Describe the principles and apply mass spectrometry and infrared (IR) spectroscopy to determine the molecular mass, molecular formula, structural features and functional groups of organic compounds

What this dot point is asking

QCAA wants you to describe how mass spectrometry (MS) and infrared (IR) spectroscopy work at a conceptual level, interpret the diagnostic peaks in each spectrum, and combine the data to determine the molecular formula and functional groups of an unknown organic compound. The dot point is the highest-yield IA3 secondary-data interpretation item and recurs every year in EA Paper 2 short response.

The answer

Spectroscopy lets a chemist identify an unknown compound non-destructively, using radiation-matter interactions to probe different aspects of structure. Mass spectrometry gives molecular mass and a fragmentation pattern; IR spectroscopy gives functional-group presence. Together they often suffice to identify a small organic molecule, and they pair with NMR (covered in the next dot point) for unambiguous identification.

Mass spectrometry (MS)

Principle. A sample is vaporised, ionised by an electron beam, accelerated through an electric field, and deflected in a magnetic field. The deflection radius depends on mass-to-charge ratio (m/z); the detector records intensity vs m/z.

What the spectrum shows.

• The x-axis is m/z (mass-to-charge ratio, in atomic mass units for singly-charged ions).
• The y-axis is relative abundance, normalised so the most intense peak (the base peak) is 100 percent.
• Each peak is one fragment ion.

The molecular ion (M+). The first ionisation event removes one electron from the parent molecule, giving the molecular cation M+. Its m/z equals the molecular mass (Mr) of the original molecule. The M+ peak is usually the highest m/z peak in the spectrum (occasionally with weak M+1 isotope peaks from 13C). Identifying M+ is the first and most important step in interpretation.

Fragmentation peaks. The molecular ion has high internal energy and fragments before reaching the detector, producing smaller cations and radical species. The radicals are not detected; the cations are. Common fragment losses for organic molecules:

The pattern of losses indicates which substituents the parent molecule carries.

Molecular formula determination. The Mr from M+ narrows the possibilities. For example, Mr = 60 is most commonly C2H4O2 (carboxylic acid or ester) or C3H8O (alcohol or ether), depending on which IR absorptions are present.

Infrared (IR) spectroscopy

Principle. Each chemical bond vibrates (stretches and bends) at a characteristic frequency. When IR radiation of that frequency passes through a sample, the bond absorbs energy and the transmission drops. The transmission vs wavenumber spectrum has dips (absorptions) at the resonant frequencies, identifying which bonds are present.

What the spectrum shows.

• The x-axis is wavenumber (cm^-1, usually plotted high to low from left to right by convention).
• The y-axis is percent transmittance (so absorption appears as a downward dip).
• Each absorption band identifies a bond or functional group.

The diagnostic regions (Unit 4 syllabus subset).

The C=O absorption around 1700 to 1750 cm^-1 is the single most useful diagnostic in Unit 4 spectra. Position shifts slightly: aldehydes around 1720, ketones around 1715, carboxylic acids around 1710, esters around 1740, amides around 1660.

Fingerprint region (600 to 1500 cm^-1). Below 1500 cm^-1, IR spectra have many small peaks that are characteristic of the whole molecule. QCAA does not expect you to interpret fingerprint peaks individually, only to use them to confirm an identification by matching to a reference spectrum.

Combining MS and IR: a worked identification

A typical IA3 / EA stimulus gives you both spectra and asks you to propose a structure. The systematic approach:

1. Read M+ from the MS. This is the Mr.
2. List molecular formulas consistent with that Mr. Common formulas for low Mr can be memorised (Mr 46: ethanol C2H6O or methanoic acid CH2O2; Mr 60: C3H8O or C2H4O2; Mr 74: C4H10O or C3H6O2; Mr 88: C4H8O2 or C5H12O).
3. Use IR to identify functional groups present.
   • Broad 3200 to 3550 cm^-1? Alcohol.
   • Very broad 2500 to 3300 cm^-1? Carboxylic acid.
   • Sharp 1700 to 1750 cm^-1? Carbonyl. Position narrows aldehyde / ketone / acid / ester.
   • Both C=O and broad O-H? Almost certainly carboxylic acid.
   • C=O without O-H, and no N? Aldehyde, ketone, or ester.
4. Use MS fragmentation to narrow further.
   • Loss of 17 (OH)? Likely carboxylic acid or alcohol.
   • Loss of 29 (CHO)? Likely aldehyde.
   • Loss of 43 (CH3CO)? Likely methyl ketone or methyl ester.
   • Fragment at m/z = 45 (COOH+)? Carboxylic acid.
5. Propose the structure and check that all spectral features are accounted for.

The worked example in the 2023 past question above follows this routine: Mr = 60, broad O-H present, C=O present at 1715 cm^-1, MS fragments at 45 and 43, no aldehyde C-H at 2700 to 2800. Diagnosis: ethanoic acid.

Strengths and limitations

MS strengths. Determines Mr exactly. Fragmentation pattern identifies many functional groups indirectly.

MS limitations. Cannot distinguish isomers with identical fragmentation (rare for simple compounds, more common for larger ones). Requires sample volatilisation; not all compounds are stable to ionisation.

IR strengths. Fast, non-destructive, identifies functional groups directly. Routine in industry and IA2 / IA3 contexts.

IR limitations. Does not give Mr. Cannot count carbons or hydrogens. Cannot distinguish chain isomers (propan-1-ol vs 2-methylpropan-2-ol both show O-H and C-O). Pair with MS and NMR for unambiguous identification.

Common traps

Misidentifying M+. Some compounds fragment so completely that M+ is very weak or absent. Look for the highest reasonable m/z; if there are isotope peaks (M+1, M+2), the M+ is the lowest of the cluster.

Confusing O-H (alcohol) with O-H (acid) IR absorptions. Alcohol O-H is broad and centred around 3300 to 3500 cm^-1. Acid O-H is much broader and extends down to 2500 cm^-1, often overlapping the C-H region. The breadth and position discriminate.

Treating the fingerprint region as interpretable peak by peak. QCAA expects the diagnostic-region absorptions (above 1500 cm^-1) to be interpreted; fingerprint peaks confirm an identification but are not individually assigned at Unit 4 level.

Forgetting to check both spectra before committing. A structure consistent with MS alone may be wrong if IR contradicts it (or vice versa). Use both consistently.

Citing "MS gives Mr; IR gives functional groups" without applying it. QCAA marks for the application, not the slogan. The marks come from explicit feature-by-feature reasoning that closes on a single named structure.

In one sentence

Mass spectrometry gives the molecular mass (from the M+ peak) and fragmentation pattern (which identifies likely substituents) of an organic compound; infrared spectroscopy gives the functional groups present (from diagnostic absorptions: O-H around 3200 to 3550 cm^-1, C=O around 1700 to 1750 cm^-1, C-H of aldehyde around 2700 to 2800 cm^-1); combining the two spectra typically lets you propose a unique structure for an unknown Unit 4 organic compound.