Part 3: Combining NMR, infrared and mass spectra

Solve original AQA-style structural problems by checking molecular formula, IR, carbon environments, proton integration and splitting together.

Build a chain of evidence

Start with the molecular formula or deduce it from composition and molecular mass. Identify plausible functional groups from IR. Count the sample 13C and 1H environments, use shifts and areas, then join fragments using splitting. End by checking the proposed structure against every supplied observation.

A molecular ion from a singly charged organic species can give the nominal molecular mass. Do not automatically treat the tallest peak, which may be a fragment, as the molecular ion. High-resolution mass can distinguish possible formulas with the same nominal mass.

IR and NMR answer different questions: an IR carbonyl absorption indicates a bond type, whereas the carbonyl 13C shift helps distinguish an acid/ester environment from an aldehyde/ketone environment. A formula with two oxygens and a carbonyl but no OH does not universally prove an ester; check alternatives such as an ether plus a carbonyl against the full data.

Original formula calculation

Suppose elemental analysis gives C 54.55%, H 9.09%, O 36.36%. In a hypothetical 100 g sample, divide by relative atomic masses: 54.55/12 ≈ 4.546; 9.09/1 = 9.09; 36.36/16 ≈ 2.273. Divide by the smallest to obtain 2 : 4 : 1, giving empirical formula C2H4O.

The empirical formula mass is 44. A molecular ion consistent with Mr 88 therefore gives molecular formula C4H8O2. Using the supplied exact masses C = 12.000000, H = 1.007825 and O = 15.994915 gives 4 × 12 + 8 × 1.007825 + 2 × 15.994915 = 88.05243 for this formula. Use the mass conventions and precision stated in the question.

Unknown A: assemble the evidence

The following are original, idealised teaching data, not a reproduced experimental spectrum. Unknown A has formula C4H8O2. Its IR spectrum contains a strong C=O absorption near 1740 cm−1, without the broad O–H absorption of an alcohol or carboxylic acid. There are four 13C signals near δ 9, 28, 52 and 174 ppm.

The carbonyl carbon near 174 ppm supports an acid or ester. Together with the formula, absence of an acid OH feature and the proton data, an ester is a consistent assignment. The four carbon signals fit four different carbon positions.

Unknown A: methyl propanoate

The triplet integrating to 3 H and the quartet integrating to 2 H form an ethyl fragment. The quartet at 2.35 ppm places CH2 beside the carbonyl. The 3 H singlet at 3.75 ppm is consistent with an OCH3 group, with no ordinary vicinal proton coupling across the ester linkage.

Join these fragments as CH3CH2COOCH3, methyl propanoate. The structure has four carbons, eight hydrogens and two oxygens; one ester carbonyl; and four distinct carbon environments. All eight hydrogens are assigned.

The 13C assignments are terminal CH3 ≈ 9, CH2 adjacent to C=O ≈ 28, OCH3 ≈ 52 and ester carbonyl ≈ 174 ppm. Ethyl ethanoate has the same molecular formula but its OCH2 group would give the oxygen-shifted quartet instead of the oxygen-shifted methyl singlet.

Unknown B: symmetry resolves the carbonyl

Original example: B has formula C3H6O, an IR carbonyl absorption, two 13C signals including one near 205 ppm, and one proton singlet integrating to all six hydrogens near 2.1 ppm.

The carbonyl shift supports an aldehyde or ketone. Two carbon environments require equivalence among the three carbons. Propanone, CH3COCH3, gives equivalent methyl carbons and one carbonyl carbon; its six equivalent H give a singlet because the adjacent carbonyl carbon has no H.

Propanal does not fit: it has three different carbon environments and an aldehyde proton. The conclusion follows from all the data, rather than from the IR carbonyl absorption alone.

When a unique structure is not justified

If two structures fit the supplied formula and all measured features, state what additional evidence would distinguish them. Examples include a missing integration, an aldehyde test or a better-resolved spectrum. A weak absent signal should not be treated as definitive unless the question provides a complete idealised spectrum.

AQA June 2022 Paper 2 Q06.1 uses a levels-of-response scheme for combined IR and NMR reasoning. Its report emphasises putting the evidence into a logical sequence. June 2023 Q05 also shows that overlap and missing integration can limit deductions. A correct final structure with no explained assignments may not demonstrate the reasoning requested.

A useful final check is to label every atom environment on your structure, predict the count and pattern, and cross off each observation only when it is explained. If one signal remains unexplained, reconsider the structure rather than ignoring the inconvenient data.

Quick checks

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

Sources