Stereochemistry of Hydroxylation of Isomers of Dicarboxylic Acid

To determine the products formed from compounds $P$ and $Q$, we analyze their structures and the stereochemistry of the hydroxylation reaction step-by-step:

1. Identification of Compounds $P$ and $Q$

The molecular formula $\text{C}_4\text{H}_4\text{O}_4$ has a degree of unsaturation (DU) equal to: $\text{DU} = C + 1 - \frac{H}{2} = 4 + 1 - \frac{4}{2} = 3$ Since $P$ and $Q$ are dicarboxylic acids, two degrees of unsaturation belong to the two carbonyl groups ($\text{-COOH}$), and the remaining one degree of unsaturation corresponds to a carbon-carbon double bond ($\text{C=C}$).

• Compound $P$ forms a cyclic anhydride upon heating, which requires the two carboxyl groups to be in close proximity (cis-configuration). Thus, $P$ is maleic acid (cis-butenedioic acid).
• Compound $Q$ is the geometrical isomer of $P$, which is fumaric acid (trans-butenedioic acid).

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2. Nature of the Reaction

Treatment with dilute alkaline $\text{KMnO}_4$ (Baeyer's reagent) results in $syn$-hydroxylation across the alkene double bond, adding two $-\text{OH}$ groups from the same face.

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3. Stereochemical Outcome of the Reaction

Reaction of $P$ (cis-isomer):

• Applying the stereochemical rule for additions: $\text{Cis-alkene} + \text{Syn-addition} \longrightarrow \textit{Meso}\text{-compound}$
• The syn-hydroxylation of maleic acid ($P$) yields meso-tartaric acid, which corresponds to structure $S$:

$$S = \begin{array}{c|c} \text{COOH} \\ \text{H} - \text{OH} \\ \text{H} - \text{OH} \\ \text{COOH} \end{array}$$

• Structure $S$ contains a plane of symmetry bisecting the $\text{C2-C3}$ bond and is internally compensated. Therefore, $S$ is optically inactive.

Reaction of $Q$ (trans-isomer):

• Applying the stereochemical rule for additions: $\text{Trans-alkene} + \text{Syn-addition} \longrightarrow \text{Pair of Enantiomers (Racemic mixture)}$
• The syn-hydroxylation of fumaric acid ($Q$) yields an equimolar pair of enantiomers, $(2R, 3R)$-tartaric acid and $(2S, 3S)$-tartaric acid, represented by structures $T$ and $U$:

$$T = \begin{array}{c|c} \text{COOH} \\ \text{H} - \text{OH} \\ \text{HO} - \text{H} \\ \text{COOH} \end{array} \quad \text{and} \quad U = \begin{array}{c|c} \text{COOH} \\ \text{HO} - \text{H} \\ \text{H} - \text{OH} \\ \text{COOH} \end{array}$$

• Since an equal mixture of $T$ and $U$ is formed, the resulting pair $(T, U)$ constitutes a racemic mixture, which is optically inactive due to external compensation.

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Conclusion

• Reaction of $P$ produces optically inactive $S$.
• Reaction of $Q$ produces optically inactive pair ($T, U$).

Thus, the correct option is B.