Which The Following Optically Lively Alcohol Is Handled With Hbr, A Racemic Combination Of Alkyl Bromides

(S)-2-butanol will endure an SN2 response with HBr to supply a racemic combination of alkyl bromides. Here choice B is the proper reply.

When optically lively alcohol is handled with HBr, the response follows an SN1 or SN2 mechanism. In the case of SN1, a carbocation intermediate is fashioned, and in SN2, a bottom assault by the nucleophile happens. The stereochemistry of the product is determined by the configuration of the intermediate and the path of assault.

In the case of (S)-2-butanol, the hydroxyl group is hooked up to the second carbon atom, which makes it a major alcohol. When handled with HBr, it undergoes an SN2 response, the place the hydroxyl group is changed by the bromine atom. The nucleophile assaults from the bottom of the molecule, resulting in an inversion of configuration.

This leads to the formation of a racemic combination of alkyl bromides, as each enantiomers have an equal probability of being attacked from both facet. On the opposite hand, (R)-2-butanol, being the enantiomer of (S)-2-butanol, can even endure the identical response and produce the identical racemic combination of alkyl bromides.

In the case of (R)-1-phenyl ethanol and (S)-1-phenyl ethanol, they’re secondary alcohols and may endure both SN1 or SN2 reactions relying on the response circumstances. However, the response mechanism will result in the formation of a combination of diastereomers, relatively than a racemic combination of enantiomers.

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Complete query:

Which of the next optically lively alcohols, when handled with HBr, leads to a racemic combination of alkyl bromides?

a) (R)-2-butanol

b) (S)-2-butanol

c) (R)-1-phenyl ethanol

d) (S)-1-phenyl ethanol