Chemistry 9701 · AS & A Level · States of matter

States of matter — practice question

Piperine is the compound that gives black pepper its hot flavour. It is an amide and can be split into piperic acid and piperidine.
(a)[1]

Suggest suitable reagents and conditions for this transformation.

(b(i))[1]

How many stereoisomers have the same structural formula as piperic acid, including piperic acid itself?

(b(ii))[2]

Draw the skeletal formula of a stereoisomer of piperic acid that is not the same as the one shown above.

(b(iii))[2]

Suggest structures for the compounds that would be produced when piperic acid is treated with excess hot concentrated acidified $\text{KMnO}_4$.

(c(i))[1]

Write down the expression for $K_w$.

(c(ii))[1]

Using your expression and the $K_w$ value from the Data Booklet, calculate the pH of $0.150\ \text{mol dm}^{-3}\ \text{NaOH(aq)}$.

(c(iii))[1]

The pH of a $0.150\ \text{mol dm}^{-3}$ solution of piperidine is 11.9. Suggest why this value is different from your answer in (c)(ii).

(c(iv))[2]

How would you expect the basicity of piperidine to compare with that of ammonia? Explain your answer.

(d(i))[1]

A $20.0\ \text{cm}^3$ volume of $0.100\ \text{mol dm}^{-3}\ \text{HCl}$ was added slowly to $10.0\ \text{cm}^3$ of $0.150\ \text{mol dm}^{-3}$ piperidine. Calculate the number of moles of $\text{HCl}$ left at the end of the addition.

(d(ii))[1]

Hence calculate $[\text{H}^+]$ and the pH at the end of the addition.

(d(iii))[2]

On the axes below, sketch the change in pH as a total of $20.0\ \text{cm}^3$ of $0.100\ \text{mol dm}^{-3}$ HCl is added. Make the end point clear.

(d(iv))[2]

From the list of indicators below, place a tick in the box next to the one most suitable for this titration.

Worked solution & mark scheme

This 17-mark question has a full step-by-step worked solution and mark scheme. One marking point: Warm in dilute HCl(aq) or H$_2$SO$_4$(aq)

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