The meso-form among the given stereoisomers is the (R,S)-isomer of II, i.e. meso‑2,3‑dichlorobutane.​

Introduction

Understanding how to identify the meso form of 2‑bromo‑3‑chlorobutane and 2,3‑dichlorobutane is essential for mastering stereochemistry questions in competitive exams. Meso compounds have multiple chiral centres but are achiral due to an internal plane of symmetry with opposite configurations such as R and S on equivalent centres.​

Step 1: Interpret molecules I and II

• Molecule I: CH₃–CH(Br)–CH(Cl)–CH₃ = 2‑bromo‑3‑chlorobutane (non‑identical substituents at C‑2 and C‑3).​

• Molecule II: CH₃–CH(Cl)–CH(Cl)–CH₃ = 2,3‑dichlorobutane (identical substituents at C‑2 and C‑3).​

Each has two stereogenic centres (C‑2 and C‑3), so maximum stereoisomers = 22=422=4, but meso possibility must be checked using symmetry.​

Step 2: Conditions for a meso compound

For simple acyclic systems with two tetrahedral stereocentres, a meso compound must:​

• Contain at least two chiral centres.

• Possess an internal plane of symmetry, making the molecule superimposable on its mirror image.

• Have configurations of the type (R,S) (or (S,R)) on equivalent stereocentres (same set of substituents on each carbon).

2,3‑Dichlorobutane satisfies the “equivalent centre” requirement, whereas 2‑bromo‑3‑chlorobutane does not because the two stereocentres carry different substituents (Br vs Cl).​

Step 3: Analyse each option

Option (A): (R,R)-isomer of I

• In 2‑bromo‑3‑chlorobutane, assign priorities at each chiral carbon: Br > Cl > CH₂CH₃ > CH₃ at C‑2, and Cl > Br > CH₂CH₃ > CH₃ at C‑3 (order pattern differs).​

• The (R,R) and (S,S) isomers of I form an enantiomeric pair and are chiral with no internal symmetry; therefore they cannot be meso.​

• Hence, the (R,R)-isomer of I is not meso.

Option (C): (R,S)-isomer of I

• Because substituents on C‑2 and C‑3 are different, even an (R,S) configuration cannot generate a plane of symmetry through the C₂–C₃ bond.​

• The two halves of the molecule are not identical; consequently, the (R,S)‑2‑bromo‑3‑chlorobutane is chiral and appears as a diastereomer of the (R,R)/(S,S) pair, not as a meso compound.​

• Thus, (R,S)-isomer of I is not meso.

Option (D): (S,S)-isomer of II

• In 2,3‑dichlorobutane, centres C‑2 and C‑3 have the same substituents (Cl, CH₃, CH₂CH₃, H), so their R/R or S/S combinations lead to a chiral molecule lacking any mirror plane.​

• The (R,R) and (S,S) configurations of II are enantiomers and rotate plane‑polarised light, so neither is meso.​

• Hence, (S,S)-isomer of II is not meso.

Option (B): (R,S)-isomer of II → Correct

• For 2,3‑dichlorobutane, drawing the (R,S) configuration in a Fischer projection places the two chlorine atoms on opposite sides at C‑2 and C‑3, with the CH₃ and CH₂CH₃ groups arranged symmetrically.​

• This arrangement introduces an internal mirror plane passing vertically through the C₂–C₃ bond, making the molecule superimposable on its mirror image and therefore achiral despite two chiral centres.​

• Thus, (R,S)-2,3‑dichlorobutane is the meso form, and option (B) is correct.​

Key concepts summarised

• Meso form exists only when stereocentres are equivalent and carry opposite configurations (R,S) so that a plane of symmetry arises, giving optical inactivity.​

• 2‑Bromo‑3‑chlorobutane (molecule I) cannot show a meso form because its two stereocentres are not equivalent (different halogens attached), so none of its stereoisomers match the meso criteria.​

• 2,3‑Dichlorobutane (molecule II) has a meso stereoisomer with configuration (R,S), making (R,S)-isomer of II the only meso form among the given options.​