JEE Challenger
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Stereoisomeric Relationships of D Erythrose Isomers

The Fischer projection of D-erythrose is depicted below.

D-Erythrose and its corresponding isomers are designated as P, Q, R, and S in Column-I. Select the correct matching of P, Q, R, and S with D-erythrose from Column-II.

Column-IColumn-IIP, Q, R, S (see structures in figure)1. Diastereomer2. Identical3. Enantiomer\begin{array}{ll} \text{Column-I} & \text{Column-II} \\ \text{P, Q, R, S (see structures in figure)} & 1.\text{ Diastereomer} \\ & 2.\text{ Identical} \\ & 3.\text{ Enantiomer} \end{array}
Question Diagram 1
Question Diagram 2

Options

A

P→2,Q→3,R→2,S→2P \rightarrow 2, Q \rightarrow 3, R \rightarrow 2, S \rightarrow 2

B

P→3,Q→1,R→1,S→2P \rightarrow 3, Q \rightarrow 1, R \rightarrow 1, S \rightarrow 2

C

P→2,Q→1,R→1,S→3P \rightarrow 2, Q \rightarrow 1, R \rightarrow 1, S \rightarrow 3

Correct
D

P→2,Q→3,R→3,S→1P \rightarrow 2, Q \rightarrow 3, R \rightarrow 3, S \rightarrow 1

Step-by-Step Solution

To determine the correct relationships of structures P, Q, R, and S with D-erythrose, we determine the absolute configurations ((R/S)(R/S) system) of the chiral centers (C2\text{C2} and C3\text{C3}) for D-erythrose and each compound.


1. Configuration of D-Erythrose

The Fischer projection of D-erythrose is:

CHO (C1)∣H−C2−OH∣H−C3−OH∣CH2OH (C4)\begin{array}{c} \text{CHO (C1)} \\ | \\ \text{H} - \text{C2} - \text{OH} \\ | \\ \text{H} - \text{C3} - \text{OH} \\ | \\ \text{CH}_2\text{OH (C4)} \end{array}

Configuration at C2:

  • Priority of groups attached to C2:
    1. −OH-\text{OH} (Priority 1)
    2. −CHO-\text{CHO} (Priority 2, attached to (O,O,H)(\text{O}, \text{O}, \text{H}))
    3. −C3-\text{C3} (Priority 3, attached to (O,C,H)(\text{O}, \text{C}, \text{H}))
    4. −H-\text{H} (Priority 4)
  • Determining (R/S)(R/S):
    • Group 1 (−OH-\text{OH}) is on the Right.
    • Group 2 (−CHO-\text{CHO}) is at the Top.
    • Group 3 (−C3-\text{C3}) is at the Bottom.
    • Direction 1→2→31 \rightarrow 2 \rightarrow 3 goes counter-clockwise.
    • Since the lowest priority group (−H-\text{H}) is on a horizontal bond, counter-clockwise corresponds to the (2R)(2R) configuration.

Configuration at C3:

  • Priority of groups attached to C3:
    1. −OH-\text{OH} (Priority 1)
    2. −C2-\text{C2} (Priority 2, attached to (O,C,H)(\text{O}, \text{C}, \text{H}))
    3. −CH2OH-\text{CH}_2\text{OH} (Priority 3, attached to (O,H,H)(\text{O}, \text{H}, \text{H}))
    4. −H-\text{H} (Priority 4)
  • Determining (R/S)(R/S):
    • Group 1 (−OH-\text{OH}) is on the Right.
    • Group 2 (−C2-\text{C2}) is at the Top.
    • Group 3 (−CH2OH-\text{CH}_2\text{OH}) is at the Bottom.
    • Direction 1→2→31 \rightarrow 2 \rightarrow 3 goes counter-clockwise.
    • With −H-\text{H} on a horizontal bond, counter-clockwise corresponds to the (3R)(3R) configuration.

Thus, D-erythrose has the (2R,3R)(2R, 3R) configuration.


2. Analysis of Structures P, Q, R, and S

Structure P:

  • C2 Center: The −OH-\text{OH} group is on a dashed bond (pointing away) and −H-\text{H} is on a wedged bond. Evaluating priorities gives the (2R)(2R) configuration.
  • C3 Center: The −OH-\text{OH} group is on a wedged bond and −H-\text{H} is on a dashed bond. Evaluating priorities gives the (3R)(3R) configuration.
  • Configuration of P: (2R,3R)(2R, 3R), which is Identical to D-erythrose. P→2\mathbf{P \rightarrow 2}

Structure Q:

  • C2 Center: The configuration is inverted relative to P, yielding (2S)(2S).
  • C3 Center: The configuration is retained relative to P, yielding (3R)(3R).
  • Configuration of Q: (2S,3R)(2S, 3R), which has one inverted stereocenter. Therefore, it is a Diastereomer of D-erythrose. Q→1\mathbf{Q \rightarrow 1}

Structure R:

  • C2 Center: The configuration is retained relative to P, yielding (2R)(2R).
  • C3 Center: The configuration is inverted relative to P, yielding (3S)(3S).
  • Configuration of R: (2R,3S)(2R, 3S), which has one inverted stereocenter. Therefore, it is a Diastereomer of D-erythrose. R→1\mathbf{R \rightarrow 1}

Structure S:

  • C2 Center: The configuration is inverted relative to P, yielding (2S)(2S).
  • C3 Center: The configuration is inverted relative to P, yielding (3S)(3S).
  • Configuration of S: (2S,3S)(2S, 3S), which is the non-superimposable mirror image of (2R,3R)(2R, 3R). Therefore, it is the Enantiomer of D-erythrose. S→3\mathbf{S \rightarrow 3}

Conclusion

P→2,Q→1,R→1,S→3P \rightarrow 2, \quad Q \rightarrow 1, \quad R \rightarrow 1, \quad S \rightarrow 3

Thus, the correct option is C.