AS Chemistry: Infrared and Mass Spectrometry — Revision Notes

Condensed for the final weeks. For the full explanation, use the Infrared and Mass Spectrometry study guide.

Infrared spectroscopy

Bonds vibrate — stretching and bending — and absorb infrared radiation at frequencies matching their natural vibration. The frequency depends on bond strength and the masses of the atoms, so each bond type absorbs in a characteristic region.

Key absorptions:

These exact ranges are printed in the Data section supplied in the exam (syllabus 22.1.1 directs you to it), so precise memorisation isn’t required — but knowing the shape of the table saves time:

Bond	Wavenumber (cm⁻¹)	Appearance
C–O	1040–1300	—
C=C	1500–1680	Weak
C=O (amide)	1640–1690	Strong
C=O (carbonyl/carboxyl)	1670–1740	Strong and sharp
C=O (ester)	1710–1750	Strong
C≡N	2200–2250	Sharp
C–H (alkane)	2850–2950	—
O–H (carboxylic acid)	2500–3000	Very broad
N–H (amine)	3300–3500	Medium
O–H (hydroxy/alcohol)	3200–3600	Broad

The two peaks that identify most spectra are C=O and O–H, and their combination distinguishes the functional group:

• C=O present, O–H absent → aldehyde, ketone or ester.
• C=O present, very broad O–H at 2500–3000 → carboxylic acid.
• O–H broad at 3200–3600, no C=O → alcohol.

That decision tree answers the standard “identify the compound” question.

Background (beyond the 9701 syllabus, not examinable at AS): distinguishing a primary from a secondary amine by counting N–H peaks, and using the fingerprint region (below 1500 cm⁻¹) for identification by database comparison, are both useful analytical chemistry context but go beyond what 19.1 and 22.1 require — 19.1 explicitly states that classifying amines as primary/secondary/tertiary is not tested at AS, and the syllabus only requires recognising the listed characteristic absorptions, not fingerprint-region matching. Don’t confuse a nitrile’s sharp C≡N absorption at 2200–2250 with a carbonyl’s C=O — the wavenumbers are very different even though both are described as “sharp.”

Application: IR is used in breathalysers and in monitoring exhaust gases, because CO and CO₂ absorb characteristically.

Relative atomic mass from isotope abundances

A mass spectrum of an element’s own isotopes can be used to calculate its relative atomic mass, Ar, as a weighted mean:

Ar = Σ(isotopic mass × % abundance) ÷ 100

Worked example. Chlorine’s spectrum shows ³⁵Cl at 75% abundance and ³⁷Cl at 25% abundance. Ar = (35 × 75 + 37 × 25) ÷ 100 = (2625 + 925) ÷ 100 = 35.5. Note this 75 : 25 (i.e. 3 : 1) abundance split is exactly why chlorine’s M and M+2 peaks appear in a 3 : 1 ratio in a compound’s spectrum.

Mass spectrometry

Stages: ionisation → acceleration → deflection → detection (background only — the syllabus does not require you to know how a mass spectrometer works, only how to interpret the spectra it produces). Ions are separated by mass-to-charge ratio (m/e — equivalent to the more common m/z notation used elsewhere).

The molecular ion peak (M⁺) is the peak at the highest m/e (ignoring isotope peaks) and gives the relative molecular mass directly.

Fragmentation produces smaller peaks. Common losses worth recognising:

Loss	Fragment
15	CH₃
17	OH
29	CHO or C₂H₅
31	CH₂OH or OCH₃
45	COOH

The base peak is the tallest peak — the most stable, and therefore most abundant, fragment.

Isotope patterns

The M+2 peak is the giveaway for halogens, and the ratio identifies which:

• Chlorine — M and M+2 in a 3 : 1 ratio, from ³⁵Cl and ³⁷Cl.
• Bromine — M and M+2 in a 1 : 1 ratio, from ⁷⁹Br and ⁸¹Br.

Seeing two peaks two units apart in roughly equal height means bromine; a 3:1 ratio means chlorine. This is examined regularly and is quick marks.

M+1 arises from ¹³C, and its size relative to M indicates the number of carbon atoms, since each carbon contributes about 1.1%. As a rough rule, the number of carbon atoms ≈ (height of M+1 as a % of M) ÷ 1.1 — so an M+1 peak at roughly 3.3% of the height of M suggests a three-carbon compound.

Exam traps

• Confusing the broad acid O–H (2500–3000) with the alcohol O–H (3200–3600).
• Interpreting individual fingerprint-region peaks instead of comparing spectra.
• Taking an isotope peak as the molecular ion.
• Reversing the chlorine and bromine M+2 ratios.
• Forgetting that IR frequency depends on bond strength and atomic masses.
• Reading the base peak as the molecular ion.

Self-test

1. What determines the frequency at which a bond absorbs infrared?
2. How do you distinguish an alcohol, a carboxylic acid and a ketone from an IR spectrum?
3. What does the molecular ion peak tell you?
4. What M+2 ratios identify chlorine and bromine?
5. What is the fingerprint region used for?
6. (Background, beyond AS) An IR spectrum shows a medium peak with two components at 3300–3500 cm⁻¹. What functional group is present?
7. A compound’s mass spectrum shows an M+1 peak at about 6.6% of the height of M. Roughly how many carbon atoms does it contain?

Answers: 1. The strength of the bond and the masses of the atoms it joins. 2. A broad O–H at 3200–3600 with no C=O indicates an alcohol; a C=O with a very broad O–H at 2500–3000 indicates a carboxylic acid; a C=O with no O–H indicates an aldehyde or ketone. 3. The relative molecular mass of the compound. 4. Chlorine gives M : M+2 of 3 : 1; bromine gives 1 : 1. 5. Identifying a compound by comparing the whole pattern with a reference database, since it is unique to each substance. 6. An amine (N–H absorption at 3300–3500 cm⁻¹). Distinguishing primary from secondary amines by counting N–H peaks is background context beyond what AS 19.1 requires — 19.1 states that classifying amines this way is not tested at AS. 7. About six carbon atoms (6.6 ÷ 1.1 = 6).