JEE Main 2023 · 11 April · morning · Single correct · +4 / −1

L-isomer of tetrose X(C₄ H₈ O₄) gives positive Schiff's test and has two chiral carbons. On acetylation, 'X' yields triacetate. 'X' also undergoes following reactions
‘A’ ←(HNO₃) ‘X’ →(NaBH₄) ‘B’
'X' is

  1. (a)
    Option (a)
  2. (b)
    Option (b)
  3. (c)
    Option (c)
  4. (d)
    Option (d)
Answer and solution

Answer (d)

Solution

Chemistry ma’am · 4 steps

Two open-chain L-tetroses pass the early clues; the chiral NaBH₄ product is what picks threose.

  1. Read the clues. A positive Schiff's test needs a free -CHO group, so X is an open-chain aldotetrose. Such a sugar has two chiral carbons (C2 and C3) and three OH groups, which explains the triacetate. L means the OH on C3, the lowest chiral carbon, points left in the Fischer projection.
  2. Use the reduction. NaBH₄ turns CHO into CH₂OH, so both ends of the chain become identical. If the two middle OH groups are on the same side, the product has a mirror plane and is meso, which is achiral. B is chiral, so the two OH groups must be on opposite sides.
    OptionFree CHO?Chiral carbonsNaBH₄ productFits?
    (a) cyclic hemiacetalNo3not relevantNo
    (b) cyclic hemiacetalNo3not relevantNo
    (c) L-erythrose, both OH on the leftYes2erythritol, meso (achiral)No
    (d) L-threose, OH right then leftYes2threitol, chiralYes
  3. Check oxidation. HNO₃ oxidises both ends of L-threose to COOH. This gives A, an optically active tartaric acid, which is consistent with the scheme.
  4. So. X is L-threose, option (d).

The idea to keep. When reduction or oxidation makes both ends of a sugar chain identical, look for a mirror plane. Same-side OH groups give a meso, achiral product. Opposite-side OH groups give a chiral one.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 150 times in JEE Main since 2020. This question is from 2023.