ANSWERS

Five multiple choice questions with worked out answers an full explanations based on the infrared spectrum of ethyl ethanoate

These questions are an experiment of doc brown using AI to generate practice exam questions based on typical specifications of UK A-level chemistry exam boards - I have checked and re-edited the questions where necessary, if you think there is any error PLEASE email me at chem55555@hotmail.com asap.

I don't mind if students/teachers do a selected printout of these questions and answers.

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1. Ethyl ethanoate shows a very strong absorption around 1740 cm⁻¹. Which bond vibration is responsible?

A. C=C stretching in an alkene

B. O–H stretching in an alcohol

C. C=O stretching in an ester

D. C–H bending in a methyl group

Correct answer: C — C=O stretching in an ester Explanation: Esters have a characteristic carbonyl stretch slightly higher than aldehydes and ketones, typically 1735–1750 cm⁻¹. Ethyl ethanoate fits this pattern exactly. A-level examiners love this because it tests whether students know that different carbonyl-containing functional groups absorb at slightly different frequencies.

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2. Ethyl ethanoate does not show a broad absorption between 3200–3600 cm⁻¹. What does this absence tell you?

A. The molecule contains no C–H bonds

B. The molecule contains no O–H group

C. The molecule contains no C=O group

D. The molecule contains no C–O bonds

Correct answer: B — The molecule contains no O–H group Explanation: A broad, rounded peak in the 3200–3600 cm⁻¹ region is the classic signature of an O–H stretch (alcohols and carboxylic acids). Ethyl ethanoate is an ester, so it has no O–H. This is a common exam trick: students must recognise that absence of evidence is evidence of absence in IR interpretation.

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3. A strong absorption near 1050–1300 cm⁻¹ is present in the spectrum. What functional group vibration does this correspond to?

A. C–O stretching in an ester

B. C–Cl stretching in a halogenoalkane

C. N–H bending in an amine

D. C=C stretching in an aromatic ring

Correct answer: A — C–O stretching in an ester Explanation: Esters show two key IR features:

• A strong C=O stretch around 1740 cm⁻¹
• A strong C–O stretch in the 1050–1300 cm⁻¹ region This second peak helps distinguish esters from aldehydes/ketones, which lack the C–O single bond.

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4. The IR spectrum of ethyl ethanoate shows several peaks around 2850–3000 cm⁻¹. What do these represent?

A. Aromatic C–H stretching

B. Alkane C–H stretching

C. Aldehyde C–H stretching

D. Carboxylic acid O–H stretching

Correct answer: B — Alkane C–H stretching Explanation: All organic molecules containing sp³ C–H bonds show absorptions in the 2850–3000 cm⁻¹ region. Ethyl ethanoate contains two alkyl groups (ethyl + methyl), so these peaks are expected and help confirm the presence of alkyl chains.

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5. Which combination of IR absorptions would uniquely identify ethyl ethanoate as an ester rather than an aldehyde or ketone?

A. Strong peak at 1740 cm⁻¹ only

B. Broad peak at 3300 cm⁻¹ and strong peak at 1740 cm⁻¹

C. Strong peak at 1740 cm⁻¹ and strong peak at 1050–1300 cm⁻¹

D. Weak peak at 1680 cm⁻¹ only

Correct answer: C — 1740 cm⁻¹ + 1050–1300 cm⁻¹ Explanation: The carbonyl peak alone cannot distinguish esters from aldehydes or ketones. The C–O stretch in the 1050–1300 cm⁻¹ region is the crucial second feature that confirms an ester. This combination is the standard A‑level identification pattern.

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ANSWERS