl groups, because each $$-COOH$$ contains two oxygens. The simplest aromatic dicarboxylic acid with two adjacent carboxyls is phthalic acid, $$C_{6}H_{4}(COOH)2$$, whose molecular formula is exactly $$C_8H_6O_4$$. In addition, phthalic acid readily forms phthalic anhydride on heating, exactly as stated in the question. Therefore

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

Oxidation of an alkyl side-chain on a benzene ring to a carboxyl group is a standard reaction; every benzylic carbon that still possesses at least one hydrogen converts completely to $$-COOH$$. Consequently (A) must contain two one-carbon side chains (each having at least one benzylic hydrogen) located ortho to each other, so that both become carboxyl groups during oxidation.

Let us write the general form of such a molecule before oxidation:

$$C_6H_4-CH_3 \quad\text{and}\quad C_6H_4-CH_2Cl$$ placed ortho to each other.

Now we verify the molecular formula. Counting atoms in $$\text{o-(chloromethyl)toluene}$$, i.e. 1-chloromethyl-2-methylbenzene, we have

Ring carbons: $$6$$; side-chain carbons: $$1+1=2 \;\; \Rightarrow \;\; C_8.$$ Hydrogens on the ring: because two ring positions carry substituents, only four hydrogens remain $$\;(H_4)$$. Hydrogens on the methyl group: $$H_3.$$ Hydrogens on the chloromethyl group: $$H_2.$$ Total hydrogens: $$4+3+2 = 9.$$ Chlorine atoms: $$1.$$

Thus the calculated formula is indeed $$C_8H_9Cl$$, matching the given formula for (A). Moreover, the benzylic $$CH_2Cl$$ reacts with alcoholic $$AgNO_3$$ to give a white precipitate of $$AgCl$$, while both benzylic side chains oxidise to yield phthalic acid. All experimental facts are satisfied.

Therefore compound (A) is $$\text{o-(chloromethyl)toluene}$$ (1-chloromethyl-2-methylbenzene), which corresponds to option (2).

Hence, the correct answer is Option 2.