Topic 20 · page 3 of 15

Infrared and mass spectra

Infrared absorptions show which bonds are present, and a mass spectrum gives the relative molecular mass and clues to the structure from the way the molecule breaks up. Used together they separate isomers.

15 min read20.1

Read first: Tests for functional groups

Infrared: which bonds are present 20.1

Bonds vibrate, and each type absorbs infrared radiation at its own wavenumber, in cm⁻¹. The spectrum plots transmittance, so an absorption is a dip, with 4000 on the left and 500 on the right. The strongest evidence comes from a few bonds.

Bond	Where it is found	Wavenumber / cm⁻¹	Look of the dip
O–H	alcohols	3230–3550	broad
O–H	carboxylic acids	2500–3300	very broad
N–H	amines	3300–3500	medium
C–H	alkanes and alkyl groups	2850–2950	sharp
C=O	aldehydes, ketones, acids, esters	1680–1750	strong and sharp
C=C	alkenes	1620–1680	weak to medium
C–Cl	chloroalkanes	600–800
C–Br	bromoalkanes	500–600

These are the ranges used with this specification. In the exam the data are given, so practise using the table in front of you. Older booklets print slightly different limits, for example 3750–3200 for the O–H of an alcohol, so always use the one you are given.

A schematic infrared spectrum of propanoic acid with wavenumber from 4000 on the left to 500 on the right. A very broad absorption spans about 3300 to 2500 and is labelled O-H of the acid. A strong sharp absorption at 1715 is labelled C=O. Smaller absorptions lie in the fingerprint region below 1500. — A schematic spectrum, not measured data. The very broad O-H and the sharp C=O together show a carboxylic acid.

MethodReading an infrared spectrum

Look at 1680–1750. A strong, sharp dip means C=O.
Look above 2500. A broad dip means O–H: very broad from 2500 to 3300 in an acid, broad near 3350 in an alcohol. A medium dip at 3300–3500 is N–H.
Say what is absent as well as what is present.
Ignore the region below about 1500 cm⁻¹, apart from C–Cl and C–Br. It is the fingerprint region and cannot be read bond by bond.

What you see	Conclusion
C=O only, no broad O–H	aldehyde, ketone or ester
C=O and very broad O–H	carboxylic acid
broad O–H, no C=O	alcohol
C=C and nothing else above 1650	alkene

Mass spectra: the molecular ion and fragments 20.1

In the mass spectrometer a molecule loses an electron and becomes the molecular ion, M⁺. Its m/z equals the relative molecular mass, $M_\text{r}$ . It is the peak at the highest m/z, apart from a tiny peak one unit higher.

The molecular ion is unstable. It breaks into a positive fragment ion, which is detected, and a radical, which has no charge and is not detected.

\ce{[CH3COCH3]^{+.} -> CH3CO+ + CH3^.}

m/z	Fragment ion
15	CH₃⁺
29	C₂H₅⁺ or CHO⁺
31	CH₂OH⁺
43	CH₃CO⁺ or C₃H₇⁺
45	$\text{COOH}^{+}$
57	C₄H₉⁺ or C₂H₅CO⁺

Each mass is the sum of the atoms in the ion. For CH₃CO⁺: 12.0 + 3 × 1.0 + 12.0 + 16.0 = 43.0.

Isomers have the same M⁺ but different fragments

Propanone and propanal both have the formula C₃H₆O, so both show M⁺ at 58. Their fragments are different.

Two schematic mass spectra drawn as vertical lines. Propanone has a molecular ion at 58 and its tallest peak at 43, labelled CH3CO+, with a smaller peak at 15. Propanal has a molecular ion at 58, a small peak at 57, its tallest peak at 29, and smaller peaks at 28 and 27. — Schematic spectra: peak heights are only a guide. Propanone has a strong peak at 43, propanal at 29.

Compound	Characteristic fragments
propanone, CH₃COCH₃	43 (CH₃CO⁺, the tallest), 15 (CH₃⁺)
propanal, CH₃CH₂CHO	29 (C₂H₅⁺ or CHO⁺, the tallest), 57, 28, 27

Putting the two together 20.1

Worked exampleTwo isomers of C₃H₆O₂

Illustrative data. A compound has its molecular ion at m/z = 74. Its infrared spectrum has a very broad absorption from 3300 to 2500 cm⁻¹ and a strong sharp absorption at 1710 cm⁻¹. Identify it.

$M_\text{r}$ = 74: C₃H₆O₂ gives 36.0 + 6.0 + 32.0 = 74.0. Other formulae also give 74, so more data are needed.
Strong, sharp 1710: C=O is present.
Very broad 3300–2500: the O–H of a carboxylic acid.
The compound is propanoic acid, CH₃CH₂COOH. Methyl ethanoate, CH₃COOCH₃, has the same formula but no O–H, so it would show the C=O and no broad absorption above 2500. It would also give a strong peak at 43 (CH₃CO⁺) in the mass spectrum.

Check yourself

State the bond responsible for (a) a strong sharp absorption at 1715 cm⁻¹ and (b) a broad absorption at 3350 cm⁻¹.
List the absorptions above 1650 cm⁻¹ you would predict for butanone, CH₃COCH₂CH₃.
How can an infrared spectrum tell ethanoic acid from ethanol?
A compound C₂H₆O has its molecular ion at 46 and a base peak at 31. Identify the compound and write an equation for the formation of the fragment at 31.
Which peak in a mass spectrum would you use to tell propanone from propanal, and why?
A compound has a molecular ion at m/z = 74 and a strong infrared absorption at 1740 cm⁻¹ but nothing between 2500 and 3600 cm⁻¹. Suggest its functional group and a possible structure.

Answers cover these until you have tried the questions

(a) C=O. (b) O–H of an alcohol.
C–H at about 2850–2950 (sharp) and C=O at 1680–1750 (strong, sharp). There is no O–H and no N–H, so nothing broad above 3000.
Ethanoic acid has a very broad O–H (2500–3300) and a C=O at about 1700. Ethanol has a broad O–H at 3230–3550 and no C=O.
The base peak at 31 is CH₂OH⁺ and the M⁺ at 46 fits C₂H₆O, so the compound is ethanol. $\ce{[CH3CH2OH]^{+.} -> CH2OH+ + CH3^.}$ (Methoxymethane would give a strong peak at 45, CH₃OCH₂⁺.)
Use the tallest fragment peak. Propanone gives 43 (CH₃CO⁺); propanal gives 29 (C₂H₅⁺ or CHO⁺). Both have M⁺ at 58, so the molecular ion cannot separate them.
A strong C=O with no O–H rules out a carboxylic acid and an alcohol group, so an ester fits. With $M_\text{r}$ = 74 (C₃H₆O₂) a possible structure is methyl ethanoate, CH₃COOCH₃, or ethyl methanoate, HCOOCH₂CH₃. More data (NMR) would choose between them.