This may help.
http://en.wikipedia.org/wiki/Leaving_group
Why is a weaker base a better leaving group? Can someone please explain that? I understand that the more stable the leaving group is after it has left, that means the more able it was to accept an electron pair prior to leaving, but I don't understand the weaker base part?
5 answers
It tells you what is, but it doesn't tell you why that is. Perhaps it's an assumed piece of information, but I seem to have missed it.
What I don't understand is what it matters that the leaving group is a weak base or a strong base. Doesn't a base in general just want to accept protons? A weak base wants protonation, but doesn't a strong base want protonation as well?
Okay maybe I'm looking at this the wrong way. Is it because what really matters is the R in the R-X (X being the halogen and the future leaving group)? Like if X is the weaker conj base, it means the acid is very strong, and therefore is willing to give its electrons away to the leaving halogen?
Sorry if I'm not making sense, I'm trying as best as I can to understand this.
What I don't understand is what it matters that the leaving group is a weak base or a strong base. Doesn't a base in general just want to accept protons? A weak base wants protonation, but doesn't a strong base want protonation as well?
Okay maybe I'm looking at this the wrong way. Is it because what really matters is the R in the R-X (X being the halogen and the future leaving group)? Like if X is the weaker conj base, it means the acid is very strong, and therefore is willing to give its electrons away to the leaving halogen?
Sorry if I'm not making sense, I'm trying as best as I can to understand this.
Update on what's going on in my head:
I think I'm getting confused because I am thinking of a bronsted acid/base instead of lewis. The weaker the lewis base is, the stronger the lewis acid is. So once the lewis base donates its electrons, the halogen of the lewis acid falls out as the leaving group. The leaving group is now the conjugate lewis acid, and therefore accepts electrons. The stronger this conj acid is, the more electrons it will take away from the carbon, therefore making it more favorable for the lewis base to bond with this carbon.
Hope someone can let me know if I'm in the right direction, and if not explain the right stuff to me :)
I think I'm getting confused because I am thinking of a bronsted acid/base instead of lewis. The weaker the lewis base is, the stronger the lewis acid is. So once the lewis base donates its electrons, the halogen of the lewis acid falls out as the leaving group. The leaving group is now the conjugate lewis acid, and therefore accepts electrons. The stronger this conj acid is, the more electrons it will take away from the carbon, therefore making it more favorable for the lewis base to bond with this carbon.
Hope someone can let me know if I'm in the right direction, and if not explain the right stuff to me :)
If we look at the halogens and consider that weak bases do not share their electrons well because their electrons are farther away from the nucleus,
then the order of reactivity is strong base F->Cl->Br->I- weak base
making it easier for the lower halogens bonds to be broken,
so if we have an equilibrium reaction
B- + R-A<->R-B + A-
the equilibrium position will be on the side where R-A or R-B forms the stronger bond, for example if B=Cl- and A=I- then the equilibrium position will be on the right hand side.
does this help?
then the order of reactivity is strong base F->Cl->Br->I- weak base
making it easier for the lower halogens bonds to be broken,
so if we have an equilibrium reaction
B- + R-A<->R-B + A-
the equilibrium position will be on the side where R-A or R-B forms the stronger bond, for example if B=Cl- and A=I- then the equilibrium position will be on the right hand side.
does this help?
Yes, I think that helps. Thank you!
2 answers