Description of how mass spectra of organic compounds are used to determine the Mr and how fragment peaks enable the analytical chemist to build up a picture of the molecule
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6:22 why is ch2oh a radical????....... and would it also be possible if e.g. we fragment the CH3CH2 off, for the OH to get a positive charge instead. do they get the positive charge randomly????
Thank you very much for the video. How do you determine which fragment will be the positive ion and which will be the radical? Do you not do both as it could be either? Thanks
the m+1 peak represents the naturally occurring isotope of carbon which is c-13 but you'd normally ignore the m+1 peak and focus on the molecular ion peak for these questions. hope that helped and good luck if you have chem unit 2 today :)
Sorry to bother you, but I have another question: A student mixes 100 cm3 of 0.200 mol dm-3 NaCl(aq) with 100 cm3 of 0.200 mol dm-3 Na2CO3(aq). This is on my sample paper for ocr. I thought that there are 3 atoms of Na therefore on the product side there would be 3 Na ions. Next, I Calculated the mols of NaOH (or either since they have the same conc/volume) and got 0.02 mols. Then I did 0.02 x 3 (because there is 3 atoms of Na+ ions on the product side) Lastly, I did 0.06=c x 200/1000 (200 as the overall solution has 200 cm3) and rearranged to get c= 0.3 mol dm-3. The thing that's bothering me is whether I was correct in thinking there is 3 atoms of Na+ ions. what if I balanced the equation and it gave me a number in front of the Na product? Thanks, and again sorry to bother you
It depends on what type of positive fragment you see. When you split the molecular ion into its fragments, the one with the m/z value relative to the peak, is the one which takes the positive ion and the rest of the molecular then becomes a radical. So, it doesn't really matter as long as the spectrometer picks up the positive peak.
Remember that the molecule loses one electron in the ionisation chamber of the mass spectrometer giving it an initial 1+ charge. If a further bond is broken (homolytically) to produce fragments, one part will carry the 1+ charge and be a radical cation (this is detected by the spectrometer) the other will not carry the charge and be a radical (not detected). There are various ways that fragmentation can occur, involving different types of bond fission. I teach A level Chemistry which does not require such an in depth look at bond breaking so I keep it as simple as possible.
like wouldn't it be random fragmentation. And also why couldn't the CH3 become the radical instead of the CH2OH couldnt it be the other way round (5:33
BAMitsFRESHPINK I think it could be either, he might've just given the CH3 the charge because its a really common fragment. Also I think that the exam question will have a mass spectra which only detects positive charges, so you don't have to worry about which radicals are being formed
