Edexcel International A Level Chemistry: Mass spectra and infrared spectroscopy (YCH11)

This guide teaches Topic 10D, Mass spectra and IR, of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, outcomes 10.21, 10.22 and 10.23 (Core Practical 8). This part of Topic 10 is assessed in the IAS written examination for Unit 2: Energetics, Group Chemistry, Halogenoalkanes and Alcohols. Topic 15E in Unit 4 extends it; that later content is not needed here.

Afterwards, condense it with the Topic 10D revision notes and check yourself on the spectra practice set. For the whole course, see the Edexcel IAL Chemistry course page and the outcome checklist; the free diagnostics show your weak spots.

Outcomes in Topic 10D

Outcome	You should be able to
10.21	Use the m/z of the molecular ion and fragmentation patterns in mass spectra to suggest possible structures of simple organic compounds
10.22	Use infrared spectra, or IR data, to deduce functional groups, and predict IR absorptions from wavenumber data for (i) C–H in alkanes, alkenes and aldehydes, (ii) C=C in alkenes, (iii) O–H in alcohols and carboxylic acids, (iv) C=O in aldehydes, ketones and carboxylic acids, (v) C–X in halogenoalkanes, (vi) N–H in amines
10.23	Core Practical 8: analysis of some inorganic and organic unknowns

You need the functional groups from Topics 4, 5 and 10. If naming is shaky, revisit Topic 4 on naming and alkanes plus Topic 5 on alkenes. The mass spectrometer itself is Topic 2 (outcome 2.6); see these notes on the instrument and isotopes.

Mass spectra of organic compounds (10.21)

The molecular ion

In the spectrometer, a molecule M loses one electron to form a positive ion:

M(g) → M⁺(g) + e⁻

This is the molecular ion. The electron’s mass is negligible and the charge is 1+, so its m/z equals Mr. The molecular ion peak is at the highest m/z, apart from a very small M+1 peak caused by carbon-13. Do not mistake that small peak for the molecular ion.

Fragmentation

Molecular ions have excess energy and many of them break apart. A bond breaks to give one positive ion and one uncharged radical:

M⁺ → X⁺ + Y•

Only the charged piece, X⁺, is detected. The uncharged radical is not accelerated or deflected, so you work out what was lost from the difference in m/z.

Two terms to keep apart:

• The molecular ion peak gives Mr.
• The base peak is the tallest peak (100% relative abundance), often a fairly stable fragment rather than M⁺.

Common fragment ions and losses

Use integer masses: H = 1, C = 12, N = 14, O = 16.

m/z	Likely ion
15	CH₃⁺
29	C₂H₅⁺ or CHO⁺
31	CH₂OH⁺
43	C₃H₇⁺ or CH₃CO⁺
45	COOH⁺
57	C₄H₉⁺ or C₂H₅CO⁺

Common differences between M⁺ and a fragment peak:

Mass lost	Likely radical lost
15	CH₃•
17	OH•
29	C₂H₅• or CHO•
43	C₃H₇• or CH₃CO•
45	COOH•

A peak at 31 suggests a CH₂OH group, as in a primary alcohol. For 43 and 29, use the IR spectrum to decide.

Isotope patterns from chlorine and bromine

This builds on outcomes 2.6 and 2.7. Chlorine is about 3 : 1 chlorine-35 to chlorine-37; bromine is close to 1 : 1 bromine-79 to bromine-81. One halogen atom gives two molecular ion peaks two units apart:

• heights about 3 : 1 → one Cl atom
• heights about 1 : 1 → one Br atom

Fragments that keep the halogen also appear as pairs.

Worked example 1: two isomers of C₅H₁₂

A hydrocarbon gives peaks at m/z 72, 57 and 15 only (ignore the tiny M+1 peak). Suggest whether it is pentane or 2,2-dimethylpropane.

Step 1. M⁺ at 72. C₅H₁₂ = 5(12) + 12(1) = 72, so both isomers fit the molecular ion.

Step 2. Identify the fragments. 72 − 57 = 15, so m/z 57 is C₄H₉⁺ formed by loss of CH₃•. m/z 15 is CH₃⁺.

Step 3. Consider each structure. Pentane, CH₃CH₂CH₂CH₂CH₃, can break between C2 and C3 to give C₂H₅⁺ (29) and C₃H₇⁺ (43). 2,2-dimethylpropane, C(CH₃)₄, has four identical C–C bonds, each joining the central carbon to a methyl group. Breaking any of them can only give C(CH₃)₃⁺ (57) or CH₃⁺ (15).

Step 4. No peaks at 29 or 43, so the compound is 2,2-dimethylpropane.

Worked example 2: which bromopropane?

A compound gives two peaks of equal height at m/z 122 and 124, a strong peak at 43, and a pair of equal peaks at 107 and 109. There are no peaks at 93 or 95. Identify the compound.

Step 1. Two molecular ion peaks two units apart and equal in height show one bromine atom. C₃H₇⁷⁹Br = 36 + 7 + 79 = 122, and with ⁸¹Br it is 124. The formula is C₃H₇Br.

Step 2. m/z 43 is a single peak, so it has lost the bromine: 122 − 79 = 43, which is C₃H₇⁺.

Step 3. The pair at 107 and 109 still contains bromine. 122 − 107 = 15, so this is loss of CH₃•, giving C₂H₄Br⁺.

Step 4. Compare the isomers. 2-bromopropane, CH₃CHBrCH₃, loses a methyl group to give CH₃CHBr⁺ (107/109). 1-bromopropane, CH₃CH₂CH₂Br, would be expected to give CH₂Br⁺ at 93/95 by breaking the C1–C2 bond. The data fit 2-bromopropane.

Infrared spectroscopy (10.22)

What the spectrum shows

Covalent bonds vibrate. A bond absorbs infrared radiation whose frequency matches its stretching frequency, so different bonds absorb at different frequencies. Polar bonds absorb strongly.

An IR spectrum plots transmittance (%) against wavenumber (cm⁻¹), the reciprocal of wavelength. Absorptions appear as dips, with the scale usually running from about 4000 cm⁻¹ down to about 500 cm⁻¹. Below about 1500 cm⁻¹ the many overlapping absorptions form a fingerprint that can be matched against a reference spectrum.

Wavenumber data used on this page

Always use the ranges a question supplies. These typical values are used on this page and its practice set:

Bond	Compound type	Range (cm⁻¹)	Notes
C–H	alkanes	2962–2853	in almost every organic molecule
C–H	alkenes (=C–H)	3095–3010	just above 3000
C–H	aldehydes (CHO)	2900–2820 and 2775–2700	the lower band is the giveaway
C=C	alkenes	1669–1645	often weak to medium
O–H	alcohols	3550–3230	broad, due to hydrogen bonding
O–H	carboxylic acids	3300–2500	very broad, overlaps the C–H bands
C=O	aldehydes	1740–1720	strong
C=O	ketones	1720–1700	strong
C=O	carboxylic acids	1725–1700	strong
C–Cl	chloroalkanes	800–600
C–Br	bromoalkanes	600–500
C–I	iodoalkanes	about 500
N–H	amines	3500–3300	sharper than alcohol O–H

How to read a spectrum

1. Look at 1750–1680 cm⁻¹. A strong peak means C=O.
2. Above 2500 cm⁻¹: broad 3550–3230 means alcohol O–H; very broad 3300–2500 means acid O–H; sharper 3500–3300 suggests N–H.
3. C=O but no O–H: a peak at 2775–2700 cm⁻¹ means aldehyde; none means ketone.
4. Check 1669–1645 and 3095–3010 cm⁻¹ for C=C and =C–H.
5. Check the low-wavenumber end for C–X.
6. State what is absent too; absences rule groups out.

Worked example 3: predicting absorptions

Predict the absorptions you would expect in the IR spectrum of 3-bromoprop-1-ene, CH₂=CHCH₂Br.

• =C–H stretch at 3095–3010 cm⁻¹ (alkene hydrogens)
• C–H stretch at 2962–2853 cm⁻¹ (the CH₂ next to the bromine)
• C=C stretch at 1669–1645 cm⁻¹
• C–Br stretch at 600–500 cm⁻¹

There should be no broad absorption above 3200 cm⁻¹ and no strong peak at 1700–1740 cm⁻¹, because the molecule has no O–H and no C=O.

Worked example 4: following an oxidation

Butan-1-ol is oxidised with potassium dichromate(VI) in dilute sulfuric acid (outcome 10.18). Three samples are analysed. Describe the key IR features of each.

Sample	Key absorptions	Reason
Butanal (distilled off as it forms)	C=O at 1740–1720; C–H at 2775–2700; no O–H band	aldehyde
Butanoic acid (heated under reflux)	C=O at 1725–1700; very broad O–H at 3300–2500	carboxylic acid
Mixture with unreacted butan-1-ol	the product’s C=O peak plus a broad O–H at 3550–3230	some alcohol remains

Both products absorb near 1720 cm⁻¹, so the O–H and aldehyde C–H bands are what separate them.

Using both techniques together

Work in this order: functional group from IR, Mr from the molecular ion, a formula that fits both, then fragments to place the groups.

Worked example 5: an unknown carbonyl compound

Compound K contains C, H and O. Its IR spectrum has a strong peak at 1712 cm⁻¹, no broad band above 2500 cm⁻¹ and no peak at 2775–2700 cm⁻¹. Its mass spectrum shows the molecular ion at m/z 86, the base peak at 57 and a strong peak at 29. There is no peak at 43. Suggest a structure for K.

Step 1. IR: C=O present, no O–H, no aldehyde C–H. K is a ketone.

Step 2. Mr = 86. A ketone with one oxygen: CₙH₂ₙO. 14n + 16 = 86, so n = 5 and the formula is C₅H₁₀O.

Step 3. 86 − 57 = 29: loss of C₂H₅• (not CHO•, as K is a ketone), leaving C₂H₅CO⁺ at 57. The peak at 29 is C₂H₅⁺. So an ethyl group is joined to the carbonyl carbon on each side.

Step 4. Check the other ketones. Pentan-2-one, CH₃COCH₂CH₂CH₃, and 3-methylbutan-2-one, CH₃COCH(CH₃)₂, both contain CH₃CO and a C₃H₇ group, so each would be expected to give a peak at 43 (CH₃CO⁺ or C₃H₇⁺). There is none.

Step 5. A structure that fits is CH₃CH₂COCH₂CH₃, pentan-3-one. Say “suggest” and give your reasoning, because spectra support a structure rather than prove it.

Core Practical 8: analysis of unknowns (10.23)

The specification names this practical as the analysis of some inorganic and organic unknowns. Unit 3 questions may ask you to recall or interpret observations from tests for ions, gases and organic functional groups in Units 1 and 2. Spectra then help you confirm a conclusion. The tests you already know include bromine water for C=C (5.5); flame colours (8.17, 8.18); tests for carbonate, sulfate and ammonium ions (8.19); silver nitrate for halide ions (8.27); PCl₅ giving misty fumes with –OH (10.17); and, from 10.18, acidified dichromate(VI) turning orange to green with primary and secondary alcohols and aldehydes, Benedict’s or Fehling’s solution with aldehydes, and sodium hydrogencarbonate fizzing with carboxylic acids.

Link each observation to a conclusion, then check it against the spectra. An orange-to-green change shows an oxidisable compound but not which type; a broad 3550–3230 cm⁻¹ band with no C=O then confirms an alcohol.

Common errors

• Taking the base peak as the molecular ion. The molecular ion is at the highest m/z, not the tallest peak.
• Writing a fragment without its positive charge, or showing the radical as charged.
• Reading the 2500–3300 cm⁻¹ acid band as an alcohol. Alcohol O–H sits higher and is less wide.
• Saying “C=O, so it is an aldehyde” without checking 2775–2700 cm⁻¹.
• Missing the halogen pattern: two peaks two units apart in a 3 : 1 or 1 : 1 ratio.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017 (Pearson Education Limited). Mass spectra and IR is part 10D of Topic 10, outcomes 10.21 to 10.23, in Unit 2.