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Topic: Fluoride group as leaving group in the molecule  (Read 9962 times)

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Offline craken66

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Fluoride group as leaving group in the molecule
« on: October 20, 2009, 04:31:34 AM »
Hello, everyone.

I have one question!

I wonder about 'fluoride group' as leaving group in some molecules as following.

I've known that fluoride group is poor leaving group in alkyl chain, which is different from other Halogen atoms (Cl, Br, I).

Then, what do you know is 'fluoride group' when bound by aromatic group (phenyl group)? good leaving or poor?

Especially, in 4,4'-oxybis(fluorobenzene)

Thak you for your reading!




Offline jinclean

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Re: Fluoride group as leaving group in the molecule
« Reply #1 on: October 20, 2009, 06:20:56 AM »
Helo!
In my opinion,whether a group is a good leave group  depends on the bond energy of the C-X bond,for example the F have more electronegativity than ohter halogens especially the iodine,because in the periodic table ,F is in the second row,the VIIA group,it can put more nagetive charge in its control ,to construct more effective interaction. Due to its small radius,the bond between F and C is short than the C-Cl or C-I,so the C-F bond energy is more than other C-X  bond.So the fluorin isn't a good leave group.

Offline Mobius1988

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Re: Fluoride group as leaving group in the molecule
« Reply #2 on: March 15, 2012, 07:15:19 PM »
Hmm, in my chemistry group we were discussing ways of doing SNAr reactions on halogen-aryl groups and one of the suggestions the supervisor gave was to substitute a bromine for a fluorine to make the molecule more reactive.

If fluorine is a worse leaving group why is this so?

Offline Arctic-Nation

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Re: Fluoride group as leaving group in the molecule
« Reply #3 on: March 16, 2012, 07:55:16 AM »
In nucleophilic aromatic substitutions, the rate-determining step is the addition of the nucleophile to the aryl system, breaking the aromaticity in the process. Since the addition puts a full negative charge into the ring, any electronegative substituent (preferably ortho/para, obviously) will help stabilize the anionic intermediate.
Fluorine, now, is very electronegative, and will strongly accelerate the rate of addition. The next step is the expulsion of the leaving group, which is much faster than the addition of the nucleophile, even though fluoride doesn't really like to be kicked out.

So, you speed up the slow step and slow down the fast step by using fluoride, but since the rate of the reaction is determined only by the slow step, the reaction as a whole goes faster.

Offline Mobius1988

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Re: Fluoride group as leaving group in the molecule
« Reply #4 on: March 17, 2012, 08:00:39 PM »
Cheers Arctic-Nation

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