Fragmentation of Alcohols, Amines and Halogenoalkanes

Introduction

Alcohols, amines and halogenoalkanes all contain a heteroatom — oxygen, nitrogen or a halogen — bonded to carbon. When their molecular ions break apart in a mass spectrometer, the heteroatom decides where the break happens. Oxygen and nitrogen use their lone pairs to stabilise a positive charge on the neighbouring carbon, while halogens tend to leave as radicals. Learning these few patterns lets you recognise each family from its mass spectrum almost at a glance.

Core explanation

Alcohols: a weak molecular ion. The molecular ion of an alcohol is often small, and for tertiary alcohols it may be almost invisible. The ion fragments easily because two low-energy routes are available: losing water and α-cleavage.

Loss of water (M − 18). Many alcohols show a peak 18 units below the molecular ion, from loss of a neutral H₂O molecule. Butan-1-ol (M = 74) gives a clear peak at m/z 56. A peak at M − 18 is a useful hint of an alcohol, but it is not proof, because other oxygen compounds can also lose water.

α-Cleavage next to oxygen. The C–C bond next to the C–O carbon breaks. The oxygen-containing piece keeps the charge because the oxygen lone pair forms a C=O⁺ double bond, giving a resonance-stabilised oxonium ion . The alkyl group lost as a radical is not seen. The mass of the oxonium ion reveals how the carbinol carbon is substituted:

Alcohol type Typical oxonium ion m/z --- --- --- Primary, RCH₂OH CH₂=OH⁺ 31 Secondary, CH₃CH(OH)R CH₃CH=OH⁺ 45 Tertiary, (CH₃)₂C(OH)R (CH₃)₂C=OH⁺ 59

Ethanol (M = 46) has its base peak at m/z 31 after losing CH₃·; propan-2-ol (M = 60) has its base peak at m/z 45 after losing CH₃·; 2-methylpropan-2-ol (M = 74) has its base peak at m/z 59. When there is a choice, the larger alkyl radical is usually lost preferentially, because a larger radical is more stable.

Amines: the nitrogen rule and iminium ions. A molecule with one nitrogen atom has an odd nominal molecular mass, so an odd M⁺ peak immediately suggests an amine (or other nitrogen compound). Amines fragment by α-cleavage even more readily than alcohols, because nitrogen donates its lone pair more easily than oxygen. The product is an iminium ion . Primary amines with an unbranched α-carbon give CH₂=NH₂⁺ at m/z 30 , often the base peak; ethylamine (M = 45) is a good example. Note that iminium fragments from simple amines have even m/z values, the opposite of the molecular ion.

Halogenoalkanes: losing the halogen. The C–X bond is relatively weak, so the molecular ion often loses a halogen radical, leaving a carbocation. 1-Bromopropane (M = 122 and 124) gives a large peak at m/z 43 (C₃H₇⁺). Chloroalkanes can also lose HCl (M − 36). The isotope patterns are decisive: chlorine gives M and M+2 peaks in a ratio of about 3:1 , and bromine gives M and M+2 of about 1:1 . Fragments that still contain the halogen show the same doublet, while fragments that have lost it do not.

Step-by-step reasoning

To classify a spectrum of a compound that may contain a heteroatom:

1. Check whether M is odd; if so, suspect one nitrogen atom. 2. Look for M+2 doublets of 3:1 (Cl) or 1:1 (Br). 3. Look for M − 18, which suggests an alcohol. 4. Find low-mass peaks at 31, 45 or 59 (oxonium ions) or 30, 44, 58 (iminium ions). 5. Work out which alkyl radical was lost to form the main fragment.

Visual explanation

Draw the molecular ion of propan-2-ol with a dashed line through one C–C bond beside the C–OH carbon. On one side, a CH₃· radical drifts away unseen; on the other, the ion CH₃CH=OH⁺ carries the charge to the detector at m/z 45. A curly arrow from the oxygen lone pair into the C–O bond shows why this piece is stabilised.

Real-world analogy

A heteroatom acts like a hinge on a folding table. When the table is stressed, it folds at the hinge rather than breaking at a random point. Because the fold always happens beside the hinge, the size of the piece that remains tells you where the hinge was.

Real-world example

Forensic laboratories identify amphetamine-type stimulants partly from their mass spectra. These amines undergo α-cleavage beside nitrogen, and amphetamine itself gives a dominant iminium ion at m/z 44. Recognising this fragment helps analysts classify a sample quickly before confirming it against reference spectra.

Why?

Why does the charge stay on the heteroatom fragment in α-cleavage? In the oxonium or iminium ion, every atom except hydrogen has a full octet, because the lone pair forms a π bond to the positive carbon. This makes the cation much more stable than a simple carbocation, so the fragmentation leading to it is strongly favoured.

Common misconception

"The tallest peak in an alcohol spectrum is the molecular ion." For most alcohols the molecular ion is weak, and the base peak is an oxonium fragment. Always look for the highest m/z peak (ignoring small isotope peaks), not the tallest one.

Worked example

Question: A compound has a molecular ion at m/z 74, a small peak at 56 and a base peak at 59. It contains C, H and O only. Suggest its structure.

Reasoning: M − 18 = 56 suggests an alcohol losing water. m/z 59 corresponds to (CH₃)₂C=OH⁺, formed when a tertiary alcohol with two methyl groups on the carbinol carbon loses a radical. 74 − 59 = 15, so the lost radical is CH₃·. The alcohol is therefore (CH₃)₃COH.

Answer: 2-Methylpropan-2-ol, (CH₃)₃COH.

Quick check

1. Why does a primary alcohol such as ethanol often show a large peak at m/z 31? Answer: α-Cleavage removes the alkyl radical and leaves the oxonium ion CH₂=OH⁺, which is stabilised by the oxygen lone pair.

Exam focus

Learn the key numbers: M − 18 for water loss, 31/45/59 for oxonium ions, 30 for CH₂=NH₂⁺, and 3:1 or 1:1 doublets for Cl and Br. Always write fragments as positive ions with a charge, and state what neutral species was lost.

Advanced insight

α-Cleavage is a radical-driven process: the molecular ion is a radical cation, and in the favoured mechanism the unpaired electron on the heteroatom pairs with one electron from the adjacent C–C bond. The competition between α-cleavage and dehydration depends on the internal energy of the ion, which is why spectra recorded with softer ionisation methods often show a much stronger molecular ion for the same alcohol.

Summary

Heteroatoms direct fragmentation. Alcohols give weak molecular ions, M − 18 peaks from water loss and oxonium ions at m/z 31, 45 or 59 by α-cleavage. Amines have odd molecular masses and give iminium ions, typically m/z 30 for primary amines. Halogenoalkanes lose the halogen to form carbocations and show 3:1 (Cl) or 1:1 (Br) M and M+2 patterns.

Practice questions

1. Propan-2-ol has M = 60. Explain the base peak at m/z 45. Answer: α-Cleavage loses a CH₃· radical (15), leaving the oxonium ion CH₃CH=OH⁺ at m/z 45. 2. A spectrum shows its molecular ion at m/z 59. What does the odd value suggest? Answer: By the nitrogen rule, the compound probably contains one nitrogen atom, for example an amine. 3. 1-Bromopropane shows peaks of equal height at m/z 122 and 124 and a large peak at 43. Identify the species at 43. Answer: The propyl cation, C₃H₇⁺, formed by loss of a bromine radical from the molecular ion. 4. Why does the peak at m/z 43 in the spectrum of 1-bromopropane have no partner at 45 of similar height? Answer: The fragment no longer contains bromine, so it does not show the 1:1 isotope doublet.