Part 1: Accurate mass and carbon-13 NMR

Distinguish nominal mass from accurate molecular mass, then use carbon environments and symmetry to narrow the structures consistent with a formula.

What is the mass spectrum actually measuring?

A mass spectrometer separates/detects ions according to mass-to-charge ratio, m/z. In common electron-ionisation organic spectra, the molecular ion is a radical cation, M⁺•, formed by removal of one electron. For a singly charged molecular ion its m/z gives molecular-mass information using the question’s stated mass convention.

Fragment peaks come from charged fragments after bond cleavage; neutral fragments are not detected as separate mass-spectrum peaks. The base peak is the most intense peak and is not necessarily the molecular ion. A weak or absent molecular ion and isotope peaks can complicate identification, so the highest observed m/z is not automatically Mᵣ.

Topic 7 provides the fragmentation and isotope foundations. Topic 19 adds accurate masses: compounds with the same whole-number nominal mass can have detectably different masses when the supplied isotope masses are summed to four decimal places.

Original worked calculation: nominal mass 58 is not enough

Use the isotope masses supplied, not rounded periodic-table averages. For this example take ¹²C = 12.0000, ¹H = 1.0078, ¹⁴N = 14.0031 and ¹⁶O = 15.9949. The accurate relative molecular mass is the sum of the mass of every atom in the chosen isotopic composition. Keep all supplied decimal places until the end.

For C₃H₆O: 3(12.0000) + 6(1.0078) + 15.9949 = 58.0417. For C₄H₁₀: 4(12.0000) + 10(1.0078) = 58.0780. For C₂H₆N₂: 2(12.0000) + 6(1.0078) + 2(14.0031) = 58.0530. All have nominal mass 58, but the accurate values differ.

A reported value of 58.0417 consistent with the first candidate supports C₃H₆O. It cannot distinguish propanal from propanone because those structural isomers have the same formula and hence the same accurate molecular mass. Further tests or spectra must establish connectivity.

An accurate relative molecular mass is dimensionless; a mass in unified atomic mass units and a molar mass in g mol⁻¹ are related quantities but should be labelled correctly. Follow the precision and uncertainty given in the question. Do not round each atomic contribution to a whole number and then claim a four-decimal result.

Chemical shift describes a nucleus’s local environment

In a magnetic field, nuclei such as ¹H and ¹³C can interact with radiofrequency radiation. Their surrounding electrons modify the local magnetic environment, so nuclei in different chemical environments resonate at different frequencies. NMR therefore gives information about the positions/environments of atoms within the molecular structure.

Chemical shift, δ, expresses the resonance relative to a reference on a parts-per-million scale, commonly with tetramethylsilane, TMS, at δ = 0. Electron-withdrawing groups often reduce shielding and move simple nearby nuclei to larger δ, but ring currents and functional-group effects mean the supplied data chart should guide interpretation.

TMS has equivalent carbons and equivalent hydrogens, giving a convenient single reference signal in each spectrum, and it is relatively unreactive and readily removed. A deuterated solvent reduces unwanted proton signals, although residual solvent/water signals may still be visible. Do not count a stated solvent or reference peak as part of the compound. MRI uses related nuclear-magnetic-resonance physics; detailed instrument engineering is not required here.

Count equivalent positions before counting peaks

In a routine proton-decoupled ¹³C NMR spectrum, each chemically distinct carbon environment normally gives one signal. Symmetry can make several carbons equivalent. Count sets of equivalent atoms, not simply the number of carbon atoms and not just the number of labels CH₃, CH₂ and C=O.

Propan-2-ol, CH₃CH(OH)CH₃, has two carbon environments: equivalent end methyls and the central C–OH carbon. Propan-1-ol has three because its two end carbons are different and its CH₂ groups occupy different surroundings. Pentan-3-one has three sets: two equivalent terminal CH₃, two equivalent CH₂ and one C=O. Pentan-2-one has five different carbons.

A useful procedure is to label one carbon A, compare the complete environment of every other carbon with it, and give equivalent atoms the same letter. Molecular symmetry or reversing identical arms can establish equivalence. Simply drawing atoms on opposite sides of a page does not make them different.

Ordinary decoupled ¹³C signal heights/areas are not normally interpreted as carbon counts in this course, and the proton (n+1) splitting rule is not applied to these decoupled carbon signals. Accidental overlap or weak signals can make a real spectrum less simple than the ideal count.

Use shift and count as two separate constraints

A high-δ ¹³C signal in the carbonyl region supports a C=O carbon; aldehyde/ketone carbonyls generally lie further downfield than acid/ester/amide carbonyls. Alkene/aromatic carbon signals lie in another characteristic region, while saturated carbons are generally at lower δ. A saturated carbon bonded to O usually appears further downfield than a similar carbon bonded only to C/H.

Use the reference chart for the actual categories and ranges in the question. Broad approximate regions overlap, so a shift assignment is supporting evidence rather than a complete structure. Carbon NMR shows carbonyl carbons that have no attached H; these do not appear as direct proton signals in a ¹H spectrum.

Original interpretation: a C₃H₆O compound has only two ¹³C signals, one at an illustrative δ 30 and one at δ 206, and gives a positive DNPH test. The carbonyl signal supports aldehyde/ketone, while two equivalent methyls explain a single low-δ signal. Propanone fits; propanal would normally have three carbon environments.

A ring can hide several equivalent carbons

Benzene has one carbon environment because all six ring positions are equivalent. Methylbenzene has five overall: four ring environments (ipso, paired ortho, paired meta, para) plus its methyl carbon. A phenyl ring is therefore not automatically one carbon environment after substitution.

For the three positional isomers of methylbenzoic acid, a para arrangement with two different opposite substituents has four ring environments plus methyl and COOH carbon, giving six in total. The ortho and meta isomers normally give six different ring carbons plus those two, giving eight. This prediction assumes resolved signals and the stated candidate family.

The key is to relate each signal to an environment and use symmetry across the whole structure. Merely saying “six peaks means six carbons” would miss the eight-carbon para structure.

Quick checks

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

Sources