Infrared spectra of aromatic rings

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: Para-cresol: 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: [...]... 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 Ketal Acetal Ketals give... 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... frequency of vinyl and aryl ethers compared to alkyl ethers can be explained by resonance: Amine infrared spectra Butylamine Dibutylamine Tributylamine N-methylaniline 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 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... 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 . 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. 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 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