i have the mass spectrum for the molecule but-3-en-1-ol. The Mr of this is 72

There are four peaks that are labelled which I have to name the species responsible for forming that peak.

The peaks are:
m/z = 66
m/z = 67
m/z = 71
m/z = 73 - this peak is labelled as (M + H)^+

My answer so far...
The peak at m/z=73 is the but-3-en-1-ol with one added hydrogen, somewhere to the double bond?
For the other peaks am I just removing hydrogens as the mass only decreases by 8 in total so i couldn't remove a carbon or the oxygen.

Does this mean that the peaks are:
m/z=71 is the removal of 1H
m/z=67 is the removal of another 4H (5H in total)
m/z=66 is the removal of another 1H (6H in total)

is this is correct? Which hydrogens are removed first and what does each species look like for each peak, and are there any charges on the species produced. Also is there a special name for this sort of fragmentation, e.g alpha cleavage?

  • $\begingroup$ 73 peak, if it has low intensity (~4-5% of molecular peak) may be a result of presence of C13. Same for other peaks: most of them should have low-intensity pairs one atomic mass unit heavier, coming from replacement of one atom with C13. For some heavier molecules those peaks may dominate. $\endgroup$
    – permeakra
    Commented Oct 19, 2014 at 6:06
  • 2
    $\begingroup$ Here is the mass-spectrum webbook.nist.gov/cgi/cbook.cgi?ID=C627270&Mask=200 and I don't see 66/67 m/z ions. Could you recheck and clarify? $\endgroup$
    – permeakra
    Commented Oct 19, 2014 at 7:36

1 Answer 1


As you have pointed out, the peak at M = 73 is M+H, protonation is however more likely to occur at the oxygen atom.

MS fragmentation of butenol

In my opinion, peaks at M = 66 and 67, i.e. with a rather small mass difference to $\mathrm{M^+}$, are negligible.

Common sense and the link in permeakra's comment suggest that the major fragmentation of but-3-en-1-ol occurs along the red dotted line, yielding an allylic and a hydroxymethyl fragment.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.