Animated Solution for Chemistry - Organic Chemistry: Compound (A), ${\text{C}}_8{\text{H}}_9\text{Br}$ gives a white precipitate when warmed with alcoholic ${\text{AgNO}}_3$. Oxidation of (A) gives an acid (B), ${\text{C}}_8{\text{H}}_6{\text{O}}_4$. (B) easily forms anyhydride on heating. Identify the compound (A).

This problem is a classic organic chemistry puzzle that tests your ability to connect chemical tests, oxidation reactions, and structural geometry to deduce an unknown molecule. Let's break down the clues step-by-step to reveal the identity of Compound A.

The first piece of evidence is that Compound A, with the molecular formula ${\text{C}}_8{\text{H}}_9\text{Br}$, gives a white precipitate when warmed with alcoholic ${\text{AgNO}}_3$.

What does this signify? The white precipitate is silver bromide ($\text{AgBr}$). For an organic halide to react with silver nitrate and release a halide ion, the carbon-halogen bond must be relatively weak and capable of forming a stable carbocation intermediate.

If the bromine atom were attached directly to the benzene ring (an aryl halide), the partial double bond character arising from resonance would make the $\text{C}-\text{Br}$ bond too strong to break under these conditions. Therefore, the bromine atom must be located on a side chain, making it a benzylic halide. This means we have a $-{\text{CH}}_2\text{Br}$ group attached to the ring.

Next, we are told that oxidation of Compound A yields an acid, Compound B, with the formula ${\text{C}}_8{\text{H}}_6{\text{O}}_4$.

When alkylbenzenes are subjected to strong oxidation (like with ${\text{KMnO}}_4$), any carbon chain attached to the benzene ring that possesses at least one benzylic hydrogen is completely oxidized down to a carboxyl group ($-\text{COOH}$).

The formula ${\text{C}}_8{\text{H}}_6{\text{O}}_4$ corresponds perfectly to a benzene dicarboxylic acid, ${\text{C}}_6{\text{H}}_4(\text{COOH}{)}_2$. This is a massive revelation! It confirms that our original Compound A must have exactly two separate carbon-containing side chains attached to the benzene ring.

The final and most definitive clue is that Compound B easily forms an anhydride upon heating.

Anhydride formation involves the loss of a water molecule between two carboxyl groups. Geometrically, this reaction is highly favored only if the resulting cyclic structure is stable—typically a five- or six-membered ring.

For a benzene dicarboxylic acid, this proximity is only achieved if the two carboxyl groups are adjacent to each other, which is the ortho position. This specific isomer is known as phthalic acid. If the groups were meta (isophthalic acid) or para (terephthalic acid), they would be too far apart to form a cyclic anhydride easily.

Now, let's put all the pieces together.

Since Compound B is ortho-phthalic acid, the two side chains in our starting Compound A must also be ortho to each other. We already established from the ${\text{AgNO}}_3$ test that one of these side chains is a bromomethyl group ($-{\text{CH}}_2\text{Br}$).

To satisfy the overall molecular formula of ${\text{C}}_8{\text{H}}_9\text{Br}$, the second side chain must simply be a methyl group ($-{\text{CH}}_3$).

Therefore, Compound A is ortho-methylbenzyl bromide. Looking at the given visual options, option (d) perfectly represents this structure, with the $-{\text{CH}}_2\text{Br}$ and $-{\text{CH}}_3$ groups situated at the 1,2-positions on the benzene ring.