JEE Advanced 2021 · Paper 1, Chemistry question 11

The reaction. $\ce{HNO3}$ oxidises both ends of an aldohexose to $\ce{-COOH}$. D-Glucose gives D-glucaric acid, $[\alpha]_D=+52.7^\circ$. The product with $-52.7^\circ$ is its enantiomer, L-glucaric acid.

Which sugars give L-glucaric acid? Once both ends are COOH, the chain can be read from either end. So two aldohexoses give L-glucaric acid:

• L-glucose;
• D-gulose (L-glucose with its two ends swapped).

The options, read from the Fischer projections:

• (A) D-mannose gives D-mannaric acid.
• (B) L-gulose gives D-glucaric acid ($+52.7^\circ$).
• (C) L-glucose gives L-glucaric acid.
• (D) D-gulose gives L-glucaric acid.

(C), (D)