L-isomer of a compound 'A'($$C_4H_8O_4$$) gives a positive test with $$[Ag(NH_3)_2]^+$$. Treatment of 'A' with acetic anhydride yields triacetate derivative. Compound 'A' produces an optically active compound (B) and an optically inactive compound (C) on treatment with bromine water and $$HNO_3$$ respectively. Compound (A) is :

Explanation: Compound $$A$$ has the molecular formula $$C_4H_8O_4$$.

A positive Tollens’ test confirms the presence of an aldehyde group $$-CHO$$.

Formation of a triacetate derivative shows the presence of three hydroxyl groups $$-OH$$.

Thus, Compound $$A$$ is an aldotetrose with the structure:

$$CHO-CH(OH)-CH(OH)-CH_2OH$$

Since the compound is an $$L$$-isomer, the $$-OH$$ group on the bottom-most chiral carbon must be on the left side in the Fischer projection.

This eliminates the $$D$$-isomers.

On oxidation with nitric acid $$HNO_3$$, both the aldehyde group and primary alcohol group are converted into carboxylic acids.

The resulting compound is optically inactive, meaning it must be a meso compound possessing an internal plane of symmetry.

For Option A, the oxidized product has the $$-OH$$ groups on opposite sides, so no plane of symmetry exists and the product remains optically active.

For Option B, the oxidized product has both $$-OH$$ groups on the same side, producing a meso compound with an internal plane of symmetry.

Thus, the nitric acid oxidation product from Option B is optically inactive.

Oxidation with bromine water converts only the aldehyde group into a carboxylic acid while leaving the terminal $$CH_2OH$$ group unchanged.

Because the two ends of the molecule are now different, the product cannot possess symmetry and remains optically active.

Therefore, Option B satisfies all the given conditions.