[diablo]

Hi folks,

I've got following problem: I cant clearly describe how the ß-elimination of FMOC works ^^ --> i know that the base abstracts the Proton (H),

then somehow a double bound is formed and after restructure the result is what is shown after the second arrow.. can anyone please help me to formulate the reaction between the first and second arrow ?^^



thx
Diablooo

[macman104]

I guess I'm not sure where the confusion is.  The arrow pushing shows exactly what is happening, I can't see any steps that aren't show...

[diablo]

I guess I'm not sure where the confusion is.  The arrow pushing shows exactly what is happening, I can't see any steps that aren't show...

Then how would u describe (in your own words) the reaction between first and second arrow ? --> starting with the negatively charge . why a double bound is formed etc...

greets diabloo

[macman104]

The negative charge pushes into the C-C bond to form a double bond.  This carbon now would have 5 bonds, so it breaks it's bond with oxygen.  And those electrons form a double bond with the C-O bond.  You are also relocating a negative "C" charge to an oxygen, which is more electrophilic, so that is another favorable aspect of the elimination as well.

I guess my initial question was:  "Do you not understand what is happening" Or "Do you not understand why it's happening".  The picture very clearly shows what bonds form where, which is why I was confused by your initial question.

If that didn't completely answer what you were looking for, feel free to post a follow-up question, still trying to figure out what part needs more explaining.

[diablo]

ok thx, its much clearer now ! 

greets diabloo

[MrDax]

Hello.

New to the site, so be gentle.

I can think of two reasons driving the deprotonated FMOC forming the double bond and ejecting the protected peptide: greater stability of the product and peptide as a good leaving group.

First is the deprotonated FMOC - it has 14 pi electron, making the entire molecule aromatic (Hückels rule: 4 x 3 + 2 = 12 + 2 = 14 --> one aromatic system). However, the the pi-electrons of an external double bond do not add electrons to the aromatic system. Twelve electrons do not make an aromatic system (4n+2 = 12 --> n = 5/2), so the diphenyl-part are seperate. Now the molecule has two aromatic groups instead of one. Ie. formation of the double bond makes the molecule more stable.

The second reason is that protected peptide is a good leaving group, as macman104 explained before. Also, the negative charge is delocalized among the COO^--system.

Hopefully I cleared something up