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.

Examples in context

Example 1. Tetrahydrocannabinol screening at NSW Forensic Services. The NSW Forensic and Analytical Science Service uses FTIR to confirm tetrahydrocannabinol (THC) in seized cannabis extracts. THC's structure shows two diagnostic IR signatures: a broad O-H stretch near 3450 cm$^{-1}$ from the phenol group and a sharp aromatic C=C signal near 1620 cm$^{-1}$. The fingerprint region below 1500 cm$^{-1}$ matches a NIST library reference uniquely. Comparison takes seconds against a stored spectrum and meets the evidentiary standard required for prosecutions under the NSW Drug Misuse and Trafficking Act. The HSC framework of "identify functional groups by characteristic ranges and confirm with fingerprint" is the same workflow forensic analysts use in court.

Example 2. Quality control of aspirin synthesis in NSW HSC depth study. Stage 6 students who synthesise aspirin record an FTIR spectrum to confirm conversion of salicylic acid (broad O-H around 3200, C=O at 1660 cm$^{-1}$) to aspirin (sharp ester C=O at 1750 cm$^{-1}$ plus separate carboxylic acid C=O at 1690 cm$^{-1}$). The disappearance of the broad O-H from the salicylic-acid phenol is the diagnostic indicator of complete esterification. Markers reward students who annotate the new aspirin spectrum with the wavenumber shifts and explain the functional-group change. This depth-study task is examined as a 5 to 7 mark Section II question almost every year.

Try this

Q1. State the wavenumber ranges associated with each of: O-H of an alcohol, O-H of a carboxylic acid, C=O of a carbonyl. [3 marks]

• Cue. Alcohol O-H: 3200 to 3550 cm$^{-1}$ (broad); acid O-H: 2500 to 3300 cm$^{-1}$ (very broad); C=O: 1670 to 1750 cm$^{-1}$ (sharp, strong).

Q2. A compound has molecular formula $C_3H_6O$ and shows a sharp strong absorption at 1720 cm$^{-1}$ but no absorption above 3000 cm$^{-1}$. Calculate the degree of unsaturation and identify the compound. [3 marks]

• Cue. Degree of unsaturation $= (2 \times 3 + 2 - 6) / 2 = 1$; no O-H rules out alcohol; C=O present; the compound is propanone ($CH_3COCH_3$).

Q3. A student records IR spectra of three unknowns X, Y, Z. X shows broad O-H at 3300, no C=O. Y shows sharp C=O at 1720, no O-H. Z shows broad O-H at 3000 and sharp C=O at 1710. (a) Identify the functional group in each. (b) Predict a possible compound for each. (c) Explain why the carboxylic acid O-H is broader than the alcohol O-H. [3+2+2 marks]

• Cue. (a) X alcohol, Y ketone or aldehyde, Z carboxylic acid. (b) X ethanol, Y propanone, Z ethanoic acid. (c) Carboxylic acid O-H forms strong hydrogen-bonded dimers in solution, broadening the absorption.