[susus]

Sorry I\m writing with capital letters, I have a problem in the keyboard haha .

Well, in my question is , organic chemistry ,

In sn1 and sn2 , we know that leaving group order is like this
I- > br- > cl- > f-
I- is best leaving group

Is that in all kind of solvents . Or it's just in neutral solvent ?

You can explain it deeply , I learn org. Chemistry

[Unknown008]

In either reactions, you have to have a particle which can act as nucleophile. Otherwise, there is no reaction at all.

In SN2, you need to have a nucleophile strong enough to displace the anion.

[susus]

How is that related to solvents .

[Unknown008]

Water is a solvent which is also a nucleophile, while butane cannot act as a nucleophile...

[DrBob1]

The leaving group order is a consequence of the carbon to halide bond strength. (C-F is stronger than C-I) Any effect of the solvent on the leaving group would be due to solvation of the ion -- more solvation would make a more stable product and hence could increase the reaction rate. But at this point the reaction is nearly over and I would think the effect would be of relatively little importance. If you look at the energy profile of the reaction you will see that this solvation would lower the energy of the products thereby icreasing the overall exotherm of the reaction but the kinetics are dominated by the activation energy at the beginning stages of the reaction.
So, while solvent effects can completely reverse the NUCLEOPHILICITY of the halogens I would think there would be little effect on their function as leaving groups.

[DrBob1]

1. These more concern the formation of by products, rather than the order of reactivity.
2 Please, Pentane as a solvent - at least you can keep this in a test tube! Can't do that with butane.

[Unknown008]

Okay, thanks for the clear ups DrBob. I was speculating as per what I'm sure about in that area.

But that said, would a solvent containing a lot of halogen ions not affect the dissociation of the molecule as per the equilibrium constant?

Also, I thought that butane was used a solvent for bromine and assumed that it could be used again for other solutes... :confused:

[DrBob1]

The reason I got huffy about butane as a solvent is that it's a GAS at room temperature. By using pentane (petroleum ether) you can work in open glassware. Hydrocarbon solvents are not real good for handling bromine (halogens in general) because there can be photochemical reactions with them. Think how many moles of hydrogen atoms are there. CCl4 is the usual solvent of choice.
I'm not sure about the question about solvents containing a lot of halogen ions. This means salty water and the primary effect will be to make an organic solute insoluble. The added ions sound like competitors for both SN1 abd SN2 reactions. You do not want them there.

[Unknown008]

Okay I wanted to make sure, and finally, the site allowed me to rep you! :p

[DrBob1]

I appreciate it.
By the way, there is no equilibrium constant involved in these reactions: in an SN1 reaction each dissociation (ionization) gives a reaction and, of course, in SN2 there is no ionization at all.

[Unknown008]

I was referring to:



Isn't that what happens? :confused:

[DrBob1]

No, I'm afraid not. You have shown a primary alkyl halide; these react solely by an SN2 reaction mechanism, The primary carbocation is simply too high in energy to participate.
Tertiary alkyl halides ionize as you show, this is why a chiral substrate gives a racemic product. I think each ionization leads to the product, not to an equilibrium. If equilibration were the case we would find racemazition of the starting material; I have never seen a mention of this. I doubt that this is solely because no one bothered to look for it. (But take a shot at it Ucky, you could become fameous!)

[Unknown008]

Oops, I actually meant a tertiary alkyl halide and forgot the H atoms, my bad :(

But I no more have my notes so I ended up completely dependent on the net to get the parts I want to confirm, and I found this on wikipedia:

SN1 reaction - Wikipedia, the free encyclopedia

The second reaction is apparently reversible, (it is very slow though) and that's what made me think of the above reaction I wrote (but unfortunately forgot the R' and R'' parts and put H2 instead). Maybe it's because of the yield, the racemization occurs over a long time.

And unfortunately, for the time being, I'm more heading towards accounting than more science... maybe in my late years :o

[DrBob1]

The "second reaction" is (CH3)3CBr ------> (CH3)3C+ + Br-. It is indeed reversible (it is the final step in the addition of HBr to methylpropene). But as I said, I do not think this happens in the practice of an SN1 reaction. That would lead to the racemization of a chiral starting material, not just to the formation of a racemic product. The carbcation must be made to react quickly or else side reactions (rearrangements, loss of H+, etc) will take place. So I still think that it CAN BE reversible, but ISN'T.
The Wikipedia article says this is a slow step; that is correct. But remember that it is slow in comparison to the other steps. It is the "rate determining step". The overall reaction can be quite fast in practical terms.
I think you show great promise as a scientist, Ucky, but you havr to follow your dreams. Good luck.

[Unknown008]

Aww, I wish I could send you another rep, but for that, I need to go to the 'Go' skin if I want to bypass the distribution of greenies. That'll be for another time :)

Thank you!

Hm... so if there were no nucleophile at all, that might be possible after a long long time, understanding that the reaction is reversible..

[DrBob1]

When you're working with carbocations there is no "long time". They are exremely reactive and if nothing comes along to which they can form a bond, they will lose a proton and become an alkene. Before or after rearranging.