Leaving group E1 reaction?
Why is Cl a better leaving group than F in this specific example?
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Because the chloride ion is better stabilized (larger radius) and because of the low binding energy this bond can be broken more easily (consider the high EN value of fluorine)
Why does dle atomic size also play a role?
I explained that: The larger radius means a larger surface over which the negative charge of the ion (as LG) can be distributed.
If you have a protic solvent, the resulting carbocation is stabilized by this, a sterically demanding base will let the whole run after E2, an aprotic solvent, if necessary according to SN2.
E1 occurs when after leaving the AG, a carbocation arises that can be stabilized somehow. It depends on the circumstances, an elimination with base runs differently than one without!
At the same time, the quality of the AG is also decisive for the speed and how everything is going. If you have a strong base here, an E1cb is also possible and an anion is created!
Look at the picture
http://snap.ashampoo.com/F2YKAerXvEyKdbG2VqtnvGiOQxGBDOXAdNWazXEJKVYkX74AvSNTeJLDuaqlaTBE
Perhaps you will realize what problems there are and what role the solvent plays.
Always note that the quality of the AG decides what happens (or not). Fluor is a very bad AG. The negative charge can only be poorly compensated/distributed over the atom. Iod is much bigger! Chlorine is also larger than fluorine!
The lower mechanism that was drawn by you will run STETS as the cation on the sec. C atom is better stabilized (or a solvent can optionally stabilize this charge than with a primary carbon atom. So here E1 will run. If you use a (strong) base (which should not be a nuc – why?) the reaction will proceed according to E2 [abstraction of a proton].
read already;) You weren’t fast enough.
Can you answer BITTE’s next question?
Why does this depend on the solvent? I mean, what else should the lion do except to leave the molecule with full shells?
So, could there really be a carboanion here, but then the fluorine would go without a negative charge?
When I have time, I answer questions, not when you ask me 😉 and you say “BITTE” – with courtesy you keep coming here
Could you answer my question, too?
Okay. This means that if we look at the reaction above more precisely, that a fluoride ion would arise here and consequently a carbocation ?
jep, so it’s… consider yourself whether it’s energetically favorable when many electrons are too close to narrow space (repulsion forces)
I can’t start with the word “stabilized” here in the context? Does that mean that the electrons are distributed over a larger space or what exactly?