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(C) The reaction conditions $(H_2O, \Delta)$ favor $S_N1$ and $E1$ mechanisms. For the starting material $X$ to yield two stereoisomeric alcohols ($A$

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$III$

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(C) The reaction conditions $(H_2O, \Delta)$ favor $S_N1$ and $E1$ mechanisms. For the starting material $X$ to yield two stereoisomeric alcohols ($A$ and $B$) that are not enantiomers (diastereomers) and two enantiomeric alkenes ($C$ and $D$),the substrate must be a chiral molecule that forms a carbocation intermediate. Structure $III$ is $1$-bromo-$1$-methyl-$4$-methylcyclohexane. Upon ionization,it forms a carbocation at the $C1$ position. The attack of $H_2O$ from the top or bottom face of the planar carbocation yields two diastereomeric alcohols ($cis$ and $trans$ isomers),which are $A$ and $B$. The elimination of a proton from the carbocation can occur in two ways to form alkenes. Due to the presence of the chiral center at $C4$,the resulting alkenes $C$ and $D$ are enantiomers. Thus,structure $III$ fits all the given criteria.

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