L- isomer of a compound 'A' (C4H8O4) gives a positive test with [Ag(NH3)2]+. Treatment of 'A' with acetic anhydride…
Biomolecules · Class 12 · JEE Main Previous Year Question
L- isomer of a compound 'A' gives a positive test with . Treatment of 'A' with acetic anhydride yields triacetate derivative. Compound 'A' produces an optically active compound (B) and an optically inactive compound (C) on treatment with bromine water and respectively. Compound (A) is:
- a✓
Option a
- b
Option b
- c
Option c
- d
Option d
Option a
Step 1: Identifying Functional Groups — A positive test shows 'A' is an aldose. Triacetate formation indicates three hydroxyl groups, so 'A' is an aldotetrose ().
Step 2: Analyzing Oxidation Products — Bromine water gives an optically active aldonic acid (B), while gives an optically inactive saccharic acid (C). Inactivity in (C) implies a symmetrical meso compound.
Step 3: Conclusion — For a tetrose to yield a meso-dicarboxylic acid, it must be erythrose. Structure (a) represents the of this configuration.
Key Point: Symmetry in the dicarboxylic acid (tartaric acid derivative) differentiates erythrose from threose.
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