If I have a chiral substrate, should I always assume that both enantiomers are present? 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?
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?