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Considering the Sn1 reaction of (S)-2,3-dimethyl-3-pentanol, knowing that this molecule has a chiral carbon, should I always assume that both enantiomers are present, and therefore the products of a reaction will always have $2^n$ products for $n$ stereocenters on the substrate, assuming that each stereocenter has equal chance of reacting? Is there two products of this Sn1 reaction due to the fact that the molecule is chiral, and therefore its enantiomer is present and has equal chance of reacting?

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In this SN1 reaction there would be two products even if only one enantiomer reactions.

Enantiomerically pure (S)-2,3-dimethyl-3-pentanol reacts with HBr to produce both (S)-3-bromo-2,3-dimethylpentane and (R)-3-bromo-2,3-dimethylpentane in equal proportions as a racemic mixture.

ALSO, enantiomerically pure (R)-2,3-dimethyl-3-pentanol reacts with HBr to produce both (S)-3-bromo-2,3-dimethylpentane and (R)-3-bromo-2,3-dimethylpentane in equal proportions as a racemic mixture.

When this kind of phenomenon happens, there must be an achiral intermediate that can react to produce either enantiomer with equal propensity. In this case, what is the structure (geometry and everything) of the carbocation formed when the leaving group leaves?

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    $\begingroup$ To add to this fine answer, reactions of this sort will not generate racemic mixtures in general, but rather only when both faces of the prochiral intermediate are equally attractive for attack. If one side of the substrate is harder to access than the other (for instance, due to steric blocking), some degree of enantiomeric excess (E.E.) will result (i.e. one enantiomeric product being produced in greater quantities than the other) $\endgroup$
    – Richard Terrett
    Dec 3, 2012 at 2:36
  • $\begingroup$ So, basically the stereocenter is rearranged to sp2 hybridization in the carbocation, and both sides of the unhybridized p orbital on the carbon are equally electrophilic. $\endgroup$
    – Leonardo
    Dec 3, 2012 at 4:31
  • $\begingroup$ @Leonardo, yes, that this the explanation generally given. $\endgroup$
    – Ben Norris
    Dec 3, 2012 at 11:04

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