Isotope Patterns: M+1 and M+2 Peaks

Introduction

Look closely at almost any molecular ion peak and you will see a small partner one unit to its right. In compounds of chlorine or bromine there is a much larger partner two units to the right. These are not impurities: they are the same molecules containing heavier isotopes. Because each element has a fixed natural isotopic mixture, the heights of these peaks form a signature that reveals how many carbon atoms are present and whether chlorine or bromine is in the molecule.

Core explanation

Carbon-13 and the M+1 peak. Natural carbon is about 98.9% ¹²C and 1.1% ¹³C . In a molecule with one carbon atom, 1.1% of molecules contain ¹³C and appear one unit higher. With n carbon atoms, the chance that any one of them is ¹³C is roughly n × 1.1%, so the M+1 peak is approximately n × 1.1% of the height of the M peak. A long-chain molecule therefore has a noticeably taller M+1 peak than a small one. (Hydrogen-2 and nitrogen-15 contribute a little, but carbon dominates for organic compounds.)

Chlorine and the M+2 peak. Chlorine has two isotopes, ³⁵Cl (about 75%) and ³⁷Cl (about 25%), a ratio of roughly 3:1 . A molecule with one chlorine atom therefore produces two molecular ion peaks two units apart, with heights in the ratio 3:1. Chloroethane, C₂H₅Cl, shows M at m/z 64 and M+2 at m/z 66.

Bromine. Bromine is about 50.7% ⁷⁹Br and 49.3% ⁸¹Br, close to 1:1 . A compound with one bromine atom shows two peaks of almost equal height separated by two units. Bromoethane gives peaks at m/z 108 and 110.

Two halogen atoms. Probabilities multiply. For two chlorine atoms, with fractions 3/4 and 1/4:

- both ³⁵Cl: 3/4 × 3/4 = 9/16 (M) - one of each: 2 × 3/4 × 1/4 = 6/16 (M+2) - both ³⁷Cl: 1/4 × 1/4 = 1/16 (M+4)

giving 9:6:1 . For two bromine atoms, the same logic with ½ and ½ gives 1:2:1 for M, M+2 and M+4.

Fragments carry the pattern too. Any fragment that retains the halogen shows the same doublet. If a fragment loses the halogen, its doublet disappears, which helps locate the halogen in the structure.

Sulfur has a smaller M+2 contribution (³⁴S, about 4.2%), giving an M+2 peak of roughly 4% for each sulfur atom — a useful but subtler clue.

Formulae

Number of carbon atoms: n ≈ (height of M+1 ÷ height of M) × 100 ÷ 1.1

Step-by-step reasoning

To read an isotope pattern:

1. Find the molecular ion M. 2. Measure the M+1 height as a percentage of M; divide by 1.1 to estimate carbon atoms. 3. Check the M+2 peak. About one-third of M suggests one Cl; about equal to M suggests one Br. 4. If there is an M+4 peak, consider two halogens and compare with 9:6:1 or 1:2:1. 5. Check fragment ions for the same doublets to see which pieces contain the halogen.

Visual explanation

Picture three molecular-ion clusters side by side. A hydrocarbon: one tall line with a tiny neighbour. A chloroalkane: two lines two units apart, the right one a third the height of the left. A bromoalkane: two lines two units apart of equal height, like a pair of twin towers.

Real-world analogy

Imagine a bag of marbles in which one in every four is slightly heavier. If you pick one marble repeatedly and weigh it, three times out of four you get the lighter mass and once the heavier. Pick two marbles each time and the combinations come out 9:6:1 — exactly the dichloro pattern.

Real-world example

Environmental chemists screening water for persistent pollutants such as polychlorinated biphenyls look for clusters of M, M+2, M+4 and M+6 peaks. The distinctive heights show how many chlorine atoms each molecule carries, identifying contaminants among thousands of other compounds.

Why?

Why do chlorine and bromine give large M+2 peaks while oxygen and nitrogen do not? Chlorine and bromine have heavier isotopes that are very common (25% and 49%), whereas ¹⁸O is only about 0.2% of oxygen and ¹⁵N about 0.4% of nitrogen. Only abundant heavy isotopes produce prominent peaks.

Common misconception

"Two peaks of equal height at m/z 122 and 124 mean two different compounds." For a single pure substance, equal peaks two units apart point to one bromine atom; both are the same molecule with different bromine isotopes.

Worked example

Question: A compound with no significant M+2 peak shows M at m/z 58 (relative height 50.0) and M+1 at m/z 59 (height 1.65). Is it butane, C₄H₁₀, or propanone, C₃H₆O?

Reasoning: n ≈ (1.65 ÷ 50.0) × 100 ÷ 1.1 = 3.3 ÷ 1.1 = 3.0. Butane has four carbons and would give M+1 of about 4.4% of M; propanone has three and gives about 3.3%.

Answer: Three carbon atoms, so the compound is propanone, C₃H₆O.

Quick check

1. A molecular ion cluster shows peaks at m/z 112 and 114 in a 3:1 height ratio. Which halogen is present, and how many atoms? Answer: One chlorine atom, because ³⁵Cl and ³⁷Cl occur in roughly a 3:1 ratio.

Exam focus

Learn the three key signatures: M+1 ≈ 1.1% per carbon; M:M+2 = 3:1 for one Cl; M:M+2 = 1:1 for one Br. Show the probability working for two halogens. When asked to identify M in a halogen compound, state that the M+2 peak is an isotope peak, not a heavier molecule.

Advanced insight

The exact isotope distribution is binomial. The ratio of M+1 to M stays close to n × 1.1%, but for large molecules the M+2, M+3 and higher peaks also grow, because molecules with two or more ¹³C atoms become common. For a molecule with 100 carbon atoms, the M+1 peak is actually taller than M, and the monoisotopic peak is no longer the most abundant. Protein chemists therefore quote average and monoisotopic masses separately.

Summary

Heavier isotopes give peaks beyond the molecular ion. The M+1 peak comes mainly from carbon-13 (1.1% per carbon) and gives the number of carbon atoms. One chlorine atom gives M:M+2 of 3:1; one bromine atom gives about 1:1. Two chlorines give 9:6:1 and two bromines 1:2:1. Fragments keep the pattern if they keep the halogen.

Practice questions

1. A compound has M at height 60.0 and M+1 at height 4.0. Estimate the number of carbon atoms. Answer: (4.0 ÷ 60.0) × 100 ÷ 1.1 = 6.1, so six carbon atoms. 2. Predict the m/z values and approximate height ratio of the molecular ion peaks of bromomethane, CH₃Br. Answer: m/z 94 (with ⁷⁹Br) and m/z 96 (with ⁸¹Br), in a ratio of about 1:1. 3. Show that dichloromethane, CH₂Cl₂, gives molecular ion peaks at m/z 84, 86 and 88 in the ratio 9:6:1. Answer: 12 + 2 + 70 = 84 with two ³⁵Cl; one ³⁷Cl gives 86 and two give 88. Probabilities are 9/16, 6/16 and 1/16, a ratio of 9:6:1. 4. In the spectrum of 1-chloropropane, a fragment appears at m/z 43 with no partner at m/z 45. What does this show? Answer: The fragment, C₃H₇⁺, has lost the chlorine atom, so it has no chlorine isotope pattern.