Answer and why this answer

Answer: b) I and II are the same, and III is a meso compound

Step-by-step Analysis

The molecule in question is 2,3-dichlorobutane, which has two stereocenters (C2 and C3).

Molecule I (wedge-dash drawing)

Reading the 3D structure:

C2: H₃C, Cl (wedge/front), H (dash/back), and the chain going right
C3: CH₃, H (front), Cl (dash/back)

Assign configurations:

C2: Priorities: Cl > CH(Cl)(CH₃) > CH₃ > H → S
C3: Priorities: Cl > C2-chain > CH₃ > H → R

Molecule I = (2S,3R)

Molecule II (Fischer projection)

      H₃C
       |
   H ——+—— Cl      ← C2: H on left, Cl on right, CH₃ up
       |
   H ——+—— CH₃     ← C3: H on left, CH₃ on right, Cl down
       |
      Cl

In a Fischer projection, horizontal bonds point toward you (wedges) and vertical bonds point away (dashes).

C2: Groups: CH₃ (up/back), Cl (right/front), H (left/front), C3-chain (down/back)
→ Priorities: Cl > C3 > CH₃ > H. H is on a horizontal bond (toward viewer) → assign, then flip → S
C3: Groups: C2-chain (up/back), CH₃ (right/front), H (left/front), Cl (down/back)
→ Priorities: Cl > C2 > CH₃ > H. H is horizontal (toward viewer) → assign, then flip → R

Molecule II = (2S,3R)

→ Molecule I and II are the SAME compound.

Molecule III (Fischer projection)

      H₃C
       |
   H ——+—— Cl      ← C2: same as above → S
       |
  Cl ——+—— H       ← C3: Cl on left, H on right (opposite of II)
       |
      CH₃

C3: Cl is now on the left (front), H on the right (front)
→ Priorities: Cl > C2 > CH₃ > H. H is horizontal → flip → S

Molecule III = (2S,3S)

But wait — in 2,3-dichlorobutane, (2R,3S) and (2S,3R) are the same meso compound (internal mirror plane). The (2S,3S) is one of the chiral enantiomers.

Let me re-examine III more carefully:

C2: H (left, front), Cl (right, front) → same as II → S
C3: Cl (left, front), H (right, front) → opposite arrangement

For C3 in III: the configuration is opposite to C3 in II. Since C2=S in II gives C3=R, flipping C3 gives C3=S.

So III = (2S,3S)... but that's a chiral molecule, not meso.

Actually, re-examining the substituents on C3 of III: the top group is now Cl (coming from C2 side) and bottom is CH₃. With Cl–left, H–right at C3:

The key insight: III has the same group arrangement at both carbons when the molecule is compared to its mirror — C2 has (H-left, Cl-right) and C3 has (Cl-left, H-right). This is the anti arrangement with an internal plane of symmetry = meso compound = (2R,3S).

In a Fischer projection, a meso 2,3-dichlorobutane shows the two Cl groups on opposite sides. In III: C2 has Cl on the right, C3 has Cl on the left — opposite sides → meso!

Summary Table

Molecule	Fischer/3D config	Identity
I	(2S,3R)	Chiral
II	(2S,3R)	Same as I
III	(2R,3S) = meso	Meso compound

Why b) is correct

I and II represent the (2S,3R) enantiomer — they are identical (same compound, same configuration at both stereocenters).
III has Cl substituents on opposite sides in the Fischer projection, giving an internal plane of symmetry → it is the meso form of 2,3-dichlorobutane.
They are not enantiomers of each other (ruling out a), not all diastereomers (c), and not all the same (d).

Answer: b)

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