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Topic: 1,4 -Diphenyl - 1,3- butadiene Mass Spectrum  (Read 749 times)

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Offline xshadow

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1,4 -Diphenyl - 1,3- butadiene Mass Spectrum
« on: October 08, 2020, 04:16:09 AM »
 Hi

I have to discuss the mass spectrum of  diphenyl butadiene  that I got from a wittig  synthesis


I find very hard to discuss it because I've studied only the fragmentation of easy molecule (with 1 functional group)

Here I have 2 double bond and two phenyl all conjugated and I don't  understand the possible fragmentation


 I link the MS:



And here my doubts...

I can't do the classical fragmentation for aromatic ring, for example (because in benzylic position I have a double bond)

Also the molecular ion ...where is the positive charge? In the aromatic ring or in the C=C?? Could be located in both of them? (so different type of fragmentation)

-The 91 peak should be tropyliuml cation...bit I don't know the correct fragmentation in order to get it..
It need a  benzylic -CH2 for the rearrangment (and in order to have 91 as m/z)but here I have only 1 H benzylic due to the C=C  near the ring.  Is a Hydrogen trasposition   needed?


-I also see the loss of 1H  atom from the molecular ion. Which Hydrgon is it? A H bonded to C=C or one bonded to the ring?

Thanks
« Last Edit: October 08, 2020, 05:59:11 AM by xshadow »

Offline GabrielTojo

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Re: 1,4 -Diphenyl - 1,3- butadiene Mass Spectrum
« Reply #1 on: October 23, 2020, 08:00:04 PM »
To form the molecular ion, an electron has been removed from a molecular orbital, giving a radical cation delocalized throughout the system. Numerous resonant forms can be drawn showing the cation and the radical on any carbon.

From the molecular ion, any hydrogen can be lost giving the M-1 peak.

Regarding the peak of 91, obviously a transfer of hydrogen to a benzylic carbon is necessary, but it is difficult to imagine. I only could suggest that such migration should be easy as electron impact produces radical cations with an energy exceeding quite a lot the energy of a C-H bond.

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