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

What this dot point is asking

NESA wants you to describe how a mass spectrometer ionises and separates molecules by mass-to-charge ratio, read a mass spectrum to find the molecular ion (M+), the base peak (tallest, set to 100%) and fragment ions, recognise common fragment losses, and identify chlorine and bromine from M and M+2 patterns.

The answer

How a mass spectrometer works

The instrument has five stages:

1. Vaporisation. The sample is heated and admitted to a high-vacuum chamber as a gas.

2. Ionisation. A beam of high-energy electrons (about 70 eV) collides with each molecule, knocking out one electron to form a radical cation:

   $$M + e^- \rightarrow M^{+ \bullet} + 2e^-$$

   This is the molecular ion. Some molecular ions break apart into smaller cations and neutral radicals; these are the fragment ions.

3. Acceleration. Positive ions are accelerated through a potential difference, gaining the same kinetic energy. Lighter ions reach higher velocities.

4. Deflection. A magnetic field bends each ion's path. Lighter (and more highly charged) ions deflect more. The radius of deflection depends on the mass-to-charge ratio, $m/z$.

5. Detection. Ions hit a detector; their abundance at each $m/z$ is recorded.

The result is a mass spectrum: a bar chart of relative abundance (0 to 100%) against $m/z$. The base peak is set to 100%.

Reading a mass spectrum

Molecular ion (M+)

The peak at the highest $m/z$ (excluding the small isotope satellites) is the molecular ion. Its $m/z$ is the molecular mass of the compound.

Base peak

The tallest peak, set to 100% relative abundance. Often a fragment, not the molecular ion. The base peak indicates the most stable cation formed in fragmentation.

Fragment peaks

Other peaks. The difference between $M^+$ and a fragment is the mass of the neutral radical lost.

Common fragment losses

So a loss of 17 ($OH$) suggests an alcohol; a loss of 45 ($COOH$) suggests a carboxylic acid; a loss of 29 ($CHO$) suggests an aldehyde.

Common diagnostic fragments

Isotope patterns

Chlorine

$^{35}Cl$ : $^{37}Cl$ = 3 : 1. A compound with one $Cl$ shows M : M+2 = 3 : 1. Two chlorines give M : M+2 : M+4 = 9 : 6 : 1.

Bromine

$^{79}Br$ : $^{81}Br$ $\approx$ 1 : 1. A compound with one $Br$ shows M : M+2 of roughly equal height. Two bromines give M : M+2 : M+4 = 1 : 2 : 1.

Carbon

$^{12}C$ : $^{13}C$ = 98.9 : 1.1, so the M+1 peak is about 1.1% per carbon atom. A 10-carbon molecule shows an M+1 peak about 11% of M; this can be used to count carbons.

Worked logic for an unknown

Suppose a mass spectrum shows:

• $m/z = 88$ molecular ion (small peak).
• $m/z = 73$ (loss of 15, $CH_3$).
• $m/z = 43$ (base peak, $CH_3CO^+$).
• $m/z = 15$ ($CH_3^+$).

Molecular formula at 88 is consistent with $C_4H_8O_2$. Base peak at 43 ($CH_3CO^+$) and loss of $CH_3$ from M+ both point to an acetyl group. Loss of 45 (88 to 43) is $C_2H_5O$, i.e. an $OC_2H_5$ ethoxy group. The compound is ethyl ethanoate $CH_3COOC_2H_5$.

Strengths and limits

Strengths. Gives the exact molecular mass and structural fragments. Picks out chlorine, bromine and sulfur by isotope pattern. Sensitive to nanograms of sample. Used routinely in forensics, drug testing, environmental analysis (often coupled to gas chromatography, GC-MS).

Limits. Destroys the sample during analysis. Cannot distinguish stereoisomers. Cannot always distinguish structural isomers if they fragment similarly (propan-1-ol and propan-2-ol have very similar spectra). Best combined with NMR and IR for unambiguous structural assignment.

Examples in context

Example 1. Drug testing at the Forensic and Analytical Science Service Lidcombe. FASS uses GC-MS to confirm methamphetamine in seizures and biological samples. Methamphetamine has a molecular ion at $m/z = 149$. The base peak at $m/z = 58$ arises from loss of $C_6H_5CH_2$ (91), corresponding to $\alpha$-cleavage adjacent to the nitrogen. A second diagnostic fragment at $m/z = 91$ confirms the benzyl cation. The NIST mass spectral library match is the legal standard for confirmation in NSW courts. The HSC framework of "molecular ion, base peak, characteristic fragment losses" is the same one FASS analysts use, just at higher resolution.

Example 2. Identifying ethanol versus methanol in NSW HSC depth study. Stage 6 candidates given two unlabelled mass spectra must distinguish methanol ($M = 32$) from ethanol ($M = 46$). Methanol shows molecular ion at $m/z = 32$, with $m/z = 31$ from loss of H. Ethanol shows molecular ion at $m/z = 46$ and base peak at $m/z = 31$ from loss of $CH_3$ (15), the diagnostic alpha-cleavage of a primary alcohol. The HSC question tests whether the candidate correctly identifies the molecular ion as the highest m/z signal and the base peak as the tallest. Markers reward students who explicitly cite the fragment loss value as the diagnostic indicator.

Try this

Q1. Describe the five stages of a mass spectrometer in order. [3 marks]

• Cue. Ionisation (electron impact), acceleration (electric field), deflection (magnetic field), detection (electron multiplier), recording (data system).

Q2. A compound shows a molecular ion at $m/z = 88$, a base peak at $m/z = 73$ and another peak at $m/z = 43$. Identify the fragment losses and deduce the compound (a saturated organic ester of $C_4H_8O_2$). [3 marks]

• Cue. Loss of 15 = $CH_3$ from $m/z$ 88; loss of 45 = $COOH$ to give $m/z$ 43 ($C_3H_7^+$); the compound is methyl propanoate.

Q3. A mass spectrum shows a 3:1 isotope cluster at $m/z = 78$ and 80. (a) Identify which element is present. (b) Calculate the molecular mass of a hypothetical methyl halide containing this element. (c) Predict the corresponding M and M+2 m/z values. [2+2+1 marks]

• Cue. (a) Chlorine ($^{35}Cl$ and $^{37}Cl$ in 3:1 ratio). (b) Methyl chloride $CH_3Cl$, $M = 50.5$. (c) $m/z = 50$ and 52.