Generally, a molecule containing one Chlorine atom would give molecular ions M+ and M+2 in a 3:1 ratio, because Chlorine contains 3 times as much of the 35Cl isotope as of the 37Cl isotope.

In your case (2 Chlorine atoms), one would expect to get M+, M+2 and also M+4, because now there's more possible combinations of the aforementioned isotopes. I think your M-1 comes from something else. Do you have any aromatic structures with for example ether-, aldehyde- or amino functions or perhaps a benzylic hydroxyl?

Regards,

JR

Chlorine has two isotopes 35Cl and 37Cl. The intensity of M+2 intensity would be 2/3 height of the M. If M is 100% then M+2 is 66.67%.

If you have two atoms you have to choose the first one. There is a 3/4 possibility of choosing a 35Cl

Then you have to choose the second atom. Once again there is 3/4 chance of choosing a 35Cl

So the overall probability of choosing a 35Cl35Cl molecule is 3/4 x 3/4 = 9/16 (M)

Likewise the probability of getting a 37Cl37Cl is 1/4 x 1/4 = 1/16 (M+4)

Hence the probability of 37Cl35Cl = 6/16 (M+2)

So the ratio is 9 : 6 : 1

In mass spectrum, M-1 at atomic mass 34 is due to either HCl, H2S, etc......

Perhaps is too simple but otherwise don't mind.... did you consider the average atomic weight?

If you consider in the calculation of the molecular weight the "average weight" of a chlorine atom i.e 35.45 then is simply explained why you get get a molecular weight effectively 0.9 units less.

Mass spectrometry is indeed capable to isolate single isotopes, this of course means for exapmel forCH2Cl2 that the calculate average weight is 84.9, while you get on the mass spectrometer a spectrum displaying the ion at m/z 84 (all 35 Cl), plus a very small ion at m/z 85 (all 35 Cl + 13C), an ion at m/z 86 (about 66% of the m/z 84 ion abundance due to 35Cl+ 37Cl) and finally an ion at m/z 88 (about 10-15% of the m/z 84 ion abundance ) .

if this is your case remember just that in a mass spectrum you can practically Always are capable to separe individual isotopes, and "averaged weights" in the Periodic table are practically meaningless. otherwyse you should take into consideration Hydorogen losses due to peculiar substituents as suggested by our collegues.