22. Analytical techniques

Infrared spectroscopy measures absorption of bond vibrations. Functional groups give characteristic absorption regions, so compare peaks with the supplied data rather than relying on one isolated feature.

Use the fingerprint region to support identity, but prioritise diagnostic absorptions such as O–H, C=O, C–H and N–H. Peak position and broadness both carry information.

A broad absorption around 3200–3600 cm⁻¹ suggests O–H; a strong peak near 1700 cm⁻¹ suggests C=O. Together they can distinguish an alcohol from a carbonyl compound.

IR identifies functional groups, not usually the complete carbon skeleton. Do not assign a group from a single noisy peak without checking the rest of the spectrum.

A mass spectrum plots relative abundance against m/e. Peaks may represent molecular ions, fragments or isotopic variants; the tallest peak is the base peak, not necessarily the molecular ion.

Use the isotope pattern and peak spacing to support identity. The working of the instrument is not required, but the distinction between abundance and mass-to-charge value is essential.

A peak at m/e 43 can be a common fragment, while a higher-mass molecular-ion peak gives the parent mass if it is present. Chlorine and bromine give distinctive M:M+2 patterns.

The base peak is not automatically M⁺, and a fragment’s m/e is not the relative molecular mass of the whole compound.

Relative atomic mass is the weighted mean of an element’s isotopic masses: Ar = Σ(isotopic mass × percentage abundance)/100.

Use fractional or percentage abundances consistently and check that they sum to 100%. The result should lie between the lightest and heaviest isotope masses.

For 75% isotope 35 and 25% isotope 37, Ar = (35×75 + 37×25)/100 = 35.5.

Do not take a simple arithmetic mean unless abundances are equal, and do not confuse Ar with one isotope’s mass number.

The molecular ion M⁺ is formed when the molecule loses one electron without fragmenting. Its m/e, usually with charge +1, gives the relative molecular mass of the intact molecule.

First identify a plausible high-mass parent peak, then check isotope patterns and fragments. The molecular ion can be weak or absent, so use other evidence if necessary.

If the molecular-ion peak is at m/e 46 for a singly charged molecule, Mr is approximately 46. A peak at 31 may instead be a fragment.

Do not choose the tallest peak automatically or treat a doubly charged ion as if m/e equalled Mr without considering charge.

Fragment ions form when a molecular ion breaks into smaller charged pieces. A proposed fragment must have a plausible formula, m/e value and neutral partner that together reconstruct the parent.

Use common stable cations and the molecular structure as constraints. Fragment abundance reflects stability and pathway, not simply the number of atoms.

A peak at m/e 29 may be C₂H₅⁺. Pairing it with a neutral radical or molecule must preserve the parent formula and electron/charge accounting.

Any formula matching a mass is not automatically the fragment identity; check connectivity and complementary loss.

The M+1 peak mainly reflects molecules containing one ¹³C instead of ¹²C. For a simple organic compound, n ≈ 100 × abundance(M+1)/(1.1 × abundance(M)).

Use the ratio of peak abundances, not their m/e positions. The result should be close to a whole number and is an estimate subject to other minor isotopes.

If M+ has abundance 50 and M+1 is 5.5, n ≈ 100×5.5/(1.1×50)=10 carbon atoms.

Do not use the base peak in place of M+, and do not round a physically implausible result without checking peak assignment.

A compound containing chlorine or bromine can show an M+2 peak because ³⁷Cl and ⁸¹Br are two mass units above the common isotopes. The relative M:M+2 pattern carries the evidence.

One chlorine gives an approximate 3:1 M:M+2 pattern, while one bromine gives about 1:1. Multiple atoms create larger isotope clusters, so inspect the whole pattern.

A molecular ion with nearly equal M and M+2 peaks suggests one bromine atom; a 3:1 pair suggests one chlorine atom.

M+2 does not prove which halogen from mass difference alone. Use abundance ratios and consider more than one halogen.