Infrared spectra of aromatic rings

Infrared spectra of alcohols and phenols O-H stretch: The free O-H stretch is a small sharp peak at 3650-3600.. The strong C-O stretch at 1260-1000 can be used to determine the structure

Infrared spectra of aromatic rings

Aromatic IR spectra are messy and difficult They show many small bands of no diagnostic value, but some are useful

While Aromatic and Alkene C-H stretches both occur just over 3000, the C=C aromatic stretches appear

between 1600 and 1450, outside the usual range for alkenes which is near 1650 Aromatic C=C stretches

are often in pairs, with one at 1600 and one at 1475

Like substituted alkenes, there are out-of-plane C-H bends between 900 and 690 These result in an overtone/combination band between 2000 and 1667, which is often used to assign how the aromatic molecule is substituted I assume this because the overtone/combination region is less likely to be

obscured by other peaks

The shaded boxes in the table above actually correspond to out-of-plane C=C bends

Below are some spectrum examples:

Infrared spectrum of toluene

Infrared spectrum of ortho-diethylbenzene

Infrared spectrum of meta-diethlbenzene

Infrared spectrum of ortho-diethylbenzene

Infrared spectrum of para-diethylbenzene

Infrared spectrum of styrene

The in-plane bending occurs between 1000 - 1300, but these bands are not useful since they overlap with stronger bands in this region

The original C-H out-of-plane bands can also be used to assign substitution directly in some cases, rather then using the overtone/combination region It is suggested to use the the 900 - 690 region only for certain substituents:

Reliable substituents:

Alkyl-Alkoxy-Halo-Amino-Carbonyl-Unreliable:

Nitro-Carboxylic acid and derivatives Sulfonic acids and derivatives

Monosubstituted: Always gives a strong band at 690, but in a halocarbon solvent this may be obscured by

a C-X stretch A second strong band usually appears at 750

Orthosubstituted: One strong band near 750

Metasubstituted: One band near 690 and one near 780 A third band is often found at 880

Parasubstituted: One strong band between 800 and 850

Ester infrared spectra Vinyl acetate

Methyl benzoate

Methyl methacrylate

Methyl salicylate

Ethly butyrate

The C=O of an ester appears near 1750-1735 which can overlap with some ketone C=O stretches One can usually eliminate ketones by considering the strong and broad C-O peak at 1300-1000 A ketones absorptions will have a weaker and narrower bands Compare the ethyl butyrate ester above to the ketone below:

Esters can also conjugate on the side of the single-bonded oxygen:

Apparently this form of conjugation interferes with resonance of the carbonyl group, hence increasing the arbsorption frequency of the C=O bond

α-keto esters: One might expect two C=O peaks due to the two different carbonyls In practice, it usually manifests as a shoulder on the main C=O peak or a single broadened band

β-keto esters: These give a strong intensity doublet for the C=O stretches Since they do not tautomerize

to the same extent as β-diketones, one cannot generally observe the OH stretch from the enol form

Infrared spectra of alcohols and phenols

O-H stretch: The free O-H stretch is a small sharp peak at 3650-3600

The hydrogen-bonded O-H stretch is a broad strong peak at 3400-3300

For the neat liquid, all the alcohol is considered hydrogen bonding, so the broad peak is the only one of the two visible As the alcohol is dissolved in a non-hydrogen bonding solvent, the free O-H peak becomes more visible:

The right-side picture is of an extremely diluted alcohol, or a gas

C-O-H bend: Broad weak peak at 1440-1220, often obscured by CH3 bends

C-O: Stretch occurs at 1260-1000

1-Hexanol:

2-Butanol:

The free O-H stretch occurs near 3640, 3630, 3620, 3610 for primary, secondary, tertiary, and phenolic alcohols respectively But resolving frequency differences this small requires careful calibration, so these distinctions are of no use

Intramolecular hydrogen bonding usually shifts the O-H stretch to a lower frequency For example, for methyl silicate it is situated about 3200, around 150 less then the average phenol

The strong C-O stretch at 1260-1000 can be used to determine the structure of the alcohol, notably more easy than with the O-H stretch:

Carboxylic acid infrared spectra

The obvious way to recognize an acid is by noting both a OH and a C=O stretch The C=O stretch for the monomer is at 1760-1730, but carboxylic acids form dimers even in dilute solutions, the dimer C=O stretch is at 1730-1700 As usual for carbonyls, conjugation lowers it and a halogen on the α-carbon increases it

The C-O stretch for an acid is near 1260

O-H out-of-plane bending produces a broad, medium-weak band around 930

Isobutyric acid

Benzoic acid

Aldehyde infrared spectra

The normal C=O stretch for aldehydes is 1725, since for ketones the normal stretch is 1715, it is difficult

to distinguish the two using a C=O stretch

To distinguish aldehydes from other carbonyl-containing compounds, look for the doublet in the C-H region, near 2850 and 2750 The 2750 is more useful, because the 2850 can overlap with other C-H stretches

Nonanal The peak at 1460 is due to bending of CH2 Methylene groups often absorb stronger when attached to a cabonyl

Crotoaldehyde

Benzaldehyde

The reason the C-H splits into two peaks is due to fermi-resonance, from a coupling of the C-H stretch with the first overtone of the aldehyde C-H bend at 1400-1350

Amide infrared spectra

Amides show a very strong C=O peak at 1680-1630

Primary amines give two N-H stretch peaks, one near 3350 and one near 3180, from asymmetric and symmetric vibrations respectively

Secondary amines give one N-H stretch peak at 3300

N-H bending at 1640-1550 for both secondary and primary amides

N-methyl acetamide

Note how the C=O partially overlaps the N-H bend

For dilute solutions the C=O is found at 1690 This is for the same reason as carboxylic acids - hydrogen bonding normally lowers the C=O frequency

Like other carbonyl compounds, the C=O stretch frequency increases with decreasing ring size, by about

40 per carbon removed

Ether Infrared spectra

The obvious way to know a molecule is an ether is to see a C-O peak, but no C=O or O-H, since the absence of a C=O or O-H stretch confirms it is not an ester, acid, or alcohol

The C-O stretch is found between 1000 and 1300

Aliphalic ethers give one strong asymmetric stretch around 1120, and a very weak symmetric stretch around 850

Aryl alkyl ethers give two bands around 1250 and 1040, symmetric and asymmetric respectively

An example of each is below:

Dibutyl ether

Anisole c6h5och3

Vinyl ethers: These give a strong asymmetric stretch near 1220, and a very weak symmetric stretch near 850

The increase in asymmetric stretch frequency of vinyl and aryl ethers compared to alkyl ethers can be explained by resonance:

Amine infrared spectra

Dibutylamine

Tributylamine

N-H stretch: At 3500-3300 This is split into two bands for a primary, one for a secondary, and is

nonexistent for a tertiary amine The intensity can be small for aliphatics, often vanishingly small for secondary aliphatic amines For aromatics the stretch can be quite large

The higher frequency band is from a symmetric vibration, the lower from asymmetric vibration

In dilute solutions (non hydrogen-bonded) the peaks are shifted higher

N-H in-plane-bend: A broad peak at 1640-1560 for primary amines Secondary amines absorb near 1500 Can overlap a C=C peak

N-H oop bend: Sometimes observed near 800, seen more easily in aliphatic amines

C-N stretch: Occurs at 1350-1000, varying intensity 1250-1000 for aliphatics, 1350-1250 for aromatics The higher frequency in aromatics is from resonance increasing double-bond character between the ring and attached nitrogen

For a neat liquid, the N-H peak is typically weaker and sharper than O-H:

Ketone infrared spectra

In addition to the obvious C=O stretch, ketones also have a C-CO-C bend:

Aromatic ketones have this stretch at the higher end, aliphatic ketones have this stretch at the lower end Methylene and methyl groups next to the carbonyl tend to absorb with greater intensity than usual The 1715 C=O stretch produces a very weak overtone band at 3430 Avoid confusing it with O-H absorptions, which are much more intense

The ketene above can be considered an extreme example of a strained ring S character of the double bond increases with ring strain, and this reaches a maximum when attached to a sp-hybridized carbon in a ketene

1,2-diketones, also called α-ketones, have one strong absorption about 1716 As expected with normal carbonyls, this peak is lower when the diketone is conjugated Diketones may also produce a narrowly-spaced doublet due to the asymmetric and symmetric absorptions (eg the carbonyls might stretch

together, or one might stretch while the other contracts)

1,3-diketones, also called β-ketones, show more complicated patterns due to tautomerization to the enol form:

The enol form is particularly stable for diketones due to the hydrogen bond, so spectra often show both C=O peaks

The shifted C=O stretch in the enol can be explained by the hydrogen bond, and by resonance:

The relative intensities of the two types of C=O can provide a method for seeing which form is more prevalent

Ketone examples:

Cyclopentanone

Mesityl oxide

Acetophenone

Cyclopentanone

Ketals and acetals infrared spectra

Acetal

Ketals give four strong bands at 1200-1020

Acetals give five strong bands at 1200-1020

C=C stretching vibrations

The C=C stretch vibration appears between 1670 and 1640 Frequency increases as substituents are added to the double bond.

Monosubstituted ones are about 1640

Disubstituted about 1650

Tri and tetra absorb around 1670

Trans substituents absorb about 10 higher than cis.

The C=C is typically a weak absorption It may be completely obscured by a shifted C=O stretch It won't absorb at all if the attached groups are symmetric Cis alkenes have less symmetry than trans, so will have a more intense absorption, despite being at a lower frequency.

Rings and terminal alkenes typically have high and low symmetry respectively, so we would expect terminal alkenes to have

a higher absorption than cyclic alkenes.

Conjugation with another C=C or a C=O will increase single bond character, decreasing the frequency.

Conjugated systems often have two closely spaced peaks, resulting from possible conformations.

C-H bending region