Why is the product of HBr with (R)-3-methyl-3-hexanol going to be 3-brom-3-methylhexane?

A racemic mixture (#50":"50#, #"R:S"#) suggests attack on either face of a planar carbocation intermediate, which indicates an #S_N1# reaction.

(R)-3-methyl-3-hexanol is a tertiary alcohol, which is bulky. Thus, at ordinary temperatures, it tends to undergo #S_N1# reactions because it can't easily be back-side attacked (too much steric hindrance).

It also follows that formation of a carbocation intermediate is easier (because more hyperconjugation stabilizes the central cationic carbon).

The loss of the water molecule (the leaving group) is the slow step in the mechanism (with NO participation of #"Br"^(-)#), whereas the nucleophile can easily attack the planar intermediate quickly.

This is why the mechanism is first order with respect to the substrate, i.e. #S_N1#.

Since the nucleophilic attack of the intermediate is fast, the product that forms is almost entirely dependent on the planarity of the intermediate.

In other words, either face could be attacked, and the umbrella flip is equally as likely to go either way (either retaining or inverting the stereochemistry), thus giving a 50/50 mixture.

This is contrary to the #S_N2# mechanism, where no carbocation intermediate forms, and only a transition state exists (if briefly).

Since no carbocation intermediate forms, only a one-directional back-side attack can occur, which umbrella-flips to only invert the stereochemistry in a second-order reaction with respect to the substrate.