Analytical Techniques: Question 9

Worked solution

Calculating the mass of the fragment lost

The molecular ion peak is at $m/z = 86$ and the fragment ion peak is at $m/z = 71$, so the neutral fragment lost from the molecular ion has mass:

$86 - 71 = 15$

A mass of 15 corresponds to a methyl radical, $\text{CH}_3^\bullet$ ($12 + 3(1) = 15$).

Why this fits a ketone

Ketones characteristically fragment by alpha-cleavage: the carbon–carbon bond immediately next to the carbonyl carbon breaks, losing an alkyl radical from one side and leaving a positively charged acylium ion, $\text{R-CO}^+$, on the other. Losing a methyl radical (mass 15) by this route from compound A ($M_r = 86$) gives an acylium fragment at:

$m/z = 86 - 15 = 71$

which matches the observed peak exactly, and is a mechanism specific to a carbonyl-containing compound such as a ketone.

Why the other options are wrong

• B proposes a mass loss of 18 (water), but $86 - 71 = 15 \neq 18$, so the arithmetic does not fit; a ketone also has no O–H group available to lose as water.
• C proposes a mass loss of 29 ($\bullet\text{CHO}$), but $29 \neq 15$; this fragment is typical of an aldehyde losing its terminal $\text{CHO}$ group, not a ketone.
• D proposes a mass loss of 17 ($\bullet\text{OH}$), but $17 \neq 15$; this fragment is typical of a carboxylic acid, which has an O–H bond to lose, unlike a ketone.

Final answer

A, loss of $\text{CH}_3^\bullet$ (mass 15) by alpha-cleavage, forming an acylium ion at $m/z = 71$.