Suggest major fragment ions without rearrangement

A fragment peak requires a charged formula and nominal mass, not only a named neutral loss.

GCE A-Level H3 Chemistry 9813-2027

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Learning objectives

Suggest major fragment ions without rearrangement

Assign formula before mass

A fragment peak requires a charged formula and nominal mass, not only a named neutral loss.

CH₃CO⁺ has m/z 43; C₂H₅CO⁺ has m/z 57.

Work from a displayed structure, not from the peak number. Mark a possible bond cleavage, decide which product retains positive charge, and calculate its m/z only after writing its formula.

Preserve connectivity

Fragments must be obtainable by cleaving bonds in the supplied structure while retaining charge on one product.

Do not assemble atoms from disconnected ends without an allowed rearrangement.

Simple α-cleavage breaks a bond next to a functional group or charged centre. The resulting ion can be relatively abundant when its charge is stabilised, but the neutral partner must still be included in the atom balance.

Use the no-rearrangement boundary

For this outcome, test simple bond cleavage unless rearrangement is explicitly supplied.

Balance the complementary neutral radical for an odd-electron parent.

For an odd-electron molecular ion, homolytic cleavage can send one electron from the broken bond with each product, giving an even-electron cation plus a neutral radical. Fishhook arrows show these single-electron movements when the mechanism is requested.

A mass difference is a useful check, not a mechanism by itself. Several neutral species can share nominal mass, so confirm that the proposed loss exists as a connected part of the parent structure.

Fragment butan-2-one

Cleavage beside C=O gives CH₃CO⁺ (m/z 43) + C₂H₅• or C₂H₅CO⁺ (m/z 57) + CH₃•.

Both channels preserve connectivity and atom balance.

Butan-2-one has two different C–C bonds beside C=O. Cleaving one side gives CH₃CO⁺ at m/z 43 plus C₂H₅•; cleaving the other gives C₂H₅CO⁺ at m/z 57 plus CH₃•.

Try this

Starting from the butan-2-one molecular ion, write both α-cleavage equations, identify the charged fragment in each and calculate both m/z values.

Check your answer

The two balanced channels are C₄H₈O⁺• → CH₃CO⁺ (m/z 43) + C₂H₅• and C₄H₈O⁺• → C₂H₅CO⁺ (m/z 57) + CH₃•. Each breaks a bond adjacent to C=O and conserves atoms and charge.

Fragment propan-1-ol

α-cleavage can give CH₂OH⁺ at m/z 31 plus C₂H₅•.

A proposed C₂H₅O⁺ at m/z 45 must be checked against the actual broken bond and parent connectivity.

For propan-1-ol, mark the C–C bond next to the carbon bearing oxygen. The oxygen-containing fragment CH₂OH⁺ has nominal mass 12 + 3 + 16 = 31, leaving C₂H₅• neutral.

Try this

Show the α-cleavage of the propan-1-ol molecular ion that produces the m/z 31 peak, including the neutral radical and nominal-mass calculation.

Check your answer

Cleavage beside the oxygen-bearing carbon gives CH₂OH⁺ + C₂H₅•. The charged fragment has m/z 12 + 3 + 16 = 31; only that charged product is detected in this channel.

Cross-check neutral loss

For diethyl ether, CH₃CH₂OCH₂CH₃, M⁺• = 74 and a fragment at 59 give a neutral loss of 15, consistent with CH₃•.

Mass difference supports but does not alone prove a formula; connectivity must agree.

For M⁺• = 74 and a fragment at 59, a loss of 15 is consistent with CH₃• only if a methyl group can be cleaved from the supplied parent. Write both products to demonstrate that check.

Try this

Diethyl ether, CH₃CH₂OCH₂CH₃, gives M⁺• at m/z 74 and a fragment at m/z 59. Propose the bond that breaks, write the fragmentation equation and verify the loss of 15 using both the mass difference and the parent's connectivity.

Check your answer

74 − 59 = 15, consistent with loss of CH₃•. Breaking the C–C bond of one ethyl group, next to the carbon bonded to oxygen, gives CH₃CH₂OCH₂CH₃⁺• → CH₃CH₂OCH₂⁺ + CH₃•. The charged fragment C₃H₇O⁺ has m/z 36 + 7 + 16 = 59 and is still one connected piece of the parent. C₂H₃O₂⁺ also has nominal mass 59, but it needs two oxygen atoms and diethyl ether has only one.

Common mistake: mass-only guessing

Different ions can share nominal m/z, so formula and precursor structure matter.

A peak 43 is not automatically CH₃CO⁺ in every molecule.

m/z 43 is a common acylium-ion value, not a universal identity. Hydrocarbon ions and other formulas can also have nominal mass 43; the parent structure decides which assignment is defensible.

Try this

Correct this statement: ‘A peak at m/z 43 always proves that CH₃CO⁺ is present.’

Check your answer

m/z 43 fixes only the mass-to-charge value, not a unique formula. CH₃CO⁺ is a sensible assignment when the parent contains a carbonyl group and a connected α-cleavage can produce it; otherwise consider other connected ions with the same nominal mass.

Combine the spectrum

For butan-2-one, M⁺• 72 with 43 and 57 supports two α-cleavages.

Next: stereochemical projections.

For each important peak, show parent → charged fragment + neutral partner, label the broken bond, calculate m/z and check total atoms and charge. This is the standard of evidence expected for a structural assignment.

Try this

For butan-2-one, connect the molecular ion at m/z 72 to fragments at m/z 57 and 43. Write both balanced cleavage equations.

Check your answer

C₄H₈O⁺• → C₂H₅CO⁺ (m/z 57) + CH₃• and C₄H₈O⁺• → CH₃CO⁺ (m/z 43) + C₂H₅•. Both are α-cleavages next to C=O, and each equation conserves atoms and overall charge.