← Spectroscopic data

7 Solving structures: bringing it all together

Combined, these data sets enable the identification of many compounds, and this problem-solving process is the focus of this topic’s assessment.

What each technique contributes

• Mass spectroscopy tells the reader the molar mass and some of the atoms present, and indicates possible structures.
• IR indicates the functional groups present (OH, NH or C=O).
• NMR gives information about the arrangement of the molecule’s structure, as well as some of the functionality.

Drawing information together

• The strategy for a problem depends on the individual problem. The most important thing to do, with any problem in chemistry, is to read the question fully. Sometimes part of the formula or other information is provided in addition to spectral data. One common example is that the empirical formula is provided, which you use with the mass spectrum to evaluate the molecular formula. Typically, questions will provide a selection of potential structures.
• When you start to analyse multiple spectra, start by taking the most robust and easily interpreted information from each spectrum.
• From MS: the first thing to take away is the molecular mass (from the molecular ion), and whether it indicates the presence of nitrogen (odd mass → 1 or 3 or 5… nitrogens; even mass → 0 or 2 or 4…).
• From ¹³C NMR: the first thing to take away is the number of chemical environments, and if any of these are carbonyl carbons and/or alkene carbons.
• From IR: you should be able to make the main judgments regarding the presence of C=O and the presence or absence of OH/NH signals. From here you should have significantly reduced the possibilities, and your further analysis should be directed. Based on what is needed to differentiate the possible structures.
• Lastly, once you have established facts about the molecule (has a C=O, has an OH, has at least 5 carbons, etc.) you can subtract the mass of these parts from the molar mass and find out what is left. That way you can evaluate what remaining possibilities there are.

Worked example A straight forward example Six candidate structures are given, along with a ¹³C NMR spectrum, an IR spectrum, a mass spectrum and a zoomed view of the highest m/z region. Which structure produces all of these spectra? Starting with the ¹³C NMR there are three signals, none of them have double bonds or are carbonyls. This eliminates 1, 5 and 6 as possibilities. The IR also tells us there are no carbonyls, but also tells us there are no OH’s or NH’s. This confirms 1 and 6 is not a possibility and further adds 3 to the list of not appropriate. The mass spectrum has a molecule ion 150 and 152 in near equal heights. This tells us there is a Br atom present, and as the mass is even, there are an even number of nitrogens present. ✓ Only molecule 2 could produce all of the spectra given. Notice the shape of the argument: each spectrum is used to strike out candidates, and the answer is whatever survives all three. Full worked example A full example: a full description An IR, a mass spectrum (with a zoomed molecular-ion region) and a ¹³C NMR are given. Deduce the structure. NOTE: the workbook flags that this analysis goes well beyond what can be expected in an assessment. It is included to show the level of detail a careful, systematic approach can reach. From the mass spectrum: the molar mass is 102, and therefore there is an even number (0 or 2 or 4) of nitrogens. From the IR: there are no NH or OH bonds, but there is at least one carbonyl. The ¹³C NMR indicates there are five environments, and therefore at least five carbon atoms, with one of them being a carbonyl and the rest saturated (or in a triple bond). The position of the carbonyl indicates it is not an aldehyde or ketone, so it is in a carboxylic acid, acyl chloride, ester or amide. Combining this with the IR (no N–H) gives: an ester, an acyl chloride or a tertiary amide. More detail from the mass spectrum: no chlorines, bromines or iodines (no isotope peaks, and the mass is less than 127), so it is not an acyl chloride. We know there is a C=O and 4 other carbons, totalling mass 76. So the rest of the molecule has mass 102 − 76 = 26. Since 26 is smaller than two nitrogens (28), the nitrogen rule tells us there are no nitrogens, so it is not an amide either. The molecule is an ester. The remaining mass = 26 − 16 = 10, and only one atom fits this mass: 10 hydrogen atoms. So the molecular formula is C₅H₁₀O₂. DBE = ½(5 × 2 + 2 − 10) = 1, which is due to the C=O, so there are no cyclic features or other double or triple bonds. Now use the ¹³C in detail: δ 60 is the C–O on the ester; δ 28 is most likely CH/CH₂; and the remaining peaks at δ 9 and 14 (both below 20) are CH₃’s not bonded to an electronegative element or carbonyl. Examining all nine possible five-carbon esters: numbers 2, 4, 6 and 9 have fewer than five chemical environments. Numbers 1, 5 and 8 only have one CH₃ not bonded to an O or C=O. The remaining two are harder to differentiate, but the tables say 0–15 ppm is a CH₃ bonded to a CH₂. In structure 3, one of the CH₃’s is bonded to a CH and would therefore have a chemical shift above 15 ppm. ✓ The structure is number 7: ethyl propanoate, C₅H₁₀O₂. ℹ

How the internal is marked

Straight from the workbook’s sample assessment. The tasks give you a set of IR, ¹³C NMR and MS spectra relating to a single compound, and these compounds contain one functional group.

• Achieved, one piece of information from each spectra is correctly identified.
• Merit, two spectra are correctly interpreted and linked to possible structures.
• Excellence. The correct structure is identified with full justification linked to the integration of information provided by all three spectra.

Read those three lines carefully, because they tell you exactly how to write. Achieved is a list of observations. Merit connects observations to candidate structures. Excellence requires you to say why the structure you picked fits every spectrum, and, in the sample assessment’s own words, “why each of the other structures do not fit the spectra provided”. ✓

The sentence pattern that earns Excellence

Every claim should name its evidence. Compare:

❌ “It is an ester.”
✅ “The IR shows a strong C=O at 1740 cm⁻¹ with no broad absorption at 3000–3500, so there is a carbonyl but no O–H or N–H. The ¹³C shows that carbonyl at δ 172, which is in the 160–185 band rather than 180–220, so it is an acid, acyl chloride, amide or ester rather than a ketone or aldehyde. Combining the two, the absence of N–H rules out a primary or secondary amide and the absence of O–H rules out the acid, so it is an ester.”

Then do the same in reverse for each rejected structure. That is what “integrating” means in the criteria.

Practise just this