= Solution
\Image[/past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-3-q39-sidechain-assignments.png]
{title=Proton letters and <carbon> numbers assigned to the two side chains}
On the ester-containing side chain, read outward from the <cyclohexene>:
$$
\mathrm{ring{-}\underset{11;\ F,F}{CH_2}{-}O{-}
\underset{1}{C}(=O){-}
\underset{9;\ D}{CH}{=}
\underset{2;\ A}{CH}{-}
\underset{3\ (ipso)}{Ph}}.
$$
A is the downfield beta-vinylic proton and D the alpha-vinylic proton next to the <ester>.
The phenyl ortho pair is \b[B on <carbon> 7]; the meta pair is \b[C on <carbon> 6]; its para proton is the remaining \b[C on <carbon> 5]. Thus the C envelope integrates to three protons, while B integrates to two. <Carbon> 3 is the unprotonated ipso site.
In the other substituent, \b[10] is the quaternary C-OH <carbon>. The methyl assignments are \b[N on 13] and \b[O on 16], reading the very closely spaced <HSQC> cross peaks together with the positive methyl responses in <DEPT-135>. Their mutual long-range correlations and correlations to 10 support the gem-dimethyl group. The two methyl carbons are distinguishable in this chiral molecular environment; their exact stereotopic names are not determined here.
Back to article page
