Statement I: (S)-2-Bromobutane reacts with SH− ions in DMSO at 50°C to give a product with inverted configuration.
Statement II: The reaction of (S)-2-bromobutane with SH− ions follows the SN2 mechanism.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I.
In an SN2 reaction the nucleophile attacks the carbon from the side opposite to the halogen, and the new bond forms as the C–X bond breaks, in a single step.
\( \displaystyle \ce{CH3CH2CH(Br)CH3 + SH^- -> CH3CH2CH(SH)CH3 + Br^-} \)
Statement I: (S)-2-Bromobutane reacts with SH− ions in DMSO at 50°C to give a product with inverted configuration. — True. SH− enters on the side away from Br, so the product, butane-2-thiol, has the inverted configuration.
Statement II: The reaction of (S)-2-bromobutane with SH− ions follows the SN2 mechanism. — True. SH− displaces Br− by a one-step backside attack, which is the SN2 path.
Why Statement II explains Statement I?
Because the reaction goes by SN2, the nucleophile can only attach on the side opposite to the leaving bromine.
The three other groups turn over like an umbrella in the wind, so the configuration of the optically active carbon is inverted. An SN1 path would instead give a racemic mixture.
NCERT (2026-27 edition):
6.7.1 Reactions of Haloalkanes – “In case of optically active alkyl halides, the product formed as a result of SN2 mechanism has the inverted configuration as compared to the reactant. This is because the nucleophile attaches itself on the side opposite to the one where the halogen atom is present. … Thus, SN2 reactions of optically active halides are accompanied by inversion of configuration.”