JEE Advanced 2021 Paper 1 Q11 Chemistry Biomolecules & Polymers Carbohydrates Hard

JEE Advanced 2021 Paper 1 · Q11 · Carbohydrates

Given: D-Glucose (Fischer projection: CHO at top; then H-C-OH, HO-C-H, H-C-OH, H-C-OH; CH$_2$OH at the bottom) on reaction with HNO$_3$ gives a product P with [$\alpha$]$_D$ = +52.7°.

The compound(s), which on reaction with HNO$_3$ will give the product having degree of rotation, [$\alpha$]$_D$ = -52.7°, is(are):

  1. A. [Figure: option A — Fischer projection of an aldohexose: CHO at top; from C2 to C5 the configurations are HO-C-H, HO-C-H, H-C-OH, H-C-OH; CH$_2$OH at the bottom. (i.e. left, left, right, right from C2.)]
  2. B. [Figure: option B — Fischer projection of an aldohexose: CHO at top; from C2 to C5 the configurations are HO-C-H, HO-C-H, H-C-OH, HO-C-H; CH$_2$OH at the bottom. (i.e. left, left, right, left.)]
  3. C. [Figure: option C — Fischer projection of an aldohexose: CHO at top; from C2 to C5 the configurations are HO-C-H, H-C-OH, HO-C-H, HO-C-H; CH$_2$OH at the bottom. (i.e. left, right, left, left.)]
  4. D. [Figure: option D — Fischer projection of an aldohexose: CHO at top; from C2 to C5 the configurations are H-C-OH, H-C-OH, HO-C-H, H-C-OH; CH$_2$OH at the bottom. (i.e. right, right, left, right.)]
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Correct answer:C, D

Solution

HNO$_3$ oxidises both ends of the hexose (CHO → COOH and CH$_2$OH → COOH) to give a saccharic (aldaric) acid. D-glucose yields D-glucaric acid with [$\alpha$]$_D$ = +52.7°. We need a hexose whose aldaric acid is the enantiomer of D-glucaric acid, i.e. L-glucaric acid with [$\alpha$]$_D$ = -52.7°.

Two sugars give L-glucaric acid: - The L-enantiomer of glucose itself (left, right, left, left from C2), which is option (C). - L-gulose (right, right, left, right from C2): oxidation to its aldaric acid followed by a 180° in-plane rotation gives the same dicarboxylic acid as L-glucaric acid (since both ends become COOH and the molecule is then symmetrical with respect to end-flip). This is option (D).

Option (A) is D-mannose configuration which yields mannaric acid (meso, optically inactive). Option (B) is D-galactose configuration which yields meso galactaric (mucic) acid. Neither gives [$\alpha$]$_D$ = -52.7°. Hence the correct answers are (C) and (D).

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