the principles and interpretation of mass spectrometry (molecular ion peak, fragmentation pattern, M+1 isotope peaks) and infrared (IR) spectroscopy (characteristic absorption bands of functional groups) for the identification of organic compounds

What this dot point is asking

VCAA wants you to interpret a mass spectrum (identify the molecular ion, use isotope patterns to spot Cl/Br/C, interpret common fragmentation losses) and an infrared spectrum (recognise characteristic absorption bands for major functional groups), and to combine the two to identify or distinguish organic compounds.

The answer

Mass spectrometry (MS) of organic compounds

A mass spectrometer ionises the molecule (usually with electron impact), accelerates the ions in a vacuum, separates them by mass-to-charge ratio (m/z), and detects them. The output is a stick spectrum of relative abundance against m/z. Three things to look at in any organic mass spectrum:

1. Molecular ion (M+) peak. The peak at the highest m/z (ignoring the small isotope peaks above it) corresponds to the whole molecule with a single electron removed. Its m/z equals the molecular mass (in u). For an unknown C3H8O, M+ appears at m/z = 60.

The M+ peak may be weak or absent if the molecular ion fragments easily (alcohols often have very small M+ peaks).

2. Isotope peaks (M+1, M+2). A small peak at one mass unit above the molecular ion (the M+1 peak) is due to natural ^13C (about 1.1% abundance). Its relative intensity is roughly 1.1% per carbon atom, which can be used to estimate the number of carbons.

A peak two mass units above the molecular ion (the M+2 peak) of comparable intensity to M+ is the signature of chlorine (3:1 ratio of M:M+2, because Cl-35 and Cl-37 occur in roughly 3:1 abundance) or bromine (1:1 ratio of M:M+2, because Br-79 and Br-81 are roughly equally abundant). The intensity ratios are diagnostic:

• 3:1 ratio at M and M+2: one Cl atom.
• 1:1 ratio at M and M+2: one Br atom.
• 9:6:1 ratio at M, M+2, M+4: two Cl atoms.

3. Fragmentation pattern. The high-energy collisions break the molecular ion into smaller cations. Common diagnostic losses:

Look for both the loss (M+ - fragment) and the fragment (the peak itself). A peak at m/z = 43 might be C3H7+ (propyl cation) or CH3CO+ (acylium ion); the context (alcohol or ketone) tells you which.

Infrared spectroscopy (IR)

IR radiation excites bond vibrations. Each bond has a characteristic stretching frequency (cm^-1) determined by bond strength and atomic masses. The spectrum plots % transmittance (y, with peaks pointing down) against wavenumber (x, cm^-1, conventionally running 4000 to 400 from left to right).

The key bands you must recognise:

The fingerprint region (below ~1500 cm^-1) is too complex to interpret peak-by-peak but is unique to each compound and useful for matching against a database.

The IR-MS workflow

Given an unknown spectrum and an MS:

1. From MS: read the molecular ion peak to get the molecular mass.
2. Check the isotope peaks for Cl, Br or unusual heteroatoms.
3. Use the degree of unsaturation formula (DoU = (2C + 2 + N - H - X) / 2) to find rings/multiple bonds.
4. From IR: identify the functional groups present (broad O-H, sharp C=O, etc.). The presence and position of bands gives the answer faster than the absence of a band.
5. Use fragments to confirm: the loss of 17 (OH) is consistent with an alcohol; the loss of 18 (H2O) is alcohol dehydration; a peak at 31 is CH2OH^+ from a primary alcohol; a peak at 77 is phenyl.
6. Propose a structure and cross-check both spectra are consistent with it.

Carboxylic acid vs ester vs ketone

A common VCE puzzle is to distinguish three carbonyl species:

• Carboxylic acid: very broad O-H around 2500 to 3300 cm^-1 + strong C=O around 1715. In MS, loss of 45 (COOH).
• Ester: no O-H. Strong C=O around 1740 and strong C-O around 1200. In MS, loss of OR' from the alkyl side.
• Ketone: no O-H. Strong C=O around 1715. No strong C-O peak. Symmetric fragmentation around the carbonyl.

Examples in context

Example 1. Identifying contaminants in Yarra River sediments. Researchers at EPA Victoria Centre for Applied Sciences in Macleod use GC-MS and FTIR to characterise organic contaminants in dredged Yarra sediment. A sample shows $M^{+}$ at m/z $= 254$ with $M+2$ at $254 \times 0.33 \approx 84$ relative units, indicating one chlorine and another $M+4$ at $\sim 5$ units suggesting bromine absent. Fragment at $219$ (loss of $35 = \text{Cl}$) confirms aryl chloride. IR: strong band $1685 \, \text{cm}^{-1}$ ($\text{C=O}$ conjugated), $3070 \, \text{cm}^{-1}$ aromatic C-H, $1600 \, \text{cm}^{-1}$ ring stretch. Degree of unsaturation $= (2 \times 12 - 9 + 2) / 2 = 8.5$, rounded to $8$: a benzene ring (4) plus 4 more, consistent with PCB-related chlorinated biphenyl, likely legacy industrial contamination.

Example 2. Doping control on Bondi triathlete urine samples. ASDTL chemists analyse a positive sample showing $M^{+}$ at $288$ matching testosterone ($\text{C}_{19} \text{H}_{28} \text{O}_2$). IR absorptions at $1717 \, \text{cm}^{-1}$ (cyclohexenone $\text{C=O}$), $3400 \, \text{cm}^{-1}$ broad (-OH), $2950 \, \text{cm}^{-1}$ (sp$^3$ C-H), and no broad acid -OH at $2500$ to $3300 \, \text{cm}^{-1}$ confirm an alcohol-ketone, not a carboxylic acid. Fragments at $272$ (loss of -CH$_4$? unusual) and $246$ (loss of $42 = \text{C}_3 \text{H}_6$) follow the testosterone fragmentation library. Confirmation requires three ions within $0.1 \%$ of reference. The urinary metabolite ratio versus epitestosterone exceeds $4:1$, triggering a sanction under WADA rules.

Try this

Q1. State the position (in $\text{cm}^{-1}$) of the IR absorption that would distinguish a carboxylic acid from an ester. [2 marks]

• Cue. Carboxylic acid: very broad O-H from $2500$ to $3300 \, \text{cm}^{-1}$; ester lacks this. Both have C=O around $1700$ to $1735 \, \text{cm}^{-1}$.

Q2. A compound has $M^{+} = 88$, $M+1 = 4.4\%$ of $M$, no $M+2$. IR shows strong absorption at $1730 \, \text{cm}^{-1}$ and $1200 \, \text{cm}^{-1}$, no broad O-H. (a) Calculate carbons. (b) Suggest molecular formula. (c) Suggest a structural class. [4 marks]

• Cue. (a) $4.4 / 1.1 = 4$ C. (b) $\text{C}_4 \text{H}_8 \text{O}_2$. (c) Ester (C=O and C-O without O-H); for example methyl propanoate or ethyl ethanoate.

Q3. A compound has $M^{+} = 122$, $M+2$ equal in height to $M$. IR: $1690 \, \text{cm}^{-1}$, $3050 \, \text{cm}^{-1}$, $1600 \, \text{cm}^{-1}$. (a) State what the $M+2$ pattern implies. (b) Identify the C=O environment. (c) Propose a structure. [2+2+2 marks]

• Cue. (a) One Br atom (1:1 isotope ratio). (b) Conjugated/aromatic C=O. (c) For mass $122$ minus $79$ (Br) leaves $43$, a CH$_3$-CO group; a bromoaromatic with no extra OH; possibly bromoacetophenone, $\text{BrC}_6 \text{H}_4 \text{COCH}_3$.