Consider the series of reactions given and identify the final product [Z].
Show Hint
Nucleophilic addition of any Grignard reagent ($\text{R-MgX}$) to a cyclic ketone always produces a tertiary alcohol where both the hydroxyl group ($-\text{OH}$) and the new alkyl/aryl group ($\text{R}$) are attached to the exact same carbon atom ($1$-position), yielding a 1-substituted cyclohexanol.
Understanding the Concept:
This question traces a series of functional group interconversions:
Acid-catalyzed dehydration of a cyclic secondary alcohol to form an alkene.
Hydroboration-oxidation of the alkene to yield an alcohol via anti-Markovnikov addition.
Oxidation of a secondary alcohol to a ketone using chromium trioxide ($\text{CrO}_3$).
Nucleophilic addition of a Grignard reagent to a carbonyl group to form a tertiary alcohol.
Step 1: Track intermediate transformations from starting material to [Y].
Heating cyclohexanol with $20%\,\text{H}_3\text{PO}_4$ undergoes dehydration to form cyclohexene [W].
Treating cyclohexene [W] with $(\text{BH}_3)_2$ followed by alkaline $\text{H}_2\text{O}_2$ undergoes hydroboration-oxidation. For a symmetric cyclic alkene, this returns the structure to cyclohexanol [X].
Oxidizing cyclohexanol [X] with $\text{CrO}_3$ (Jones reagent conditions) converts the secondary alcohol into a ketone, yielding cyclohexanone [Y].
Step 2: React cyclohexanone [Y] with the Grignard reagent to find [Z].
Cyclohexanone [Y] contains a highly polarized carbonyl carbon. The nucleophilic phenyl group from phenylmagnesium bromide ($\text{C}_6\text{H}_5\text{MgBr}$) attacks this carbonyl carbon:
\[
\text{Cyclohexanone [Y]} + \text{C}_6\text{H}_5\text{MgBr} \rightarrow \text{Intermediate Complex} \xrightarrow{\text{H}_2\text{O}} \text{1-Phenylcyclohexanol [Z]}
\]
This addition transforms the carbonyl group into a tertiary alcohol bearing an attached phenyl group, yielding 1-phenylcyclohexanol [Z].
