JEE Main · 2022 · Shift-ImediumBIO-029

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

Question

L- isomer of a compound 'A' \ce(C4H8O4)\ce{(C4H8O4)} gives a positive test with [Ag(NH3)2]+{[Ag(NH_3)_2]^+}. 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 \ceHNO3\ce{HNO3} respectively. Compound (A) is: image

Options
  1. a

    Option a

  2. b

    Option b

  3. c

    Option c

  4. d

    Option d

Correct Answera

Option a

Detailed Solution

Step 1: Identifying Functional Groups — A positive Tollen’s\text{Tollen's} test shows 'A' is an aldose. Triacetate formation indicates three hydroxyl groups, so 'A' is an aldotetrose (C4C_4).

Step 2: Analyzing Oxidation Products — Bromine water gives an optically active aldonic acid (B), while \ceHNO3\ce{HNO3} 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 L-isomer\text{L-isomer} of this configuration.

Key Point: Symmetry in the dicarboxylic acid (tartaric acid derivative) differentiates erythrose from threose.

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L- isomer of a compound 'A' (C4H8O4) gives a positive test with [Ag(NH3)2]+. Treatment of 'A' with… (JEE Main 2022) | Canvas Classes