Chemistry 9701 · AS & A Level · Analytical techniques

Analytical techniques — practice question

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

Explain why $\text{CDCl}_3$ is preferred as a solvent instead of $\text{CHCl}_3$.

(b)[1]

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

(c)[1]

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

(d(i))[2]

Describe and explain the splitting patterns for the peaks at $\delta = 1.4$ and $\delta = 4.3$. splitting pattern observed at $\delta = 1.4$ reason for the splitting pattern at $\delta = 1.4$ splitting pattern observed at $\delta = 4.3$ reason for the splitting pattern at $\delta = 4.3$

(d(ii))[1]

In each molecule of compound $E$, five protons give 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]

The mass spectrum of compound $E$ contains fragment ions with $m/e$ values of $29$ and $77$. Give the formulae of these fragment ions. fragment ion with $m/e = 29$ fragment ion with $m/e = 77$

Worked solution & mark scheme

This 9-mark question has a full step-by-step worked solution and mark scheme. One marking point: CDCl$_3$ gives no signal

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