Ir Spectral Interpretation Assignment

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Infrared Spectroscopy

1. Theory

Vibrational modes within a molecule can be described using the anharmonic oscillator model. This model assumes that the two masses (with known weight) are connected with a spring (with known strength). With the help of quantum mechanical calculations (Schroedinger equation) you can find the frequencies of basic stretching and bending modes:


masses of involved atoms
F=force constant

From this equation, one can deduce some basic trends can be deducted:

a. If the force constant F (= bond strength) increases, the stretching frequency will increase as well (in cm-1)

CC bondtripledoublesingle
(in cm-1)210016501130
functional groupalkynealkenealkane

b. If the masses of the involved atoms increase, the peak will shift to lower wavenumbers e.g. H/D-exchange in labeling experiments although the bond strength remains the same.

C-X bondC-HC-D
(in cm-1)~3000~2200

Based on the equation, one would always expect a sharp lines at a well defined wavenumber. Unfortunately, the change in vibration modes is always accompanied with change in rotational mode (Stokes and Anti-Stokes). The required energy for this process is much smaller (1-5 cm-1) and causes together with some other effects the broading of the 'lines'.

The number of basic stretching and bending modes expected for a molecule increases with the number of atoms in the molecule. For non-linear molecules 3N-6 (2N-5 bending, N-1 stretching) vibrations are observed (e.g. CH2Cl2). For linear molecules e.g. CO2 one expect to find 3N-5 (3*3-5=4) modes. If there is no symmetry in the molecule most of them will be observed the IR spectrum; the remaining modes will be observed in a Raman spectrum. The more complicated the molecule is (the more atoms it possesses and the lower the symmetry), the more peaks can be observed in the IR spectrum (see example 3 and 4).

A signal is only observed in the IR spectrum, if the dipole momentum of the molecule changes during the interaction with the electromagnetic radiation. This is very likely for groups which already possess a significant dipole momentum to start with e.g. C-O, C-Cl, O-H, etc. (strong peaks). Groups with a small difference in electronegativity e.g. C-H, C-C, C=C, etc., will usually show weak or medium sized peaks in the IR spectrum.

An interpretation of an IR spectrum should include a detailed assignment of the peaks: exact wavenumber from the spectrum (integer), the intensity (w/m/s/br) and which functional group it represents, and maybe in addition the corresponding literature value. However, it is not necessary to interpret every little peak in the IR spectrum. Also, be cautious when you compare spectra which were obtain with different techniques (solution, Nujol mull, KBr pellet). The actual number of your vibration changes quite at bit, especially for highly polar compounds (Why ?).

2. Interpretation of the spectrum

The IR-spectrum can be divided into five ranges major ranges of interest for an organic chemist:

a. From 2700-4000 cm-1 (E-H-stretching: E=B, C, N, O)

In this range typically E-H-stretching modes are observed. The C-H-stretching modes can be found between 2850 and 3300 cm-1, depending on the hydrization. The range from 2850-3000 cm-1belongs to saturated systems (alkanes, sp3, example 1), while the peaks from 3000-3100 cm-1 indicate an unsaturated system (alkenes, sp2,example 2; aromatic ring, example 3,4). Latter ones are usually weak or medium in intensity. The CH-function on a C-C-triple bond (alkynes) will appear as a sharp, strong peak around 3300 cm-1. The differences in wavenumbers are mainly due to different hybridization (=bond strength, the s-character in the bond, the stronger the bond is) and number of ligands on the carbon atom.

The O-H-stretching modes from alcohols and phenols (example 5) are mostly broad and very strong (3200-3650 cm-1) The O-H-peaks due to carboxylic acids (example 6) show a very broad and less intense peak between 2500 and 3500 cm-1. The change in peak shape is a result of the different degree of hydrogen bonds in alcohol and carboxylic acids. These peaks change significantly with the polarity of the solvent. They usually become sharper if the polarity of the solvent increases e.g. camphor in CCl4 and CH2Cl2 (see reader).

Peaks that are due to N-H-stretching modes are sharper than O-H-peaks (3300-3500 cm-1). Primary amines (example 7) have two peaks (sym./asym. vibration) in this range, while secondary amines (example 8) have only one peak. There are no peaks in this area for tertiary amines. Why?

Two small or medium peaks at ~2750 and ~2850 cm-1 are a result of an aldehyde (example 15).

b. From 2000 - 2700 cm-1 (E-X-triple bonds: E=X=C, N, O)

This range covers mainly the triple bond stretching modes. The C-C-triple bond of alkynes (2130-2150 cm-1) is usually fairly weak, if observed at all. The C-N-triple bond of nitriles (example 10) (2100-2160 cm-1). In most cases a peak (with varying intensity) around 2349 cm-1 (together with 667 cm-1). This is due to the CO2 in the beam (poor background correction).

c. From 1500 - 2000 cm-1(E-X-double bonds: E=X=C, N, O)

This is the most important range in the entire IR spectrum for organic chemists. If there is a very strong peak between 1640 and 1850 cm-1, there is most likely a carbonyl function in the molecule. Analysis of the exact peak position will reveal further what type of carbonyl function is present. The general rule is the more reactive the carbonyl compound is, the further to the right (=higher wavenumber) the C=O stretching frequency will be. The following sequence is observed:

acid chlorides > anhydrides > ester > aldehydes > ketones > carboxylic acids > amides

In order to identify a specific group additional information is needed from the other ranges of the IR spectrum:

Carbonyl derivativeC=O-vibration (cm-1)ExampleComment
acid chloride1780-1820example 11
carboxylic acid1700-1725example 6broad peak between 2500 and 3500 cm-1
anhydride ~ 1760 and ~1810example 12
ester1730-1750example 13medium or strong peak between 1000 and 1300 cm-1
aldehyde1720-1740example 15weak or medium peaks peaks at 2750 and 2850 cm-1
ketone1705-1725example 14
amide1630-1680example 16peaks at 3300-3500 cm-1 for primary or
secondary amides

Ring strain usually increases the C=O stretching frequency. Conjugation of teh carbonyl group with other double bonds (aromatics, alkenes) or the formation of hydrogen bonds decreases the bond strength (shift: 15-60 cm-1 to lower wavenumbers) as the following examples of cyclic ketones demonstrate.

For all 'anhydride type systems' (O=C-X-C=O, X=O, NR, S, CH2) two peaks are observed in this region due to a symmetric and an asymmetric stretching mode.

In addition to the dominant C=O-mode, the C-C-double bond (example 2 -4 )is also located in this area (1600-1660 cm-1). However, if there is a carbonyl group present, it might be difficult to locate a weak or medium sized peak right next to it. Sometimes it is only observed as a shoulder. If there is a coupling between a C=C-group and other double bonded systems e.g. C=O or aromatic systems, the intensity will increase due to the increase in dipole momentum in the double bond. A medium or strong peak in this area corresponds to aromatic ring.

A nitro group shows two very intense peaks in the range between 1300-1400 cm-1(sym.) and 1500-1600 cm-1 (asym.) (example 17).In a good spectrum, it might be possible to deduct the substitution pattern on an aromatic system from the overtone combination vibrations in the range from 1660-2000 cm-1 e.g. four peaks with increasing intensity are characteristic for a monosubstitution on your ring (example 4 and table for Arenes below).

d. From 1000-1500 cm-1 (E-X-single bonds, deformation, rocking modes)

A strong peak around 1450 cm-1 indicates the presence of methylene groups (CH2), while an additional strong peak about 1375 cm-1 is caused by a methyl group (CH3) (examples 1, 8-10). A symmetric 'doublet' with medium intensity around 1370 cm-1 is characteristic for an isopropyl group, while an asymmetric 'doublet' between 1365 and 1390 cm-1 is often due to a t-Bu-group.

The C-O-C-functions of ethers and esters are typically found as strong peaks in the range between 1000 and 1300 cm-1 (example 13). Generally, assignments in this area have to be done with extreme care, because there are a lot of ring absorbances in this ‘fingerprint area’.

e. Below 1000 cm-1 (out of plane modes, C-X: X=Cl, Br, I, heavier atoms)

This range belongs to the ’fingerprint area’, where assignments are a little bit uncertain. In some cases you can find information about the substitution pattern of alkenes or aromatic ring systems, if the range between 700-900 cm-1 is analyzed correctly (oop bending).


Substitution patternPeaks at cm-1Intensity
monosubstitution895-915, 985-995strong, strong
cis-1,2-subst.680-730 strong
1,1-disubst855-885 strong
trisubst.790-830 medium


Substitution pattern Peaks at cm-1Intensity
mono680-710, 720-760strong, strong
ortho (1,2)725-760strong
meta (1,3)670-710, 750-805, 870-900strong, strong, medium
para (1,4)800-860strong
1,2,3680-715, 750-780medium, strong
1,3,5670-700, 830-910medium, strong
1,2,4785-830, 870-900strong, medium

Besides this information you may find strong C-Cl-vibrations around 740 cm-1 e.g. CH2Cl2.

3. Conclusion

Nobody is expected to memorize all those numbers. A lot of details can be deducted if a basic understanding of general trends is there. Although IR spectroscopy can reveal a lot of information about a molecule, but it is not enough to identify an unknown compound for sure. It is also important to keep in mind that an attempt to get too many details out of the ‘fingerprint area’ can be very misleading. In order to determine the nature of an unknown compound correctly, other techniques such as NMR (1H,13C, HETCOR, DEPT), Mass spectrometry, the chemical reactivity (solubility tests, sometimes derivatives) and physical properties (melting point, boiling point, refractive index) are required as well. Based on all these data, a structural suggestion for an unknown compound can be made.

4. Examples

Example 1: Alkane - Undecane

Frequency in cm-1Assignment
2924, 2850C-H (sp3, stretch)
1467CH3 (sp3, bend)
1378CH2, CH3 (sp3, bend)
721for chain deformation (n>4)

Example 2: Alkene - 1-Pentene

Frequency in cm-1Assignment
3080C-H (sp2, stretch)
2965C-H (sp3, s)tretch
1642C=C (stretch)
992, 911=C-H (oop bending)

Example 3: Arene - Benzene

Frequency in cm-1Assignment
3035, 3060, 3090C-H (sp2, aromatic)
1479C=C (stretch)
1036=C-H (bending in ring plane)
673=C-H (oop bending)

Example 4: Substituted Arene - Ethylbenzene

Frequency in cm-1Assignment
3030, 3050=C-H (sp2, stretch)
2966C-H (sp3, stretch)
1730-1940combination overtones
(for a monosubstitution)
1605, 1496C=C (stretch)
1450CH2, CH3 (bend)
746, 697 C-H (oop bending, monosubst.)

Example 5: Alcohols - Cyclohexanol

Frequency in cm-1Assignment
3600-3100O-H (stretch)
2930, 2850CH2 (sp3, stretch)
1067C-O (stretch)

Example 6: Carboxylic Acid - Hexanoic Acid

Frequency in cm-1Assignment
3500-2400O-H (stretch, O-H bridges)
2932C-H (sp3, stretch)
1711C=O (stretch)

Example 7: Primary Amine - Propylamine

Frequency in cm-1Assignment
3367, 3280N-H (stretch, two peaks !)
2958C-H (sp3, stretch)
1607 N-H (broad, scissoring)
816 N-H (broad, oop bending)

Example 8: Secondary Amine - Diethylamine

Frequency in cm-1Assignment
3280N-H (stretch, one peak only)
2964C-H (sp3, stretch)
1452CH3 (bend)
1378CH2, CH3 (bend)
776N-H (oop bending)

Example 9: Tertiary Amine - Triethylamine

Frequency in cm-1Assignment
2980C-H (sp3, stretch)
1460CH3 (stretch)
1380CH2, CH3 (bend)

Example 10: Nitrile - Butyronitrile

Frequency in cm-1Assignment
2973C-H (sp3, stretch)
2250C=N (stretch)
1465CH3 (stretch)
1386CH2, CH3 (bend)

Example 11: Carboxylic Acid Chloride - Valeryl chloride

Frequency in cm-1Assignment
3560overtone from C=O peak
2963C-H (sp3, stretch)
1801C=O (stretch)
753C-Cl (stretch)

Example 12: Carboxylic Acid Anhydrides - Propionic anhydride

Frequency in cm-1Assignment
2987C-H (sp3, stretch)
1819C=O (asym. stretch)
1751C=O (sym. stretch)
1094C-O (stretch)

Example 13: Ester - Butyl acetate

Frequency in cm-1Assignment
2962C-H (sp3, stretch)
1743C=O (stretch)
1243, 1031C-O-C (stretch)

Example 14: Ketone - Cyclohexanone

Frequency in cm-1Assignment
3405overtone from C=O peak
2938C-H (sp3, stretch)
1714C=O (stretch)

Example 15: Aldehyde - Butyraldehyde

Frequency in cm-1Assignment
2950C-H (sp3, stretch)
2820, 2720C-H (sp2, stretch, aldehyde)
1720C=O (stretch)
1460CH3 (stretch)
1388CH2, CH3 (bend)

Example 16: Amide - Propionamide

Frequency in cm-1Assignment
3362, 3200N-H (stretch, two peaks for
a primary amide)
2920C-H (sp3, stretch)
1661C=O (stretch)

Example 17: Nitro compounds - 1-Nitropropane

Frequency in cm-1Assignment
2978C-H (sp3, stretch)
1554NO2 (asym. stretch)
1387NO2 (sym. stretch)

Spectrum 1: o-Nitrotoluene (R1, C3)


n(CH, sp2)


n(CH, sp3)

1524, 1613

n(C=C, aromatic)

1384, 1524



d(CH3, bend)


oop, ortho-subst. arene

Spectrum 2: Benzamide (R3, C3)

3177, 3369

n(NH2, prim. amide)


n(CH, sp2)


n(C=O, amide)

1495, 1618

n(C=C, aromatic)


d(NH2, amide II band)

Spectrum 3: Terephthalic acid (R1, C4)


n(OH, carboxylic acid)


n(C=O, carboxylic acid)

1511, 1575

n(C=C, aromatic)


n(C-OH, acid)


oop, para-subst. arene

Spectrum 4: Bis(o-tolyl) sulfoxide (R4, C3)


n(CH, sp2)


n(CH, sp3)

1474, 1590

n(C=C, aromatic)

1382, 1453

d(CH3, bend)


n(S=O, sulfoxide)


oop, ortho-subst. arene

Spectrum 5: p-Anisidine (R2, C2)

3348, 3423

n(NH2, prim. amine)

1511, 1618

n(C=C, aromatic)


d(CH3, bend)

1033, 1236

n(COC, ether)


oop, para-subst. arene

Spectrum 6: m-Methoxybenzaldehyde (R1, C1)


n(CH, sp2)


n(CH, sp3)

2733, 2840

n(CH, aldehyde)


n(C=O, aldehyde)

1485, 1597

n(C=C, aromatic)

1385, 1450

d(CH3, bend)

1040, 1264

n(COC, ether)

684, 774, 901

oop, meta-subst. arene

Spectrum 7: 2,6-Dimethylphenol (R3, C2)


n(OH, phenol


n(CH, sp2)


n(CH, sp3)

1478, 1615

n(C=C, aromatic)

1379, 1446

d(CH3, bend)


n(C-OH, phenol)

671, 757

oop, 1,2,3-subst. arene

Spectrum 8: Bibenzyl (R1, C4)


n(CH, sp2)


n(CH, sp3)

1492, 1600

n(C=C, aromatic)



699, 752

oop, mono-subst. arene

Spectrum 9: (+)-Camphoric anhydride (R4, C1)


n(CH, sp3)

1761, 1802

n(C=O, anhydride)

1386, 1456

d(CH2, CH3, bend)

994, 1044

n(COC, anhydride)

Spectrum 10: Acetic acid, 3-buten-1-yl ester (R5, C4)


n(CH, sp2)


n(CH, sp3)


n(C=O, ester)


n(C=C, alkene)

1383, 1478

d(CH2, CH3, bend)

1024, 1242

n(COC, ester)

918, 996

oop, mono-subst. alkene

Note: The designation behind the name is the row and column number the molecule is located in the table.

General comments:

1. The average grade for the assignment is ~36.2 points (out of 40). If you scored around or less than 30 points, I would advise you to see your TA or the instructor to seek some help. There is still some time to practice and do better in the final exam.

2. One of the most common mistakes was not to analyze the n(C-H) region correctly to determine what type of compound (alkane, alkene, aromatic) is present. Just writing "sp3" or "C-H" is not sufficient here.

3. The n(OH) peaks for carboxylic acids and alcohols look very different and can be easily identified by looking at them (see reader). However, in some cases intramolecular hydrogen bonding broadens and shifts (to lower wavenumbers) the peak of an alcohol.

4. Carbonyl peaks can be shifted to lower wavenumbers when the carbonyl function is conjugated to another p-system i.e., double bond or aromatic ring.

5. The presence of two peaks in the carbonyl range (1630-1850 cm-1) indicates either the presence of an "anhydride type" of function or two different carbonyl functions (i.e., conjugated and saturated ketone (see exampke 1)). These peaks can vary greatly in terms of their relative intensities.

6. Peaks due to amine, alkyne or alcohol functions are very different in appearance. Alkyne peaks (CH stretch) are very sharp and fairly intense due to an "isolated" motion. Amine peaks are usually a little broader and less intense. The presence of two peaks suggests a primary amine (NH2) while secondary amines show only one peak in this range. Alcohols (OH-stretch) usually show a rounded peak in this area. There is also a difference between amine and amides in terms of width and intensity of the peaks.

8. If the spectrum shows a lot of peaks (independent from their size), the molecule has to have a lot of atoms and also cannot be very symmetric.

9. Amide and amine are not the same functionality! Many students mislabeled the peaks in these peaks.

10. A significant number of assignments were rather sloppily done: very poor labeling (i.e., "C-H", "aromatic", "sp2/sp3", etc.) structures that are not a choice, benzene rings without the double bonds, five-bonded carbon or nitrogen atoms, etc.

11. In many cases, groups were presented in the labeling that are not part of the final moelcule i.e., CH2 group if the molecule did not have one. Students that did this repeatedly received a deduction due to the inconsistancy of the data and the proposed structure.

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