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(C) The reaction of a tertiary alcohol with $HBr$ proceeds via an $S_N1$ mechanism.$1$. The hydroxyl group is protonated to form a good leaving group

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Racemization (loss of configuration) occurs at the carbon undergoing substitution

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(C) The reaction of a tertiary alcohol with $HBr$ proceeds via an $S_N1$ mechanism.$1$. The hydroxyl group is protonated to form a good leaving group $(-OH_2^+)$.$2$. The leaving group departs to form a stable carbocation intermediate $(Ph-C^+(CH_3)(CD_3))$.$3$. This carbocation is planar (sp$^2$ hybridized).$4$. The nucleophile $(Br^-)$ can attack the planar carbocation from either side with equal probability,leading to the formation of both enantiomers.$5$. Therefore,racemization occurs at the chiral center.

Which describes the best stereochemical aspects of the following reaction?
$Ph-C(CH_3)(CD_3)(OH) + HBr \rightarrow \text{Product}$

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Which describes the best stereochemical aspects of the following reaction?$Ph-C(CH_3)(CD_3)(OH) + HBr \rightarrow \text{Product}$