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Topic: N-arylation of a sulfoximine with a substituted benzaldehyde  (Read 1834 times)

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Offline GregRC

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N-arylation of a sulfoximine with a substituted benzaldehyde
« on: August 28, 2013, 03:23:00 PM »


I'm going to be doing something like this soon. This is a pretty noob question, but how exactly are percents and equivalents worked out? If I were to start with 100mg of the benzaldehyde, where would I go from there? Do I use the MW somehow? If someone could explain this to me it would be very helpful!

Offline kriggy

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Re: N-arylation of a sulfoximine with a substituted benzaldehyde
« Reply #1 on: August 28, 2013, 03:52:20 PM »
Percents you mean yield?
Ok, so to get 1 mole of product 3 you need 1 mole of reagent 1 and 2 moles of reagent 2. So from 370grams of 2 (2 moles) of you get, theoreticaly, 255 grams of 3 (1mole). If you use lesser amount of reagents you get lesser amount of product.
When you do your reaction you only get lets say 150 grams of 3 that is 150/255=0,588 -> 58% yield.

Equivalent means same molar amount of reagent. So if its 2 eq its twice that much.

I hope I answered your question and calculated molecular weights correctly

Offline GregRC

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Re: N-arylation of a sulfoximine with a substituted benzaldehyde
« Reply #2 on: August 29, 2013, 02:36:00 PM »
Alright, but what about 5% catalyst and 1.8 equivalent Cs2CO3? Do the percentages depend on how much toluene is used or? I figure the equivilent is figured out by how much benzaldehyde is used?

Offline kriggy

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Re: N-arylation of a sulfoximine with a substituted benzaldehyde
« Reply #3 on: August 29, 2013, 04:05:33 PM »
I dont know. Its probably 5% solution It should be noted in experimental section of that paper

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