 Solution

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.