Chemistry 9701 · AS & A Level · Organic synthesis

Organic synthesis — practice question

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

Suggest reagents and conditions needed for this reaction.

(b(i))

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

(b(ii))

Draw the skeletal structure of a stereoisomer of piperic acid that is different from the one shown above.

(b(iii))[4]

Suggest structures for the compounds formed when piperic acid is treated with an excess of hot concentrated acidified $\text{KMnO}_4$.

(c(i))

Write down the expression for $K_w$.

(c(ii))

Use your expression together with the value of $K_w$ in the Data Booklet to calculate the pH of $0.150\ \text{mol dm}^{-3}\ \text{NaOH(aq)}$.

(c(iii))

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

(c(iv))[5]

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

(d(i))

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

(d(ii))

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

(d(iii))

On the axes below, sketch how the pH changes during the addition of a total of $20.0\ \text{cm}^3$ of $0.100\ \text{mol dm}^{-3}$ HCl. Clearly indicate the end point.

(d(iv))[6]

From the list of indicators below, place a tick in the box beside the indicator that is most suitable for this titration.

Worked solution & mark scheme

This 16-mark question has a full step-by-step worked solution and mark scheme. One marking point: Heat with dilute $\text{HCl(aq)}$ or $\text{H}_2\text{SO}_4\text{(aq)}$

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