Infrared Spectroscopy

AS · 11 min

Test-tube reactions tell you which functional groups are present by what you see. Infrared (IR) spectroscopy does the same job instrumentally, faster and with a few milligrams of sample: each type of bond absorbs infrared radiation at its own characteristic frequency, so the absorptions in a spectrum reveal which bonds, and therefore which functional groups, the molecule contains. At AS you need to read a simple IR spectrum using the table of absorptions in the Data Booklet and identify the functional groups present. IR questions appear in Paper 2, usually combined with mass spectra, test-tube results or a synthesis.

How infrared absorption works

Covalent bonds are not rigid. They behave like springs, continually vibrating: stretching (the bond length changing) and bending (the bond angle changing). Each vibration has a natural frequency that depends on:

  • the strength of the bond (a stronger bond vibrates at a higher frequency: C≡N above C=O above C–O);
  • the masses of the atoms (bonds to light hydrogen atoms vibrate at high frequency: O–H, N–H and C–H all absorb above 2500 cm−12500\ \text{cm}^{-1}).

When infrared radiation of exactly that frequency passes through the sample, the bond absorbs energy and vibrates more strongly. Only vibrations that change the dipole moment of the molecule absorb infrared, which is why polar bonds such as O–H and C=O give strong absorptions.

An IR spectrum plots transmittance (the percentage of radiation that passes through) against wavenumber (in cm−1\text{cm}^{-1}, proportional to frequency). The wavenumber axis runs from about 4000 cm−14000\ \text{cm}^{-1} on the left to 500 cm−1500\ \text{cm}^{-1} on the right. Each absorption appears as a dip (often called a "peak") pointing downwards.

4000 3000 2000 1500 1000 500 100 0 propan-1-ol, CH₃CH₂CH₂OH O–H (broad) C–H C–O 4000 3000 2000 1500 1000 500 100 0 propanone, CH₃COCH₃ C–H C=O 4000 3000 2000 1500 1000 500 100 0 propanoic acid, CH₃CH₂COOH O–H (very broad, 2500–3300) C=O C–O wavenumber / cm⁻¹ (transmittance / % on the vertical axis)
Simplified infrared spectra. Each dip is an absorption. An alcohol shows a broad O–H band near 3300 cm⁻¹; a ketone shows a strong, sharp C=O band near 1715 cm⁻¹; a carboxylic acid shows both the C=O band and a very broad O–H band from about 2500 to 3300 cm⁻¹ that overlaps the C–H absorptions.

The table of characteristic absorptions

Key result
bondfunctional groups containing the bondabsorption range / cm−1\text{cm}^{-1}appearance
C–Ohydroxy, ester1040–1300strong
C=Caromatic compound, alkene1500–1680weak (unless conjugated)
C=Oamide1640–1690strong
C=Ocarbonyl (aldehyde, ketone), carboxyl1670–1740strong
C=Oester1710–1750strong
C≡Calkyne2150–2250weak (unless conjugated)
C≡Nnitrile2200–2250weak
C–Halkane2850–2950strong
N–Hamine, amide3300–3500weak
O–Hcarboxyl2500–3000strong and very broad
O–Hhydroxy (alcohol)3200–3650strong (broad when hydrogen bonded)

These are the ranges used in the Cambridge Data Booklet. In the exam, always use the table printed in your Data Booklet.

The three most useful absorptions to recognise on sight:

  1. C=O at about 1700 cm−11700\ \text{cm}^{-1}: a strong, sharp dip. Present in aldehydes, ketones, carboxylic acids and esters.
  2. Alcohol O–H at about 32003200–3650 cm−13650\ \text{cm}^{-1}: strong and broad (hydrogen bonding between molecules gives a range of O–H bond strengths, which broadens the band).
  3. Carboxylic acid O–H at 25002500–3000 cm−13000\ \text{cm}^{-1}: very broad, often stretching from about 2500 to 3300 cm−13300\ \text{cm}^{-1} and overlapping the C–H absorptions. The very strong hydrogen bonding in acid dimers makes it broader and shifts it lower than the alcohol O–H.

Almost every organic compound shows C–H absorptions near 2900 cm−12900\ \text{cm}^{-1}, so these tell you little.

The fingerprint region

The region below about 1500 cm−11500\ \text{cm}^{-1} contains many overlapping absorptions from bending vibrations and from single bonds such as C–C and C–O. The pattern is unique to each compound, like a fingerprint. It is hard to interpret bond by bond, but comparing it with a database of known spectra can identify a compound exactly (or confirm that a product is pure, by showing no extra absorptions).

Reading a spectrum

Method

Identifying functional groups from an IR spectrum

  1. Look first at 16401640–1750 cm−11750\ \text{cm}^{-1}. A strong absorption means C=O is present.
  2. Look at 25002500–3650 cm−13650\ \text{cm}^{-1} for O–H:
    • very broad, centred near 3000 cm−13000\ \text{cm}^{-1} (2500–3000), together with C=O: carboxylic acid;
    • broad at 32003200–3650 cm−13650\ \text{cm}^{-1}, no C=O: alcohol.
  3. C=O present but no O–H: aldehyde, ketone or ester. A strong C–O absorption at 10401040–1300 cm−11300\ \text{cm}^{-1} as well suggests an ester.
  4. Check the other diagnostic regions: 22002200–2250 cm−12250\ \text{cm}^{-1} (C≡N), 33003300–3500 cm−13500\ \text{cm}^{-1} weak (N–H), 15001500–1680 cm−11680\ \text{cm}^{-1} weak (C=C).
  5. Combine the evidence with the molecular formula and any other data to suggest a structure.
  6. Note what is absent as well as what is present: no O–H rules out alcohols and acids.
compound typeO–H (3200–3650, broad)O–H (2500–3000, very broad)C=O (1670–1750)C–O (1040–1300)
alcoholyesnonoyes
aldehyde or ketonenonoyesno
carboxylic acidnoyesyesyes
esternonoyes (1710–1750)yes

IR spectroscopy cannot easily tell an aldehyde from a ketone at AS level (both absorb in the same C=O range); use Tollens' or Fehling's test for that (see Tests for carbonyl compounds).

Tip

Infrared absorption is also behind the greenhouse effect: carbon dioxide, water vapour and methane absorb infrared radiation emitted by the Earth's surface, because their bond vibrations change their dipole moments. Breathalysers use IR too, measuring the C–H absorption of ethanol in breath.

Worked examples

Routine: reading absorptions

The IR spectrum of a compound shows strong absorptions at 2950 cm−12950\ \text{cm}^{-1} and 1715 cm−11715\ \text{cm}^{-1}, and nothing between 3200 and 3650 cm−13650\ \text{cm}^{-1} or between 2500 and 3000 cm−13000\ \text{cm}^{-1} apart from the sharp C–H absorption. Which bonds are present, and what type of compound could it be?

Solution

2950 cm−12950\ \text{cm}^{-1}: C–H (alkane part). 1715 cm−11715\ \text{cm}^{-1}: C=O (carbonyl). No broad O–H absorption, so not an alcohol or a carboxylic acid.

The compound is an aldehyde or ketone (an ester is also possible if there is a strong absorption at 10401040–1300 cm−11300\ \text{cm}^{-1}).

Routine: distinguishing three oxidation products

Propan-1-ol can be oxidised to propanal or propanoic acid. Explain how IR spectra would distinguish propan-1-ol, propanal and propanoic acid.

Solution

Propan-1-ol: broad O–H absorption at 32003200–3650 cm−13650\ \text{cm}^{-1}; no C=O absorption.

Propanal: strong C=O absorption at 16701670–1740 cm−11740\ \text{cm}^{-1}; no O–H absorption.

Propanoic acid: C=O absorption at 16701670–1740 cm−11740\ \text{cm}^{-1} and a very broad O–H absorption at 25002500–3000 cm−13000\ \text{cm}^{-1}.

Standard: following a reaction

A student oxidises ethanol with acidified potassium dichromate(VI), distilling the product. Describe how the IR spectrum of the distillate would show (a) that oxidation has occurred, (b) whether the product is ethanal or ethanoic acid, (c) whether any ethanol remains.

Solution

(a) A strong C=O absorption at 16701670–1740 cm−11740\ \text{cm}^{-1} appears, which ethanol does not have.

(b) If the product is ethanal, there is no O–H absorption at all. If it is ethanoic acid, there is a very broad O–H absorption at 25002500–3000 cm−13000\ \text{cm}^{-1}.

(c) A broad absorption at 32003200–3650 cm−13650\ \text{cm}^{-1} (alcohol O–H) would show unreacted ethanol.

Standard: an ester or an acid?

Compound V, CX4HX8OX2\ce{C4H8O2}, has strong IR absorptions at 29802980, 17401740 and 1240 cm−11240\ \text{cm}^{-1} and no absorption between 2500 and 3650 cm−13650\ \text{cm}^{-1} other than C–H. Deduce the type of compound and suggest two possible structures.

Solution

1740 cm−11740\ \text{cm}^{-1}: C=O, in the ester range (1710–1750). 1240 cm−11240\ \text{cm}^{-1}: C–O. No O–H, so not a carboxylic acid (which would show a very broad band at 25002500–3000 cm−13000\ \text{cm}^{-1}).

V is an ester, CX4HX8OX2\ce{C4H8O2}. Possible structures: ethyl ethanoate, CHX3COOCHX2CHX3\ce{CH3COOCH2CH3}; methyl propanoate, CHX3CHX2COOCHX3\ce{CH3CH2COOCH3} (also propyl methanoate or 1-methylethyl methanoate).

Exam-hard: combining IR with chemical tests

Compound K, CX3HX6O\ce{C3H6O}, decolourises bromine water. Its IR spectrum has a broad absorption at 3350 cm−13350\ \text{cm}^{-1} and a weak absorption at 1645 cm−11645\ \text{cm}^{-1}, but nothing between 1670 and 1750 cm−11750\ \text{cm}^{-1}. It reacts with sodium, giving a gas. Identify K, explaining each piece of evidence.

Solution

Broad absorption at 3350 cm−13350\ \text{cm}^{-1}: O–H of an alcohol (confirmed by the gas, hydrogen, with sodium).

No absorption at 16701670–1750 cm−11750\ \text{cm}^{-1}: no C=O, so not propanal or propanone (the obvious CX3HX6O\ce{C3H6O} isomers).

Weak absorption at 1645 cm−11645\ \text{cm}^{-1}: C=C (range 1500–1680, weak); confirmed by decolourising bromine water.

An alcohol with a C=C and three carbons, CX3HX6O\ce{C3H6O}: prop-2-en-1-ol, CHX2=CHCHX2OH\ce{CH2=CHCH2OH}.

Exam-hard: checking each step of a synthesis

Propanone is converted to 2-hydroxy-2-methylpropanenitrile, which is then hydrolysed to 2-hydroxy-2-methylpropanoic acid. For each of the three compounds, list the absorptions (bond and range) you would expect in the region above 1500 cm−11500\ \text{cm}^{-1}, and explain how IR shows that each step has gone to completion.

Solution

Propanone, CHX3COCHX3\ce{CH3COCH3}: C–H 28502850–29502950; C=O 16701670–1740 cm−11740\ \text{cm}^{-1}.

2-hydroxy-2-methylpropanenitrile, (CHX3)X2C(OH)CN\ce{(CH3)2C(OH)CN}: O–H (hydroxy) 32003200–36503650, broad; C–H 28502850–29502950; C≡N 22002200–2250 cm−12250\ \text{cm}^{-1} (weak). No C=O.

2-hydroxy-2-methylpropanoic acid, (CHX3)X2C(OH)COOH\ce{(CH3)2C(OH)COOH}: O–H (hydroxy) 32003200–36503650; O–H (carboxyl) 25002500–30003000, very broad; C=O 16701670–1740 cm−11740\ \text{cm}^{-1}. No C≡N.

Step 1 is complete when the C=O absorption of propanone has disappeared and the C≡N and O–H absorptions have appeared. Step 2 is complete when the C≡N absorption has gone and the very broad carboxyl O–H and the C=O absorption have appeared.

Watch out
  • Reading the axis the wrong way. Wavenumber decreases from left to right. An absorption at 1700 cm−11700\ \text{cm}^{-1} is to the right of one at 3000 cm−13000\ \text{cm}^{-1}.
  • Calling every broad band "alcohol". A very broad band centred near 3000 cm−13000\ \text{cm}^{-1} together with C=O is a carboxylic acid. The alcohol O–H is at higher wavenumber (32003200–3650 cm−13650\ \text{cm}^{-1}).
  • Using C–H to identify a compound. Nearly all organic compounds have C–H absorptions near 2900 cm−12900\ \text{cm}^{-1}; they are not diagnostic.
  • Expecting IR to give the whole structure. IR identifies bonds and functional groups. It cannot by itself distinguish isomers with the same groups (propanal and propanone, or butan-1-ol and butan-2-ol), except by fingerprint comparison with known spectra.
  • Quoting a single number. Give the range from the Data Booklet (for example "C=O, 16701670–1740 cm−11740\ \text{cm}^{-1}") and name the bond, not just the functional group.
Exam tip
  • A typical question gives a spectrum (or a list of absorptions) and asks you to "identify the bonds responsible for the absorptions at X and Y". Answer with the bond (O–H, C=O) and, if asked, the functional group, using the Data Booklet ranges.
  • "Use the IR spectrum to show that compound A is a carboxylic acid and not an alcohol": point to the broad O–H at 25002500–3000 cm−13000\ \text{cm}^{-1} and the C=O at about 1700 cm−11700\ \text{cm}^{-1}.
  • When a question combines IR with a molecular formula, use the formula to list possible structures first, then use IR to eliminate them.
  • State the absence of an absorption as evidence too: "no absorption at 16701670–1750 cm−11750\ \text{cm}^{-1}, so no C=O".
Summary
  • Bonds vibrate at characteristic frequencies and absorb infrared radiation of matching frequency; vibrations that change the dipole moment absorb.
  • Spectra plot transmittance against wavenumber (4000→500 cm−14000 \to 500\ \text{cm}^{-1}); absorptions are dips.
  • Key absorptions: C=O 16701670–17501750 (strong, sharp); alcohol O–H 32003200–36503650 (broad); carboxyl O–H 25002500–30003000 (very broad); C–O 10401040–13001300; C≡N 22002200–22502250; N–H 33003300–35003500; C=C 15001500–16801680; C–H 28502850–2950 cm−12950\ \text{cm}^{-1}.
  • Alcohol: O–H, no C=O. Aldehyde/ketone: C=O, no O–H. Acid: C=O + very broad O–H. Ester: C=O + C–O, no O–H.
  • The fingerprint region (below 1500 cm−11500\ \text{cm}^{-1}) identifies a compound by comparison with known spectra.

Practice

Question
  1. Explain why molecules absorb infrared radiation at particular frequencies.
  2. State the absorption ranges, from the Data Booklet, for the O–H bond in an alcohol and in a carboxylic acid, and explain why the carboxylic acid O–H absorption is so broad.
  3. Which bond is responsible for a weak absorption at 2230 cm−12230\ \text{cm}^{-1}? Name a compound type that contains it.
  4. List the absorptions above 1000 cm−11000\ \text{cm}^{-1} that you would expect in the IR spectrum of ethyl ethanoate, and state one absorption that would be absent.
  5. A spectrum shows a strong absorption at 1710 cm−11710\ \text{cm}^{-1} and a very broad absorption from 25002500 to 3300 cm−13300\ \text{cm}^{-1}. Identify the functional group.
  6. Explain how IR spectroscopy could be used to distinguish butan-2-ol from butanone.
  7. What is the fingerprint region, and how is it used?
  8. Propan-1-ol is heated with concentrated phosphoric acid. Describe how the IR spectrum of the product differs from that of propan-1-ol.
  9. Compound P, CX4HX8O\ce{C4H8O}, has a strong IR absorption at 1720 cm−11720\ \text{cm}^{-1} and no absorption between 3200 and 3650 cm−13650\ \text{cm}^{-1}. It gives a pale yellow precipitate with alkaline aqueous iodine and does not react with Tollens' reagent. Identify P.
  10. Compound Q, CX3HX6OX2\ce{C3H6O2}, has strong IR absorptions at 1745 cm−11745\ \text{cm}^{-1} and 1240 cm−11240\ \text{cm}^{-1} and no absorption between 2500 and 3650 cm−13650\ \text{cm}^{-1} apart from C–H. On hydrolysis it gives methanol. (a) Identify Q. (b) Explain why the spectrum rules out the isomer propanoic acid. (c) Describe how the IR spectrum would change if Q were completely hydrolysed by dilute acid and the organic acid product isolated.
Answers
  1. Covalent bonds vibrate (stretch and bend) at natural frequencies that depend on the bond strength and the masses of the atoms. When infrared radiation of the same frequency passes through, the bond absorbs energy and vibrates more (if the vibration changes the dipole moment). Different bonds therefore absorb at different, characteristic frequencies.
  2. Alcohol O–H: 32003200–3650 cm−13650\ \text{cm}^{-1}. Carboxyl O–H: 25002500–3000 cm−13000\ \text{cm}^{-1}, very broad. Carboxylic acid molecules are strongly hydrogen bonded (as dimers), which weakens and varies the O–H bonds over a wide range of strengths, so they absorb over a wide range of frequencies.
  3. C≡N; nitriles (for example propanenitrile).
  4. C–H 28502850–2950 cm−12950\ \text{cm}^{-1}; C=O 17101710–1750 cm−11750\ \text{cm}^{-1}; C–O 10401040–1300 cm−11300\ \text{cm}^{-1}. Absent: any O–H absorption (25002500–3650 cm−13650\ \text{cm}^{-1}).
  5. Carboxylic acid (C=O and very broad carboxyl O–H).
  6. Butan-2-ol shows a broad O–H absorption at 32003200–3650 cm−13650\ \text{cm}^{-1} and no C=O. Butanone shows a strong C=O absorption at 16701670–1740 cm−11740\ \text{cm}^{-1} and no O–H.
  7. The region below about 1500 cm−11500\ \text{cm}^{-1}, which contains many complex absorptions unique to each compound. The spectrum of an unknown is compared with a database of spectra; an exact match in this region identifies the compound.
  8. The product is propene. The broad O–H absorption (32003200–3650 cm−13650\ \text{cm}^{-1}) and the strong C–O absorption (10401040–1300 cm−11300\ \text{cm}^{-1}) disappear; a weak C=C absorption appears at 15001500–1680 cm−11680\ \text{cm}^{-1}.
  9. C=O present, no O–H, so an aldehyde or ketone. No reaction with Tollens': a ketone. Positive iodoform: CHX3COX−\ce{CH3CO-}. P is butanone, CHX3COCHX2CHX3\ce{CH3COCH2CH3}.
  10. (a) C=O in the ester range and C–O, no O–H: an ester. Hydrolysis gives methanol, so the alkyl part is methyl and the acid part has two carbons: Q is methyl ethanoate, CHX3COOCHX3\ce{CH3COOCH3}. (b) Propanoic acid would show a very broad carboxyl O–H absorption at 25002500–3000 cm−13000\ \text{cm}^{-1}, which is absent. (c) The isolated product, ethanoic acid, still shows C=O (now in the 16701670–1740 cm−11740\ \text{cm}^{-1} range) and C–O, but a very broad O–H absorption at 25002500–3000 cm−13000\ \text{cm}^{-1} appears.

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