Mass spectrometry of organic compounds explained: HSC Chemistry Module 8

A 5 mark answer needs the molecular formula derived from m/z = 60, the structure deduced from the fragments, and the chemistry of each loss.

Step 1: Molecular formula from M+

$m/z = 60$ with only C, H, O. Trying $C_3H_8O$: $3(12) + 8(1) + 16 = 60$. Matches. Other options ($C_2H_4O_2$ = 60 is also valid) need to be checked against the fragmentation.

Step 2: Identify the loss from M+ to base peak

$60 - 43 = 17$, which is the loss of $OH$. Loss of $OH$ is diagnostic of an alcohol losing a hydroxyl radical to give a carbocation.

Step 3: Identify the m/z = 15 fragment

This is $CH_3^+$ (methyl cation), formed by cleavage of a C-C bond.

Step 4: Deduce structure

Loss of OH plus presence of $CH_3^+$ is consistent with $CH_3CH_2CH_2-OH$ (propan-1-ol) or $CH_3CH(OH)CH_3$ (propan-2-ol). Both are $C_3H_8O$, both can lose $OH$, and both can lose $CH_3$. Propan-2-ol is more likely the answer because the secondary carbocation $CH_3CH^+CH_3$ at $m/z = 43$ is more stable than the propan-1-ol fragment, so the base peak at 43 is expected to dominate. The unknown is most consistent with propan-2-ol.

Step 5: Origin of each peak

• $m/z = 60$: molecular ion $[CH_3CH(OH)CH_3]^+$, formed by loss of one electron.
• $m/z = 43$: $[CH_3CHCH_3]^+$ (loss of OH, 17). Stable secondary carbocation, so the base peak.
• $m/z = 15$: $[CH_3]^+$ (loss of $CH_3CHOH$, 45). Methyl cation.

Markers reward (1) the molecular formula at 60, (2) loss of 17 identified as $OH$ and the alcohol implication, (3) loss of 15 ($CH_3$) and identification of the methyl cation, (4) a self-consistent structure, (5) stability argument for the base peak.

The signature of a 3:1 ratio between M and M+2 is one chlorine atom.

Natural chlorine is 75% $^{35}Cl$ and 25% $^{37}Cl$ (a 3:1 ratio). Any compound with one chlorine therefore shows two molecular ion peaks, the lighter one with $^{35}Cl$ at M and the heavier one with $^{37}Cl$ at M+2, in a 3:1 ratio.

For $m/z = 78$: a compound containing one chlorine has a non-chlorine part of mass $78 - 35 = 43$, which is $C_3H_7$. The unknown is therefore $C_3H_7Cl$, chloropropane (either 1-chloropropane or 2-chloropropane, indistinguishable from this evidence alone).

A 1:1 M to M+2 ratio would indicate bromine ($^{79}Br$ and $^{81}Br$ are roughly equal). Two chlorines give a 9:6:1 pattern at M : M+2 : M+4.

Markers reward (1) identifying chlorine from the 3:1 ratio, (2) calculating the rest of the molecular formula, (3) noting the contrast with bromine.

(a) Mass lost

$46 - 31 = 15$.

(b) Fragment

A loss of 15 corresponds to $CH_3$ (a methyl radical).

(c) Functional group

Loss of $CH_3$ from a molecular ion of 46 is consistent with a primary alcohol undergoing alpha-cleavage next to the $OH$ group; the compound is ethanol, $CH_3CH_2OH$ ($M = 46$), giving $CH_2OH^+$ at $m/z = 31$.

Marking criteria: 1 mark for the correct mass difference, 1 mark for naming the correct fragment ($CH_3$), 1 mark for a functional group/structure consistent with both the molecular ion and the loss.

(a) Halogen

Bromine.

(b) Expected ratio

M : M+2 $\approx$ 1 : 1, because $^{79}Br$ and $^{81}Br$ occur in almost equal natural abundance.

(c) Why not a second compound

The M+2 peak arises from molecules that contain the heavier $^{81}Br$ isotope instead of $^{79}Br$; both peaks belong to the SAME compound (they are isotopologues), not an impurity or a second substance, because every other atom in the molecule is identical between the two peaks.

Marking criteria: 1 mark per part; part (c) must explicitly use the word isotope or isotopologue and reject the "second compound" interpretation to gain the mark.

(a) Moles of 1-bromopropane.

(b) Number of molecules.

Rounded to 3 significant figures:

(c) Molecular ion m/z values and ratio.

With $^{35}Cl$ replaced by $^{79}Br$/$^{81}Br$: the two molecular ions appear at $m/z = 122$ ($C_3H_7\,^{79}Br$) and $m/z = 124$ ($C_3H_7\,^{81}Br$), in an approximately 1 : 1 height ratio because $^{79}Br$ and $^{81}Br$ are almost equally abundant in nature.

Marking criteria: 1 mark for the correct moles calculation to 4 s.f., 1 mark for correct use of $N = nN_A$, 1 mark for the final molecule count to 3 s.f. with units, 1 mark for both correct m/z values (122 and 124), 1 mark for the correct approximate 1 : 1 ratio with a stated reason.

Reading the spectrum. Four peaks are labelled: a weak molecular ion at $m/z = 88$, a small peak at $m/z = 73$, the base peak at $m/z = 43$, and a small peak at $m/z = 15$.

Fragment losses.

• $88 \rightarrow 73$: loss of $88 - 73 = 15$, i.e. loss of $CH_3$ (a methyl radical cleaves from the ethoxy end).
• $88 \rightarrow 43$: loss of $88 - 43 = 45$, i.e. loss of $OC_2H_5$ (the ethoxy group leaves as a radical), giving the stable acylium cation $CH_3CO^+$ at $m/z = 43$. Because $CH_3CO^+$ is a resonance-stabilised acylium cation, it is the most abundant ion and forms the base peak.
• $m/z = 15$: the $CH_3^+$ cation, formed by further fragmentation or directly from the methyl end of the molecule.

Structure justification. A base peak from loss of 45 ($OC_2H_5$) plus a peak from loss of 15 ($CH_3$) is only consistent with an ester having a $CH_3CO-$ acyl group and an $-OC_2H_5$ alkoxy group either side of the carbonyl oxygen, i.e. $CH_3COOC_2H_5$ (ethyl ethanoate), matching the $M = 88$ molecular ion ($C_4H_8O_2$: $4(12) + 8(1) + 2(16) = 88$).

Marking criteria: 1 mark for each correctly identified loss (88 to 73, and 88 to 43), 1 mark for correctly naming both fragments ($CH_3$ and $OC_2H_5$), 1 mark for the stability argument explaining why $m/z = 43$ is the base peak, 1 mark for a structure consistent with all four labelled peaks and the molecular formula check.

This is a 6-mark PROPOSE AND EVALUATE: markers reward reasoned use of the spectrum plus a balanced judgement on the technique's reliability.

Reasoning to confirm identity.

• The molecular ion at $m/z = 149$ fixes the molecular mass of the suspected compound.
• The base peak at $m/z = 58$ (loss of 91 from M+) and the fragment at $m/z = 91$ (a benzyl-type cation, $C_6H_5CH_2^+$, commonly from alpha-cleavage next to a nitrogen atom) together form a fragmentation "fingerprint" specific to the compound's structure, not just its mass.
• A chemist compares the FULL pattern of peaks (m/z values and their relative abundances) against a certified reference library spectrum (e.g. NIST) run under identical ionising conditions (about 70 eV electron impact); a match across the molecular ion, base peak and at least one further diagnostic fragment is treated as confirmatory, since it is extremely unlikely that two different compounds fragment identically at every position.

Evaluation of strengths and limitations.

• Strength: mass spectrometry gives an exact molecular mass and a highly specific fragmentation pattern, and (coupled to GC) can identify a compound from a complex mixture at nanogram sensitivity, which is why GC-MS library matching is accepted as legal-standard confirmatory evidence.
• Limitation: mass spectrometry alone cannot distinguish stereoisomers (e.g. two enantiomers of the same drug give identical spectra), and closely related structural isomers can sometimes give near-identical fragmentation patterns, so an m/z match is necessary but not always sufficient on its own.
• Judgement: for legal-standard forensic confirmation, mass spectrometry should be combined with a second independent technique (commonly IR or NMR, or a second chromatographic retention time) so that structural features invisible to fragmentation (like stereochemistry) are also confirmed; used alone, a mass spectrum match is strong but not absolute proof of identity.

Marking criteria: 2 marks for correctly reasoning through the specific peaks given (m/z 149, 91, 58) as a fingerprint rather than just a mass, 2 marks for a correctly justified strength AND limitation, 2 marks for an explicit, reasoned judgement about combining techniques (not just asserting mass spectrometry is "good" or "bad").

This is a 7-mark ASSESS: markers reward a judgement backed by a specific worked comparison and correct use of the equilibrium/collision reasoning requested.

Band 6 PLAN.

• Thesis: mass spectrometry alone is NOT sufficient to fully determine structure in every case, because structural isomers that fragment via similar bond cleavages can give near-identical spectra; propan-1-ol and propan-2-ol are a worked counter-example to the claim.
• Worked comparison: both are $C_3H_8O$, $M = 60$. Propan-2-ol (secondary alcohol) readily loses $CH_3$ (15) to give the relatively stable secondary carbocation $CH_3CH^+OH$ at $m/z = 45$ as a strong/base peak. Propan-1-ol (primary alcohol) also loses $H$ or $CH_2OH$-type fragments and can show a peak near $m/z = 31$ ($CH_2OH^+$) and around 43/45 from different cleavages; the two spectra can look similar enough near $m/z = 43$-$45$ that the fragmentation pattern alone is not always conclusive at the level of intensities available on a standard classroom-referenced instrument.
• Ionisation energy reasoning: electron-impact ionisation is not an equilibrium process governed by Le Chatelier's principle in the usual reversible-reaction sense (there is no reverse reaction converting fragments back to M+ inside the source), so raising the ~70 eV electron energy does not "shift" anything towards revealing more structural detail; it simply supplies more excess energy to the molecular ion, increasing the amount and randomness of fragmentation. This makes the molecular ion peak weaker and can destroy fine structural distinctions rather than resolving them, so more energy makes the ambiguity worse, not better.
• Judgement: mass spectrometry gives the molecular mass and a strong first hypothesis about functional groups and connectivity, but for isomers that fragment similarly it must be combined with $^1H$/$^{13}C$ NMR (which directly reports the chemical environment and number of each type of hydrogen/carbon) or IR, so the claim that mass spectrometry ALONE is always sufficient is false in general, though true for many simpler unknowns with distinctive fragmentation.

Model paragraph (excerpt). The claim that mass spectrometry alone can always determine structure fails for isomer pairs like propan-1-ol and propan-2-ol, both of $M = 60$, because both are primary/secondary alcohols capable of losing similar small radicals to give carbocations in a similar mass range; without an independent technique such as NMR to directly distinguish the number and environment of hydrogen atoms, a chemist cannot be certain which isomer produced a given spectrum. Increasing the electron-impact energy beyond the standard 70 eV does not help, since fragmentation in the ion source is a one-way process driven by excess internal energy rather than a reversible equilibrium; more energy only produces more extensive, more random fragmentation, which can further wash out the specific structural clues the chemist is trying to read.

Marker's note: top-band answers (1) correctly identify that mass spectrometry is not always sufficient and support this with the propan-1-ol/propan-2-ol example specifically, (2) correctly explain that ionisation is not a Le Chatelier equilibrium and reason instead in terms of excess energy driving fragmentation, (3) end with an explicit, reasoned judgement rather than a neutral list of pros and cons, and (4) name at least one complementary technique (NMR or IR) that resolves the ambiguity.