Spectroscopy — OCR A-Level Chemistry

This row is a course heading for the NMR spectroscopy work in section 6.3.2, not an assessed answer target. Treat it as a map of what you must be able to do: interpret a carbon-13 NMR spectrum of an organic molecule and use it to make predictions, and interpret a proton (hydrogen-1) NMR spectrum, including the use of chemical shift, integration (relative peak areas) and splitting patterns to deduce environments of carbon atoms and hydrogen atoms. Read the specification sub-statements under this heading, then work through your textbook and class notes so each technique is practised on real spectra. Use the heading to organise revision: build a checklist of the skills named beneath it, and check you can apply each one to unfamiliar molecules rather than only to memorised examples.

Carbon-13 NMR tells you how many different carbon environments a molecule contains. Each distinct environment gives one peak, so count the peaks to predict the number of non-equivalent carbon atoms; a molecule with a plane of symmetry will show fewer peaks than its total carbon count. Use the chemical shift of each peak, read against a correlation table, to predict the type of carbon environment, for example a carbonyl carbon or a carbon attached to oxygen. Remember that carbon-13 has low natural abundance and no nuclear spin splitting is normally resolved, so peaks appear as single lines. Combine peak count and shift values with the molecular formula and any other data to propose or check a structure, and state your reasoning explicitly.

Carbon-13 NMR tells you how many distinct carbon environments a molecule contains. Each set of equivalent carbons gives one signal, so the number of signals equals the number of carbon environments. To count them, look for symmetry: carbons related by a mirror plane or a rotation axis are equivalent and share a signal. For example, ethanol, CH₃CH₂OH, has two carbon environments (the CH₃ carbon and the CH₂ carbon), so its ¹³C NMR spectrum shows two signals. Butan-2-ol, CH₃CH(OH)CH₂CH₃, has four distinct carbons and gives four signals, whereas butan-1-ol has four too but with different shifts. A molecule such as benzene has one environment because all six carbons are equivalent. Count signals, not atoms, and check for symmetry before deciding.

The chemical shift of each ¹³C NMR signal, measured in parts per million (ppm) relative to tetramethylsilane (TMS) at 0 ppm, indicates the type of carbon environment. The shift depends on the local electronic environment: carbons near electronegative atoms or in unsaturated groups are deshielded and appear at higher ppm. Typical ranges help identify environments: alkyl carbons appear around 10–50 ppm, carbons adjacent to oxygen in alcohols and ethers around 50–90 ppm, alkene carbons around 100–150 ppm, and carbonyl carbons of aldehydes, ketones, esters and carboxylic acids around 160–220 ppm. For example, a signal near 170 ppm suggests a carbonyl carbon, while one near 60 ppm suggests a carbon bonded to oxygen. Match each observed shift to these ranges to assign the type of environment.

This criterion asks you to use combined spectroscopic evidence to choose between candidate structures for an organic molecule. Work through the data systematically: use the molecular formula from high-resolution mass spectrometry to fix the molecular formula and degree of unsaturation, then use infrared absorptions to identify or exclude functional groups such as C=O, O–H, N–H and C≡N. Turn to proton NMR: count the number of environments from the number of peak groups, use integration ratios to fix the relative numbers of equivalent hydrogens, and use chemical shift to place each group (for example an aldehyde proton near 9–10 ppm or an O–H proton that exchanges). Apply splitting patterns to establish which hydrogens are on adjacent carbons, and use the coupling constant where given to distinguish coupling partners.

This criterion requires you to analyse a high-resolution proton NMR spectrum and use it to predict features of the molecule. High resolution resolves splitting, so begin by counting the peak groups to establish the number of hydrogen environments. Read the integration trace or the given ratios to fix the relative numbers of equivalent hydrogens in each environment, scaling to the molecular formula where one is supplied. Use chemical shift to assign each environment to a type of proton, for example alkyl near 0.9–1.5 ppm, a proton on a carbon adjacent to a carbonyl, an aldehydic proton near 9–10 ppm, or an exchangeable O–H proton. Apply the n+1 rule to each group: a signal split into n+1 peaks indicates n equivalent hydrogens on adjacent carbons, and the coupling constant confirms which groups couple.

A proton environment is a set of hydrogen atoms that are chemically equivalent, meaning they sit in the same structural position and experience the same magnetic surroundings. To count environments, look for symmetry: hydrogens on the same carbon of a CH₃ group are equivalent, and a plane of symmetry can make two separate groups equivalent too. For example, ethanol CH₃CH₂OH has three environments: the CH₃ hydrogens, the CH₂ hydrogens and the O–H hydrogen. Propan-2-ol (CH₃)₂CHOH has three as well, because the two CH₃ groups are equivalent by symmetry. Each distinct environment gives one peak in a low-resolution proton NMR spectrum, so the count equals the number of peaks.

Chemical shift values, measured in parts per million (ppm) on the δ scale, tell you what kind of proton environment is present because each type of hydrogen absorbs at a characteristic shift. Use a data table of typical shifts: alkyl CH₃ hydrogens appear around δ 0.9–1.7, hydrogens on a carbon next to a C=O or arene ring appear higher, an aldehyde hydrogen (RCHO) appears near δ 9–10, a carboxylic acid O–H near δ 10–12 and an alcohol O–H is variable around δ 1–5. For example, a peak near δ 9.8 signals an aldehyde proton, while a peak near δ 2.1 fits a CH₃ next to a carbonyl. Match each observed shift to the closest entry in the table to assign the environment type.

In a low-resolution ¹H NMR spectrum, each set of chemically equivalent protons gives one peak, and the area under that peak is proportional to how many protons produce it. The integration trace is a stepped line whose step heights are in the same ratio as the peak areas, so you read the relative numbers of each type of proton from the step heights or from ratio numbers printed beside the peaks. For example, steps of 3, 2 and 1 show three types of proton in the ratio 3:2:1, which for a molecule with six protons could be a CH₃ group, a CH₂ group and an OH proton. Convert the ratio to whole numbers by dividing by the smallest value, and check the total against the molecular formula. Integration gives relative numbers only, not absolute numbers, so a ratio of 3:2:1 is equally consistent with 6:4:2.

High-resolution ¹H NMR spectra show spin–spin splitting because a proton's signal is influenced by the magnetic states of non-equivalent protons on adjacent carbon atoms. Apply the n + 1 rule: if n non-equivalent protons sit on the neighbouring carbon atom(s), the signal splits into n + 1 peaks. One adjacent proton gives a doublet, two give a triplet, three give a quartet and none gives a singlet. Only protons that are non-equivalent to the observed proton and are on adjacent atoms cause splitting; equivalent protons do not split each other. For example, in CH₃CH₂OH the CH₂ protons are adjacent to three CH₃ protons, so the CH₂ signal is a quartet, while the CH₃ signal is a triplet from the two CH₂ protons.

This outcome asks you to move from spectral data to a structure. Combine the molecular formula with infrared absorptions, carbon-13 NMR chemical shifts and proton NMR integration, splitting and chemical shift to decide which candidate structure fits. For example, a formula C₄H₈O₂ with an infrared absorption near 1700 cm⁻¹, a carbon-13 signal near 170 ppm and a proton NMR spectrum of a triplet, a quartet and a singlet in ratio 3:2:3 points to ethyl ethanoate rather than a carboxylic acid or an alcohol. Work by eliminating candidates: check the number of carbon environments against the carbon-13 signals, check the number of proton environments and their integration against the proton NMR, and check splitting patterns against neighbouring hydrogens. Confirm that every piece of evidence is consistent before selecting the option.

This outcome asks you to work in the reverse direction: given a structure, predict what its carbon-13 or proton NMR spectrum would show. For carbon-13 NMR, identify each distinct carbon environment, remembering that symmetry makes some carbons equivalent, and predict the number of signals. For proton NMR, identify each distinct hydrogen environment, predict the relative integration from the number of hydrogens, and predict splitting using the n+1 rule based on hydrogens on adjacent carbon atoms. For example, propan-1-ol shows three carbon environments and proton environments including a CH₃ split into a triplet by the adjacent CH₂. Use chemical shift ranges to place signals in the correct region. Select the option that matches your predicted number of signals, integrations and splitting patterns.

In ¹H NMR, chemical shift is measured relative to a reference so that values from different machines and samples can be compared. Tetramethylsilane, Si(CH₃)₄, is that standard: its twelve equivalent protons give one sharp signal, set at δ = 0 ppm. TMS is chemically inert, volatile (so it is easily removed) and soluble in organic solvents, and it absorbs at a shift beyond that of most sample protons, so its peak rarely overlaps theirs. A small amount is added to the sample, and every other peak is reported as a shift from the TMS line. Because all twelve protons are equivalent, TMS produces a single reference peak rather than a cluster.

A ¹H NMR spectrum detects protons, so the solvent must not contribute its own hydrogen signal that would swamp or confuse the sample peaks. Deuterated solvents such as CDCl₃ replace hydrogen with deuterium (²H), which does not give a signal in the ¹H NMR region, so the solvent is effectively invisible. The sample is dissolved in a large excess of solvent, so any ordinary protonated solvent would dominate the spectrum; using CDCl₃ avoids this. Deuterated solvents also dissolve a wide range of organic compounds and are commercially available. The deuterium signal can additionally be used to lock the spectrometer field, but the central reason is to prevent solvent protons appearing in the spectrum.

In a ¹H NMR spectrum, protons on oxygen or nitrogen (an O–H or N–H proton) exchange rapidly with deuterium when deuterium oxide, D₂O, is added. Deuterium has spin quantum number 1, not ½, so a deuterated proton no longer splits or resonates in the ¹H spectrum in the same way; the O–H or N–H signal therefore disappears from the spectrum. To identify such a proton, record the spectrum, add a few drops of D₂O, shake, re-record, and compare: a peak that vanishes belonged to an exchangeable O–H or N–H proton. A peak that remains belongs to a C–H proton, which does not exchange. This is a diagnostic test, not a way of counting all protons, and it distinguishes exchangeable from non-exchangeable protons rather than identifying the whole molecule on its own.

This row sits in Module 6, section 6.3.2 Spectroscopy, where you bring together the spectroscopic and spectrometric methods you have met across the course. Treat 'combined techniques' as the skill of selecting and sequencing evidence rather than a single instrument. In practice you read a set of data, decide which technique answers which question, and build one consistent structure. For example, an infrared spectrum shows which functional groups absorb, a mass spectrum gives the molecular ion and fragmentation pattern, and proton and carbon-13 NMR give hydrogen and carbon environments with splitting and integration. Use each result to check the others, and reject a proposed structure that contradicts any one piece of evidence.

This criterion asks you to combine several analytical techniques to work out a full organic structure, not to interpret one spectrum in isolation. Typical data sets give an elemental analysis or molecular formula, an infrared spectrum, a mass spectrum (including the molecular ion and fragmentation) and a proton NMR spectrum with chemical shifts, integration and splitting. Work systematically: use the molecular ion to fix Mr, use infrared to identify functional groups such as O–H, C=O or C=C, use integration to count protons in each environment and splitting to deduce neighbouring hydrogens, then assemble a structure that satisfies every piece of evidence. Check your candidate against the molecular formula and the number of environments before committing.

Elemental analysis gives the percentage by mass of each element in a compound, from which you calculate the empirical formula. Divide each percentage by the relative atomic mass of that element to get the mole ratio, then divide through by the smallest value to obtain the simplest whole-number ratio. For example, a compound containing 40.0% carbon, 6.7% hydrogen and 53.3% oxygen gives mole ratios of 3.33 : 6.7 : 3.33, which simplify to CH₂O. Combine the empirical formula with the relative molecular mass from the mass spectrum to obtain the molecular formula. This links back to the mole calculations in 2.1.3 c.

Mass spectrometry ionises a sample, accelerates the ions and separates them by mass-to-charge ratio, m/z. In organic chemistry the key skill is reading a mass spectrum to find the molecular ion, M⁺, which gives the relative molecular mass, and to interpret fragmentation peaks that reveal structural features. The molecular ion forms when a molecule loses one electron, so its m/z equals the relative molecular mass for a singly charged ion. A peak at M−15 often signals loss of a methyl group, CH₃, and M−29 can indicate loss of an ethyl group, C₂H₅. The base peak is simply the most abundant ion, not necessarily M⁺. This links back to 4.2.4 f–g, where you met ionisation and the mass spectrometer, so apply that instrumental knowledge here to deduce structure.

Infrared spectroscopy identifies bonds by the frequencies at which they absorb IR radiation, because each bond vibrates at characteristic wavenumbers measured in cm⁻¹. You read an IR spectrum by matching absorption peaks to functional groups: a broad O–H absorption around 2500–3300 cm⁻¹ indicates a carboxylic acid, a broad O–H around 3200–3600 cm⁻¹ indicates an alcohol, a sharp C=O around 1680–1750 cm⁻¹ indicates a carbonyl, and C–H stretches appear just below 3000 cm⁻¹. The fingerprint region below about 1500 cm⁻¹ gives a pattern useful for identifying a specific compound by comparison with a database. This builds on 4.2.4 d–e, where you met the IR spectrometer and the idea of absorption by bonds, so apply that instrumental background to assign peaks and to distinguish, for example, an alcohol from a carboxylic acid.

NMR spectroscopy identifies hydrogen environments in a molecule. In ¹H NMR, the number of signals equals the number of chemically non-equivalent proton environments; the chemical shift (δ, in ppm) indicates the environment, the integration (relative peak area) gives the number of protons in that environment, and splitting follows the n + 1 rule, where n is the number of non-equivalent protons on adjacent carbon atoms. A singlet means no neighbouring protons, a doublet one neighbour, a triplet two. The O–H and N–H protons are often exchangeable and may appear as broad signals. ¹³C NMR gives one signal per non-equivalent carbon environment. Use a chemical-shift data table to assign environments, then combine the evidence to deduce a structure.