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(reference, second source here)

I've found the bigger peaks like m/z = 57, and 70 but how do I get 101? I've tried everything I could think of, like inductive cleavage or McLafferty rearrangement. Since it's 101, using rule of 13 I get the most likely formula to be C5H9O2. But the alcohol-chain has 11 hydrogens, how do I take 2 hydrogens away and end up with 5 carbons?

  • 1
    $\begingroup$ M/z=101 is ~ 4 times less intense than m/z=43. Minor fragmentation between C2-C3 of the alcohol moiety gives these two cations CH3CH2CO2CH2CH2+ (101) and CH3CHCH3+ (43). $\endgroup$
    – user55119
    Oct 9 '20 at 20:28
  • 1
    $\begingroup$ I should add that m/z=43 is more intense because it is a secondary cation rather than primary (m/z=101). $\endgroup$
    – user55119
    Oct 10 '20 at 2:12

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