Stereochemistry of Bromine Addition to Cis and Trans But-2-ene

To determine the correct statements, we analyze the mechanism and stereochemistry of the bromination reaction for both alkenes:

1. Reaction Mechanism and Stereospecificity

The addition of bromine ($\text{Br}_2$) to an alkene in an inert solvent like $\text{CHCl}_3$ occurs via electrophilic addition. It involves the formation of a cyclic bromonium ion intermediate, followed by an $S_N2$ nucleophilic attack by the bromide ion ($\text{Br}^-$) from the opposite face.

This leads to an anti-addition (also referred to as $trans$-addition) of the two bromine atoms across the double bond. Therefore, Statement (D) is correct.

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2. Stereochemical Analysis of Reaction (i)

• Reactant: $trans$-but-2-ene
• Addition mode: $trans$-addition (anti-addition)

According to the stereochemical mnemonic for addition reactions: $\text{trans-alkene} + \text{anti-addition} \longrightarrow \text{meso-product}$

Thus, the reaction produces the optically inactive meso compound, $(2R, 3S)\text{-2,3-dibromobutane}$. Hence, $\mathbf{M}$ and $\mathbf{N}$ represent the same meso molecule, meaning $\mathbf{M} \equiv \mathbf{N}$.

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3. Stereochemical Analysis of Reaction (ii)

• Reactant: $cis$-but-2-ene
• Addition mode: $trans$-addition (anti-addition)

According to the stereochemical mnemonic: $\text{cis-alkene} + \text{anti-addition} \longrightarrow \text{racemic mixture}$

Thus, the reaction produces a pair of enantiomers:

• $\mathbf{O} = (2R, 3R)\text{-2,3-dibromobutane}$
• $\mathbf{P} = (2S, 3S)\text{-2,3-dibromobutane}$

Since $\mathbf{O}$ and $\mathbf{P}$ are non-superimposable mirror images (enantiomers), they are not identical. Therefore, Statement (A) is incorrect.

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4. Comparison of Product Relationships

• $\mathbf{M}$ (or $\mathbf{N}$) is the meso stereoisomer: $(2R, 3S)\text{-2,3-dibromobutane}$.
• $\mathbf{O}$ is $(2R, 3R)\text{-2,3-dibromobutane}$.
• $\mathbf{P}$ is $(2S, 3S)\text{-2,3-dibromobutane}$.

Since stereoisomers that are not mirror images of each other are diastereomers:

• The pair ($\mathbf{M}$ and $\mathbf{O}$) consists of the $(2R, 3S)$ and $(2R, 3R)$ forms, which are diastereomers.
• The pair ($\mathbf{N}$ and $\mathbf{P}$) consists of the $(2R, 3S)$ and $(2S, 3S)$ forms, which are also diastereomers.

Therefore, Statement (B) is correct and Statement (C) is incorrect.

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Final Conclusion

The correct statements are B and D.