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.

Worked example

Unknown: M+ at m/z = 88, peaks at 73, 60, 45, 43, 29. IR has strong absorption at 1735 cm^-1 and at 1180 cm^-1; no peak above 3000 cm^-1.

Mass 88 with no O-H suggests an ester (C4H8O2). The strong 1735 is a C=O; 1180 is C-O of an ester. Fragments:

• m/z = 73 (loss of 15): loss of CH3.
• m/z = 60 (loss of 28): loss of CO or C2H4.
• m/z = 45: COOH^+ or CHO2^+, suggesting the ester.
• m/z = 43: CH3CO^+ (acetyl cation), strongly suggesting the acid part is acetic acid.
• m/z = 29: CHO^+ or C2H5^+.

A likely structure: ethyl propanoate (CH3CH2COOCH2CH3, M = 102, NO) or methyl propanoate (CH3CH2COOCH3, M = 88, YES) or propyl methanoate (HCOOCH2CH2CH3, M = 88). The CH3CO^+ at 43 is more consistent with methyl ethanoate (CH3COOCH3, M = 74, NO), so reconsider. With M = 88 and a CH3CO^+ peak: methyl propanoate gives CH3CH2CO^+ at 57, not 43. Propyl methanoate (HCOOC3H7) gives HCO^+ at 29 and C3H7^+ at 43. So propyl methanoate is consistent. Best structure: propyl methanoate, HCOOCH2CH2CH3 (or its isopropyl analogue).

(Exam-grade answers always include the working and acknowledge multiple consistent structures when the data does not fully resolve.)

Common traps

Reading the IR axis the wrong way. Wavenumber on the x-axis runs right to left (high to low cm^-1). Transmittance on the y-axis dips downward for absorbance peaks.

Confusing the broad O-H of an alcohol with the broad O-H of a carboxylic acid. Alcohol O-H is around 3300 cm^-1, often above the C-H region. Acid O-H is much broader and often centred around 3000 cm^-1, overlapping with C-H.

Missing the M+2 isotope clue. A strong M+2 peak the same height as M+ means one bromine atom. A 3:1 ratio means one chlorine.

Forgetting that the molecular ion can be weak or absent. For some compounds, M+ is barely visible. Use the next highest peak and an even-electron loss to back-calculate.

Calling every C=O a ketone. A C=O alone could be an aldehyde, ketone, ester, amide or acid. Combine with the O-H or C-O or N-H to identify the family.

Skipping the degree of unsaturation. A DoU of 1 implies one C=C or one ring; 4 implies a benzene ring.

In one sentence

Mass spectrometry gives the molecular mass from the molecular ion, the M+1 isotope peak counts carbons (about 1.1% per C), M+2 spots Cl (3:1) and Br (1:1), and fragmentation patterns show common losses (15 CH3, 17 OH, 18 H2O, 29 CHO, 45 COOH); IR identifies functional groups by absorption bands (broad O-H around 3300, N-H near 3400, sharp C=O around 1700, C-O around 1100), and the two spectra together let you propose a complete organic structure.