Mass spectrum of the unknown compound "MS88"

I'm given the above spectrum. $M^+ = 102$, probably $\ce{C_6H_14O}$ or $\ce{C_5H_10O_2}$.

I'm thinking of ethyl propionate. (that picture also shows the fragments I expect). I can account for the following peaks: m/z = 73: Loss of ethyl from acyl side (expected lowish intensity) m/z = 57: Loss of alkoxy to form $\ce{CH3CH2CO+}$ (moderate intensity) m/z = 45: Alkoxy ion (weak) m/z = 29: ethyl carbocation, moderate intensity.

I can't account for 74, since according to my book the alkoxy side would not proceed through normal Mclafferty, but the Mclafferty + 1 rearrangement... That should yield m/z=75.

I proposed methyl butyrate, and of course that's not right. My instructor wrote to pay attention to the m/z = 57 (M-45) and m/z = 85 (M-17). This fits well with carboxylic acids, but again cannot account for the m/z = 74. McLafferty of carboxylic acids would result in a fragment of m/z=60...

Can someone help me out, I've come to a dead end for hours now.


2 Answers 2


I was on to the wrong structure. My main issue with the carboxylic acid was the fragment m/z=60 that it would produce by McLafferty. I didn't have that, so I thought that left me with t-Bu-COOH, which was the only structure with no $\gamma$-hydrogens. Besides I couldn't explain the m/z=74 peak with an acid. On the other hand I couldn't explain the M-17 peak by ethylpropanoate.

But of course, if the carboxylic acid has a methyl substituent at the $\alpha$-carbon, then that methyl group would also be part of the McLafferty fragment, therefore resulting in m/z=74!

The spectre is of: 2-methyl-butanoic acid

M-17 and M-45:

Alpha cleavage of carbonyl => loss of hydroxyl radical or carboxyl radical

m/z = 74:

Loss of ethene by McLafferty rearrangement

m/z = 45:

$\ce{COOH^+}$ fragment

M-15 and M-29:

Loss of methyl radical or ethyl radical

m/z = 29:

Ethyl carbocation


For m/z 74 it could be methyl ester: in the loss of a $\ce{CH2=CH2}$ by McLafferty rearrangement.


  • $\begingroup$ This doesn't seem likely to me, especially considering the intensity. Do you suggest a mechanism :-)? $\endgroup$
    – user1160
    Commented Nov 9, 2013 at 20:21
  • $\begingroup$ May be I'm wrong, but I've thought in the lost of a CH2=CH2 by a McLafferty rearrangement. $\endgroup$
    – RFG
    Commented Nov 9, 2013 at 20:57
  • $\begingroup$ Okay I just read it again. It is the ethyl ester that fits with many of the peaks, but I was troubled because I thought the book said it would only undergo McLafferty+1 rearrangement (with double proton transfer, M/Z=75), but apparently that's just an additional mode of fragmentation to the normal McLafferty arrangement. This gives [CH3CH2C(OH)O]+ m/z=74. I would like to explain the m/z=85 though, now that my instructor pointed my attention to it. hmm $\endgroup$
    – user1160
    Commented Nov 10, 2013 at 7:29
  • 1
    $\begingroup$ Possibilty: Loss of a hydroxyl group (m = 17). $\endgroup$
    – RFG
    Commented Nov 10, 2013 at 10:01

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