Part 2: Mass spectra and combined structure deduction

Distinguish molecular ions, isotope peaks and charged fragments, then make all analytical evidence fit one candidate.

The molecular ion reveals Mr when present

A molecular ion is the intact molecule after losing an electron, commonly written [M]⁺•. In the singly charged organic spectra used here, its m/z is numerically the molecular mass. It may be small or absent; the tallest peak is the base peak, not necessarily the molecular ion.

A small M+1 peak arises mainly from molecules containing one ¹³C atom instead of ¹²C. It is not a molecule that has gained a hydrogen. Therefore “choose the furthest-right peak” is unsafe without recognising isotope peaks. OCR AS spectra here are limited to the relevant C, H and O compounds; organic halogen isotope patterns are not an AS requirement.

Give a structure with a positive charge

Fragmentation can split a molecular ion into a positive fragment ion and a neutral radical. The mass spectrum detects the ion, not the uncharged radical. Check that the proposed fragment can be built from the parent’s atoms and show the positive charge.

For butanone, CH₃COCH₂CH₃, Mr = 72. A fragment [CH₃CO]⁺ has m/z = 2(12) + 3(1) + 16 = 43; [CH₃CH₂]⁺ has m/z 29. A peak at 43 can also arise from a different ion such as C₃H₇⁺ in another molecule, so mass alone does not prove one connectivity.

Worked cleavage: [CH₃COCH₂CH₃]⁺• → [CH₃CO]⁺ + CH₃CH₂•. Both atoms and total charge balance. If asked for a fragment structure, a bare m/z number or an uncharged formula is incomplete.

Worked deduction from three techniques

An illustrative compound contains 54.5% C, 9.10% H and 36.4% O. Dividing by Ar gives 4.54 : 9.10 : 2.275 ≈ 2 : 4 : 1, so empirical formula C₂H₄O. A molecular ion at m/z 44 makes the molecular formula the same.

IR shows C=O absorption and no broad O–H; within these AS candidate families this supports ethanal, CH₃CHO. A plausible CHO⁺ fragment has m/z 29. Check the formula contains exactly two C, four H and one O and that a carbonyl is present. This is constructed teaching data, not a measured spectrum.

Use a workflow: elemental analysis → empirical formula; molecular ion → molecular formula; IR → functional groups; fragments → compatible substructures; finally audit all evidence. Absence of a weak peak is less decisive than a clear contradictory strong feature.

Know when the evidence is insufficient

C₃H₆O with a carbonyl absorption and Mr = 58 could be propanal or propanone. A peak at m/z 43 may help with supplied fragmentation context but is not automatically unique. If both candidates remain possible, state the limitation and request distinguishing data rather than inventing a certainty.

Do not add NMR interpretation to the AS requirement: it is developed in Year 13. An unfamiliar AS question can still combine the techniques here with chemical reactions or supplied information, so revise synthesis and analysis together.

Worked empirical-to-molecular formula calculation

Constructed teaching data: a compound contains 66.7% carbon, 11.1% hydrogen and 22.2% oxygen, with a molecular ion at m/z 72. On a 100 g basis, relative amounts are 66.7/12 = 5.558…, 11.1/1 = 11.1 and 22.2/16 = 1.3875. Divide by the smallest to obtain approximately 4:8:1, giving C₄H₈O.

The empirical-formula mass is 4(12) + 8(1) + 16 = 72. Since the singly charged molecular ion also has m/z 72, the molecular formula is C₄H₈O. The small deviations from exact integer ratios reflect rounding of the supplied percentages.

If the molecular mass were twice the empirical-formula mass, all subscripts would double. Do not change just carbon or force a multiplier that is inconsistent with the data. Check that the selected peak is the molecular ion rather than a fragment or M+1 isotope peak.

A plausible fragment must come from the parent structure

For pentan-3-one, CH₃CH₂COCH₂CH₃, Mr = 86. Cleaving a bond beside the carbonyl can give [CH₃CH₂CO]⁺ with m/z = 3(12) + 5(1) + 16 = 57 and a neutral ethyl radical of mass 29. The masses add to 86 and total charge remains +1.

A nominal mass can fit more than one formula: C₄H₉⁺ also has m/z 57. Use the parent connectivity and other evidence to decide which suggestion is plausible. Do not simply assemble any collection of atoms of the right mass and claim the spectrum proves that fragment structure.

An ion must be positively charged in these spectra; an uncharged fragment is not the species assigned to the peak. Where a radical is the neutral partner, show its dot. The molecular ion is usually a radical cation, so its + and dot convey separate information.

Combine chemical and spectral evidence to resolve an ambiguity

Suppose an unknown is made by oxidising a secondary alcohol with an unchanged four-carbon skeleton. Its molecular ion is at m/z 72, and IR shows C=O without a broad alcohol O–H band. The oxidation history narrows the carbonyl family to a ketone under the ordinary conditions, and the four-carbon ketone is butan-2-one, CH₃COCH₂CH₃.

A supplied fragment assigned as [CH₃CO]⁺ at m/z 43 is consistent with that conclusion. It strengthens the account, but a peak at 43 alone would not prove the structure because different ions can share that nominal mass. Each piece of evidence answers a different question: mass constrains formula, IR identifies bonding, and the reaction history constrains connectivity.

Without the secondary-alcohol starting information or other distinguishing evidence, C₄H₈O plus C=O would also permit aldehydes. State exactly which evidence removes each alternative rather than presenting a guess as a unique deduction.

Do not confuse tallest, heaviest and intact

The base peak is simply the most intense peak, defined as 100% relative abundance. It can be a stable fragment ion. The molecular ion is the intact ionised molecule and can be much smaller in intensity. An isotope peak can lie just to its right, so neither “tallest” nor “furthest right” is an unconditional identification rule.

For an ordinary C/H/O molecule, a small M+1 signal commonly comes mainly from molecules containing one ¹³C in place of ¹²C. This adds one mass unit without adding a new carbon atom or a hydrogen. Use the pattern and information supplied; estimating carbon number from M+1 intensity is not required here.

Some molecular ions are weak or absent. If the question supplies only incomplete fragment data, be honest about the resulting limitation. The AS task is to interpret the evidence provided, not to invent an unseen parent peak or bring in unsupported NMR data.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official OCR A mark allocations.

Sources