AS Chemistry: Infrared and Mass Spectrometry — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Infrared and Mass Spectrometry revision notes

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Section A

1. State what determines the frequency at which a bond absorbs infrared radiation. [2]

2. State what the molecular ion peak indicates. [1]

3. (Background — beyond the 9701 specification, not examinable at this level.) Explain what the fingerprint region is used for.

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Section B

4. Three compounds have the formula C₃H₆O₂, C₃H₆O or C₃H₈O. Their IR spectra show:

• A: strong sharp peak at 1715 cm⁻¹, no peak above 3000 cm⁻¹
• B: strong peak at 1710 cm⁻¹ and a very broad peak from 2500–3000 cm⁻¹
• C: broad peak at 3350 cm⁻¹, no peak near 1700 cm⁻¹

(a) Identify the functional group present in each. [3]

(b) Explain how B is distinguished from C. [2]

(c) Suggest a structure for B. [1]

5. A mass spectrum shows peaks at m/e 108 and 110 (m/e — equivalent to the more common m/z notation used elsewhere) in a ratio of approximately 1 : 1, plus a fragment peak at m/e 29.

(a) Identify the halogen present, with a reason. [2]

(b) Suggest the identity of the fragment at m/e 29. [1]

(c) Deduce the relative molecular mass of the compound. [1]

6. Another spectrum shows M and M+2 peaks in a ratio of 3 : 1.

(a) Identify the halogen and explain the ratio. [3]

(b) Explain what causes an M+1 peak, and what its relative height indicates. [3]

7. A mass spectrum shows fragment peaks formed by the loss of 15 and 45 mass units from the molecular ion.

(a) Suggest the identity of each fragment lost. [2]

(b) State the functional group these two losses together suggest is present. [1]

8. An IR spectrum shows a sharp peak at about 2240 cm⁻¹ and a medium-intensity peak, with two components, at about 3350–3400 cm⁻¹.

(a) Identify the bond responsible for each absorption. [2]

(b) State the functional group each absorption is consistent with. [1]

(c) (Background — beyond the 9701 specification; AS 19.1 states that classifying amines as primary/secondary/tertiary will not be tested.) Suggest why this N–H absorption appears as two components rather than one.

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Answers

1. The strength of the bond [1] and the masses of the atoms joined by it [1].

2. The relative molecular mass of the compound [1].

3. (Background only — not AS-examinable.) It is unique to each compound, so a spectrum can be matched against a database of reference spectra to identify the substance. Individual peaks in this region are not interpreted.

4. (a) A: ketone or aldehyde, C₃H₆O (C=O only, one degree of unsaturation) [1]. B: carboxylic acid, C₃H₆O₂ (C=O plus very broad O–H) [1]. C: alcohol, C₃H₈O (broad O–H, no C=O) [1] — C must be saturated, since it shows no C=O and no evidence of a C=C or ring, so it takes the third formula rather than sharing C₃H₆O with A.

(b) B’s O–H is very broad and at lower wavenumber (2500–3000), characteristic of a carboxylic acid [1], and B also has a C=O peak, which C lacks entirely [1].

(c) CH₃CH₂COOH (propanoic acid) for B [1]. (C, the saturated alcohol, is propan-1-ol, CH₃CH₂CH₂OH, or propan-2-ol, CH₃CH(OH)CH₃ — either is consistent with the spectrum.)

5. (a) Bromine [1], because the M and M+2 peaks are of approximately equal height, reflecting the near 1 : 1 natural abundance of ⁷⁹Br and ⁸¹Br [1].

(b) C₂H₅⁺ (or CHO⁺) [1].

(c) 108 [1] — the lower of the two peaks is the molecular ion containing ⁷⁹Br.

6. (a) Chlorine [1]. ³⁵Cl and ³⁷Cl occur in a natural abundance ratio of about 3 : 1 [1], so molecules containing the lighter isotope are three times as common, giving M three times the height of M+2 [1].

(b) The ¹³C isotope, present at about 1.1% natural abundance [1]. Its height relative to M indicates the number of carbon atoms in the molecule [1] — roughly, the percentage height divided by 1.1 [1].

7. (a) Loss of 15: CH₃ [1]. Loss of 45: COOH [1]. Both are standard losses worth recognising directly from the fragment mass, without working out a full structure first. (b) A carboxylic acid group [1] — losing 45 (COOH) directly from the molecular ion is a strong indicator on its own, and losing 15 (CH₃) as well narrows the structure further, consistent with a short-chain carboxylic acid such as propanoic or ethanoic acid.

8. (a) The sharp peak at ~2240 cm⁻¹ is a C≡N bond [1]; the peak at ~3350–3400 cm⁻¹, appearing as two components, is an N–H bond, distinct from the broader O–H absorptions seen elsewhere in the spectrum [1]. (b) C≡N is consistent with a nitrile [1]; N–H is consistent with an amine [1]. (c) (Background only.) The two components arise from the symmetric and asymmetric stretching of the two N–H bonds in a primary amine, –NH₂ (a secondary amine has only one N–H bond and shows a single peak) — useful analytical context, but distinguishing primary from secondary amines is not tested at AS.

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Where marks are usually lost

• Confusing the broad acid O–H (2500–3000) with the alcohol O–H (3200–3600).
• Interpreting individual fingerprint-region peaks (background knowledge only; not AS-examinable).
• Reversing the chlorine and bromine M+2 ratios.
• Taking an isotope peak as the molecular ion.
• Reading the base peak (tallest) as the molecular ion.
• Confusing a nitrile’s sharp C≡N absorption (~2220–2260) with a carbonyl’s C=O (~1680–1750) — very different wavenumbers, easy to mix up if only skimming for “a sharp peak.”
• Forgetting that a primary amine’s N–H shows as two peaks (symmetric and asymmetric stretching of the two N–H bonds), while a secondary amine shows only one — useful background, though classifying amines this way is not tested at AS.
• Guessing a fragment’s identity from its mass alone without checking it against the standard loss table (15, 17, 29, 31, 45).

Questions 7 and 8 draw on the fragmentation-loss table and the amine/nitrile absorptions from the Infrared and Mass Spectrometry revision notes, material the earlier questions on this page don’t reach, since they focus on the carbonyl/hydroxyl decision tree and halogen isotope patterns instead.