Analytical techniques — OCR A-Level Chemistry

This row is course guidance, not an assessed answer target. Read it as an instruction to learn how infrared (IR) spectroscopy identifies bonds in organic molecules. Absorbed IR radiation excites bond vibrations; the wavenumber absorbed depends on the bond and its environment. Learn to read a spectrum in the 4000–400 cm⁻¹ range and to match characteristic absorptions: a broad O–H stretch around 2500–3300 cm⁻¹ for a carboxylic acid, a sharp C=O stretch around 1650–1750 cm⁻¹, and C–H stretches just below 3000 cm⁻¹. Use the fingerprint region below about 1500 cm⁻¹ to confirm identity by comparison with a database spectrum. Practise annotating spectra and linking each peak to a specific bond.

Infrared radiation carries energy that a molecule can take up only when the energy matches a gap between its vibrational energy levels. In a covalent bond the atoms behave as if joined by a spring, so the bond can stretch (symmetric and asymmetric) or bend. Absorbing an IR photon raises the bond from one vibrational level to a higher one, so the bond vibrates with greater amplitude and the radiation is absorbed rather than transmitted. Because different bonds need different energy gaps, each bond type absorbs at a characteristic wavenumber, which is why IR spectra identify O–H, C=O and C–H environments. A bond must change its dipole during the vibration to absorb IR strongly, so symmetric stretches of symmetrical molecules absorb weakly.

Atmospheric gases that contain C=O, O–H or C–H bonds absorb infrared radiation at characteristic wavenumbers. Carbon dioxide absorbs through its C=O bonds, water vapour through its O–H bonds and methane through its C–H bonds. When these gases absorb outgoing infrared radiation emitted by the warmed Earth, they gain vibrational energy and re-emit radiation in all directions, so some returns towards the surface and the atmosphere retains more energy. This greenhouse effect is the basis of the suspected link between rising concentrations of such gases and global warming. Because the effect is linked to energy use, reducing reliance on fossil fuels and improving efficiency are presented as responses that limit further emissions.

Infrared spectroscopy passes infrared radiation through a sample and records which wavenumbers are absorbed; the resulting spectrum is a plot of absorbance (or transmittance) against wavenumber in cm⁻¹, usually read from about 4000 cm⁻¹ down to 500 cm⁻¹. An organic compound absorbs at wavenumbers that match the vibrational energies of its bonds, so particular absorption peaks act as fingerprints for particular bond types. To identify a compound you read the position of each significant peak, decide which bond vibration it corresponds to, and then assemble the functional groups those bonds imply. For example, a broad absorption centred near 3300 cm⁻¹ points to an O–H bond, while a sharp strong peak near 1700 cm⁻¹ points to a C=O bond; finding both suggests a carboxylic acid rather than an alcohol or a ketone.

An alcohol contains an O–H bond, and that bond absorbs infrared radiation over a broad range, typically quoted as roughly 3200–3600 cm⁻¹ for hydrogen-bonded O–H in a liquid or concentrated sample. The absorption is broad because hydrogen bonding varies the O–H bond strength across many molecules, spreading the absorption over a range of wavenumbers. To identify an alcohol from a spectrum, locate a broad absorption in that region, read its wavenumber in cm⁻¹, and confirm that no other feature contradicts the assignment. A sharp, narrow absorption near 3300 cm⁻¹ is more typical of a terminal alkyne C–H stretch, and a C=O absorption near 1700 cm⁻¹ would indicate a carbonyl compound instead.

Infrared spectroscopy identifies functional groups by the wavenumbers at which bonds absorb. The C=O bond of an aldehyde or a ketone gives a strong, sharp absorption in the region around 1680–1750 cm⁻¹. To decide that a compound is an aldehyde or a ketone, you locate that peak and confirm that no broad O–H absorption of a carboxylic acid appears around 2500–3300 cm⁻¹ and no O–H of an alcohol appears around 3230–3550 cm⁻¹. Aldehydes and ketones share the same C=O absorption, so infrared alone cannot usually separate them; further tests such as Tollens' or Fehling's reagent distinguish an aldehyde from a ketone. Read the wavenumber axis, note the peak position and shape, then eliminate other groups before concluding.

A carboxylic acid shows two diagnostic infrared features together. The C=O bond absorbs strongly and sharply around 1680–1750 cm⁻¹, and the O–H bond of the carboxyl group absorbs broadly around 2500–3300 cm⁻¹ because of hydrogen bonding. The breadth of that O–H absorption is the key clue: it is much wider than the sharper O–H stretch of an alcohol, which appears around 3230–3550 cm⁻¹. To identify a carboxylic acid, confirm both peaks are present and note the broad shape of the O–H signal. A spectrum showing only a C=O peak suggests an aldehyde, ketone, ester or acid chloride instead, so the pair of absorptions together is what supports the identification.

Infrared spectroscopy identifies bonds by the wavenumber at which they absorb. Use the supplied data table rather than memory. A broad absorption around 3200–3600 cm⁻¹ indicates an alcohol O–H bond; a very broad absorption around 2500–3300 cm⁻¹ indicates the O–H of a carboxylic acid, often with a C=O near 1680–1750 cm⁻¹. A sharp absorption near 1640–1750 cm⁻¹ indicates C=O, and a sharp absorption near 1620–1680 cm⁻¹ indicates C=C. Absorptions around 2850–3100 cm⁻¹ indicate C–H stretching. To interpret, list the absorptions present, assign each using the table, then combine the assignments into a structure. To predict, work from a proposed structure and state which absorptions it must show and which it must lack. An unfamiliar substance is handled the same way: read the supplied data, assign peaks, and deduce functional groups.

Infrared spectroscopy detects gases because each gas absorbs at characteristic wavenumbers, and the strength of absorption indicates concentration. Carbon monoxide and nitrogen monoxide from car emissions absorb infrared radiation at their own characteristic wavenumbers, so an infrared detector can monitor their concentration in exhaust or roadside air and check compliance with limits. In a modern breathalyser, ethanol vapour in exhaled breath absorbs infrared radiation at a characteristic wavenumber; the amount of absorption is proportional to the ethanol concentration, which gives a measure of alcohol in the breath. The method is non-destructive and can give a continuous reading, which suits monitoring pollution and roadside testing.

This row is a course heading, not an assessed answer target. Treat it as the umbrella for the analytical technique of mass spectrometry as taught in OCR A Level Chemistry (H432), section 4.2.4. Read it as directing you to the technique itself: how a sample is vaporised and ionised, how ions are accelerated and separated by mass-to-charge ratio, and how the resulting spectrum is detected and displayed. Use the heading to organise your notes around the instrument's stages, the meaning of the x-axis (mass-to-charge ratio, m/z) and y-axis (relative abundance), and the difference between the molecular ion and fragment ions. Link forward to the specific sub-statement on identifying the molecular ion peak, and check your class notes and textbook for the exact depth your centre expects.

In electron impact ionisation a vaporised organic sample is bombarded with high-energy electrons. Each molecule that loses one electron forms a radical cation, written M⁺• (often shown simply as M⁺), with a 1⁺ charge and the same relative isotopic composition as the intact molecule. The molecular ion appears as a small peak at the highest m/z value in the spectrum, and that m/z value equals the relative molecular mass Mr of the compound. Fragmentation then produces lower-mass ions, so the tallest peak is usually a fragment, not the molecular ion. For example, ethanol (Mr = 46) shows a molecular ion peak at m/z 46 and fragment peaks at m/z 45, 31 and 29. Read the highest significant m/z, check it is consistent with a plausible formula, and use it to confirm Mr.

In electron impact ionisation a molecule loses an electron to form a radical cation, M⁺•, which can break into a charged fragment and a neutral radical. Only charged fragments appear as peaks, so the molecular ion peak gives the relative molecular mass and each fragment peak corresponds to a charged piece of the structure. To analyse a spectrum, subtract each fragment m/z value from the molecular ion m/z value to find the mass of the neutral group lost, then match that difference to a structural fragment. For example, a peak 15 units below M⁺• is consistent with loss of a methyl group, CH₃•, leaving a charged remainder. Use accurate m/z values and the molecular formula to decide which bonds can break and which fragments are plausible.

This section covers infrared spectroscopy and mass spectrometry and how they are used together to identify organic compounds. Infrared spectroscopy reveals which functional groups are present through absorption at characteristic wavenumbers, while mass spectrometry gives the molecular mass from the molecular ion peak and structural clues from the fragmentation pattern. Combining the two techniques lets you propose a structure: IR narrows the functional groups, and the mass spectrum fixes the molecular formula and supports the fragments. Read the specification to see which techniques are required here and which come later in the course, and use each spectrum for the information it actually provides rather than expecting one technique to give a full structure alone.

Deduction means combining several spectra into one consistent structure rather than reading a single spectrum in isolation. Work systematically: use the molecular ion from mass spectrometry to fix the molecular formula, then use the infrared spectrum to identify functional groups by their absorption ranges, and finally use the proton NMR spectrum to place the groups. In NMR, count the number of environments from the number of peaks, read the integration to give the relative numbers of hydrogen atoms in each environment, and use chemical shift to identify the type of proton. Splitting patterns reveal neighbouring hydrogens: an ethyl group shows a CH₃ triplet integrating to 3H and a CH₂ quartet integrating to 2H, a 3:2 ratio. Cross-check every proposed structure against all the data and reject any that contradicts one piece of evidence.

Elemental analysis determines the percentage by mass of each element in a compound, from which the empirical formula is calculated. Divide each percentage by the relative atomic mass of that element to obtain the mole ratio, then divide through by the smallest value to give 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. The empirical formula combined with the relative molecular mass from mass spectrometry gives the molecular formula. This links to the earlier work on empirical and molecular formulae in section 2.1.3c.

A mass spectrum plots relative abundance against m/z (mass-to-charge ratio). In the OCR A Level course you use it to find the molecular mass from the molecular ion peak, M⁺, and to infer fragments from other peaks. For example, ethanol gives a molecular ion at m/z = 46 and a prominent fragment at m/z = 31 from loss of CH₃, plus m/z = 45 from loss of a hydrogen atom. The M⁺ peak sits at the highest m/z of the main cluster (ignoring small M+1 isotope peaks). Fragmentation follows the weakest bond, so the spectrum reveals structural pieces. You interpret peaks, not calculate them from first principles, and you link each fragment to a plausible structure.

An infrared spectrum plots percentage transmittance against wavenumber (cm⁻¹). You use it to identify functional groups from absorption bands and to fingerprint a compound by matching its whole spectrum to a database. A broad absorption around 3200–3600 cm⁻¹ indicates an O–H group in an alcohol; a sharp band near 1700 cm⁻¹ indicates a C=O group; a C–H stretch appears just below 3000 cm⁻¹. The fingerprint region below about 1500 cm⁻¹ is complex and is used for matching rather than for assigning individual bonds. You read the position and shape of bands, then link them to bonds present in the structure.