A-level Chemistry/WJEC/Module 2/Analysis

There are three different techniques for structural determination on the WJEC syllabus - mass spectroscopy (MS), infrared spectroscopy (IR) and nuclear magnetic resonance spectroscopy (NMR). Each has its own purpose and all are valuable in identifying compounds. The three put together are often enough to identify a molecule.

Mass spectrometry

Much of the theory behind mass spectrometry is covered at in unit 1, but aimed at identifying atoms rather than molecules. To summarize, the sample chemical to be tested is vaporised and then ionised by an electron beam. The resultant positive ions are then accelerated by charged plates and deflected by an electromagnet of variable strength. When the magnet is adjusted correctly, the ions are detected. The signal is recorded as the electromagnet strength is varied to calculate a mass to charge ratio usually represented as "m/z". Because the charge is almost always +1, we can take m/z = M_r.

Mass spectrum of pyridine, C₅H₅N. Notice that the most common signal is labelled as "100 %" and the other signals are measured in proportion to this "100 %" value. The M_r of the molecule is 79, as shown by the largest major peak. There is a small fragment with a mass of 80, which is due to a small fraction of C₅H₆N⁺ ions forming, and the presence of the isotopes ¹⁵N and ¹³C. The fragment with a mass of 52 is 27 mass units lighter than the C₅H₅N molecule. This is probably due to some molecules losing the fragment CHN⁺ and forming C₄H₄⁺ ions.

Structure of pyridine.

The general idea behind use with molecules is the same - a mass to charge ratio is detected and peaks can appear at varying values of m/z due to the component atoms having a range of isotopes. However, the matter is complicated by fragmentation of the molecule on ionisation. Groups of atoms break off the molecule to produce many more peaks of varying mass on the resulting spectrum.

This bond fission produces a mixture of fragment ions which can be used to suggest the functional groups in a molecule. Spectra from molecular compounds are more complex than for elements, which means the spectra are more difficult to interpret, but the spectra offer more information.

The main features to watch out for are:

• The heaviest major peak (the one on the right of the spectrum) is caused by the whole molecular ion.
• In organic molecules, a small peak one value above the molecular ion caused by ¹³C and by the molecule acquiring an extra H atom during ionisation
• Major peaks at values for known functional group masses:

A peak at m/z = 15 suggests a methyl group (CH₃⁺)

A peak at m/z = 29 suggests an ethyl group (CH₃CH₂⁺)

(As a rule, adding 14 to the mass of a fragment gives the mass of a similar fragment with an extra CH₂).

A peak at m/z = 30 suggests an amine group (H₂NCH₂⁺)

A peak at m/z = 31 suggests an alcohol group (HOCH₂⁺)

Peaks at m/z = 35 and m/z = 37, in a 3:1 ratio, are chlorine atoms (³⁵Cl⁺ and ³⁷Cl⁺)

A peak at m/z = 77 suggests a phenyl group (C₆H₅⁺)

Peaks at m/z = 79 and m/z = 81, in a 1:1 ratio, are bromine atoms (⁷⁹Br⁺ and ⁸¹Br⁺)

• Similar numbers are involved if you consider the difference between the molecular ion and lighter ions in the spectrum. For example, a peak with a mass of 59, in a spectrum that also shows a molecular ion with a mass of 88, suggests that the molecules often lose a fragment with a mass of 29 i.e. probably an ethyl group.

Infrared spectroscopy

Molecules can convert energy from electromagnetic radiation into vibrations in the covalent bonds between their atoms. Different bonds vibrate at different frequencies and therefore by passing infrared radiation through a sample of a substance it can be observed which wavelengths are absorbed. When compared to known data, frequencies which are absorbed can be used to analyse the functional groups present in a compound.

The frequency of infrared radiation is conventionally measured in wavenumbers (ṽ) which have units of cm^-1 ("per centimetre"). A wavenumber is the number of waves that would fit into a given length, usually 1 cm. It would be an unusual question to ask, but to convert wavenumber to frequency or wavelength you need to do this:

Example: Wavenumber ṽ = 2000 cm^-1 = 200 000 m^-1

(If there are 2000 waves per centimetre, there will be 200 000 waves per metre)

Wavelength λ = 1 / ṽ = 1/200 000 = 5 x 10^-6 m = 5 μm

Frequency f = c x ṽ = 3.00 x 10⁸ m s^-1 x 200 000 m^-1 = 6.00 x 10¹³ Hz

Ethanol IR Spectrum. The broad signal at 3358 cm^-1 is from the O-H bond. The sharp signals at 2974, 2927 and 2887 cm^-1 are from C-H bonds. Signals below 1500 cm^-1 are "fingerprint" signals which come from various possible sources. The signal at 1050 cm^-1 is from the C-O bond, but with other fingerprint signals (such as the one at 1090 cm^-1) we cannot say for certain that this signal proves that the C-O bond is present.

The process produces a graph with (usually) percentage transmittance on the y-axis and wavenumber along the x-axis. The trace is high and dips where an area of high absorption is present. Data is supplied in the Data Booklet (provided in every exam) for the frequencies where each group absorbs energy - for example the N-H bond in amines forms a trough from around 3100 – 3500 cm⁻¹.

Below around 1500 cm⁻¹, most graphs become very complex and with a lot of interacting troughs - this is known as the fingerprint region and is unique for any compound. This is the region used to conclusively identify a sample if it is pure and fingerprint region data is available.

Nuclear Magnetic Resonance Spectroscopy

This is perhaps the most complex information you will be required to interpret in the exam. The basic principle is some atomic nuclei (e.g. ¹H i.e. protons or ¹³C) have a magnetic spin. In a magnetic field, the nuclei can absorb energy from radio frequency radiation to move into higher energy states. In the higher energy state, the spin of the nucleus is no longer parallel to the applied magnetic field. When the nucleus loses this energy it emits radio frequency energy which gives emission peaks in the spectrum.

Electrons around a nucleus shield it from the applied field and so the frequencies of emission vary depending on the environment that the nucleus is in. This is affected by even small variances in electronegativity of nearby atoms in the molecule, and this is measured in terms of 'chemical shift' (usually represented by δ) - the standard NMR spectrum appears as a trace with emission on the y axis and decreasing δ on the x axis, with spikes at points of high emission where resonance occurs.

¹H NMR spectrum of vanillin (CH₃OC₆H₃OHCHO). The signal at 3.959 ppm is from the 3 H atoms on CH₃O-. The signal at 6.39 ppm is from the H atom on -OH. The signal at 9.823 ppm is from the H atom on -CHO. The other three signals are from the H atoms on C₆H₃. The CH₃O- signal is three times as intense as the other signals, because it is generated by 3 H atoms, and the other signals come from one H atom each.

¹H NMR spectrum of 1,3-dibromopropane. The signal at 3 ppm is from the 4 H atoms on carbons 1 and 3. The signal at 2.3 ppm is from the central 2 H atoms on carbon 2. With three carbon atoms but only two signals, the spectrum shows that the molecule must be symmetrical. The 3 ppm signal is twice as intense as the 2.3 ppm signal, because it is generated by 4 H atoms, and the other signal comes from 2 H atoms.

¹³C NMR spectrum of 1,3-dibromopropane. The signal at 40.7 ppm is from carbons 1 and 3. The signal at 39.8 ppm is from carbon 2. With three carbon atoms but only two signals, the spectrum shows that the molecule must be symmetrical. With ¹³C NMR the signal intensities do not reliably give the number of carbon atoms generating each signal.

As with IR spectrometry, data is supplied on your Data Booklet for chemical shifts relating to the hydrogen and carbon atoms in specific chemical groups.