Identify each compound below from its molecular formula and its ${}^{\mathbf{13}}\mathbf{C}$NMR spectrum.

(a) ${\mathbf{C}}_{\mathbf{11}}{\mathbf{H}}_{\mathbf{22}}\mathbf{O}$

(b) ${\mathbf{C}}_{\mathbf{8}}{\mathbf{H}}_{\mathbf{9}}\mathbf{Br}$

(c) ${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{10}}\mathbf{O}$

(d) ${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{12}}$

There are five signals but eleven carbon atoms, which means that others are similar, i.e., thecompound may be symmetrical. The signal at $210\mathrm{ppm}$represents a carbonyl carbon of a ketone group. Thus, it can be concluded that the compound is a symmetrical ketone with similar five-carbon alkyl groups. So the structure of the compound is given below.

Compound (a)

There are only four signals to denote the benzene ring, which is between $110\mathrm{ppm}$and role="math" localid="1652277756125" $117\mathrm{ppm}.$.This suggests that only four carbon atoms are similar. Others are substituted ones. Moreover, these substitutions are different. So, the structure of the compound is given below.

Compound (b)

Since there is a signal at$212\mathrm{ppm}$ , it indicates a carbonyl carbon of a ketone group. There are three signals present in the spectrum. There are five carbon atoms other than carbonyl carbon. So, it represents a symmetrical cyclic ketone. The structure of the compound is given below.

Compound (c)

The molecular formula indicates the formula of an alkene. It may be a double bond or ring. There are ${\mathrm{sp}}^2$hybridized carbon atoms which are indicated by the peak at $130\mathrm{ppm}$. So it is confirmed that the compound has a double bond. There are six carbon atoms, and they show only three signals. Cis or trans cannot be confirmed through${}^{13}\mathrm{C}$ NMR spectrum but can be confirmed by ${}^1\mathrm{H}$NMR spectrum using coupling constants.

So the structure of the compound is given below.

Compound (d)