An IR spectrum plots of the percentage of radiation that passes through a sample (transmittance) vs. the wavenumber of the radiation transmitted. If a sample absorbs a particular frequency / energy / wavenumber the transmission decreases and a downward peak or absorption band shows up on the spectrum.
The peaks are characteristic of the frequency of radiation needed to increase the vibrational energy of specific bonds (so that the bond is vibrating more strongly).
We can use an IR spectrum to identify bonds in a molecule and this tells us about the functional groups a molecule contains. For example, if we were expecting our sample to be carboxylic acid, we would be looking for confirmation of C=O, C-O and O-H bonds in the IR spectrum.
We also see peaks arising from whole molecule vibrations, particularly below 1500cm-1 which is known as the fingerprint region. We can use a database / software to identify a molecule definitely from its IR spectrum assuming the spectrum was recorded under exactly the same conditions as the reference, but the technique is most commonly used to identify chemical bonds.
The wavenumber of a peak depends on the type of vibration, the bond enthalpy and the mass of the atoms in the bond:
- short, strong bonds vibrate at higher wavenumbers than longer, weaker bonds e.g. C=C bond at 1650cm-1, C-C bond at 1000cm-1; C=O bond at 1700cm-1, C-O bond at 1150cm-1
- bonds involving atoms with a low mass vibrate at a higher wavenumber e.g. C-H bond at 3000cm-1, C-O bond at 1150cm-1, C-Cl bond at 700cm-1
- generally speaking, vibrations of polar bonds result in a more intense peak (see the C=O peaks in the spectra of butanal and butanone below)
- the peak for an O-H bond is strong and generally a broad peak (see the IR spectrum for ethanoic acid above). Hydrogen bonding between O-H groups changes the strength of the O-H bond to differing degrees in every molecule and hence the specific wavenumber of radiation absorbed. The broad peak in the spectrum is an amalgamation of all these slightly different absorptions.
The precise position of peaks in an IR spectrum depends on the environment of bonds in a molecule so a carbonyl bond (C=O) in an aldehyde is not in the same environment as a carbonyl bond in an ester, for example. This means we need to read the position of a peak off a spectrum carefully and read the data sheet even more carefully!
So what exactly do we need to be able to do?
- Identify key bonds from their peaks and give a precise wavenumber for the peak
- Link the bonds to a specific functional group
- Explain the significance of the fingerprint region
If we are making a prediction for where we would expect to see a peak on a spectrum, then we need to give the range quoted from the data sheet e.g. there would be a peak for an O-H bond (not OH / OH group / hydroxyl) in an alcohol at 3200 – 3600 cm-1. How you write it is key!
Let’s have a look at a couple of examples …
The C-H peak is strong and seen in all IR spectra, but it is often confused for an O-H peak (usually wishful thinking!). The peak for an O-H bond is strong and generally a broad peak.
If we didn’t know this spectrum was for butanal all we could deduce is that the compound contained a carbonyl group characteristic of an aldehyde. What we would write is …
‘there is a sharp peak at 1730cm-1 for C=O in an aldehyde’
The key points are that we have read the position of the peak carefully off the x-axis and linked it to a BOND (we would get no marks for saying the peak was for a carbonyl / CO group) and then a functional group. This is the biggest mistake students make!
The IR spectra for butanal and butanone are very similar – the obvious difference is the fingerprint region which is specific to a molecule and could be used to identify one over the other when compared to a reference. However, IR spectroscopy is pretty much always used in conjunction with other analytical techniques such as mass spectrometry, elemental analysis, 1H NMR and 13C NMR to get a full picture (especially if our molecule is unknown).
Amines are particularly hard to identify from an IR spectrum as the peak for N-H is either very small or masked by the O-H absorption in molecules with multiple functional groups. The absorption for C-N is 1021-1250cm-1 but as it falls into the fingerprint region it is impossible to assign a peak with any certainty.
The only peak in the fingerprint region worth looking for is C-O in an alcohol, ester, ether or carboxylic acid at 1000-1300cm-1 as it is a strong, sharp signal and can be used in conjunction with a C=O or O-H absorption to confirm the presence of a functional group (see the IR spectrum for ethanol at the top of the page).
Practice questions
- This question is about propanoyl chloride, CH3CH2COCl, whose infrared spectrum is shown below:
(a) Identify the key peaks in the spectrum, with reference to the data sheet.
(b) The sample of propanoyl chloride is left exposed to moisture in the air for a few hours. Following this, a new infrared spectrum is recorded which shows a broad peak at 3050cm-1 and a strong, sharp peak at 1710cm-1. Explain these observations and suggest a structure for the compound responsible.
(c) Propanoyl chloride reacts with ethanol to form an ester, ethylpropanoate. Predict the wavenumbers for the key peaks in the infrared spectrum of the ester with reference to the data sheet.
2. An alcohol (X) was refluxed with acidified potassium dichromate (VI) and the resulting oxidation product gave the following IR spectrum:
Identify which of the alcohols shown below could be alcohol X, with reason, and write an equation for the oxidation reaction that occurred.
3. The IR spectrum for 2-bromopropane is shown below, with the absorption band for the C-Br bond labelled.
The haloalkane was refluxed with aqueous sodium hydroxide. Explain how IR spectroscopy could be used to show that a hydrolysis reaction occurred.
4. Explain how the fingerprint region of an IR spectrum could be used to distinguish between (CH3)2CHCH2CHO and (CH3)3CCHO.
Answers
- (a) The sharp peak at 2950cm-1 is for C-H. The sharp / strong peak at 1800cm-1 is for C=O in an acyl chloride.
(b) Acyl chlorides react with water to form carboxylic acids. The broad peak at 3050cm-1 is for O-H in a carboxylic acid and the sharp peak at 1710cm-1 is for C=O in a carboxylic acid.
(c) There would be peak at 2850-2950cm-1 for C-H, a peak at 1735-1750cm-1 for C=O in an ester and a peak at 1000-1300cm-1 for C-O in an ester.
2. Molecule A is a tertiary alcohol and cannot be oxidised. Molecule C is a primary alcohol and would be oxidised to an aldehyde and then a carboxylic acid. There is no peak / absorption band between 2500-3300cm-1to signify the presence of an O-H bond in a carboxylic acid, so the molecule cannot be C.
Molecule B is a secondary alcohol and would be oxidised to a ketone, with a C=O peak at 1705-1725cm-1. There is an intense peak at 1705cm-1. The sharp peak at 2950cm-1 is for the C-H stretch. Molecule B is alcohol X.
3. Hydrolysis of 2-bromoalkane would give a propan-2-ol. The IR spectrum for propan-2-ol would not show a peak at 720cm-1 for a C-Br bond, but would show a peak at 3200-3600cm-1 for an O-H stretch in an alcohol. There would also be a peak at 1000-1300cm-1 in the fingerprint region for a C-O stretch.
4. The fingerprint region of an IR spectrum (below 1500cm-1) is unique / specific to an individual molecule. You can use software or a database to compare spectra and identify a molecule.
Ready to take it to the next level? Download a booklet of exam style questions with perfectly structured, fully explained answers and exam tips. Preparation is the key to success!