Biomolecules JEE 2021: Aldaric Acid PYQ Solution Guide

What is the hard Biomolecules JEE 2021 aldaric acid question?

The Biomolecules JEE 2021 problem is a hard, multi-correct JEE Advanced 2021 Paper 1 question on nitric-acid oxidation and Fischer-projection equivalence. Its stated expected solving time is 180 seconds. The safest method is to track the four OH positions as a sequence.

D-glucose has CHO at the top, CH₂OH at the bottom and the C2-to-C5 OH pattern R,L,R,R. Oxidation with HNO₃ produces aldaric acid P with the optical rotation shown below. The question asks which option patterns produce an aldaric acid with the same magnitude of rotation but the opposite sign. $[α]_{D} = + {52.7}^{\circ}$ $[α]_{D} = - {52.7}^{\circ}$

The option patterns from C2 to C5 are:

Under the same conditions, enantiomers have optical rotations of equal magnitude and opposite sign. The required product must therefore be the enantiomer of P. Every equivalent Fischer representation of that enantiomer must be identified.

How does HNO₃ oxidise the aldose in this question?

HNO₃ oxidises both terminal groups of an aldose to produce an aldaric acid. The aldehyde at the top and the primary alcohol at the bottom become carboxylic acids. The configurations at the four internal stereocentres, C2, C3, C4 and C5, remain unchanged.

The two terminal transformations are: $C H O \to C O O H$ $C H_{2} O H \to C O O H$

D-glucose therefore carries its internal pattern directly into product P:

R, L, R, R \to R, L, R, R

After oxidation, the molecule has identical COOH groups at both ends: $H O O C - (C H O H)_{4} - C O O H$

This creates an additional Fischer-projection equivalence when the sequence is compared from opposite ends. A 180° rotation produces another valid drawing of the same aldaric acid.

Which options form the enantiomer of P?

Options C and D form the required enantiomeric aldaric acid. The official method uses two separate operations. First, identify every representation of P by valid rotation. Then obtain P’s mirror pattern by interchanging left and right at every stereocentre.

For a 180° rotation of an aldaric-acid Fischer projection:

Reverse the order of the OH sequence.
Interchange left and right at every position.

Apply the rule to P: $R, L, R, R$

Reverse the order: $R, R, L, R$

Interchange left and right: $L, L, R, L$

These are two representations of the same product P:

R, L, R, R \equiv L, L, R, L

Both have the same optical rotation: $[α]_{D} = + {52.7}^{\circ}$

Now obtain the enantiomer by interchanging every side in the original pattern:

R, L, R, R \to L, R, L, L

Its optical rotation has the opposite sign: $+ {52.7}^{\circ} \to - {52.7}^{\circ}$

Rotate this enantiomer using the same official sequence method. First reverse the sequence:

L, R, L, L \to L, L, R, L

Then interchange left and right:

L, L, R, L \to R, R, L, R

The enantiomeric aldaric acid therefore has two equivalent representations:

L, R, L, L \equiv R, R, L, R

The complete option audit is:

A, L,L,R,R: A different stereoisomer. It is neither P nor its enantiomer.
B, L,L,R,L: P itself after a valid 180° rotation. It gives positive rotation.
C, L,R,L,L: The direct mirror pattern of P.
D, R,R,L,R: The 180°-rotated representation of P’s enantiomer.

Official answer: C and D

Why does option B still give positive optical rotation?

Option B has the pattern L,L,R,L, but a Fischer projection that looks different from R,L,R,R is not automatically an enantiomer. The valid 180° rotation test shows that B is another representation of P. Molecular identity must be checked before assigning the sign of optical rotation.

Start with P: $R, L, R, R$

Reverse its sequence: $R, R, L, R$

Now interchange every side: $L, L, R, L$

This is exactly option B. A valid 180° rotation does not change molecular identity or optical rotation. Therefore, B still gives: $[α]_{D} = + {52.7}^{\circ}$

It does not give the required negative value.

Use this test:

Generate the full mirror pattern by swapping left and right at every stereocentre.
Only then use the 180° reverse-then-interchange operation to find equivalent drawings.

Partial matching by appearance is unsafe in a multi-correct question.

Which related carbohydrate stereochemistry questions should you practise?

Practise these three transfer problems to apply the same method to meso compounds, reagent selection and equivalent Fischer projections. For every rotation, reverse the OH sequence and then interchange left and right. Do not decide molecular identity from visual appearance alone.

How do you test whether an aldaric acid is meso?

Compare the 180°-rotated pattern with the full mirror pattern. Consider an aldose with the C2-to-C5 pattern R,L,L,R after HNO₃ oxidation. If its rotated representation and mirror pattern are identical, the aldaric acid is meso and optically inactive.

Reverse the original pattern:

R, L, L, R \to R, L, L, R

Interchange sides:

R, L, L, R \to L, R, R, L

The full mirror pattern is also L,R,R,L. The aldaric acid is therefore identical to its mirror image, so it is meso and optically inactive.

Which reagent oxidises both ends of an aldose?

HNO₃ oxidises both the CHO group and the terminal CH₂OH group. Bromine water, under standard carbohydrate chemistry conditions, selectively oxidises the aldehyde and leaves the terminal primary alcohol unchanged. The two reagents therefore produce different classes of carbohydrate acids.

HNO₃: Aldose to aldaric acid
Bromine water: Aldose to aldonic acid

What are the equivalent and enantiomeric patterns of R,R,L,R?

The equivalent Fischer pattern is L,R,L,L. Reverse R,R,L,R to R,L,R,R, then interchange left and right to obtain L,R,L,L. The full mirror pattern is L,L,R,L, whose rotated representation is R,L,R,R.

R, R, L, R \to R, L, R, R \to L, R, L, L

Therefore:

R, R, L, R \equiv L, R, L, L

The enantiomeric representations are:

L, L, R, L \equiv R, L, R, R

For more questions of this type, use the free past-paper archive to search JEE Advanced papers by year, subject or chapter and check each worked solution against the same reverse-then-interchange audit.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Frequently asked questions

What is the answer to the Biomolecules JEE 2021 aldaric acid question?

The correct options are C and D. Their OH patterns, L,R,L,L and R,R,L,R, are equivalent Fischer representations of the enantiomer of the aldaric acid formed from D-glucose.

Why is option B not an enantiomer?

Option B, L,L,R,L, is the same aldaric acid as the original R,L,R,R product after a valid 180° Fischer rotation. It therefore retains the positive optical rotation instead of giving the required negative value.

How do you rotate an aldaric acid Fischer projection?

Reverse the order of the OH sequence and then interchange left and right at every stereocentre. This operation is valid here because the aldaric acid has identical COOH groups at both ends.

Which reagent converts an aldose into an aldaric acid?

HNO₃ oxidises both the aldehyde group and the terminal primary alcohol of an aldose, producing an aldaric acid. Bromine water oxidises only the aldehyde under standard conditions and forms an aldonic acid.

How do you check whether an aldaric acid is meso?

Compare the compound’s full mirror pattern with an equivalent pattern obtained by 180° Fischer rotation. If they represent the same molecule, the aldaric acid is meso and optically inactive.

Biomolecules JEE 2021: Aldaric Acid Stereochemistry