R3.4 Exam Practice

A mass spectrum shows peaks at m/z: 15, 29, 43, 58 (base peak at 43).

(a) State what the molecular ion peak represents. [1]

(b) Deduce the molecular formula from M⁺ = 58. [1]

(c) Identify the fragment responsible for m/z = 43 and m/z = 15. [2]

(d) Suggest a structure consistent with this spectrum. [1]

Show Mark Scheme

(a) The intact molecule with one electron removed (M⁺) giving the relative molecular mass [1]

(b) C₃H₆O (Mr = 58) [1]

(c) m/z 43 = CH₃CO⁺ (loss of CH₃, 58−15) [1]; m/z 15 = CH₃⁺ [1]

(d) Propanone (CH₃COCH₃) [1]

An unknown compound C₃H₆O₂ has an IR absorption at 1715 cm⁻¹ and a broad peak at 2500–3300 cm⁻¹. ¹H NMR shows two peaks: δ 1.1 (3H, triplet) and δ 2.4 (2H, quartet).

(a) Identify the functional group from the IR data. [1]

(b) Explain what the number of peaks in ¹H NMR tells you. [1]

(c) Deduce the structure of the compound. [1]

(d) Explain the splitting pattern (triplet and quartet). [2]

Show Mark Scheme

(a) Carboxylic acid: C=O stretch at 1715 + broad O–H at 2500–3300 [1]

(b) Two peaks = two different hydrogen environments [1]

(c) Propanoic acid (CH₃CH₂COOH) [1]

(d) Triplet: CH₃ has 2 neighbouring H atoms (n+1 = 3) [1]; quartet: CH₂ has 3 neighbouring H atoms (n+1 = 4) [1]