Chemistry 9701 · AS & A Level · States of matter

States of matter — practice question

When answering this question, assume that all the hydrogen atoms in a benzene ring together give one unsplit peak at $\delta = 7.2$ in a proton ($^1\text{H}$) NMR spectrum. The structures of five ketone isomers, P, Q, R, S and T, are shown.
(a)[1]

Locate every chiral carbon atom in the structures above. Put an asterisk (*) beside each chiral carbon atom.

(b(i))[2]

Fig. 8.1 shows the proton ($^1\text{H}$) NMR spectrum of one of the five isomers, P, Q, R, S or T. Identify which compound produces this spectrum. Draw the displayed formula of the identified compound. On the structure you have drawn, identify the protons that give the peaks at $\delta = 3.7$, $\delta = 2.5$ and $\delta = 1.0$.

(b(ii))[1]

Name the splitting pattern of the peak at $\delta = 3.7$. Explain why it has that pattern.

(c(i))[1]

From the letters P, Q, R, S and T, choose the two compounds that each show a doublet peak in the proton ($^1\text{H}$) NMR spectrum.

(c(ii))[1]

From the letters P, Q, R, S and T, choose the compound that has only three peaks in its proton ($^1\text{H}$) NMR spectrum.

(d)[1]

Suggest a suitable solvent to use when obtaining the spectrum shown in Fig. 8.1.

(e)[1]

The proton ($^1\text{H}$) NMR spectrum of compound T is compared with and without $\text{D}_2\text{O}$. Describe any difference between the two spectra. Explain your answer.

(f)[2]

Complete Table 8.1 to state the number of peaks in the carbon-13 NMR spectrum of each compound.

Worked solution & mark scheme

This 10-mark question has a full step-by-step worked solution and mark scheme. One marking point: Only the CH carbon of T is chiral and marked

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