Analytical techniques

Every objective in this topic is printed under "AS Level subject content" in the 9701 syllabus and carries the tier "AS", so it is examined on Paper 1 and Paper 2 as well as at A Level. Do not skip it as A Level material; the topic number is high, but the tier is not.

It is thinly examined on Paper 1. Across the papers in the bank on this site, five multiple-choice questions come from this topic, and all five are mass spectrometry: two on using the M+1 peak to count carbons, one on the M+2 peak distinguishing bromine from chlorine, one on identifying a fragment at m/e 43, and one on calculating the abundance of the M+1 peak. That is a narrow target and this note is weighted towards it.

Syllabus points

22.1 Infrared spectroscopy

• Analyse an infrared spectrum of a simple molecule to identify functional groups (see the Data section for the functional groups required).

22.2 Mass spectrometry

• Analyse mass spectra in terms of m/e values and isotopic abundances (knowledge of the working of the mass spectrometer is not required).
• Calculate the relative atomic mass of an element given the relative abundances of its isotopes, or its mass spectrum.
• Deduce the molecular mass of an organic molecule from the molecular ion peak in a mass spectrum.
• Suggest the identity of molecules formed by simple fragmentation in a given mass spectrum.
• Deduce the number of carbon atoms, n, in a compound using the [M+1]⁺ peak and the formula n = (100 × abundance of [M+1]⁺ ion) / (1.1 × abundance of M⁺ ion).
• Deduce the presence of bromine and chlorine atoms in a compound using the [M+2]⁺ peak.

Infrared spectroscopy

Covalent bonds vibrate, by stretching and by bending, at frequencies that fall in the infrared region. A bond absorbs infrared radiation whose frequency matches its own natural vibration frequency, so an infrared spectrum records which bonds are present.

The spectrum is plotted as percentage transmittance against wavenumber, in cm⁻¹, with the wavenumber running from about 4000 on the left down to 400 on the right. Absorption therefore appears as a downward trough, not a peak, which is worth fixing in mind before you read one.

You are given the absorption ranges in the Data Booklet, so nothing here needs memorising. What is examined is reading the spectrum, and there are three things to do:

1. Look for the C=O trough. A strong, sharp absorption near 1700 cm⁻¹ means a carbonyl group: an aldehyde, ketone, carboxylic acid or ester.
2. Look for the O-H trough. A broad absorption around 3200 to 3600 cm⁻¹ means an alcohol; a very broad one from about 2500 to 3300 cm⁻¹, overlapping the C-H region, means a carboxylic acid. The breadth comes from hydrogen bonding, and it is the clearest visual signature in the whole spectrum.
3. Combine the two. Broad O-H and C=O together means a carboxylic acid. C=O with no O-H means an aldehyde, ketone or ester. O-H with no C=O means an alcohol.

Everything below about 1500 cm⁻¹ is the fingerprint region. It is too complicated to interpret bond by bond, but it is unique to a compound, so it is used to confirm identity by matching against a database spectrum.

A worked reading. A spectrum shows a broad trough from 2500 to 3300 cm⁻¹ and a strong sharp trough at 1710 cm⁻¹, and the compound has Mr 60. The two troughs mean a carboxylic acid, and Mr 60 fits ethanoic acid, CH₃COOH.

Mass spectrometry: reading the spectrum

You are not required to know how the instrument works. What you must do is read the output.

A mass spectrum plots relative abundance against m/e, the mass to charge ratio. Because almost every ion carries a single positive charge, m/e is effectively the mass of the ion.

• The molecular ion peak, M⁺, is the peak at the highest m/e apart from the small isotope peaks just beyond it. Its m/e value is the relative molecular mass of the compound.
• The peaks at lower m/e are fragment ions, produced when the molecular ion breaks up.
• The base peak is the tallest peak, set to 100 per cent, and it is simply the most stable fragment. It is not necessarily the molecular ion.

Relative atomic mass from isotopic abundances

Relative atomic mass is the weighted mean of the isotope masses:

A_r = Σ (isotope mass × abundance) / Σ abundances

Worked example. Chlorine has two isotopes, chlorine-35 at 75.0 per cent and chlorine-37 at 25.0 per cent.

A_r = (35 × 75.0 + 37 × 25.0) / 100 = 35.5

Worked example with peak heights rather than percentages. A mass spectrum of an element shows peaks at m/e 63 with height 30.9 and at m/e 65 with height 14.1.

Total height:

30.9 + 14.1 = 45

A_r = (63 × 30.9 + 65 × 14.1) / 45 = 63.63

so A_r is 63.6, which is copper. Note that the heights do not have to add to 100; dividing by their total handles that.

Fragmentation

The molecular ion is unstable and breaks apart, and each fragment that keeps the positive charge gives a peak. The useful fragments are the ones whose masses are distinctive:

Two of these carry a warning that the bank tests directly. m/e 43 has two common identities, the propyl ion C₃H₇⁺ and the acylium ion CH₃CO⁺, and so does m/e 29, the ethyl ion and the CHO⁺ of an aldehyde. A question that asks which of several compounds could give a peak at 43 is asking you to remember both.

Worked example, from a served question. Which of ethanal, propan-1-ol and propan-2-ol give a fragment at m/e 43?

• Ethanal, CH₃CHO, Mr 44. Losing the hydrogen atom from the carbonyl leaves CH₃CO⁺, mass 43. Yes.
• Propan-1-ol, CH₃CH₂CH₂OH, Mr 60. Losing OH leaves C₃H₇⁺, mass 43. Yes.
• Propan-2-ol, (CH₃)₂CHOH, Mr 60. Losing OH again leaves C₃H₇⁺, mass 43. Yes.

All three, and the answer depends entirely on knowing that 43 is not exclusively one ion.

The way to work these out is always the same: take the relative molecular mass, subtract the mass of the fragment lost, and check the remainder is a group the molecule actually contains. A loss of 15 is a methyl group, a loss of 17 is OH, a loss of 18 is water, a loss of 29 is CHO or C₂H₅, and a loss of 45 is COOH.

The M+1 peak and counting carbons

Carbon is 98.9 per cent carbon-12 and about 1.1 per cent carbon-13. In a molecule with n carbon atoms, the chance that one of them is a carbon-13 is roughly n × 1.1 per cent, and any such molecule appears one mass unit higher. That is the M+1 peak.

n = (100 × abundance of [M+1]⁺) / (1.1 × abundance of M⁺)

Worked example, from a served question. The M and M+1 peaks of a compound are in the ratio 13 : 1. How many carbon atoms does it have?

n = 100 × 1 / (1.1 × 13) = 6.99

so 7 carbons. Of the options offered in that question, only 3,3-dimethylpentan-1-ol has seven: a pentan skeleton of five carbons plus two methyl groups.

Worked example, the calculation run the other way. Vitamin C is C₆H₈O₆, and its molecular ion peak has a relative abundance of 7.0 per cent. What is the abundance of the M+1 peak?

Rearranging, the M+1 abundance is the M abundance multiplied by 1.1n/100. With n = 6:

6 × 1.1 / 100 = 0.066

7.0 × 0.066 = 0.462

so 0.462 per cent. Both directions of this calculation appear in the bank, so practise the rearrangement rather than memorising one form.

Round n sensibly. The answer must be a whole number, and a value of 6.99 or 7.04 both mean seven carbons.

The M+2 peak: chlorine and bromine

Two elements have a second isotope two mass units heavier and abundant enough to produce a large M+2 peak, and their ratios are different enough to tell them apart at a glance:

So:

• M+2 about one third the height of M: one chlorine atom.
• M+2 equal in height to M: one bromine atom.

Worked example, from a served question. A compound has M and M+2 peaks of equal height, and M is fifteen times the height of M+1. Identify it from a list.

Equal M and M+2 means bromine, which eliminates every chlorine-containing option immediately.

Then use the M+1 ratio for the carbon count:

n = 100 × 1 / (1.1 × 15) = 6.06

so 6 carbons, and the answer is 3-bromo-2,2-dimethylbutane, which is C₆H₁₃Br. Note how the two ratios do different jobs: M+2 identifies the halogen, M+1 counts the carbons. Questions of this shape give you both on purpose.

With two chlorine atoms the pattern becomes M : M+2 : M+4 in the ratio 9 : 6 : 1, and with two bromines it is 1 : 2 : 1. Neither is required, but recognising a three-peak cluster as two halogen atoms is useful.

Common mistakes

• Treating this topic as A Level material because of its position in the syllabus. Every objective in it is tiered AS.
• Reading an infrared spectrum as though absorption gave an upward peak. Transmittance falls, so absorption is a trough.
• Confusing the broad O-H of an alcohol with the much broader O-H of a carboxylic acid, or missing that a carboxylic acid shows both O-H and C=O.
• Trying to assign individual bonds in the fingerprint region.
• Taking the tallest peak in a mass spectrum as the molecular ion. The tallest peak is the base peak; the molecular ion is the one at the highest m/e apart from isotope peaks.
• Assuming m/e 43 must be C₃H₇⁺, or that m/e 29 must be C₂H₅⁺. Each has two common identities.
• Inverting the M+1 formula, so that the M abundance ends up on top.
• Forgetting the factor of 1.1 altogether, which gives a carbon count about a hundred times too large.
• Reading an M+2 peak of equal height as chlorine. Equal means bromine; one third means chlorine.
• Using percentages in a relative atomic mass calculation when the data are peak heights that do not sum to 100, without dividing by their actual total.