IR

Pass infrared light through a sample of acetone and most of it comes out the other side. The main exception is a narrow band near 1715 cm⁻¹ (the units are explained below), which acetone absorbs almost completely. That absorption comes from its C=O bond. Every simple ketone absorbs near 1715 too.

Infrared (IR) spectroscopy records which infrared frequencies a compound absorbs. Each kind of bond absorbs in its own range, so the pattern of absorptions tells you which functional groups the molecule contains. IR is the first and quickest of the structure methods, so it is usually where you start. Mass spectrometry, the next section, gives the molecular weight and formula. NMR, which maps the carbon skeleton, comes later in the course, in NMR Spectroscopy.

What IR measures

Picture a bond as a spring with a ball on each end. The balls are the two atoms. They move apart and back together, over and over, and each bond does this at its own natural frequency. This motion is a stretching vibration, or a stretch for short.

Infrared light has longer wavelengths than visible red light. Light arrives in packets of energy called photons, and a longer wavelength means less energy per photon. An infrared photon has too little energy to lift an electron into a higher orbital, but enough to make a bond vibrate harder. A molecule absorbs an infrared photon when the photon's frequency matches the natural frequency of one of its bonds.

IR spectra give positions as a wavenumber, in units of cm⁻¹ (read "reciprocal centimeters"). A wavenumber counts how many waves of the light fit into one centimeter, so it equals 1 divided by the wavelength in centimeters. Acetone's C=O band has a wavelength of about 5.83 micrometers (millionths of a meter), which is 0.000583 cm. One divided by 0.000583 cm gives 1715 cm⁻¹.

More waves per centimeter means a shorter wavelength, a higher frequency and more energy. So a bigger wavenumber always means a faster vibration, which is why chemists use it.

Not every stretch shows up in a spectrum. A vibration absorbs infrared light only if it changes the molecule's dipole moment, the measure of charge separation defined in Bond polarity. Acetone's C=O is polar, with δ+ on carbon and δ− on oxygen. Each stretch moves the two charges farther apart and back again, so the dipole moment grows and shrinks, and the band is strong.

The C=C of trans-but-2-ene is the opposite case. Its two halves are identical and point in opposite directions, so the molecule has no dipole moment. When the C=C stretches, both halves move out and back together, and the dipole moment stays at zero throughout. The stretch absorbs no infrared light at all. Chemists call a vibration like this IR-inactive.

Two variables set the frequency

Compare three carbon–carbon bonds. A C–C single bond absorbs near 1000 cm⁻¹, a C=C double bond near 1650, and a C≡C triple bond near 2150. Each extra bond between the same two carbons makes the bond stronger, and a stronger bond is a stiffer spring. A stiffer spring snaps back faster, so it vibrates at a higher frequency. The spectrum measures the difference in bond strength that Bonding described.

The mass of the atoms matters too. A C–H bond absorbs at 2850–2960 cm⁻¹, far above the C–C single bond at 1000. Hydrogen is the lightest atom, and a light ball on a spring moves back and forth faster than a heavy one. That is why every bond to hydrogen (C–H, N–H and O–H) absorbs high, between about 2500 and 3600 cm⁻¹.

Deuterium shows the mass effect on its own. It is the isotope of hydrogen with one neutron in its nucleus (Atomic structure), so it makes the same bonds but weighs twice as much. A C–D bond absorbs near 2100–2200 cm⁻¹, well below a C–H bond, although the two bonds are almost equally strong.

The absorptions worth recognizing on sight

Those two rules divide a spectrum into four regions. Reading from high wavenumber to low: bonds to hydrogen fall between 4000 and 2500 cm⁻¹, triple bonds between 2500 and 2000, and double bonds between 2000 and 1500. Below 1500 comes a crowded region called the fingerprint region, which a later section covers.

The table gives the bands to know. Strong, medium and weak describe how much light a band absorbs. A broad band spreads over a wide range of wavenumbers, and a sharp one is narrow.

Printed spectra follow two conventions that look backwards at first. The wavenumber axis runs from high on the left, at 4000, to low on the right. The vertical axis shows percent transmittance, the share of the light that passes through the sample. A reading of 100% means nothing was absorbed, so the baseline sits at the top and every absorption dips down from it. Chemists still call these dips peaks, or bands.

An alcohol's O–H band is broad and strong, centered near 3350. Alcohol molecules hydrogen-bond to one another (Bond polarity), and at any instant each O–H is hydrogen-bonded a little differently from the next. Each one therefore vibrates at a slightly different frequency, and the many overlapping bands merge into one wide band.

A carboxylic acid's O–H band is broader still. It sprawls from about 3300 down to 2500 and often swallows the C–H bands near 3000. Two acid molecules hold each other through a pair of strong hydrogen bonds, which weakens each O–H further and spreads its band even wider.

An N–H band sits between 3300 and 3500. It is sharper than an O–H band, because N–H bonds form weaker hydrogen bonds. It is also weaker, because nitrogen is less electronegative than oxygen, so an N–H stretch changes the dipole moment less.

The number of N–H bands tells primary and secondary amines apart. A primary amine has an NH₂ group. Its two N–H bonds can stretch in step, or in turn, one lengthening while the other shortens. The two motions absorb at slightly different wavenumbers, so a primary amine shows two N–H bands. A secondary amine has one N–H bond and shows one band. A tertiary amine has no N–H at all, so it shows none.

The line at 3000 cm⁻¹ sorts C–H bonds by the hybridization of their carbon. C–H bands just below 3000 come from sp³ carbons. Bands just above it, from 3000 to 3100, come from sp² carbons, so they mean an alkene or an aromatic ring. (The aldehyde C–H is the one exception: its carbon is sp², but it absorbs below 3000, as shown further down.) The order follows bond strength. An sp² orbital has more s-character than an sp³ orbital (Hybridization), so it holds the bonding electrons closer to carbon and makes a shorter, stronger C–H bond.

An sp carbon has more s-character again, and its C–H band sits highest, at 3300. This ≡C–H band belongs to a terminal alkyne, one whose triple bond sits at the end of the chain. It is sharp and strong, which sets it apart from a broad O–H in the same place.

The same band separates a terminal alkyne from a nitrile. Both have a triple-bond band between 2100 and 2260, but only the alkyne has the band at 3300.

A C≡C in the middle of a chain gives only a weak band. If the groups on its two ends are the same, the stretch leaves the dipole moment at zero, just as in trans-but-2-ene, and the band disappears.

One more C–H is worth knowing. The hydrogen on an aldehyde's carbonyl carbon gives a pair of weak bands near 2720 and 2820. Almost nothing else absorbs at 2720, so that band is a good sign of an aldehyde.

The carbonyl region in detail

Four compounds share the CH₃–C(=O)– piece: acetyl chloride, methyl acetate, acetone and acetamide. Their C=O bands fall in that order, from about 1800 cm⁻¹ for the acid chloride down to 1650–1690 for the amide. The C=O band is strong and sharp and sits where little else absorbs, and its exact position tells you which kind of carbonyl you have.

The order comes from resonance. In an ester or an amide, the atom next to the carbonyl has a lone pair it can donate into the C=O. Donation gives a resonance contributor with a C–O single bond, so the real C=O has some single-bond character. A partly single bond is weaker, so it absorbs at a lower wavenumber. Nitrogen donates most, so the amide absorbs lowest. (Esters & amides, later in the course, builds the reactivity of these compounds on the same donation.)

Chlorine sits at the other end. Its 3p lone pairs overlap poorly with carbon's 2p orbital, so it donates very little. It also pulls electron density away through the σ bond. The acid chloride's C=O keeps its full double-bond character and absorbs highest, near 1800.

An ester sits a little above a ketone, even though its oxygen donates. Oxygen is more electronegative than carbon, so it also pulls electron density through the σ bond, and for the C=O stretch that pull slightly outweighs the donation.

An anhydride has two C=O groups joined through one oxygen. Like the two N–H bonds of an NH₂ group, they can stretch in step or in turn. The two motions absorb at different wavenumbers, so an anhydride shows two C=O bands, near 1820 and 1760.

Acetophenone, which has a benzene ring on its carbonyl, absorbs near 1685 instead of a ketone's usual 1715. A C=O bonded directly to a C=C or a benzene ring is conjugated with it: their p orbitals line up side by side, so the π electrons spread over both. Conjugation lowers a C=O band by about 30 cm⁻¹. (Conjugated systems, in the next chapter, explains conjugation in full.) In but-3-en-2-one, a C=C sits next to the C=O. One resonance contributor moves the C=O's π electrons onto oxygen and leaves a positive charge on the end carbon of the C=C. The C=O gains some single-bond character, just as it does in an amide.

Ring size moves the band the other way. Cyclohexanone absorbs at 1715, like any other ketone, but cyclopentanone absorbs at 1745 and cyclobutanone at 1785.

The smaller the ring, the more it squeezes the C–C–C angle at the carbonyl carbon below the 120° that an sp² carbon prefers. To make a narrow angle, the two ring bonds from that carbon use orbitals with extra p-character. The carbon has only so much p-character to share out. The orbital it uses for the σ part of the C=O is left with extra s-character, which makes a shorter, stiffer bond with a higher wavenumber.

The fingerprint region

Below about 1500 cm⁻¹ the spectrum turns into a dense cluster of overlapping bands. Some are stretches of single bonds such as C–C, C–O and C–N. Others are bending vibrations, in which a bond angle opens and closes instead of a bond lengthening. This region is the fingerprint region, and it is too crowded to assign band by band.

Its value is that two different compounds almost never give the same pattern there. To confirm that an unknown is a particular compound, compare its fingerprint region with a reference spectrum of that compound. Above 1500, read the functional groups; below 1500, compare with a known spectrum instead.

One pattern in the fingerprint region is worth reading. A benzene ring with a single substituent, a monosubstituted ring, gives two strong bands near 750 and 690 cm⁻¹. They come from the ring's C–H bonds bending out of the plane of the ring.

How to read a spectrum in thirty seconds

Ask four questions, in this order.

• Question 1: is there a strong band between 1650 and 1850? Then there is a C=O, and its exact position says which kind.
• Question 2: is there a band above 3200? A broad one is an O–H, and one or two sharper ones are N–H (a single sharp, strong band at 3300 is the ≡C–H of question 4).
• Question 3: is there a C–H band just above 3000, or a cluster of bands between 1450 and 1600? Then there is a C=C, most likely in an aromatic ring.
• Question 4: is there a sharp band between 2100 and 2260? Then there is a C≡N or a C≡C. If a sharp, strong band at 3300 comes with it, the C≡C is a terminal alkyne.

The four answers give you the functional groups.

Read the empty regions as carefully as the bands. A spectrum with no absorption between 1650 and 1850 rules out every carbonyl compound at once: aldehyde, ketone, ester, acid, amide, acid chloride and anhydride. On a multiple-choice question, that one check often removes most of the options.

A clean baseline above 3200 works the same way. It rules out an alcohol, a carboxylic acid, and any amine or amide that carries an N–H. An absence is often the stronger evidence: an impurity can add a small band, but it cannot take one away.

What carries forward

IR is the first step in every structure problem in the course. It pairs naturally with the degrees of unsaturation from Alkene structure: that count says how many rings and π bonds to look for, and IR says what kind they are. Mass spectrometry, next, supplies the molecular formula, and the NMR methods, later, the carbon skeleton.