Identifying an ester from hydrolysis tests
A colourless liquid A has the molecular formula C4H8O2 and a pleasant smell. It does not react with sodium carbonate solution and gives no colour with bromine water in the dark. When A is heated under reflux with dilute sodium hydroxide and the mixture is distilled, two products are obtained. The first product B effervesces with sodium carbonate. The second product C gives a yellow precipitate with alkaline iodine solution. A student claims that A could equally be methyl propanoate and that the tests cannot distinguish the two esters. The laboratory has only the following reagents: ethene, water, acidified potassium dichromate(VI), concentrated sulphuric acid, sodium hydroxide, and standard laboratory glassware.
1. Deduce the structure of A and decide whether the student is correct that methyl propanoate would give the same tests. Explain each test result.
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Degree of unsaturation of C4H8O2: (2C + 2 – H)/2 = (8 + 2 – 8)/2 = 1. This is consistent with one C=O and no C=C. The absence of a reaction with bromine water supports the absence of a C=C bond. No effervescence with sodium carbonate shows that A is not a carboxylic acid; the pleasant smell and hydrolysis products indicate an ester.
Hydrolysis gives B, a carboxylic acid (effervescence with carbonate), and C, an alcohol. The iodoform test on C requires a CH3CO– group or a CH3CH(OH)– group. Ethanol, CH3CH2OH, is oxidised by alkaline iodine to ethanal, CH3CHO, and gives the yellow precipitate CHI3. Methanol cannot give this test because it has no CH3CH(OH)– unit and is not oxidised to ethanal under these conditions.
If C is ethanol, B must contain the remaining two carbons: ethanoic acid, CH3COOH. Therefore A is ethyl ethanoate, CH3COOC2H5.
If A were methyl propanoate, hydrolysis would give propanoic acid and methanol. Methanol would give no yellow precipitate with alkaline iodine, so the tests do distinguish the two esters. The student is incorrect.
2. Using only ethene, water, acidified potassium dichromate(VI), concentrated sulphuric acid, sodium hydroxide and glassware, propose a three-step route from ethene to A. Give essential conditions and explain why the final step cannot give 100% yield even with a catalyst.
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Step 1: convert ethene to ethanol by catalytic hydration. C2H4 + H2O → C2H5OH Conditions: phosphoric acid catalyst, about 300 °C and high pressure (approximately 60 atm).
Step 2: oxidise ethanol to ethanoic acid. C2H5OH + 2[O] → CH3COOH + H2O Use acidified potassium dichromate(VI) and heat under reflux. The dichromate is reduced from orange to green. Reflux ensures that the intermediate ethanal is further oxidised to the acid.
Step 3: esterify ethanol with ethanoic acid. CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O Use concentrated sulphuric acid as catalyst and heat under reflux. Distil the ester as it forms.
The final yield cannot be 100% because esterification is reversible. At equilibrium all four substances are present; the acid catalyst increases the rate of attainment of equilibrium but does not change the equilibrium constant or the position of equilibrium. Yield can be improved by removing water or distilling the ester, but some losses occur during work-up, and side reactions are possible.
3. A sample of A is contaminated with a little ethanoic acid. Describe one quick test to show the contamination and explain whether this contamination would affect the iodoform conclusion.
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Add a small amount of solid sodium carbonate or sodium hydrogencarbonate to a sample of the liquid. If ethanoic acid is present, carbon dioxide is produced and effervescence is seen. Pure A, an ester, does not react in this way.
The iodoform test is carried out after hydrolysis, when the ester has been converted into ethanol and ethanoic acid. The contaminating ethanoic acid does not itself give a yellow precipitate, so it does not create a false positive. A small amount should not prevent ethanol from giving the positive iodoform result. However, if the contamination is large, the added acid could neutralise part of the alkaline iodine reagent and make the test less reliable; more reagent would then be needed. The contamination therefore does not change the structural conclusion for a small amount, but it must be allowed for in the test.