Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 1 / a

A is 3-Chlorobutan-2-one, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}(=\mathrm{O})\mathrm{CH}(\mathrm{Cl})\mathrm{C}{\mathrm{H}}_3$. Counting chlorine as one hydrogen in the index of index of hydrogen deficiency gives $(2\times 4+2-7-1)/2=1$. A single carbonyl accounts for that unsaturation; the resonance at $203.2$ ppm identifies a ketone, while the $59.0$ ppm methine bears chlorine.

The carbon-13 NMR assignments are ${\delta }_C=203.2$ for $\mathrm{C}=\mathrm{O}$, $25.7$ for $\mathrm{C}{\mathrm{H}}_3\mathrm{CO}$, $59.0$ for $\mathrm{CHCl}$ and $20.1$ ppm for $\mathrm{CH}(\mathrm{Cl})\mathrm{C}{\mathrm{H}}_3$. The four distinct carbon environments agree with the unsymmetrical structure. The small signal near 77 ppm belongs to the usual deuterated chloroform solvent, rather than to a fifth carbon of A.

The strong $1724\mathrm{c}{\mathrm{m}}^{-1}$ absorption is the carbonyl stretching vibration of a saturated ketone. Its exact wavenumber depends on substitution and measurement conditions; the important discrimination is that it is far below the high-frequency acyl chloride band in B and is accompanied by a carbonyl carbon-13 NMR resonance near 200 ppm.

Write the proton-labelled structure as $\mathrm{C}{\mathrm{H}}_3^{\mathrm{a}}\mathrm{C}(=\mathrm{O})\mathrm{C}{\mathrm{H}}^{\mathrm{b}}(\mathrm{Cl})\mathrm{C}{\mathrm{H}}_3^{\mathrm{c}}$. The assignments areThe NMR quartet is centred at $(4.371+4.302)/2=4.3365$ ppm; each adjacent line differs by $0.023$ ppm. The NMR doublet has the same spacing, $1.618-1.595=0.023$ ppm. At 300 MHz,Three equivalent methyl protons split $b$ into a $1:3:3:1$ NMR quartet; the single methine proton splits $c$ into a $1:1$ NMR doublet. No resolved long-range scalar coupling splits the acetyl methyl, giving its NMR singlet. The integrals total seven protons.