Given D-Glucose The compound(s), which on reaction with $\mathrm{HNO}_{3}$ will give the product having degree of rotation, $[\alpha]_{\mathrm{D}}=-52.7^{\circ}$ is (are)

(C) (see structure); (D) (see structure)

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Biomolecules

Given

D-Glucose
The compound(s), which on reaction with

HNO_{3}

will give the product having degree of rotation,

[α]_{D} = - 52. 7^{\circ}

is (are)

(A)(see structure)
(B)(see structure)
(C)(see structure)
✓ Correct
(D)(see structure)
✓ Correct

Answer: (C), (D)

Solution

D-Glucose on oxidation with dilute HNO₃ gives D-glucaric acid with [α]_D = +52.7°. The enantiomer of this product (L-glucaric acid) has [α]_D = -52.7°.

L-glucaric acid is obtained by oxidizing L-glucose or L-gulose with HNO₃. Looking at the Fischer projections:

Option (C) represents L-gulose
Option (D) represents L-glucose

Both of these, when oxidized with HNO₃, will give the same aldaric acid (L-glucaric acid) with [α]_D = -52.7° because the aldehyde and primary alcohol groups are both oxidized to carboxylic acids.

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · One or more correct options

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JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Biomolecules

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