QUESTION 154 (4 marks)
The simplified mass spectrum is from one of these compounds: ethanol, ethane, chloroethane or 1-chloropropane. Its molecular-ion pair is at m/z = 64 and 66. Chlorine has isotopes with masses 35 and 37 in an approximate 3:1 abundance ratio. Use C = 12, H = 1 and O = 16. The infrared spectrum has no O–H absorption.
(a) Identify the compound and give two mass-spectral reasons for your choice. [3 marks]
(b) State whether the absence of an O–H absorption is sufficient, by itself, to identify the compound from the four candidates. Explain. [1 marks]
Practice marking scheme
Answer
The compound is chloroethane, C₂H₅Cl. Its light-isotope molecular mass is 64 and one Cl atom gives a 64/66 pair in a 3:1 ratio. The missing O–H absorption alone excludes ethanol but does not distinguish the other three candidates.
Working
For C₂H₅³⁵Cl, molecular mass is 2(12) + 5 + 35 = 64. Replacing ³⁵Cl by ³⁷Cl gives 66, with the supplied isotope ratio explaining the paired peak intensities. Ethanol has molecular mass 46, ethane 30 and 1-chloropropane 78/80. The IR observation is consistent with chloroethane but is not unique among the non-alcohol candidates.
Marking criteria
- Identifies chloroethane. [1 mark]
- Uses C₂H₅³⁵Cl molecular mass 64 to distinguish its carbon skeleton. [1 mark]
- Explains the two-unit separation and approximate 3:1 intensities using one chlorine atom. [1 mark]
- Explains that absence of O–H alone leaves three possible non-alcohol candidates. [1 mark]
Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
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