📋 Structured Questions
Complete each question on paper, then check your answers against the mark scheme.
Question 1: Identification by Infrared Spectroscopy
4 marksA student oxidizes propan-1-ol. They wish to determine whether the product is propanal or propanoic acid using infrared spectroscopy.
(a) State the wavenumber range of the absorption that would be present in the infrared spectrum of propanoic acid but absent in the spectrum of propanal. Identify the bond responsible. [2]
(b) State the wavenumber range of an absorption that is present in both propanal and propanoic acid, and identify the bond responsible. [2]
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(a)
- Wavenumber range: 2500 to 3000 cm-1 [1]
- Bond: O-H bond (specifically of a carboxylic acid) [1]
(b)
- Wavenumber range: 1680 to 1750 cm-1 [1]
- Bond: C=O bond (carbonyl group) [1]
Question 2: Chemical Test Distinction
4 marksDescribe how you could use simple test tube reactions to distinguish between samples of cyclohexene, cyclohexanol, and cyclohexyl chloride. Include reagents and expected observations for each compound. [4]
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- Test for Cyclohexene: Add bromine water. Only cyclohexene will decolourise the orange-brown solution. [1]
- Test for Cyclohexanol: Add acidified potassium dichromate(VI) (K2Cr2O7/H2SO4) and warm. Only cyclohexanol (a secondary alcohol) will turn the solution from orange to green. [1]
- Test for Cyclohexyl chloride: Warm with NaOH (to hydrolyse the halogenoalkane), neutralise with nitric acid, then add silver nitrate solution. A white precipitate of silver chloride (AgCl) will form. [2] (1 mark for reagents, 1 mark for observation).
Question 3: High-Resolution Mass Spectrometry
3 marksTwo organic compounds, A and B, both have an approximate relative molecular mass (Mr) of 58.
Compound A is butane (C4H10) and Compound B is propanone (C3H6O).
The exact relative isotopic masses are: H = 1.0078, C = 12.0000, O = 15.9949.
(a) Calculate the precise molecular mass (to four decimal places) of butane and propanone. [2]
(b) Explain how a student could use these values and a high-resolution mass spectrometer to identify a sample containing one of these compounds. [1]
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(a)
- Butane (C4H10): (4 x 12.0000) + (10 x 1.0078) = 58.0780 [1]
- Propanone (C3H6O): (3 x 12.0000) + (6 x 1.0078) + 15.9949 = 58.0417 (or 58.0418 depending on rounding) [1]
(b)
- The mass spectrometer measures the m/z value of the molecular ion peak (M+) to four decimal places. The student compares the measured peak value directly with the calculated exact masses to identify the compound. [1]
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