What this chapter covers22

Every compound you have met so far was named for you. In a real laboratory nobody hands you a label: you have a colourless liquid in a vial and have to find out what it is. Two instruments do most of that work. An infrared spectrometer tells you which bonds are in the molecule, and so which functional groups it has. A mass spectrometer tells you how heavy the molecule is, how it breaks apart, and — from small companion peaks — how many carbon atoms it has and whether it contains chlorine or bromine. Put the two together and most small molecules give themselves away.

The verbs tell you how the marks are earned. Almost every outcome says analyse, deduce or suggest: you are given a spectrum and asked what it shows. You are not asked to explain how either instrument works, and 22.2.1 says so outright for the mass spectrometer. What you do need is a fluent way of reading each kind of spectrum and a set of numbers you can work with quickly: the wavenumber ranges in the data table, the masses of common fragments, the 1.1 in the carbon formula, and the 3 : 1 and 1 : 1 patterns of chlorine and bromine.

You will be given the infrared data table in the exam, so the wavenumbers themselves are not something to memorise. What wins marks is knowing which rows to look at, how to tell the two O–H rows apart by the shape of the absorption, and what a spectrum with no peak in a region tells you. The mass spectrometry calculations are the opposite: nothing is given except the formula in 22.2.5, so the method has to be yours.

The page is in three parts. The first is infrared spectroscopy: why bonds absorb infrared at all, which vibrations do, how to use the data table, and a method for reading a spectrum that works on any simple molecule. The second applies mass spectrometry to elements, where the peaks are isotopes and the job is to calculate a relative atomic mass. The third applies it to organic compounds: the molecular ion, fragmentation, the [M + 1] peak and the [M + 2] peak, ending with problems that need an infrared spectrum and a mass spectrum together.

What an infrared spectrum shows22.1.1

Infrared radiation is the part of the electromagnetic spectrum just beyond red light: longer wavelength and lower energy than anything you can see. When infrared passes through a sample of an organic compound, most of it goes straight through, but radiation of some particular frequencies is absorbed. Which frequencies are absorbed depends on which bonds the molecules contain. An infrared spectrum is a record of that pattern.

Two things about the axes catch people out, because both run the opposite way to most graphs you draw.

• The vertical axis is transmittance, the percentage of the radiation that gets through. 100% is at the top. An absorption therefore shows as a dip down from the top line, and it is still called a peak.
• The horizontal axis is wavenumber, in cm⁻¹, which is the number of waves per centimetre: wavenumber = 1 ÷ wavelength in cm. A larger wavenumber means a shorter wavelength, a higher frequency and a higher energy. The axis is drawn with the largest value on the left, usually 4000 cm⁻¹, falling to about 500 cm⁻¹ on the right.

To convert, 1 cm is 10⁷ nm, so a wavenumber of 1700 cm⁻¹ is a wavelength of 10⁷ ÷ 1700 ≈ 5900 nm, about ten times the wavelength of red light. The range an infrared spectrum covers, 4000 to 500 cm⁻¹, is 2500 to 20 000 nm. Chemists quote wavenumbers rather than wavelengths because wavenumber rises with energy, so "a higher wavenumber" means "a stiffer bond or a lighter atom" without any turning upside down in your head.

Why bonds absorb infrared22.1.1

A covalent bond is not rigid. The two atoms it holds are constantly moving towards and away from each other, as if joined by a spring, and a group of three or more atoms can also flex, changing the angle between its bonds. These movements are called vibrations. The two basic kinds are:

• stretching — the bond length increases and decreases, the atoms moving along the line of the bond;
• bending — the bond angle opens and closes, the atoms moving across the line of the bond.

Each vibration has its own natural frequency. When infrared radiation of exactly that frequency arrives, the bond can absorb a photon and vibrate more vigorously. Radiation of any other frequency passes by. So every absorption in a spectrum marks a frequency at which some bond in the molecule naturally vibrates.

The natural frequency is set by two things, exactly as for a mass on a spring:

• the stiffness of the bond — a stronger bond is a stiffer spring and vibrates faster. A triple bond vibrates faster than a double bond, and a double bond faster than a single bond between the same two atoms;
• the masses of the atoms — lighter atoms vibrate faster. Any bond to hydrogen, the lightest atom of all, vibrates at a high frequency.

Those two rules explain the whole layout of the data table. Every bond to hydrogen (O–H, N–H, C–H) absorbs at the left-hand end, above 2800 cm⁻¹. The triple bond C≡N comes next, around 2200. The double bonds C=O and C=C come next, between 1500 and 1750. Single bonds between heavier atoms — C–O, C–C, C–Cl — come last, below 1300 cm⁻¹. Stretching needs more energy than bending the same bond, so bends appear further to the right, mostly in the region below 1500 cm⁻¹.

Which vibrations absorb: the dipole rule22.1.1

Having a natural frequency is not enough. Radiation is an oscillating electric field, and it can only push on a vibration that moves electric charge about. So a vibration absorbs infrared only if it changes the dipole moment of the molecule. A vibration that does not change the dipole moment is infrared inactive: it happens, but it leaves no mark on the spectrum.

The clearest case is a molecule made of two identical atoms. In N₂, O₂ or Cl₂ both atoms have the same electronegativity, the bond has no dipole, and stretching it creates none. These molecules do not absorb infrared at all. That is why air, which is mostly nitrogen and oxygen, does not appear in an infrared spectrum, and why a spectrometer can be run open to the air. A molecule of two different atoms, such as HCl or CO, has a polar bond; stretching the bond changes the size of the dipole, so it absorbs.

Carbon dioxide shows that the rule is about the whole molecule, not each bond. Each C=O bond is polar, but the molecule is linear, so the two bond dipoles cancel. In the symmetric stretch both oxygens move out together: each bond dipole grows, but they still cancel, so the molecule's dipole stays zero. That vibration is inactive. In the asymmetric stretch one bond lengthens while the other shortens, so the dipoles no longer cancel; and in the bend the molecule stops being straight, so they no longer cancel either. Both of those absorb.

For the organic molecules in 22.1 the rule has one practical consequence. A bond with a large dipole, such as C=O or O–H, gives a strong absorption, because stretching it moves a lot of charge. A bond with little or no dipole gives a weak one. The C=C bond in a symmetrical alkene is the extreme case: its absorption is often weak, and in a perfectly symmetrical molecule it can vanish altogether. So a missing C=C peak is not proof that there is no C=C, whereas a missing C=O peak is good evidence there is no C=O.

The data table and the two regions22.1.1

You are given a table of characteristic absorptions in the exam, and 22.1.1 limits the functional groups you are expected to recognise to the ones in it. It lists each bond, the functional groups containing it, and the range of wavenumbers where it absorbs.

Learn what the group names in the middle column mean, because the table uses them rather than the family names you are used to. Hydroxy is the –OH of an alcohol. Carboxyl is the –COOH of a carboxylic acid. Carbonyl is the C=O of an aldehyde or ketone. Ester, amide, amine and nitrile mean what they say. So an alcohol is found on the "hydroxy" rows, and a carboxylic acid on the "carboxyl" rows — including its own separate O–H row.

A spectrum divides into two regions, and you use them differently.

• Above about 1500 cm⁻¹ is the functional-group region. It contains few peaks, each one belonging to a particular bond, and this is where you identify functional groups.
• Below about 1500 cm⁻¹ is the fingerprint region. It is crowded with overlapping absorptions from single bonds and bending vibrations of the whole molecule. You cannot assign most of these peaks, but the pattern as a whole is unique to one compound, like a fingerprint. Matching it against the spectrum of a known sample identifies a compound exactly; two different compounds never share a fingerprint region.

The C–O row sits inside the fingerprint region, which is why it is the least reliable row in the table: there are usually several peaks between 1040 and 1300 whether or not the molecule has a C–O bond. Use it only to confirm what another peak has already told you.

The O–H absorptions: alcohols and acids22.1.1

The O–H bond is the easiest to spot in any spectrum, because it gives the broadest absorption there is. The reason is hydrogen bonding. In a liquid or solid alcohol, the O–H groups of neighbouring molecules are hydrogen bonded to one another, and every O–H bond is held in a slightly different arrangement. Each arrangement changes the stiffness of the bond a little, so instead of one sharp wavenumber there is a spread of them, and the peak is smeared into a wide, rounded trough.

The table gives two O–H rows, and they are told apart by position and by shape:

The acid's O–H is lower and broader because the hydrogen bonding is stronger. Two acid molecules link into a pair, each O–H hydrogen bonded to the other molecule's C=O. Stronger hydrogen bonding weakens the O–H bond itself a little, so it vibrates at a lower wavenumber, and the spread of environments is wider.

The C=O absorption and the carbonyl families22.1.1

If the O–H is the easiest peak to see, the C=O is the most useful. It is strong, because the bond is very polar, and it is sharp, and it falls in a part of the spectrum where little else absorbs. A strong, sharp peak between 1640 and 1750 cm⁻¹ is as close to certain proof of a C=O bond as infrared gives.

The table has three C=O rows, and they overlap: esters at 1710–1750, aldehydes, ketones and acids at 1670–1740, and amides at 1640–1690. So the exact position of the C=O peak cannot, on its own, tell you which of these families you have. What tells them apart is which other peaks come with it:

The table shows the method: decide the family from the combination of peaks, and use the absence of a peak as evidence as firmly as its presence. An ester and a carboxylic acid can have the same molecular formula — methyl ethanoate and propanoic acid are both C₃H₆O₂ — and both have a C=O and a C–O. Only the acid has the broad O–H, so that one region settles it.

N–H, C≡N, C=C and C–O22.1.1

The remaining rows of the table each have a characteristic place and appearance.

N–H, 3300–3500 cm⁻¹ (amines and amides). This overlaps the alcohol's O–H range. The N–H peak is usually weaker and sharper than an O–H, because the N–H bond is less polar and hydrogen bonding between amine molecules is weaker. A primary amine, –NH₂, often shows two small spikes, one from each of its stretching vibrations; an O–H is always one rounded trough. If a peak in this region is broad and strong, think O–H; if it is narrow, or a pair, think N–H.

C≡N, 2200–2250 cm⁻¹ (nitriles). This is the only row in the whole middle of the spectrum. Between about 2250 and 1750 cm⁻¹ there is nothing else in the table, so any sharp peak there means a triple bond. The syllabus names only nitriles. The peak is sharp and of medium strength.

C=C, 1500–1680 cm⁻¹ (alkenes and aromatic compounds). Usually weak, for the reason given in section 4, and sometimes lost among other peaks. A benzene ring gives several peaks between 1450 and 1600. An alkene's C=C is best confirmed by chemistry — bromine water — rather than by infrared.

C–H, 2850–2950 cm⁻¹. Almost every organic molecule has C–H bonds, so this peak is almost always present and almost never informative. Its one use is as a landmark: a spectrum with nothing around 2900 either has no C–H bonds at all or has them buried under a carboxylic acid's O–H.

C–O, 1040–1300 cm⁻¹ (alcohols and esters, and also acids and ethers). Strong, but in the fingerprint region, so treat it as supporting evidence only.

Analysing a spectrum, step by step22.1.1

22.1.1 asks you to analyse the spectrum of a simple molecule. The method below works for every compound the syllabus can ask about, and it has the advantage that you write down evidence as you go, which is what the mark scheme rewards.

1. Look above 3000 cm⁻¹. Is there a broad, strong trough? That is O–H of an alcohol (3200–3600). Narrow, weaker peaks, perhaps two? N–H (3300–3500).
2. Look between 3000 and 2500 cm⁻¹. A very broad band here is the O–H of a carboxylic acid. Sharp peaks just below 3000 are C–H.
3. Look at 2200–2250 cm⁻¹. A sharp peak means C≡N.
4. Look at 1640–1750 cm⁻¹. A strong, sharp peak means C=O. Decide which kind from steps 1–3.
5. Only then look below 1500 cm⁻¹, and only to confirm (a C–O peak for an alcohol or ester) or to match against a known spectrum.
6. Check against the molecular formula if you have one. Every group you claim must fit the atoms available: a C=O and an O–H need two oxygens, for example.

The spectrum explorer below generates a model spectrum for each molecule from its bonds, using the ranges in the data table, and labels each absorption. Change the molecule and watch which windows light up.

Now try it the other way round: a spectrum, and four candidate structures. The model picks a spectrum at random; choose the structure that fits, and it explains which peaks decide it.

Following a reaction with infrared22.1.1

Because an infrared spectrum shows which functional groups are present, it shows when one group has been turned into another. That makes it a quick check on whether a reaction has worked, and questions often present it this way: spectra of a starting material and a product, and a question about what has changed.

The standard example is the oxidation of a primary alcohol with acidified potassium dichromate(VI) (topic 16). Ethanol has a broad O–H at 3200–3600 and no C=O. Distilled out as it forms, the product is ethanal: the broad O–H has gone and a strong C=O has appeared at 1670–1740. Heated under reflux, the product is ethanoic acid: the C=O is still there, and a very broad O–H has come back, but now lower, at 2500–3000.

What a mass spectrum shows22.2.1

A mass spectrometer turns a sample into positive ions and sorts them by mass. What comes out is a mass spectrum: a bar chart with one line for every kind of ion that reached the detector. Each line tells you two things.

• Its position along the horizontal axis is the ion's mass-to-charge ratio, m/e. Almost every ion in a spectrum carries a single positive charge, and for those m/e is simply the ion's mass on the relative scale (¹²C = 12 exactly).
• Its height is the ion's relative abundance: how many of those ions arrived, compared with the others. Heights are given either as percentages of all the ions, or relative to the tallest line, which is set to 100.

The word ratio matters for one reason. An ion with a 2+ charge is deflected as though it were half its mass, so it appears at half its m/e. Neon gives a small line at m/e = 10 from ²⁰Ne²⁺. Doubly charged ions are uncommon and always small, and CIE questions mention them only when a small line would otherwise be unexplained.

Isotopes and the spectrum of an element22.2.1

Put a sample of a single element into a mass spectrometer and each isotope gives its own line. The m/e of each line is the isotope's mass number, and the height of each line is the isotope's abundance. So the mass spectrum of an element is a direct picture of its isotopic composition.

Magnesium is a typical example. It has three stable isotopes, and its spectrum has three lines:

The last two columns are the same information on two scales. To go from percentages to "relative to tallest", divide each by the largest percentage and multiply by 100; to go back, divide each height by the sum of the heights and multiply by 100. A calculation of relative atomic mass works with either scale, as the next section shows, provided you divide by the right total.

Reading the spectrum of an element therefore needs three facts from you: the number of lines is the number of isotopes; the m/e of each line is its mass number; and the tallest line is the most abundant isotope, which need not be the lightest.

Calculating relative atomic mass22.2.2

The relative atomic mass of an element is the weighted mean mass of its atoms, compared with one twelfth of the mass of a ¹²C atom. "Weighted" means each isotope counts in proportion to how common it is. From a mass spectrum:

A_r = Σ(m/e × relative abundance) ÷ Σ(relative abundance)

Multiply each m/e by its abundance, add the products, and divide by the total abundance. If the abundances are percentages the total is 100; if they are heights relative to the tallest line, the total is whatever they add up to, and forgetting to divide by it is the commonest slip.

With the model you can check one more thing. Using mass numbers gives the right answer to one decimal place for most elements, but not quite for all: copper comes out at 63.6 rather than the 63.5 in the Periodic Table, and bromine at 80.0 rather than 79.9. The reason is that an isotope's actual relative mass is not exactly its mass number — ⁶³Cu is 62.930 and ⁸¹Br is 80.916 — because the mass of a nucleus is slightly less than the sum of its protons and neutrons. Exam questions either give you the m/e values to use or expect mass numbers, and in either case you use what the question gives you.

Working backwards to abundances22.2.2

The same equation can be run in reverse. If an element has just two isotopes and you know A_r, you can find the abundance of each, because there is only one unknown: call the fraction of the lighter isotope x, and the heavier one is then 1 − x.

That final check is worth making every time. A_r always lies between the lightest and heaviest isotope masses, and nearer the more abundant one. Chlorine's 35.5 is a quarter of the way from 35 to 37, so three quarters of chlorine atoms are ³⁵Cl: 75% and 25%, the ratio of 3 : 1 that section 22 will use again and again. Bromine's 79.9 is almost exactly halfway between 79 and 81, so its two isotopes are almost exactly equally common, 1 : 1.

Diatomic elements: Cl₂ and Br₂22.2.1

Chlorine and bromine exist as diatomic molecules, and their mass spectra are a first taste of what 22.2.1 means by analysing spectra "in terms of isotopic abundances". A sample of chlorine gas gives two groups of lines.

At m/e 35 and 37 are single Cl⁺ ions, made when a molecule splits in the instrument. Their heights are in the ratio 3 : 1, the isotope ratio itself.

At m/e 70, 72 and 74 are whole Cl₂⁺ ions, each made of two atoms picked from the same 3 : 1 mixture. There are three possible combinations, and their chances multiply:

The middle line is doubled because there are two ways of making it: the heavy isotope can be either of the two atoms. Leaving out that factor of 2 is the usual error, and it gives 9 : 3 : 1, which is wrong.

Bromine works the same way with its 1 : 1 ratio: lines at 79 and 81 for Br⁺ of equal height, and lines at 158, 160 and 162 for Br₂⁺ in the ratio ½ × ½ : 2 × ½ × ½ : ½ × ½ = 1 : 2 : 1.

This counting is not a side issue. It is exactly the reasoning you need in section 22, where an organic molecule contains one, two or more chlorine or bromine atoms, and the pattern of its molecular-ion lines tells you how many.

The molecular ion peak22.2.3

When an organic compound goes through a mass spectrometer, the first thing that happens to each molecule is that it loses one electron. What is left is the molecular ion, M⁺: the whole molecule, with a single positive charge.

M(g) + e⁻ → M⁺(g) + 2e⁻

An electron has so little mass that the molecular ion has, for all practical purposes, the same mass as the molecule. Its m/e is therefore the relative molecular mass of the compound. That is 22.2.3 in one line: find the molecular ion peak, and read off M_r.

The molecular ion has an odd number of electrons, because it has lost one from a molecule in which all electrons were paired. So it is a radical as well as an ion, which some books show as M^+•. That unpaired electron is part of the reason the ion is unstable and so often falls apart, as the next section shows.

Three practical points make the molecular ion peak easier to find.

• It is at the right-hand end. Ignore any small lines one or two units beyond it, which are the [M + 1] and [M + 2] peaks of sections 21 and 22. The last substantial line is M⁺.
• It need not be tall. For compounds that fragment easily — alcohols, and branched molecules in general — the molecular ion peak can be tiny, and for some tertiary alcohols it is barely visible. A small line at the end of the spectrum is not a mistake; it is often the most important line on the page.
• Its mass is calculated from whole numbers. A mass spectrum sorts individual ions, each made of particular isotopes, so M⁺ is at the mass of the molecule built from the commonest isotope of each atom: C = 12, H = 1, O = 16, N = 14, ³⁵Cl = 35, ⁷⁹Br = 79. Chloromethane's molecular ion is at 50, not at 50.5.

Fragmentation22.2.4

A spectrum is never just one line at M_r. The electrons that ionise the molecule carry far more energy than is needed to remove an electron, and much of the surplus ends up in the molecular ion, which shakes itself apart. A covalent bond breaks, and the molecular ion splits into two pieces. One piece keeps the positive charge and the other carries the unpaired electron:

M⁺ → X⁺ + Y^•

The positive ion X⁺ is detected and gives a line at its own m/e. The radical Y^• is neutral, is not accelerated or deflected, and never reaches the detector. This process is fragmentation, and the lines below the molecular ion are fragment ion peaks.

Each bond that breaks can give two different lines, because either piece can end up carrying the charge. Breaking the CH₃–CH₂ bond in the molecular ion of propanal, CH₃CH₂CHO⁺, gives either:

[CH₃CH₂CHO]⁺ → CH₃⁺ + ^•CH₂CHO    (line at 15)

[CH₃CH₂CHO]⁺ → CH₃^• + CH₂CHO⁺    (line at 43)

The two m/e values add up to M_r: 15 + 43 = 58. That is always true of the two ions from one broken bond, and it is a quick check that you have split the molecule correctly. Breaking the next bond, CH₃CH₂–CHO, gives CH₃CH₂⁺ or CHO⁺, which happen to both be 29.

Identifying fragment ions22.2.4

22.2.4 asks you to "suggest the identity" of the species behind given lines. There are two ways in, and you should use both.

From the line itself. Find a group of atoms from the molecule whose mass equals the m/e. A small set of fragments turns up again and again, and knowing them saves a great deal of arithmetic:

From the gap to the molecular ion. Subtract the line's m/e from M_r, and the answer is the mass of what was lost. A line at M − 15 means a methyl group has gone; M − 17 an OH; M − 18 a water molecule; M − 29 an ethyl group or CHO; M − 45 a COOH group. Losses are often easier to recognise than the ions that remain, because the lost pieces are small.

Two of the commonest values are ambiguous, and the ambiguity is worth knowing. 29 can be C₂H₅⁺ or CHO⁺; 43 can be C₃H₇⁺ or CH₃CO⁺. Each pair has the same whole-number mass because CO (28) weighs the same as C₂H₄ (28). The molecule settles it: a line at 43 from a compound with no oxygen cannot be CH₃CO⁺.

The model below does the splitting for you. Choose a molecule, and it breaks each bond between two non-hydrogen atoms in turn, gives the formula and m/e of both possible ions, and checks that each pair adds up to M_r.

Which peaks are tall, and the molecules lost22.2.4

Not every possible fragment gives a tall line. The height of a fragment peak depends on how many of those ions survive the trip to the detector, and that depends on how stable the ion is. Two kinds of ion are especially stable, and they explain most base peaks.

• More highly substituted carbocations. Alkyl groups push electron density towards a positive carbon and spread its charge (the inductive effect, topic 15). A tertiary carbocation, with three alkyl groups on the positive carbon, is more stable than a secondary, which is more stable than a primary. So a molecule tends to break next to a branch, leaving the charge on the branched carbon. The base peak of 2-methylpropane is at 43, (CH₃)₂CH⁺.
• Acylium ions, RCO⁺. A positive charge on the carbon of a C=O group is spread onto the oxygen, which has lone pairs to share. Aldehydes and ketones break on either side of the carbonyl carbon, and the RCO⁺ ion is usually a tall line: CH₃CO⁺ at 43 is the base peak of propanone and of butanone.

22.2.4 also asks about the molecules formed. A molecular ion can lose a small neutral molecule instead of a radical, by a rearrangement; the ion left behind still carries the charge and gives a line, and the neutral molecule is inferred from the gap. The common ones are:

Telling isomers apart22.2.4

Isomers have the same molecular formula, so their molecular ions are at the same m/e. Their fragments are not. A different arrangement of atoms means different bonds to break, and so a different set of lines — and a line that one isomer can produce and the other cannot is proof of which you have.

Propanal and propanone, both C₃H₆O, M_r 58. Both can give 15 (CH₃⁺) and 43. But only propanal has an ethyl group joined to a CHO group, so only propanal gives 29 (C₂H₅⁺ and CHO⁺), and only propanal can lose the H from its CHO to give 57. Propanone, CH₃COCH₃, has two identical methyl groups; its spectrum is dominated by 43, CH₃CO⁺, and has nothing at 29 or 57.

The isomeric butanols, C₄H₁₀O, M_r 74. The carbon carrying the OH breaks away from its neighbours, keeping the OH and the charge:

• butan-1-ol, a primary alcohol, gives CH₂OH⁺ at 31;
• butan-2-ol, a secondary alcohol, gives CH₃CHOH⁺ at 45 as its tallest line (and C₂H₅CHOH⁺ at 59). Butan-1-ol can make a line at 45 too, CH₂CH₂OH⁺, but only a small one, because the charge there is not on the carbon carrying the OH;
• 2-methylpropan-2-ol, a tertiary alcohol, gives (CH₃)₂COH⁺ at 59, from losing a methyl group, and its molecular ion at 74 is almost invisible.

Pentan-2-one and pentan-3-one, C₅H₁₀O, M_r 86. Breaking beside the C=O gives acylium ions, which make the tall lines. Pentan-2-one, CH₃COCH₂CH₂CH₃, gives two: CH₃CO⁺ at 43 and C₃H₇CO⁺ at 71. Pentan-3-one, CH₃CH₂COCH₂CH₃, is symmetrical and has only one, C₂H₅CO⁺ at 57. The line at 43 is the clean test: pentan-3-one cannot make any ion of mass 43 by breaking one bond. (It can make a small line at 71, by losing a CH₃ from the end of a chain, so 71 on its own does not decide it.)

The [M + 1] peak and the number of carbon atoms22.2.5

Look closely at the right-hand end of any organic spectrum and there is a small line one unit beyond the molecular ion. This is the [M + 1]⁺ peak. It is caused by carbon-13. About 1.1% of all carbon atoms are ¹³C rather than ¹²C, so a small proportion of molecules contain one ¹³C atom and weigh one unit more than the rest.

The more carbon atoms a molecule has, the more chances it has of including a ¹³C. With one carbon atom, 1.1% of molecules contain ¹³C; with two, about 2.2%; with n, about n × 1.1%. So the height of the [M + 1] peak compared with the M peak is proportional to the number of carbons:

abundance of [M + 1]⁺ ÷ abundance of M⁺ ≈ n × 1.1 ÷ 100

Rearranged, this is the formula in 22.2.5:

n = (100 × abundance of [M + 1]⁺ ion) ÷ (1.1 × abundance of M⁺ ion)

The model also shows how good the 1.1 rule is. Worked out properly, the chance of a molecule with n carbons containing exactly one ¹³C is n × 0.011 × 0.989ⁿ⁻¹, and the chance of none is 0.989ⁿ, so the true ratio is n × 0.011 ÷ 0.989 — about 1.1% higher than the formula assumes. For the molecules you meet, with fewer than about fifteen carbons, that never changes the rounded answer. Hydrogen-2, oxygen-17 and nitrogen-15 add a little more to the [M + 1] line, but their contributions are far smaller and the syllabus ignores them.

The [M + 2] peak: chlorine and bromine22.2.6

Chlorine and bromine each have two isotopes two mass units apart: ³⁵Cl and ³⁷Cl in the ratio 3 : 1, ⁷⁹Br and ⁸¹Br in the ratio 1 : 1. A molecule containing one of these atoms therefore comes in two versions two units apart, and its spectrum shows a second molecular-ion line at M + 2 — the [M + 2]⁺ peak — whose height is set by the isotope ratio.

The patterns for two halogen atoms come from the same counting as Cl₂ in section 15. For one Cl and one Br, the lightest molecule needs ³⁵Cl and ⁷⁹Br: ¾ × ½ = 3/8. The heaviest needs ³⁷Cl and ⁸¹Br: ¼ × ½ = 1/8. The middle line can be made two ways — ³⁷Cl with ⁷⁹Br, or ³⁵Cl with ⁸¹Br — so it gets ¼ × ½ + ¾ × ½ = 4/8. Hence 3 : 4 : 1.

The [M + 2] line is decisive because nothing else produces one of a comparable size. A compound of C, H and O does have a tiny line at M + 2, from molecules with two ¹³C atoms or one ¹⁸O, but it is a fraction of a percent of M. A line at M + 2 a third as tall as M means one chlorine atom; one the same height as M means one bromine atom.

The same pairs appear on any fragment that still contains the halogen, and that tells you where the halogen is. Chloroethane gives lines at 49 and 51, in the ratio 3 : 1, from CH₂Cl⁺; its line at 29, C₂H₅⁺, has no partner, because the chlorine has gone.

Putting infrared and mass spectra together22.1.1, 22.2

Real problems give you more than one kind of evidence, and each piece removes some possibilities. A reliable order is to settle the formula first and the structure second.

1. M_r from the molecular ion peak (22.2.3).
2. Halogens from any [M + 2] line (22.2.6). Subtract their mass.
3. Carbons from the [M + 1] line (22.2.5). Subtract 12 for each.
4. Functional groups from the infrared spectrum (22.1.1). This tells you how many oxygen or nitrogen atoms to allow for; what is left is hydrogen.
5. Structure from the fragments (22.2.4): which pieces would give the lines you see, and which isomer could not.
6. Check the answer against every piece of evidence, including peaks that are absent.

The model below sets problems of this kind. It chooses a compound, builds its mass spectrum and its infrared spectrum from its structure, and asks you to identify it. Work through the six steps before you choose.

Self-test22

Twenty-four questions across both sub-topics. Each answer comes with the reason, so a wrong answer is worth reading as closely as a right one.

Definitions to learn22

The numbers to have ready

Data used on this page22

The infrared ranges below are the ones printed in the 9701 data section, and every model spectrum on the page places its peaks inside them. In the exam, use the table you are given. The isotopic abundances are standard reference values; they are not part of the data booklet, which gives only relative atomic masses, and other sources differ in the last figure.

Masses used for m/e values: H 1, C 12, N 14, O 16, ³⁵Cl 35, ³⁷Cl 37, ⁷⁹Br 79, ⁸¹Br 81. Abundances used for the [M + 1] and [M + 2] models: ¹³C 1.1%, ³⁵Cl : ³⁷Cl = 75 : 25 (exactly 3 : 1), ⁷⁹Br : ⁸¹Br = 50 : 50 (exactly 1 : 1), as the syllabus uses them; the isotope table above gives the more precise values. The bond-vibration model uses typical force constants, listed in its panel, and the speed of light 3.00 × 10¹⁰ cm s⁻¹.