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JEE Advanced 2020 · Paper 1, Chemistry question 6

The Fischer projection of D-erythrose is shown below.

D-Erythrose and its isomers are listed as P, Q, R, and S in Column-I. Choose the correct relationship of P, Q, R, and S with D-erythrose from Column II.

Fischer projection of D-erythrose: CHO at the top, CH₂OH at the bottom, and both OH groups on the right (H on the left) at C2 and C3. Configuration (2R,3R).

Column-I

  • P
    Zigzag (wedge-dash) structure OHC-CH(OH)-CH(OH)-CH₂OH: at C2 H is dashed and OH wedged (pointing down); at C3 HO is dashed and H wedged (pointing up). Configuration (2R,3R).
  • Q
    Zigzag (wedge-dash) structure OHC-CH(OH)-CH(OH)-CH₂OH: C2 and C3 both have their substituents drawn below the chain; at C2 HO is dashed and H wedged, at C3 H is dashed and OH wedged. Configuration (2S,3R).
  • R
    Zigzag (wedge-dash) structure OHC-CH(OH)-CH(OH)-CH₂OH: at C2 H is dashed and OH wedged (pointing down); at C3 H is dashed and OH wedged (pointing up). Configuration (2R,3S).
  • S
    Zigzag (wedge-dash) structure OHC-CH(OH)-CH(OH)-CH₂OH: both C2 and C3 have their substituents drawn below the chain; at C2 HO is dashed and H wedged, at C3 OH is dashed and H wedged. Configuration (2S,3S).

Column-II

  • 1
    Diastereomer
  • 2
    Identical
  • 3
    Enantiomer
Choose one option
Answer and worked solution

Official answer: C

D-Erythrose, with both OH groups on the right in the Fischer projection, is (2R,3R). Assign each zigzag drawing using the priorities OH > CHO > C3 at C2 and OH > C2 > CH₂OH at C3.

  • P. C2: OH wedged, H dashed, giving R. C3: H wedged, OH dashed, giving R. So P is (2R,3R): identical (2).
  • Q. C2 (H wedged, HO dashed) is S; C3 (OH wedged, H dashed) is R. So Q is (2S,3R), a threose: diastereomer (1).
  • R. C2 as in P is R; C3 now has OH wedged, so it is S. So R is (2R,3S): diastereomer (1).
  • S. C2 as in Q is S; C3 has OH dashed, so it is S. So S is (2S,3S), the mirror image: enantiomer (3).

P → 2, Q → 1, R → 1, S → 3. (C)