Analytical Techniques: Question 1

Worked solution

Reading the isotope pattern

The molecular ion peak, $\text{M}^+$, appears at $m/z = 108$. A second peak at $m/z = 110$ (that is, $[\text{M}+2]^+$) is almost as tall as $\text{M}^+$. A peak exactly two mass units heavier than $\text{M}^+$, with no fragmentation needed to explain it, is the signature of an element with two naturally occurring isotopes two mass units apart, this points straight at chlorine or bromine.

Distinguishing chlorine from bromine by relative peak height

• Chlorine occurs as $^{35}\text{Cl}$ (about 76%) and $^{37}\text{Cl}$ (about 24%), an abundance ratio of roughly 3:1. A single Cl atom therefore gives an $[\text{M}+2]^+$ peak only about a third the height of $\text{M}^+$.
• Bromine occurs as $^{79}\text{Br}$ (about 51%) and $^{81}\text{Br}$ (about 49%), an abundance ratio of roughly 1:1. A single Br atom therefore gives an $[\text{M}+2]^+$ peak almost the same height as $\text{M}^+$.

Since the data describes the $m/z = 110$ peak as “almost identical height” to the $m/z = 108$ peak, this matches the 1:1 pattern, one bromine atom, not chlorine.

Why the other options are wrong

• B describes the correct 1:1 reasoning but attaches it to the wrong element: chlorine’s isotopes occur in a 3:1 ratio, not 1:1.
• C is wrong on two counts: a single chlorine atom does still give an $[\text{M}+2]^+$ peak (just a smaller one, about a third the height of $\text{M}^+$); and two chlorine atoms would in any case give three peaks ($\text{M}^+$, $[\text{M}+2]^+$ and $[\text{M}+4]^+$, roughly in the ratio 9:6:1), not just two peaks of near-equal height.
• D is wrong because fragment ions always have a lower $m/z$ than the molecular ion (a piece of the molecule has broken off, so mass is lost), never a higher one. A peak above $\text{M}^+$ can only arise from a naturally occurring heavier isotope of an atom already present in the molecule, not from fragmentation.

Final answer

A. Compound J contains one bromine atom.