Chemistry 9701 · AS & A Level · Analytical techniques

Analytical techniques — practice question

The proton NMR spectrum for compound E, recorded in the solvent $\text{CDCl}_3$, is shown. Compound E has the molecular formula $\text{C}_9\text{H}_{10}\text{O}_2$.
(a)[1]

Explain why $\text{CDCl}_3$ is used as a solvent rather than $\text{CHCl}_3$.

(b)[1]

Explain why TMS is added so that the small peak appears at chemical shift $\delta = 0$.

(c)[1]

Compound E is hydrolysed by hot $\text{NaOH(aq)}$, producing only two organic products. One of these is ethanol. Name the functional group in compound E that is hydrolysed by hot $\text{NaOH(aq)}$.

(d(i))[2]

Describe the splitting patterns of the peaks at $\delta = 1.4$ and $\delta = 4.3$, and explain them.

(d(ii))[1]

Each molecule of compound E contains five protons that produce the peaks between $\delta = 7.0$ and $\delta = 8.5$. Identify the functional group in compound E that contains these protons.

(d(iii))[1]

Give the structural formula for compound E.

(e)[2]

In the mass spectrum of compound E, fragment ions appear at $m/e$ values of 29 and 77. Give the formulae of these fragment ions.

(c(iii))[1]

Give the structural formula for compound E.

Worked solution & mark scheme

This 10-mark question has a full step-by-step worked solution and mark scheme. One marking point: $CDCl_3$ produces no peak, unlike $CHCl_3$

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