Spectroscopy

Spectroscopy and the electromagnetic spectrum

When white light passes through a prism it spreads into a band of colours, as in the photograph at the start of this unit. Visible light is only a small part of a much larger family called electromagnetic radiation. This radiation carries energy in small packets called photons. Atoms and molecules can absorb or give out photons, and which photons they absorb depends on their structure. So the pattern of absorption tells a chemist what a substance is and how its atoms are joined.

Spectroscopy

A technique used for structural elucidation (working out the structure) of molecules. It is based on the interaction of atoms or molecules with electromagnetic radiation.

Spectrum

A graph that shows how much radiation a sample absorbs (or gives out) at each wavelength, frequency or wavenumber.

Electromagnetic spectrum

The range of all frequencies of electromagnetic radiation, from radio waves to gamma rays.

All electromagnetic waves travel at the speed of light $c$. The wavelength $\lambda$ is the distance between two neighbouring crests and the frequency $\nu$ is the number of waves passing a point per second. They are linked, and the energy $E$ of one photon depends on the frequency:

Here $c=3.00\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$, $h=6.63\times 10^{-34}\mathrm{J}\mathrm{s}$ is Planck's constant, $\lambda$ is in metres and $\nu$ is in hertz ($1\mathrm{Hz}=1{\mathrm{s}}^{-1}$). A short wavelength means a high frequency and a high photon energy.

18.1Spectroscopic techniques

Four techniques are used most often to identify organic compounds. Each answers a different question about the molecule.

Each type of molecule absorbs electromagnetic radiation only in particular regions. On absorbing a particular wavelength, the molecule undergoes a particular type of change (an electron jump, a stronger bond vibration or a nuclear spin flip), and this change is observed. Measuring which wavelengths are absorbed therefore gives the key to the structure of an organic compound.

18.2Index of hydrogen deficiency (degree of unsaturation)Often examined

Start with the alkanes. An open-chain alkane $\mathrm{C}{}_n\mathrm{H}{}_{2n+2}$ holds the largest possible number of hydrogen atoms for its number of carbon atoms, so it is called saturated. For example, ethane $\mathrm{C}{}_2\mathrm{H}{}_6$ has $2(2)+2=6$ hydrogen atoms.

Now put a double bond into ethane: two hydrogen atoms must be removed, giving ethene $\mathrm{C}{}_2\mathrm{H}{}_4$. Joining the two ends of a chain into a ring also removes two hydrogen atoms: hexane $\mathrm{C}{}_6\mathrm{H}{}_{14}$ becomes cyclohexane $\mathrm{C}{}_6\mathrm{H}{}_{12}$. So every $\pi$ bond and every ring leaves the molecule short of one pair of hydrogen atoms.

Index of hydrogen deficiency (IHD) or degree of unsaturation

The total number of rings plus $\pi$ bonds in a molecule. It is found by comparing the molecular formula with the formula of the saturated open-chain compound that contains only $\sigma$ bonds: each pair of missing hydrogen atoms counts as 1.

$C$ = number of carbon atoms, $H$ = number of hydrogen atoms, $N$ = number of nitrogen atoms and $X$ = number of halogen atoms (F, Cl, Br, I). Oxygen and sulphur atoms are ignored.

18.3, 18.3.1IR spectroscopy and its principleOften examined

Infrared (IR) spectroscopy makes use of the interaction of infrared radiation with organic molecules. Its typical range is from 4000\,\mathrm{cm^{-1}} to 625\,\mathrm{cm^{-1}}, and it is used to find the functional groups present in an organic compound. Each functional group absorbs a specific part of the IR radiation, and so it is recognised. The technique is used for both organic and inorganic compounds. It is fast (high scan speed), sensitive and gives good resolution, which makes it a very useful tool for identifying molecular structure.

Positions in an IR spectrum are given as wavenumber \bar{\nu} (read 'nu bar'), the number of waves in one centimetre. It is the reciprocal of the wavelength measured in centimetres, so its unit is \mathrm{cm^{-1}}. A larger wavenumber means a higher frequency and a higher photon energy.

For example, \bar{\nu}=4000\,\mathrm{cm^{-1}} means \lambda=\tfrac{1}{4000}\,\mathrm{cm}=2.5\,\mu\mathrm{m}, and 625\,\mathrm{cm^{-1}} means \lambda=16\,\mu\mathrm{m}.

Principle of IR spectroscopy

Think of a covalent bond as a spring joining two balls (the atoms). The bond is never still: it keeps stretching (the bond length changes) and bending (the bond angle changes). Each bond vibrates at its own natural frequency, which depends on the strength of the bond and the masses of the two atoms.

1. IR radiation falls on the sample. The electric field of the radiation interacts with the dipole moment of the bonds. Bonds to hetero atoms such as O and N are polar, so they interact strongly.
2. If the frequency of the IR radiation matches the natural vibrational frequency of a bond (or of a group of bonds), the radiation is absorbed.
3. The bond then vibrates with a greater amplitude, and its dipole moment changes.
4. The more polar the bond, the bigger the change in dipole moment and the more intense the absorption. This is why \ce{C=O} and \ce{O-H} give strong bands.
5. The frequencies absorbed are characteristic of the bonds and functional groups present. So the IR spectrum reveals the molecular structure, especially the functional groups.

The infrared spectrophotometer

The major components of an infrared spectrophotometer are: a source of IR radiation, a monochromator, a sample chamber, a detector, and a data collection and processing system.

• Source: a heated ceramic rod that gives out IR radiation of all wavenumbers.
• Monochromator: a prism or grating that separates the radiation so that one wavenumber at a time reaches the sample.
• Sample chamber: holds the sample, for example a thin liquid film between plates of sodium chloride. Glass cannot be used because it absorbs IR.
• Detector: measures how much radiation passes through the sample at each wavenumber.
• Data system: a computer plots percentage transmittance against wavenumber. This graph is the IR spectrum.

Uses of IR spectroscopy

• Identifying the functional groups in an unknown compound, for example \ce{O-H}, \ce{C=O} and \ce{N-H}.
• Confirming the identity of a compound by matching its fingerprint region (below 1500\,\mathrm{cm^{-1}}) with the spectrum of a known sample.
• Checking purity: extra bands show impurities. For example, a broad \ce{O-H} band in the spectrum of a ketone shows that water or an alcohol is present.
• Following a reaction: when ethanol is oxidised to ethanal, the broad \ce{O-H} band disappears and a strong \ce{C=O} band appears near 1720\,\mathrm{cm^{-1}}.
• Everyday analysis: breathalysers measure the IR absorption of ethanol in breath, and IR sensors measure \ce{CO2} and other gases in air.

18.3.2, 18.3.3Reading an IR spectrumOften examined

An IR spectrum is a graph of percentage transmittance T (the percentage of the radiation that passes through the sample) against wavenumber. Two things look unusual at first. The wavenumber decreases from left (4000\,\mathrm{cm^{-1}}) to right. And absorption shows as a dip downwards from the 100\% line. Each dip is called an absorption band or peak, and each corresponds to a specific bond. The pattern of bands is a fingerprint of the compound.

Two regions of the spectrum

Characteristic absorption ranges (Table 18.1)

How to read an IR spectrum

1. Identify the significant (strong or broad) peaks, mainly in the functional group region above 1500\,\mathrm{cm^{-1}}.
2. Compare these peaks with the known wavenumber ranges of Table 18.1. Check the regions in turn: 3200\text{–}3500 (broad \ce{O-H}, or sharper \ce{N-H}); about 3000 (\ce{C-H}: just below 3000 for alkyl groups, just above 3000 for alkenes and aromatics); 2100\text{–}2260 (triple bonds); 1680\text{–}1750 (strong \ce{C=O}); 1400\text{–}1680 (\ce{C=C}).
3. Use the matches to deduce the functional groups present. Note also which bands are absent: no strong band near 1700\,\mathrm{cm^{-1}} means no \ce{C=O} group.
4. Combine this with the molecular formula (and its IHD) to suggest a structure. Confirm it by comparing the fingerprint region with the spectrum of a known sample.

Deducing a structure from an IR spectrum and the molecular formula

18.4UV-visible spectroscopyOften examined

UV-visible spectroscopy is the technique in which the absorption of ultraviolet and visible light by a compound is measured. Its typical range of wavelength is 200 to 800\,\mathrm{nm}: 200\text{–}400\,\mathrm{nm} is ultraviolet and 400\text{–}800\,\mathrm{nm} is visible light. It can be used on liquids, solids and gases, and is widely used in material science and biochemistry. It is used in particular to find conjugation in unsaturated compounds.

Light of these wavelengths carries enough energy to excite electrons: an electron moves from a filled orbital to an empty orbital of higher energy. The filled orbitals are bonding orbitals (\sigma and \pi) or non-bonding lone-pair orbitals (n). The empty orbitals are antibonding orbitals (\sigma^* and \pi^*). The usual electronic transitions are \pi\to\pi^*, n\to\pi^*, \sigma\to\sigma^* and n\to\sigma^*.

These transitions absorb particular ranges of wavelength and so give information about the molecular structure. A UV-visible spectrum is a graph of absorbance against wavelength. The wavelength at the top of an absorption band is called \lambda_{\max} (lambda max). The energy gap and the wavelength absorbed are linked by

so a large energy gap means a short wavelength is absorbed, and a small gap means a long wavelength is absorbed. The absorbance is also directly proportional to the concentration of the absorbing substance.

Why some compounds are coloured

A coloured compound absorbs a small range of wavelengths in the visible region (400\text{–}800\,\mathrm{nm}) and transmits (or reflects) the rest. The colour we see is not the colour that is absorbed. It is the colour of the light that is left, which is the complementary colour of the absorbed light. For example, a complex that absorbs blue or violet light appears yellow or yellow-green.

1. From the UV-visible spectrum, find the wavelength absorbed most strongly, \lambda_{\max}.
2. Find the colour of that wavelength and, if needed, the energy of the transition, \Delta E=hc/\lambda_{\max}. Short wavelengths (high energy) are violet and blue; long wavelengths (low energy) are orange and red.
3. The compound shows the complementary colour: use Table 18.2 or the colour wheel.

Finding the concentration of a solution

Another use of UV-visible spectroscopy is to find the concentration of a solution. The absorbance is directly proportional to the concentration, which is stated by the Beer–Lambert law:

A = absorbance (no unit), c = concentration in \mathrm{mol\,dm^{-3}}, l = path length of the cell (usually 1\,\mathrm{cm}), and \varepsilon = molar absorptivity, a constant for the substance at that wavelength (unit \mathrm{dm^3\,mol^{-1}\,cm^{-1}}).

1. Prepare standard solutions of known concentration and measure the absorbance of each at \lambda_{\max}.
2. Plot absorbance against concentration. This standard curve (calibration curve) is a straight line.
3. Measure the absorbance of the unknown solution in the same way and read its concentration from the graph.

Uses of UV-visible spectroscopy

• Detecting conjugation and aromatic rings: conjugated compounds absorb above 200\,\mathrm{nm}, and the more conjugated, the longer \lambda_{\max} (next section).
• Predicting and explaining the colour of compounds, including transition metal complexes.
• Finding the concentration of a solution with a standard curve, for example iron or nitrate in water, and proteins (280\,\mathrm{nm}) or DNA (260\,\mathrm{nm}) in biochemistry.
• Following the rate of a reaction in which a coloured substance is used up or formed.
• Checking purity: a trace of benzene in ethanol shows up as a band near 254\,\mathrm{nm}, where pure ethanol does not absorb.
• Material science: studying dyes, pigments and semiconductors.

18.4.1Predicting whether a compound absorbs in the UV-visible regionOften examined

To predict whether a compound will absorb in the UV-visible region, look at its electronic structure, especially at which electronic transitions are possible. If a compound has conjugated double bonds, an aromatic ring, or a group such as a carbonyl group that allows \pi\to\pi^* or n\to\pi^* transitions, it is likely to absorb in the UV-visible region.

Chromophore

The group of atoms in a molecule that is responsible for absorbing UV or visible light, for example \ce{C=C}, \ce{C=O}, \ce{C#C}, \ce{N=N}, \ce{NO2} and the benzene ring.

Conjugated system

A chain of alternating single and double (or triple) bonds, such as \ce{C=C-C=C}. The p orbitals overlap along the whole chain, so the \pi electrons are delocalised.

\lambda_{\max}

The wavelength at which a compound absorbs most strongly.

Red shift (bathochromic shift)

A move of \lambda_{\max} to a longer wavelength, for example when conjugation is increased.

Ordinary UV-visible spectrometers work from 200\,\mathrm{nm} upwards, because below 200\,\mathrm{nm} the oxygen of the air and most solvents absorb strongly. A compound whose only absorption lies below 200\,\mathrm{nm} is therefore said not to absorb in the UV-visible region. Alkanes (only \sigma\to\sigma^*) and compounds with one isolated \ce{C=C} bond are like this. This is why hexane, ethanol and water are good solvents for UV-visible work: they are transparent above 200\,\mathrm{nm}.

How conjugation shifts the absorption

In a conjugated molecule the p orbitals of neighbouring double bonds overlap side by side, so the \pi electrons are spread (delocalised) over the whole chain. The longer this delocalised system, the closer together the highest filled \pi orbital (HOMO) and the lowest empty \pi^* orbital (LUMO) become. A smaller energy gap means that a photon of lower energy, that is, of longer wavelength, is absorbed (\lambda=hc/\Delta E). The absorption also becomes more intense.

• **Example 1: benzene (\ce{C6H6}).** Benzene has a conjugated \pi system of alternating double bonds. It shows \pi\to\pi^* transitions and absorbs in the UV region at about 254\,\mathrm{nm}.
• **Example 2: buta-1,3-diene (\ce{C4H6}).** It has a conjugated system of double bonds, \ce{C=C-C=C}. It shows a \pi\to\pi^* transition and absorbs in the UV region at about 217\,\mathrm{nm}.
• **Example 3: acetone (\ce{CH3COCH3}).** It has a carbonyl group (\ce{C=O}), whose oxygen lone pair can undergo an n\to\pi^* transition. It absorbs in the UV region at about 279\,\mathrm{nm} (a weak band).

1. Look for chromophores: \pi bonds, and lone pairs on atoms next to \pi bonds.
2. Only \sigma bonds, or a single isolated \ce{C=C}: no absorption above 200\,\mathrm{nm}.
3. Conjugated \ce{C=C} bonds, an aromatic ring or a \ce{C=O} group: absorbs in the UV (\pi\to\pi^* or n\to\pi^*).
4. The more conjugated double bonds, the longer \lambda_{\max} and the stronger the absorption. With about 8 or more conjugated double bonds, \lambda_{\max} reaches the visible region and the compound is coloured.

18.5Atomic emission spectroscopy

Atomic emission spectroscopy (AES) is a technique used to identify elements and to measure their amounts (quantify them) by measuring the particular wavelengths of light emitted by a sample whose atoms have first been excited. The source of excitation may be a flame, a plasma, an electric arc or a spark.

When an atom absorbs energy (from heat, electrical energy or light), its electrons are excited to higher energy levels. These excited electrons are unstable and soon return to lower energy levels or to the ground state. In each return the electron gives out the energy difference as a photon of light:

E_2 and E_1 are the energies of the higher and lower levels, and \lambda is the wavelength of the emitted light. The energy levels of an atom are fixed (quantized), so only certain energy differences are possible. The emitted light therefore contains only certain wavelengths, and the spectrum is a series of separate bright lines, each of one wavelength. These are called spectral lines, and they represent the quantized energy levels of the atom. Each element has its own set of energy levels, so each element has a unique emission spectrum, often called its fingerprint.

How AES works

1. The sample, usually a solution, is sprayed into a flame or plasma. Here it is turned into free atoms and the atoms are excited.
2. The excited atoms emit light as their electrons return to lower levels (after the source of energy is removed, or as they leave the hot zone).
3. A prism or diffraction grating disperses the emitted light into its separate wavelengths.
4. A detector measures the intensity of each line.
5. The wavelengths of the lines identify the element (qualitative analysis). The intensity of a line depends on the number of emitting atoms, so it gives the amount of the element (quantitative analysis).

18.6Atomic absorption spectrum

An atomic absorption spectrum is observed when free atoms in the ground state absorb particular wavelengths of light and are excited to higher energy levels. If white light (all wavelengths) passes through the atoms, the wavelengths that match the energy gaps are removed, so dark lines appear in the spectrum at those wavelengths.

Atomic emission and absorption spectra are both line spectra and give the same information. For the same element the lines appear at exactly the same wavelengths, as Fig 18.6 shows for potassium, because the same energy gaps are involved: in emission an electron falls across a gap, and in absorption it jumps up across the same gap. The only difference is the appearance: in AES bright lines appear on a dark background, while in AAS dark lines appear on a bright background.

How atomic absorption spectroscopy (AAS) is used

1. The sample solution is sprayed into a flame, which breaks it into free atoms, most of them in the ground state.
2. Light of exactly the wavelength of the element being measured passes through the flame. It comes from a lamp made of that element (a hollow cathode lamp).
3. The ground-state atoms absorb part of this light, and the detector measures how much is absorbed.
4. The absorbance is proportional to the concentration of the element. AAS is very sensitive, so it is used to measure traces of metals such as lead, mercury and cadmium in water, food and blood.

18.7, 18.7.1, 18.7.2Nuclear magnetic resonance: principle and NMR-active nucleiOften examined

Nuclear magnetic resonance (NMR) spectroscopy, also called magnetic resonance spectroscopy (MRS), is based on the re-orientation of atomic nuclei that have a non-zero nuclear spin, when they are placed in an external magnetic field. The re-orientation happens when the nuclei absorb electromagnetic radiation in the radio-frequency region, roughly 4 to 900\,\mathrm{MHz}. The exact frequency depends on the isotope and increases in proportion to the strength of the external magnetic field.

Most importantly, the resonance frequency of each NMR-active nucleus depends on its chemical environment (the atoms and electrons around it). So an NMR spectrum gives information about the functional groups in the sample and about how neighbouring nuclei are connected. Because NMR spectra are unique or highly characteristic of individual compounds, NMR is one of the most important methods for identifying the structures of organic compounds.

Principle

A spinning charged nucleus behaves like a tiny bar magnet, a nuclear magnet. Normally these nuclear magnets point in random directions and all have the same energy. When a strong external magnetic field B_0 is applied, a nucleus of spin \tfrac12, such as \ce{^1H} or \ce{^{13}C}, can take up only two orientations (spin states):

• aligned with the applied field: the lower-energy state, spin quantum number m=+\tfrac12;
• opposed to the applied field: the higher-energy state, m=-\tfrac12.

The energy difference \Delta E between the two states corresponds to radio-wave frequencies. When radio waves of exactly this energy fall on the sample, nuclei in the lower state absorb them and flip their spin to the higher state. This matching of the radio frequency with the energy gap is called resonance. When the nuclei return to the lower state they give out the absorbed energy at the same frequency. This emitted radio-frequency signal is detected and gives the NMR spectrum of the compound.

\nu is the radio frequency absorbed. A stronger magnet makes \Delta E larger, so a higher frequency is needed. Instruments are named by the frequency at which their protons resonate, for example a 400\,\mathrm{MHz} spectrometer.

NMR-active nuclei

NMR-active nuclei are those that possess spin: the charged nucleus spins about an axis and so produces its own magnetic dipole moment. Hydrogen-1 (\ce{^1H}), carbon-13 (\ce{^{13}C}), fluorine-19 (\ce{^{19}F}) and phosphorus-31 (\ce{^{31}P}) all have magnetic moments and can be observed by NMR. Organic chemists mostly use \ce{^1H} and \ce{^{13}C} NMR.

To be NMR-active, a nucleus must have a non-zero nuclear spin quantum number, I\neq0. It is this spin that lets the nucleus interact with an external magnetic field and give a signal. Whether I is zero depends on the numbers of protons and neutrons in the nucleus:

18.8, 18.8.1, 18.8.2Proton NMR: deuterated solvents and TMSOften examined

Proton NMR (\ce{^1H} NMR, also called PMR) is the application of NMR spectroscopy to the hydrogen-1 nuclei in the molecules of a substance, in order to find the structure of its molecules. It is the most used form of NMR, because almost every organic compound contains hydrogen, and 99.98\% of natural hydrogen is \ce{^1H}, which gives strong signals.

Use of deuterated solvents

The sample is dissolved in a solvent before its spectrum is recorded, and the solvent is present in far greater amount than the sample. If the solvent contained ordinary hydrogen atoms, it would give a huge peak that would swamp (hide) the small peaks of the sample.

So solvents are used in which the hydrogen atoms are replaced by deuterium (\ce{^2H}, often written D). Examples are deuterochloroform \ce{CDCl3} (the most widely used), deuterated water \ce{D2O} and deuterated acetone \ce{(CD3)2CO}. Solvents that contain no hydrogen at all, such as tetrachloromethane \ce{CCl4} and carbon disulphide \ce{CS2}, are also used. In each case there are no solvent protons to interfere with the analysis.

Tetramethylsilane (TMS) as the standard of chemical shift

Chemical shifts are measured from the signal of a reference compound, tetramethylsilane (TMS), \ce{Si(CH3)4}. A small amount is added to the sample. Its signal is defined as chemical shift \delta=0\,\mathrm{ppm}.

• One sharp signal. TMS is a symmetrical, tetrahedral molecule. All 12 hydrogen atoms are chemically equivalent, so they give a single peak.
• Highly shielded. Silicon (electronegativity 1.8) is less electronegative than carbon (2.5), so electron density is pushed towards the methyl groups. The TMS protons are more shielded than the protons of almost all organic compounds, so the TMS peak lies at the far right, clear of the other peaks. Its position is taken as \delta=0.
• Strong signal. With 12 equivalent protons, only a small amount of TMS is needed.
• Inert, soluble and volatile. TMS does not react with the sample, it dissolves in most organic solvents, and it boils at about 27^\circ\mathrm{C}, so it can be evaporated off afterwards to recover the sample.

The proton peaks of most organic compounds lie between about 0 and 12\,\mathrm{ppm} (the book gives the full possible range as +14 to -4\,\mathrm{ppm}). Proton spectra also show spin-spin coupling between neighbouring protons (18.8.4). TMS is used as the reference for \ce{^{13}C} NMR too, where its carbon atoms appear at \delta=0.

18.8.3Nuclear spin flipping and chemical shiftOften examined

Nuclear spin flipping is a quantum-mechanical process in which a nucleus absorbs energy from an external source and changes its spin state. When a nucleus is irradiated with radio waves of the proper frequency, it absorbs energy and 'flips' from the lower-energy spin state to the higher-energy spin state.

If every proton in a molecule felt exactly the same magnetic field, all of them would flip at the same frequency and every spectrum would show a single peak. This does not happen, because each nucleus is surrounded by electrons, and the electrons change the field that the nucleus actually feels.

Shielding and deshielding

In the applied field B_0, the electrons around a nucleus circulate and produce a small magnetic field of their own that opposes B_0. The nucleus therefore feels a slightly smaller field than B_0: it is shielded by its electrons. A shielded nucleus comes into resonance at a lower frequency.

Shielding

When the electron density around a nucleus is high, the opposing field of the electrons is large and the nucleus feels less of the applied field. Its peak appears at low \delta (to the right, upfield).

Deshielding

When the electron density around a nucleus is reduced, for example by a nearby electronegative atom such as O or Cl, the opposing field is smaller and the nucleus feels more of the applied field. Its peak appears at high \delta (to the left, downfield).

Chemical shift

The chemical shift \delta of a nucleus is the difference between its resonance frequency and that of a standard (TMS) in the same magnetic field. It is expressed in parts per million (ppm) of the operating frequency of the spectrometer, and it is a key property for working out a molecule's structure:

\nu_{\text{sample}} and \nu_{\text{TMS}} are the resonance frequencies of the proton and of TMS (in Hz), and \nu_{\text{spectrometer}} is the operating frequency of the instrument (in Hz). Dividing by the operating frequency makes \delta the same on every instrument, whatever the strength of its magnet.

The exact frequency needed for resonance depends on three things: the strength of the external magnetic field, the identity of the nucleus (\ce{^1H}, \ce{^{13}C} and so on), and the electronic environment of the nucleus. It is the third that makes NMR so useful.

What a proton NMR spectrum tells us

1. Number of peaks (signals) = number of different chemical environments of the hydrogen atoms.
2. Position of each peak (chemical shift \delta) = the type of environment: next to O, Cl, \ce{C=O}, a benzene ring and so on.
3. Ratio of the areas under the peaks = ratio of the numbers of hydrogen atoms in each environment. The instrument draws an integration trace whose step heights give this ratio.
4. Splitting of each peak = the number of hydrogen atoms on the neighbouring carbon atom (n+1 rule, next section).

Examples from the book

Simple molecules give simple spectra. Chloroethane (ethyl chloride), \ce{CH3CH2Cl}, gives a peak at 1.5\,\mathrm{ppm} for the three alike protons of the methyl (\ce{CH3}) group and a peak at 3.5\,\mathrm{ppm} for the two protons of the methylene (\ce{CH2}) group. The \ce{CH2} protons are nearer to the highly electronegative chlorine atom, which pulls electron density away from them. They are more deshielded, feel more of the external field and so absorb further downfield (3.5\,\mathrm{ppm}) than the \ce{CH3} protons (1.5\,\mathrm{ppm}). The peak areas are in the ratio 2:3.

Benzene, \ce{C6H6}, gives a single peak at about 7.2\,\mathrm{ppm}, because all six protons are equivalent. The peak is far downfield because of the ring current: in the applied field the delocalised \pi electrons circulate around the ring, and the field they produce adds to B_0 at the positions of the hydrogen atoms outside the ring, so these protons are strongly deshielded.

18.8.4Environment of a proton and spin-spin splitting (multiplicity)Often examined

Equivalent protons. In proton NMR, all the hydrogen atoms attached to the same carbon atom have the same environment. Hydrogen atoms on different carbon atoms can also have the same environment, if those carbon atoms are bonded to exactly the same chemical groups, so that the molecule is symmetrical. For example, the two \ce{CH3} groups of propanone, \ce{CH3COCH3}, are equivalent, and all six protons give one peak.

Why peaks split

Each proton is itself a tiny magnet. The protons on the neighbouring carbon atom can be aligned with or against B_0, so they slightly increase or decrease the field felt by the proton being observed. Each possible arrangement of the neighbouring spins gives a slightly different resonance frequency, so the signal splits into several lines. This is called spin-spin splitting (spin-spin coupling).

• No hydrogen on the adjacent atoms: the signal stays a single peak, a singlet.
• One hydrogen on the adjacent atom: it can be up or down (2 arrangements), so the signal splits into two peaks of equal size, a doublet (1:1).
• Two equivalent protons on the adjacent carbon: the arrangements are ↑↑, ↑↓ or ↓↑, and ↓↓. This gives three lines, a triplet (1:2:1). The middle line is twice as tall because it can arise in two ways.
• Three equivalent protons on the adjacent carbon: four lines, a quartet (1:3:3:1).

Here n is the number of equivalent hydrogen atoms on the neighbouring carbon atom(s). This is the **n+1 rule**. It gives valuable information about the neighbourhood of a proton. The relative heights of the lines follow Pascal's triangle (1:1, 1:2:1, 1:3:3:1, 1:4:6:4:1 ...).

This is why, in the spectrum of chloroethane, the \ce{CH2} peak is a quartet (3 neighbouring protons on the \ce{CH3} group) and the \ce{CH3} peak is a triplet (2 neighbouring protons on the \ce{CH2} group).

Worked examples: predicting proton NMR spectra

18.8.5Proton exchange by D2O (deuterium exchange)Often examined

The peaks of \ce{O-H} and \ce{N-H} protons are hard to recognise. Their chemical shift changes with concentration, solvent and temperature because of hydrogen bonding, and they usually appear as singlets that are not split. A simple test picks them out.

When an NMR sample that contains \ce{O-H} or \ce{N-H} protons is shaken with a few drops of deuterium oxide, \ce{D2O} (heavy water), the \ce{O-H} or \ce{N-H} protons exchange with deuterium atoms from \ce{D2O}:

Deuterium gives no signal in a proton NMR spectrum. So when the spectrum is recorded again, the peak of the \ce{O-H} or \ce{N-H} proton has disappeared. This shows that the peak belonged to an \ce{O-H} or \ce{N-H} proton. The method is called deuterium exchange (the '\ce{D2O} shake'). Hydrogen atoms bonded to carbon do not exchange, because \ce{C-H} bonds are not polar enough to lose \ce{H+}.

Uses of deuterium exchange

• It confirms that a compound contains an \ce{O-H} group (alcohol, phenol, carboxylic acid) or an \ce{N-H} group (amine, amide).
• It shows which peak in the spectrum is the \ce{O-H} or \ce{N-H} peak, even when its position is unusual.
• It removes an \ce{O-H} peak that overlaps a \ce{C-H} peak, so the \ce{C-H} peak can be measured properly.

18.9, 18.9.1, 18.9.2Carbon-13 NMROften examined

About 1\% (more exactly 1.1\%) of all carbon atoms are the isotope \ce{^{13}C}. The rest are \ce{^{12}C}, apart from tiny amounts of radioactive \ce{^{14}C}. Carbon-13 NMR (\ce{^{13}C} NMR, also called CMR) is based on the magnetic properties of the \ce{^{13}C} nuclei. A \ce{^{13}C} nucleus (I=\tfrac12) is NMR-active and behaves like a small magnet; a \ce{^{12}C} nucleus (I=0) does not.

Like a proton, a \ce{^{13}C} nucleus in an external magnetic field B_0 can be aligned with the field or opposed to it. The opposed alignment is less stable (higher energy). Supplying exactly the right amount of energy, with radio waves of just the right frequency, flips the nucleus from the more stable to the less stable alignment, and this is detected as a peak on a graph. The flipping of the \ce{^{13}C} nucleus from one magnetic alignment to the other by the radio waves is known as the resonance condition.

Because only about 1 carbon atom in 90 is \ce{^{13}C}, the signals are weak, and the computer adds together many scans. There is also a useful simplification: two \ce{^{13}C} atoms are hardly ever next to each other, and the spectra are normally recorded in a way that stops the attached hydrogen atoms from splitting the carbon peaks. So each different carbon environment gives one single line, as in all the \ce{^{13}C} spectra in the book.

Chemical shift in 13C NMR

TMS is again the reference (\delta=0). A peak at a chemical shift of 60\,\mathrm{ppm} is said to be downfield of TMS; the further to the left a peak is, the more downfield it is. \ce{^{13}C} chemical shifts are much bigger than proton shifts: they range up to about 200\,\mathrm{ppm} (about 220\,\mathrm{ppm} for ketones), depending on the environment of the carbon atom.

The same ideas apply as for protons. A carbon bonded to an electronegative atom (O, N, Cl) is deshielded and appears downfield. The \ce{sp^2} carbons of \ce{C=C} bonds and benzene rings appear at 100\text{–}150\,\mathrm{ppm}. The carbon of a \ce{C=O} group, joined to oxygen by a double bond, is the most deshielded of all (155\text{–}220\,\mathrm{ppm}).

Deducing the structure of a molecule with 13C NMR

The \ce{^{13}C} NMR spectrum of each different molecule is distinct. The number of peaks gives the number of different carbon environments. The position of each peak, compared with standard values, gives the type of carbon atom: alkane, alkene, alkyne, aromatic and carbonyl carbons are easily recognised from their chemical shift ranges. Symmetrical molecules are simple and give fewer peaks, because some of their carbon atoms are in identical positions.

Comparing the peaks of a spectrum with these values helps to predict the structure of the molecule. In general, one spectroscopic technique on its own does not settle a structure. The combined information from different spectroscopic techniques, together with mass spectrometry, is used to find the structural formulae of organic compounds.

Comparing proton NMR (PMR) and carbon-13 NMR (CMR)

18.9.3Predicting the number of peaks in a 13C NMR spectrumOften examined

In a \ce{^{13}C} NMR spectrum, each unique carbon environment in the molecule gives a separate peak. The number of peaks equals the number of chemically different carbon atoms.

1. Look for symmetry in the molecule. Carbon atoms that are swapped by the symmetry are equivalent and give one peak.
2. Identify each unique carbon environment. Carbon atoms in different chemical environments (different neighbouring atoms or bonds) give separate peaks.
3. Use Table 18.3 to predict where each peak appears.

Using the techniques togetherOften examined

No single technique gives the whole structure. Each one answers a different question, and together they usually fix the structure without doubt.

1. Mass spectrometry: the molecular ion peak \ce{M+} gives the relative molecular mass; with analysis data this gives the molecular formula.
2. IHD from the molecular formula: the number of rings plus \pi bonds.
3. IR spectrum: the functional groups (\ce{O-H}, \ce{C=O}, \ce{N-H}, \ce{C=C} and so on).
4. **\ce{^{13}C} NMR:** the number of carbon environments and their types, for example a \ce{C=O} carbon near 200\,\mathrm{ppm}.
5. **\ce{^1H} NMR:** the number of hydrogen environments, their ratio and their neighbours. A \ce{D2O} shake picks out \ce{O-H} and \ce{N-H} protons.
6. Put the pieces together and check that the structure explains every peak.

Formula sheet

Key terms

Spectroscopy

A technique for structural elucidation of molecules based on the interaction of atoms or molecules with electromagnetic radiation.

Spectrum

A graph of how much radiation a sample absorbs or emits at each wavelength, frequency or wavenumber.

Electromagnetic spectrum

The whole range of electromagnetic radiation, from radio waves to gamma rays.

Mass spectrometry

A technique that ionises molecules and sorts the ions by mass-to-charge ratio to find molecular masses, fragment masses and isotope abundances.

Index of hydrogen deficiency (degree of unsaturation)

The total number of rings plus \pi bonds in a molecule, found from the molecular formula.

Wavenumber

The number of waves per centimetre, \bar{\nu}=1/\lambda; unit \mathrm{cm^{-1}}.

IR spectroscopy

Spectroscopy using infrared radiation (4000 to 625\,\mathrm{cm^{-1}}), which makes bonds vibrate; used to identify functional groups.

Stretching

A vibration in which the bond length changes.

Bending

A vibration in which the bond angle changes.

Percentage transmittance

The percentage of the incident radiation that passes through the sample; the y-axis of an IR spectrum.

Functional group region

The part of an IR spectrum from 4000 to 1500\,\mathrm{cm^{-1}} where functional groups absorb at predictable wavenumbers.

Fingerprint region

The part of an IR spectrum below 1500\,\mathrm{cm^{-1}} whose complex pattern is unique to each compound.

UV-visible spectroscopy

Measurement of the absorption of UV and visible light (200 to 800\,\mathrm{nm}), which excites electrons; used to find conjugation and concentration and to explain colour.

Electronic transition

The movement of an electron from a filled orbital (\sigma, \pi or n) to an empty antibonding orbital (\sigma^* or \pi^*).

Chromophore

The group of atoms responsible for absorbing UV or visible light, such as \ce{C=C}, \ce{C=O} or a benzene ring.

Conjugation

Alternating single and multiple bonds whose p orbitals overlap so that the \pi electrons are delocalised.

\lambda_{\max}

The wavelength of greatest absorption.

Complementary colour

The colour seen when a substance absorbs light of another colour; opposite colours on the colour wheel.

Beer–Lambert law

Absorbance is directly proportional to concentration and path length: A=\varepsilon cl.

Atomic emission spectroscopy (AES)

Identification and measurement of elements from the wavelengths of light emitted by their excited atoms.

Atomic absorption spectrum

Dark lines produced when ground-state atoms absorb particular wavelengths from light passing through them.

Spectral lines

Lines of single wavelengths in an atomic spectrum; each corresponds to a jump between two quantized energy levels.

NMR spectroscopy

Spectroscopy in which nuclei with spin, placed in a strong magnetic field, absorb radio waves and change their spin orientation.

NMR-active nucleus

A nucleus with non-zero spin (I\neq0), such as \ce{^1H} or \ce{^{13}C}, which behaves like a small magnet.

Resonance

Absorption of radio waves whose energy exactly matches the gap between the two spin states of a nucleus.

Nuclear spin flipping

The change of a nucleus from the lower-energy to the higher-energy spin state on absorbing radio waves.

Shielding

Reduction of the field felt by a nucleus by the opposing field of its surrounding electrons; gives a low chemical shift (upfield).

Deshielding

Loss of electron density around a nucleus, e.g. near an electronegative atom, so it feels more of the applied field; gives a high chemical shift (downfield).

Chemical shift

The difference between the resonance frequency of a nucleus and that of TMS, in ppm of the operating frequency; symbol \delta.

Tetramethylsilane (TMS)

\ce{Si(CH3)4}, the reference compound of NMR whose signal defines \delta=0.

Deuterated solvent

A solvent in which H is replaced by deuterium, such as \ce{CDCl3}, so it gives no signal in a proton spectrum.

Equivalent protons

Hydrogen atoms in identical chemical environments; they give one signal and do not split each other.

Spin-spin splitting

Splitting of an NMR signal into several lines by the spins of protons on the neighbouring carbon atom.

n + 1 rule

A proton with n equivalent neighbouring protons gives a signal of n+1 lines.

Integration

The measurement of peak areas in proton NMR; the areas are in the ratio of the numbers of H atoms.

Deuterium exchange

Shaking a sample with \ce{D2O} so that \ce{O-H} and \ce{N-H} protons are replaced by D and their peaks disappear.

Carbon-13 NMR (CMR)

NMR of the \ce{^{13}C} nuclei (1.1% of carbon); one line per carbon environment, shifts 0 to 220\,\mathrm{ppm}.

Proton NMR (PMR)

NMR of the \ce{^1H} nuclei of a compound.

Upfield and downfield

Towards the right (low \delta, shielded) and towards the left (high \delta, deshielded) of an NMR spectrum.

FBISE 2026 questions

Questions from the FBISE HSSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

Exam focus

1. Write the principle of IR spectroscopy and one application. (short)
   Asked in 2026 (Set B Q.2 (xii) OR) and the basis of long question i.
   Go to answer
2. Identify the bonds for IR bands at 3300\text{–}3500, 1700\text{–}1725 and 1050\text{–}1150\,\mathrm{cm^{-1}}. (short)
   Asked in 2026 (Set D); needs Table 18.1 learnt by heart.
   Go to answer
3. Why are deuterated solvents used in NMR? What is TMS and why is it the standard? (long)
   Asked three times in 2026 (Set B Q.2 (xiii), Set D Q.6 OR a and b).
   Go to answer
4. Predict the number of \ce{^1H} and \ce{^{13}C} NMR peaks of simple molecules (acetone, ethanol, dimethyl ether). (short)
   Three 2026 MCQs and a 3-mark question (dimethyl ether) test peak counting.
   Go to answer
5. Compare the UV absorption of cyclohexene and buta-1,3-diene; which compound absorbs at the longest wavelength? (short)
   Conjugation was asked in the old-syllabus paper and in Set B MCQ 15.
   Go to answer
6. Calculate the index of hydrogen deficiency from a molecular formula. (numerical)
   An SLO of the unit; also the first step of every structure-deduction question.
   Go to answer
7. Explain chemical shift, shielding and the n+1 splitting rule. (short)
   Book short questions ix and x; needed to explain any NMR spectrum.
   Go to answer
8. What is the deuterium exchange method and its use? (short)
   Book short question xi; a standard 3-mark question.
   Go to answer
9. Predict the colour of a compound from its UV-visible spectrum. (short)
   SLO of the unit; book short questions iii and iv.
   Go to answer
10. Principles of AES and AAS and the differences between them. (short)
    Book short questions v and vi.
    Go to answer
11. How is PMR different from CMR? NMR as a tool for structure elucidation. (long)
    Book long questions iii and iv, likely Section C questions.
    Go to answer