Chemistry 9701 · AS & A Level · Carboxylic acids and derivatives

Carboxylic acids and derivatives — practice question

Ethanedioic acid, $\text{C}_2\text{O}_4\text{H}_2$, is present in many vegetables. Its amount in spinach may be found as follows: 40.0 g of spinach leaves are crushed and blended with distilled water; the mixture is filtered and made up to $100.0\,\text{cm}^3$ with water. A $25.0\,\text{cm}^3$ aliquot is acidified and titrated with $0.0200\,\text{mol dm}^{-3}\,\text{KMnO}_4$. The equation is $2\text{MnO}_4^- + 6\text{H}^+ + 5\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}$. During the titration, $15.20\,\text{cm}^3$ of $\text{KMnO}_4$ were needed.
(a)[3]

Calculate the percentage by mass of ethanedioic acid in the spinach leaves.

(b(i))[1]

State a suitable reagent for changing ethanedioic acid into ethanedioyl chloride: $\text{HO}_2\text{C}-\text{CO}_2\text{H} \rightarrow \text{ClOC-COCl}$.

(b(ii))[2]

For the reactions of ethanedioyl chloride below, suggest the structures of compounds J and K, and draw them in the boxes.

(c(i))[2]

Use the Data Booklet to give in the boxes below the $\delta$ values for the peaks in the spectrum that come from the carbon atoms in ethoxyethane, $\text{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_3$.

(c(ii))[1]

Explain what the remaining carbon-13 NMR spectrum shows about the structure of L.

(b)[3]

Ethanedioic acid may be converted into ethanedioyl chloride: $\text{HO}_2\text{C-CO}_2\text{H} \rightarrow \text{ClOC-COCl}$ (i) State a suitable reagent for this reaction. (ii) For the reactions of ethanedioyl chloride below, suggest the structures of compounds J and K and draw them in the boxes. Ethanedioyl chloride reacts with $\text{CH}_3\text{OH}$ to give J ($\text{C}_4\text{H}_6\text{O}_4$). Ethanedioyl chloride reacts with $\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2$ to give K ($\text{C}_4\text{H}_6\text{N}_2\text{O}_2$).

(c)[3]

When ethanedioyl chloride is treated with silver ethanedioate, $\text{AgO}_2\text{C-CO}_2\text{Ag}$, in ethoxyethane at $-30^{\circ}\text{C}$, an oxide of carbon, L, is produced. L has no overall dipole and molecular formula $\text{C}_4\text{O}_6$. The carbon-13 NMR spectrum of a solution of L in ethoxyethane, $\text{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_3$, is shown. (i) Use the Data Booklet to give in the boxes below the $\delta$ values for the peaks in the spectrum that arise from the carbon atoms in ethoxyethane. (ii) Explain what the rest of the carbon-13 NMR spectrum tells you about the structure of L.

(d)[4]

When pure L is reacted with an excess of $\text{CH}_3\text{OH}$, a mixture of three compounds is produced. $\text{L } (\text{C}_4\text{O}_6) + \text{CH}_3\text{OH} \rightarrow \text{M } (\text{C}_2\text{H}_2\text{O}_4) + \text{N } (\text{C}_3\text{H}_4\text{O}_4) + \text{O } (\text{C}_4\text{H}_6\text{O}_4)$ M is one of the products formed when either N or O is heated with aqueous acid. The table gives details of the peaks recorded in the carbon-13 NMR spectra of M, N and O. (i) Suggest the structures of M, N and O. (ii) Suggest a structure for L that is consistent with all the data given in (c) and (d).

Worked solution & mark scheme

This 19-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Accurate working for the amount of $\text{MnO}_4^-$ and $\text{C}_2\text{O}_4^{2-}$” …

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