Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 3 / c

The alcohol is oxidized to cyclohexanone through an alkyl hypochlorite intermediate. Acid converts hypochlorite into electrophilic hypochlorous acid. The alcohol oxygen attacks chlorine, with acid-assisted proton transfers and loss of water giving the O–Cl derivative ${\mathrm{R}}_2\mathrm{CH}-\mathrm{O}-\mathrm{Cl}$.

A base such as acetate then removes the hydrogen attached to the carbinol carbon. The three coupled electron-pair motions are: base → H; C–H → C–O to form C=O; and O–Cl → Cl to release chloride. The resulting carbonyl is cyclohexanone. This is an elimination-like oxidation of the hypochlorite ester: the original alcohol oxygen becomes the ketone oxygen, while loss of its O-bound H and the carbinol C-bound H effects the net oxidation.

Methyl addition gives 1-methylcyclohexanol, whose acid-catalysed dehydration yields the shown 1-methylcyclohexene. The methyl end of methyllithium attacks the cyclohexanone carbonyl carbon, and the C=O pi electrons move onto oxygen, giving a lithium alkoxide. Protonation gives the tertiary alcohol.

Protonate OH to make $\mathrm{O}{\mathrm{H}}_2^{+}$, then cleave its C–O bond with both electrons going to departing water. A base removes a beta hydrogen from either adjacent ring methylene, and that C–H pair forms the C=C bond to the tertiary cationic centre. This E1 elimination regenerates the acid catalyst. The two adjacent ring positions are equivalent, and the displayed endocyclic alkene is more substituted than the possible exocyclic methylene alternative.

D is racemic trans-2-methylcyclohexan-1-ol. E is racemic 1-methyl-2-phenylcyclohexan-1-ol with OH and Ph trans, and Me and Ph cis. Number the starting alkene carbon carrying methyl as ${\mathrm{C}}_1$ and its partner as ${\mathrm{C}}_2$ in the reaction explanation below; the product name of D renumbers the OH-bearing carbon as 1.

For D, hydroboration proceeds through a concerted four-centre transition state: the alkene pi pair donates to boron while B–H supplies the new C–H bond. Boron attaches mainly to less substituted ${\mathrm{C}}_2$ and hydrogen to ${\mathrm{C}}_1$, on the same face. Under peroxide/base conditions, hydroperoxide attacks B; migration of the C–B bond to the adjacent oxygen, with O–O cleavage, replaces C–B by C–O with retention at carbon. Hydrolysis/proton transfer releases the alcohol. Thus the newly added H and OH are syn, making the pre-existing methyl and OH trans. Attack on either alkene face gives the enantiomeric pair.

For E, the peracid transfers its terminal oxygen across the alkene in concerted epoxidation, through cyclic electron flow that also breaks O–O and transfers the acidic proton to the peracid carbonyl oxygen. The resulting epoxide retains the relative configuration of the alkene substituents. The phenyl equivalent from phenylmagnesium bromide then attacks less substituted ${\mathrm{C}}_2$ from the face opposite the epoxide oxygen. Its C–O bond breaks onto oxygen, an SN2-like ring opening; the oxygen stays attached to ${\mathrm{C}}_1$. Water workup protonates that alkoxide. Consequently OH at ${\mathrm{C}}_1$ and Ph at ${\mathrm{C}}_2$ are anti, while the methyl at ${\mathrm{C}}_1$ lies on the same relative face as Ph.