Investigate the processes used to analyse the structure of simple organic compounds, including infrared spectroscopy

What this dot point is asking

NESA wants you to explain that bonds in molecules absorb infrared radiation at specific frequencies that depend on bond strength and atomic masses, identify functional groups from characteristic absorption bands in an IR spectrum, and use the fingerprint region to compare an unknown to a reference.

The answer

Why bonds absorb infrared

A covalent bond behaves like a tiny spring connecting two atoms. It has a natural vibration frequency given (to first approximation) by:

where $k$ is the bond force constant (stiffness) and $\mu$ is the reduced mass of the two atoms. Stronger bonds and lighter atoms vibrate at higher frequencies. When the molecule is hit with IR light at exactly its natural frequency, the bond absorbs energy and vibrates more vigorously. The detector sees this as a dip in the transmitted intensity at that frequency.

IR spectra are plotted as transmittance (or absorbance) versus wavenumber $\bar{\nu}$ in $cm^{-1}$, conventionally with high wavenumber on the left and decreasing to the right. The HSC range of interest is 4000 to 500 cm$^{-1}$.

Why only some vibrations absorb

A vibration only absorbs IR if it changes the molecular dipole moment. Symmetrical stretches of nonpolar bonds (like the symmetric stretch of $CO_2$, or the stretch of $H_2$) do not appear. Polar bonds (O-H, N-H, C=O) give the strongest absorptions.

The diagnostic absorption table

Memorise this:

The single most useful peak is the C=O stretch at 1700 to 1750 cm$^{-1}$, because it is intense, narrow, and only present when there is a carbonyl.

Reading a spectrum: the four-step screen

1. Is there a strong C=O around 1700 to 1750? If yes, a carbonyl is present (aldehyde, ketone, acid, ester, amide).
2. Is there a broad O-H/N-H above 3000? Broad and centred on 3300 (alcohol O-H), very broad from 2500 (acid O-H), or sharper around 3400 (amine N-H).
3. Any C-H above 3000? If yes, the molecule has sp$^2$ C-H (alkene or aromatic) or sp C-H (terminal alkyne at 3300).
4. Fingerprint region (below 1500) for matching to a reference library.

Combining the answers narrows the structural class:

• C=O present and broad O-H below 3000: carboxylic acid.
• C=O present, no broad O-H: aldehyde, ketone, ester or amide; distinguish by the C-H doublet (aldehyde) or N-H (amide) or C-O ester pattern.
• No C=O and broad O-H: alcohol.
• No C=O, no O-H, but =C-H above 3000: alkene or aromatic.
• No C=O, no O-H, only C-H below 3000: alkane.

The fingerprint region

Below 1500 cm$^{-1}$, the spectrum is dominated by C-C, C-O and skeletal bending vibrations. The pattern is too complex to assign peak by peak, but it is unique to each compound. To confirm an identification, compare the fingerprint region to a reference spectrum: a match across both functional-group region and fingerprint region is conclusive.

Strengths and limits

Strengths. Fast (seconds per spectrum on a modern FTIR), non-destructive (the sample can be recovered), identifies functional groups directly, and works on solids, liquids and gases.

Limits. Does not give exact carbon counts (use mass spectrometry). Cannot distinguish enantiomers (use chiral chromatography or polarimetry). Heavily overlapping peaks in the fingerprint region need a reference library to resolve. Mixtures give superimposed spectra that can be hard to deconvolve.

Common traps

Reading the wavenumber axis left to right. Wavenumber decreases to the right. The O-H stretch (high wavenumber) is on the left of the spectrum.

Confusing alcohol O-H with carboxylic acid O-H. Alcohol O-H is broad but centred around 3300; acid O-H is very broad, dragging down to 2500 cm$^{-1}$, often obscuring the C-H stretches.

Forgetting the aldehyde Fermi doublet. Two small peaks at 2720 and 2820 cm$^{-1}$ are diagnostic of an aldehyde C-H.

Reading the strongest peak as the most important. C-H stretches are always strong but rarely diagnostic. Look at the diagnostic regions (carbonyl, O-H, N-H) first.

Treating IR as definitive for chirality. IR sees bonds, not chirality. Enantiomers give identical IR spectra.

In one sentence

Infrared spectroscopy identifies functional groups by their characteristic bond-vibration absorptions: a sharp strong C=O at 1700 to 1750 cm$^{-1}$ signals a carbonyl, a broad O-H at 3200 to 3550 signals an alcohol (and from 2500 to 3300 a carboxylic acid), and the fingerprint region below 1500 cm$^{-1}$ uniquely identifies the molecule by comparison to a reference.