JEE MainChemistry

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Identification of Aldotetrose Stereoisomers from Chemical and Optical Properties

L-isomer of a compound 'A' (C_4H_8O_4) gives a positive test with [Ag(NH_3)_2]^(+). Treatment of 'A' with acetic anhydride yield triacetate derivative. Compound 'A' produces an optically active compound (B) and an optically inactive compound (C) on treatment with bromine water and HNO_3 respectively, compound (A) is:

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Fischer projection of an aldotetrose with CHO at top, CH2OH at bottom, and both OH groups on the left with both H atoms on the right.

A Fischer projection with CHO at the top, CH2OH at the bottom, an upper horizontal bond of H on left and OH on right, and lower of HO on left and H on right.

Fischer projection with CHO at the top, CH2OH at the bottom, and both OH groups positioned on the right with H atoms on the left.

Fischer projection with top CHO, bottom CH2OH, upper cross showing left OH and right H, and lower cross showing left H and right OH.

Step-by-step solutionView

Correct answer

Compound A is L-erythrose, the aldotetrose having both hydroxyl groups on the left in its standard Fischer projection.

Option analysis

Why each option works or fails

A ·

This is the correct structure representing L-erythrose. L-erythrose has $- CHO$ at C1, $- {CH}_{2} OH$ at C4, and both $- OH$ groups on the left (C2 and C3). Oxidation with ${HNO}_{3}$ produces meso-tartaric acid (optically inactive), oxidation with ${Br}_{2} / H_{2} O$ gives an optically active aldonic acid, and having the lowest chiral center (C3) $- OH$ on the left defines it as an L-sugar.

B ·

Confusing an erythrose stereoisomer with a threose stereoisomer, assuming oxidation of any aldotetrose by nitric acid yields an optically inactive product. In threose, the two $- OH$ groups are on opposite sides, so oxidation with ${HNO}_{3}$ yields optically active tartaric acid ( $C_{2}$ chiral symmetry), failing the condition that compound C is optically inactive.

C ·

Identifying D-erythrose instead of L-erythrose by confusing D- and L- stereochemical descriptors. In a standard Fischer projection with $- CHO$ at the top, the configuration is L when the $- OH$ on the bottom-most stereocenter (C3) points to the left, not the right.

D ·

Choosing D-threose by confusing both the relative erythro/threo configuration and D/L configuration. Oxidation of threose by nitric acid yields optically active tartaric acid, whereas the question requires an optically inactive (meso) aldaric acid, which only forms from erythrose.

Reviewed route

Solution

StepWorking

01Identify

Compound 'A' $(C_{4} H_{8} O_{4})$ reduces Tollens' reagent $[Ag ({NH}_{3})_{2}]^{+}$ , indicating an aldehyde group (-CHO). It forms a triacetate with acetic anhydride, confirming three -OH groups. Thus, 'A' is an aldotetrose (erythrose or threose).

02Mechanism

Bromine water oxidises the -CHO group selectively to -COOH, forming an aldotetronic acid (B). Nitric acid ( ${HNO}_{3}$ ) oxidises both the terminal -CHO and $- {CH}_{2} OH$ groups to -COOH, forming a tartaric acid derivative / aldaric acid (C).

03Product

Compound C is optically inactive. Therefore, it must have an internal plane of symmetry. This means C is a meso-compound. In tartaric acid, only the meso form has an internal plane of symmetry. In this meso form, both -OH groups lie on the same side in the Fischer projection. This is the erythro configuration. Thus, compound 'A' must be an erythrose.

✓Verify

The stem specifies that 'A' is the L-isomer. In Fischer projection with -CHO at C-1 and $- {CH}_{2} OH$ at C-4, the bottom-most stereocenter (C-3) must have its -OH on the left for the L-configuration. Since it is L-erythrose, the -OH at C-2 must also be on the left (both -OH groups on the left). Inspecting the options, Option A shows both C-2 and C-3 -OH groups on the left, which is L-erythrose.

Hints that build this answer step by step

A compound with formula $C_{4} H_{8} O_{4}$ reduces Tollens' reagent and forms a triacetate with acetic anhydride. What functional class does compound A belong to?
An aldotetrose containing one aldehyde group and three hydroxyl groups
Oxidation of compound A with concentrated ${HNO}_{3}$ converts both terminal groups ( $- CHO$ and $- {CH}_{2} OH$ ) to carboxylic acids ( $- COOH$ ), forming tartaric acid (compound C). If compound C is optically inactive, what is the relative stereochemistry of the two $- OH$ groups in A?
Erythro (both $- OH$ groups are on the same side in the Fischer projection, giving meso-tartaric acid)
Compound A is specifically designated as the L-isomer. In a standard Fischer projection with $- CHO$ at the top and $- {CH}_{2} OH$ at the bottom, which side must the $- OH$ group at C3 (the highest-numbered stereocenter) be on?
On the left side

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Chemistry
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 22 September 2026

Quick checks

Students also ask

Why does erythrose give an optically inactive aldaric acid, while threose gives an optically active one?

Oxidizing both terminal groups to -COOH turns erythrose into meso-tartaric acid. Meso-tartaric acid has a horizontal plane of symmetry because both -OH groups are on the same side. In contrast, threose forms d/l-tartaric acid. This product lacks a plane of symmetry and is optically active.

Answer

Compound A is L-erythrose, the aldotetrose having both hydroxyl groups on the left in its standard Fischer projection.

Why each option works or fails

A · correct: - This is the correct structure representing L-erythrose. L-erythrose has $- CHO$ at C1, $- {CH}_{2} OH$ at C4, and both $- OH$ groups on the left (C2 and C3). Oxidation with ${HNO}_{3}$ produces meso-tartaric acid (optically inactive), oxidation with ${Br}_{2} / H_{2} O$ gives an optically active aldonic acid, and having the lowest chiral center (C3) $- OH$ on the left defines it as an L-sugar.
B: - Confusing an erythrose stereoisomer with a threose stereoisomer, assuming oxidation of any aldotetrose by nitric acid yields an optically inactive product. In threose, the two $- OH$ groups are on opposite sides, so oxidation with ${HNO}_{3}$ yields optically active tartaric acid ( $C_{2}$ chiral symmetry), failing the condition that compound C is optically inactive.
C: - Identifying D-erythrose instead of L-erythrose by confusing D- and L- stereochemical descriptors. In a standard Fischer projection with $- CHO$ at the top, the configuration is L when the $- OH$ on the bottom-most stereocenter (C3) points to the left, not the right.
D: - Choosing D-threose by confusing both the relative erythro/threo configuration and D/L configuration. Oxidation of threose by nitric acid yields optically active tartaric acid, whereas the question requires an optically inactive (meso) aldaric acid, which only forms from erythrose.

Step-by-step solution

Identify: Compound 'A' $(C_{4} H_{8} O_{4})$ reduces Tollens' reagent $[Ag ({NH}_{3})_{2}]^{+}$ , indicating an aldehyde group (-CHO). It forms a triacetate with acetic anhydride, confirming three -OH groups. Thus, 'A' is an aldotetrose (erythrose or threose).
Mechanism: Bromine water oxidises the -CHO group selectively to -COOH, forming an aldotetronic acid (B). Nitric acid ( ${HNO}_{3}$ ) oxidises both the terminal -CHO and $- {CH}_{2} OH$ groups to -COOH, forming a tartaric acid derivative / aldaric acid (C).
Product: Compound C is optically inactive. Therefore, it must have an internal plane of symmetry. This means C is a meso-compound. In tartaric acid, only the meso form has an internal plane of symmetry. In this meso form, both -OH groups lie on the same side in the Fischer projection. This is the erythro configuration. Thus, compound 'A' must be an erythrose.
Verify: The stem specifies that 'A' is the L-isomer. In Fischer projection with -CHO at C-1 and $- {CH}_{2} OH$ at C-4, the bottom-most stereocenter (C-3) must have its -OH on the left for the L-configuration. Since it is L-erythrose, the -OH at C-2 must also be on the left (both -OH groups on the left). Inspecting the options, Option A shows both C-2 and C-3 -OH groups on the left, which is L-erythrose.

Identification of Aldotetrose Stereoisomers from Chemical and Optical Properties

What feels right?

Compound A is L-erythrose, the aldotetrose having both hydroxyl groups on the left in its standard Fischer projection.

Why each option works or fails

Solution

01Identify

02Mechanism

03Product

✓Verify

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