= Solution
\b[A is <3-Chlorobutan-2-one>, $\mathrm{CH_3C(=O)CH(Cl)CH_3}$.] Counting <chlorine> as one <hydrogen> in the index of <index of hydrogen deficiency> gives $(2\times4+2-7-1)/2=1$. 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 $\mathrm{C=O}$, $25.7$ for $\mathrm{CH_3CO}$, $59.0$ for $\mathrm{CHCl}$ and $20.1$ ppm for $\mathrm{CH(Cl)CH_3}$. 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.
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