Analytical techniques

In a nutshell

Chemists identify an unknown organic compound by combining evidence from several techniques. At AS you need three: elemental analysis (gives the empirical formula), mass spectrometry (gives the M_r and pieces of the structure) and infrared (IR) spectroscopy (shows which bonds, and so which functional groups, are present).

IR also explains why some atmospheric gases warm the planet, and it is used to monitor car exhaust gases and to measure alcohol in a driver's breath.

Assumed knowledge: Amount of substance: empirical and molecular formulae; Alcohols: oxidation of alcohols to aldehydes, ketones and carboxylic acids.

Core content

Infrared spectroscopy: what IR radiation does to bonds

Infrared radiation causes covalent bonds to vibrate more and absorb energy.

• Bonds are always vibrating (stretching and bending). Absorbing IR makes them vibrate more.
• Each type of bond absorbs IR at its own characteristic wavenumber range, so the wavenumbers absorbed tell you which bonds are in the molecule.
• IR does not break bonds. The molecule is unchanged afterwards.

An IR spectrum plots transmittance (the percentage of radiation that passes through) against wavenumber in cm^-1. An absorption shows up as a dip (a trough), not a peak sticking up. Exam papers still call these dips "peaks" or "absorptions".

The OCR Data Sheet gives the wavenumber ranges, so you do not need to learn them. The ones you use at AS:

Two points that stop you misreading a spectrum:

• Almost every organic compound has a C-H absorption at about 3000 cm^-1. It tells you nothing about the functional group.
• Below about 1500 cm^-1 there are many overlapping absorptions. Do not try to assign them all. Use the clear absorptions above 1500 cm^-1 (O-H and C=O) to identify the functional group.

Still don't get it? · reading an IR spectrum

Think of shining a torch through a row of tuning forks. Each fork only "rings" at its own note, and when it rings it soaks up a little of that note. Listen to what comes out the other side, and the missing notes tell you which forks were there.

Step by step:

1. Each bond (O-H, C=O, C-H...) is like a fork with its own note. Its "note" is a particular wavenumber of IR radiation.
2. The spectrometer shines every IR wavenumber through the sample and measures how much gets through (transmittance).
3. Where a bond absorbs, less gets through, so the line dips down. A dip is an absorption.
4. The x-axis runs backwards: 4000 cm^-1 on the left, falling to about 400 cm^-1 on the right. Read the wavenumber straight down from the bottom of the dip.
5. Match that wavenumber to a bond using the Data Sheet table. Ignore the C-H dip at about 3000 (nearly every organic molecule has one) and the crowded region on the right.

Exam version: "an absorption at 1630-1820 cm^-1 shows a C=O bond". Always give the bond, the wavenumber (or the Data Sheet range) and, if asked, the functional group.

Infrared spectroscopy: greenhouse gases and global warming

The atmospheric gases CO₂, H₂O and CH₄ absorb IR because they contain C=O, O-H and C-H bonds.

The causal chain:

1. Radiation from the Sun passes through the atmosphere and warms the Earth's surface.
2. The warm surface emits infrared radiation.
3. The C=O, O-H and C-H bonds in CO₂, H₂O and CH₄ absorb this IR and vibrate more.
4. The molecules re-emit the energy in all directions, including back towards the Earth.
5. So less energy escapes to space, and the lower atmosphere warms.

The suspected link to global warming:

• Burning fossil fuels has increased the concentration of CO₂ in the atmosphere. Methane levels have also risen (for example from farming and landfill).
• More greenhouse gas means more IR is absorbed, so more warming. This is the enhanced greenhouse effect.
• Rising CO₂ levels are linked to rising global temperatures.

Resulting changes to energy usage. Because the scientific evidence for global warming has been accepted, governments have agreed international targets (for example the Kyoto Protocol and the Paris Agreement) and brought in policies to:

• switch to renewable energy supplies (wind, solar, hydroelectric) in place of fossil fuels
• use less energy and use it more efficiently, to cut CO₂ emissions

Infrared spectroscopy: identifying alcohols, carbonyls and carboxylic acids

Three functional groups have a clear IR "signature". Learn which absorptions to look for, and which must be absent.

Alcohol (propan-1-ol): the broad O-H is the giveaway, and there is no C=O.

Ketone (propanone): a strong, sharp C=O, and no O-H at all. An aldehyde looks very similar.

Carboxylic acid (propanoic acid): both the C=O and a very broad O-H that sits lower than an alcohol's and often swallows the C-H absorption.

IR cannot tell an aldehyde from a ketone at AS: both just show C=O. You need other evidence (for example fragment ions in the mass spectrum, or the reaction that made it).

Infrared spectroscopy: interpreting and predicting spectra

You may be given the spectrum of a compound you have never met, or asked to predict what a spectrum would show. Use the Data Sheet and the functional groups in this specification.

To interpret a spectrum:

1. Look for an absorption at 1630-1820 cm^-1 (C=O). Present or absent?
2. Look for O-H. Is it at 3200-3600 (alcohol) or very broad at 2500-3300 (carboxylic acid)?
3. Check anything else that fits your other data, such as C=C at 1620-1680 cm^-1 for an alkene.
4. Write each conclusion as bond + wavenumber + functional group.

To predict a spectrum, list the bonds in the molecule and give the Data Sheet range for each one that matters.

A common use is following a reaction. When an alcohol is oxidised (see Alcohols):

Infrared spectroscopy: monitoring pollution and breathalysers

IR is used wherever a gas must be identified and measured quickly without destroying the sample.

• Air pollution: IR monitors gases in car emissions, such as carbon monoxide (CO) and nitrogen monoxide (NO). Each gas absorbs at its own characteristic wavenumbers, and the amount absorbed shows how much is present.
• Breathalysers: modern breathalysers use IR to measure the concentration of ethanol in the breath. The greater the absorption by bonds in ethanol, the more ethanol is present. The reading can be used as evidence in court for drink driving.

Mass spectrometry: the molecular ion and relative molecular mass

In a mass spectrometer, organic molecules are ionised by losing one electron. The whole molecule with one electron removed is the molecular ion, M⁺.

$\text{M} \rightarrow \text{M}^{+} + \text{e}^{-}$

The ions are separated by their mass-to-charge ratio, m/z. At AS every ion has a single positive charge, so m/z equals the mass of the ion.

The molecular ion peak is the peak with the highest m/z (apart from the tiny M+1 peak, below). Its m/z value is the relative molecular mass, M_r, of the compound.

Read it like this:

• The peak at the highest m/z is 58, so M_r of butane = 58. Check: C₄H₁₀ = (4 × 12.0) + (10 × 1.0) = 58.0.
• The tallest peak (m/z 43) is not the molecular ion. It is a fragment that happens to be very stable and common.
• The tiny peak at m/z 59 is the M+1 peak.

The M+1 peak. About 1% of carbon atoms are carbon-13 rather than carbon-12. Molecules containing one ¹³C atom are 1 mass unit heavier, so they give a small peak one unit to the right of M⁺. Do not mistake it for the molecular ion.

Mass spectrometry: fragmentation peaks

Inside the spectrometer, some molecular ions break apart. A covalent bond breaks to give a positive fragment ion and a neutral radical:

$\text{M}^{+} \rightarrow \text{X}^{+} + \text{Y}^{\bullet}$

• Only the positive ion (X⁺) is detected, so only it gives a peak.
• The radical (Y•) is uncharged, so it is not detected. You can work out its mass from the difference between M_r and the fragment peak.
• (Strictly, the molecular ion also has an unpaired electron and is sometimes written M^+•. What matters at AS is that the detected fragment carries the + charge.)

Every fragment ion you write must show its + charge. Common fragment ions to recognise:

Some m/z values fit two ions with the same mass (29, 43, 57). Pick the one that is actually part of the molecule you are considering.

Example: butanone, CH₃COCH₂CH₃ (M_r = 72).

For example, the peak at m/z 43 forms like this:

CH₃COCH₂CH₃⁺ → CH₃CO⁺ + CH₃CH₂•

Still don't get it? · molecular ions and fragments

Imagine posting a model made of snap-together pieces through a machine that weighs whatever arrives at the other end, but only weighs pieces with a sticker on them. Some models arrive whole. Others snap in two on the way, and only the half that kept the sticker gets weighed.

Step by step:

1. The sticker is the positive charge. The machine (the mass spectrometer) knocks one electron off each molecule, so the whole molecule becomes a + ion: the molecular ion, M+.
2. Any M+ that arrives whole is weighed at its full mass. That is the peak furthest to the right, and its m/z is the Mr.
3. Some M+ ions snap at a bond on the way. One piece keeps the + charge (the fragment ion), and the other piece is neutral (a radical).
4. The detector only "sees" charged pieces, so only the fragment ion gives a peak. The neutral piece is invisible.
5. You can still work out what fell off: Mr minus the fragment's m/z is the mass of the missing piece. A gap of 15 means CH3 was lost; a gap of 29 means C2H5 (or CHO).

Exam version: the molecular ion peak is the peak with the highest m/z, and its m/z gives the M_r. Fragment ions are positive ions formed when the molecular ion breaks up; always write them with a + charge, for example CH₃CO⁺ at m/z 43.

Combined techniques: deducing a structure

In the exam you are often given several pieces of data and asked to deduce the structure. Each technique answers a different question:

Work through it in this order:

1. Empirical formula from the percentages: divide each % by A_r, then divide by the smallest answer.
2. Molecular formula: divide M_r (from the M⁺ peak) by the empirical formula mass, and multiply up.
3. Functional group from the IR spectrum.
4. Draw every possible isomer that has that molecular formula and that functional group.
5. Use the fragment peaks to decide between them.
6. Check that your final structure fits every piece of data.

Halogen compounds are not examined for mass spectra at this level.

Worked examples

1. Elemental analysis and the molecular ion: molecular formula

Task: an organic compound contains 54.5% carbon, 9.1% hydrogen and 36.4% oxygen by mass. Its mass spectrum has the molecular ion peak at m/z = 88. Find its molecular formula.

Step 1: divide each percentage by A_r (Data Sheet values) to get the mole ratio.

$\text{C}: \frac{54.5}{12.0} = 4.54$

$\text{H}: \frac{9.1}{1.0} = 9.1$

$\text{O}: \frac{36.4}{16.0} = 2.275$

Step 2: divide by the smallest value (2.275).

$\text{C}: \frac{4.54}{2.275} = 2.00 \qquad \text{H}: \frac{9.1}{2.275} = 4.00 \qquad \text{O}: \frac{2.275}{2.275} = 1$

$\text{empirical formula} = \text{C}_2\text{H}_4\text{O}$

Step 3: find the empirical formula mass.

$M(\text{C}_2\text{H}_4\text{O}) = (2 \times 12.0) + (4 \times 1.0) + 16.0 = 44.0$

Step 4: use the molecular ion peak. M_r = 88, so:

$\frac{88}{44.0} = 2$

$\text{molecular formula} = \text{C}_4\text{H}_8\text{O}_2$

Where marks go: the M_r comes from the molecular ion (highest m/z) peak, not the tallest peak. Do not round 2.00 and 4.00 until you have divided by the smallest value.

2. Identifying fragment ions

Task: the mass spectrum of butanone, CH₃COCH₂CH₃, has peaks at m/z 72, 57, 43 and 29. Identify the species responsible for each peak.

Step 1: confirm the molecular ion.

$M_r(\text{C}_4\text{H}_8\text{O}) = (4 \times 12.0) + (8 \times 1.0) + 16.0 = 72.0$

So m/z 72 is the molecular ion, CH₃COCH₂CH₃⁺.

Step 2: for each fragment, work out what was lost, then what is left.

$72 - 57 = 15 \;\Rightarrow\; \text{CH}_3 \text{ lost, so m/z 57 is } \text{CH}_3\text{CH}_2\text{CO}^{+}$

$72 - 43 = 29 \;\Rightarrow\; \text{C}_2\text{H}_5 \text{ lost, so m/z 43 is } \text{CH}_3\text{CO}^{+}$

$72 - 29 = 43 \;\Rightarrow\; \text{CH}_3\text{CO} \text{ lost, so m/z 29 is } \text{CH}_3\text{CH}_2^{+}$

Step 3: check each ion's mass adds up.

• CH₃CH₂CO⁺ = (3 × 12.0) + (5 × 1.0) + 16.0 = 57
• CH₃CO⁺ = (2 × 12.0) + (3 × 1.0) + 16.0 = 43
• CH₃CH₂⁺ = (2 × 12.0) + (5 × 1.0) = 29

Where marks go: each fragment needs its + charge. At m/z 43, butanone gives CH₃CO⁺, not C₃H₇⁺, because butanone has no C₃H₇ group.

3. Predicting IR changes in a reaction

Task: propan-1-ol is oxidised by acidified potassium dichromate(VI). Describe how the IR spectrum of the organic product shows whether propanal or propanoic acid has formed.

1. Propan-1-ol has an O-H absorption at 3200-3600 cm^-1 and no C=O.
2. Propanal (aldehyde, distilled off): the alcohol O-H at 3200-3600 cm^-1 disappears, and a C=O absorption at 1630-1820 cm^-1 appears. No O-H at all.
3. Propanoic acid (heated under reflux): C=O at 1630-1820 cm^-1 appears, and a very broad O-H at 2500-3300 cm^-1 appears.
4. So: C=O with no O-H means propanal; C=O with a broad 2500-3300 cm^-1 O-H means propanoic acid.

Where marks go: name the bond and the wavenumber range each time, and make clear which O-H range you mean (alcohol or acid).

4. Combining all three techniques

Task: compound Y contains 48.6% carbon, 8.1% hydrogen and 43.2% oxygen by mass. Use this and the data below to deduce its structure.

Step 1: empirical formula from elemental analysis.

$\text{C}: \frac{48.6}{12.0} = 4.05 \qquad \text{H}: \frac{8.1}{1.0} = 8.1 \qquad \text{O}: \frac{43.2}{16.0} = 2.70$

Divide by the smallest (2.70):

$\text{C}: \frac{4.05}{2.70} = 1.50 \qquad \text{H}: \frac{8.1}{2.70} = 3.00 \qquad \text{O}: \frac{2.70}{2.70} = 1$

1.50 is not a whole number, so multiply every value by 2:

$\text{C} : \text{H} : \text{O} = 3 : 6 : 2 \;\Rightarrow\; \text{empirical formula} = \text{C}_3\text{H}_6\text{O}_2$

Step 2: molecular formula from the molecular ion. The highest m/z peak is 74.

$M(\text{C}_3\text{H}_6\text{O}_2) = (3 \times 12.0) + (6 \times 1.0) + (2 \times 16.0) = 74.0$

74 ÷ 74.0 = 1, so the molecular formula is C₃H₆O₂.

Step 3: functional group from IR.

• Absorption at 1715 cm^-1: within 1630-1820 cm^-1, so C=O.
• Very broad absorption across about 2500-3300 cm^-1: O-H in a carboxylic acid.
• So Y is a carboxylic acid.

Step 4: structure. The only carboxylic acid with formula C₃H₆O₂ is propanoic acid, CH₃CH₂COOH.

Step 5: check against the fragments.

Answer: Y is propanoic acid, CH₃CH₂COOH.

Where marks go: the empirical-formula ratio 1.5 must be doubled, not rounded to 2; the M⁺ confirms the molecular formula; the IR answer needs both the C=O and the broad acid O-H; each fragment needs its + charge.

Common exam mistakes

• Saying IR "breaks" bonds. IR makes bonds vibrate more; nothing breaks. Bond-breaking happens in the mass spectrometer, not in IR.
• Using the wrong O-H range. Alcohol O-H is 3200-3600 cm^-1; carboxylic acid O-H is 2500-3300 cm^-1 and very broad. Quoting the alcohol range for an acid (or the reverse) loses the mark.
• Using the C-H absorption at about 3000 cm^-1 as evidence. Nearly every organic compound has it, so it does not identify a functional group.
• Naming a wavenumber but not the bond. "There is a peak at 1715" is not enough. Write "C=O at 1715 cm^-1 (1630-1820 cm^-1)".
• Calling a carboxylic acid an alcohol or a ketone. A carboxylic acid needs both C=O and the broad acid O-H. Seeing only one of them is not enough.
• Claiming IR distinguishes an aldehyde from a ketone. Both show only C=O. Use other evidence.
• Choosing the tallest peak as the molecular ion. The molecular ion is the peak with the highest m/z (ignoring the small M+1 peak), not the most abundant one.
• Mistaking the M+1 peak for M⁺. The tiny peak one unit above M⁺ comes from carbon-13, so the M_r is one less than it.
• Leaving the + off fragment ions. "CH₃CO" or "C₂H₅" is not a fragment ion. Write CH₃CO⁺, C₂H₅⁺.
• Giving the radical as the detected species. Only the positive ion is detected; the neutral radical gives no peak.
• Picking an ion that is not in the molecule. m/z 43 is CH₃CO⁺ in a methyl ketone but C₃H₇⁺ in a propyl compound. Choose the one that fits your structure.
• Rounding empirical-formula ratios too early. A ratio of 1.50 means multiply by 2, not round up. 1.33 means multiply by 3.
• Writing only that greenhouse gases "trap heat" or "reflect" radiation. The marks are for IR being absorbed by C=O, O-H and C-H bonds, which vibrate, and the energy being re-emitted.
• Mixing up global warming with ozone depletion. Greenhouse gases absorb infrared; ozone absorbs ultraviolet. They are different problems.
• Stopping before checking. In combined questions, show that your structure fits every piece of data, including the fragments.

Key definitions

• Wavenumber: the number of waves per centimetre, in cm^-1; the unit used for the position of IR absorptions.
• Infrared absorption: infrared radiation causes covalent bonds to vibrate more and absorb energy; each type of bond absorbs at a characteristic wavenumber range.
• Greenhouse gas: a gas (such as CO₂, H₂O or CH₄) whose C=O, O-H or C-H bonds absorb infrared radiation emitted from the Earth's surface.
• Molecular ion, M⁺: the positive ion formed when a molecule loses one electron; it has the same mass as the molecule (the mass of an electron is negligible).
• Molecular ion peak: the peak with the highest m/z in a mass spectrum (apart from the small M+1 peak); its m/z gives the relative molecular mass.
• M+1 peak: a small peak one unit above the molecular ion peak, caused by molecules containing one carbon-13 atom.
• Fragment ion: a positive ion formed when a molecular ion breaks up in a mass spectrometer.
• Fragmentation: the breaking of a molecular ion into a positive fragment ion and a neutral radical; only the ion is detected.
• Empirical formula: the simplest whole number ratio of atoms of each element present in a compound.
• Molecular formula: the number and type of atoms of each element in a molecule.

Specification

• I can state that infrared radiation causes covalent bonds to vibrate more and absorb energy. (4.2.4 a)
• I can explain how CO₂, H₂O and CH₄ absorb infrared radiation through their C=O, O-H and C-H bonds, the suspected link to global warming, and how this has changed energy usage. (4.2.4 b)
• I can use an IR spectrum to identify an alcohol (O-H), an aldehyde or ketone (C=O), and a carboxylic acid (C=O and broad O-H). (4.2.4 c)
• I can interpret and predict the IR spectrum of a familiar or unfamiliar substance using Data Sheet values. (4.2.4 d)
• I can describe how IR is used to monitor CO and NO in car emissions and to measure ethanol in breathalysers. (4.2.4 e)
• I can identify the molecular ion peak in a mass spectrum and use it to find the relative molecular mass, and I know a small M+1 peak comes from carbon-13. (4.2.4 f)
• I can suggest the structures of fragment ions, with their + charge, from peaks in a mass spectrum. (4.2.4 g)
• I can deduce the structure of an organic compound by combining elemental analysis, mass spectra and IR spectra. (4.2.4 h)

Related notes

• Atomic structure and isotopes: mass spectra of elements, m/z and isotopes
• Amount of substance: empirical and molecular formula calculations
• Basic concepts of organic chemistry: functional groups, radicals and structural isomers
• Alkanes: combustion, CO from incomplete combustion
• Alcohols: oxidation products you can track with IR
• Organic synthesis: making and purifying the compounds you then analyse
• Spectroscopy (A level): NMR, and combining all the spectroscopic techniques