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?
This may help.
http://en.wikipedia.org/wiki/Leaving_group
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.
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 :)
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?
Yes, I think that helps. Thank you!
The concept of a weaker base being a better leaving group is related to the idea of nucleophilicity and basicity. To understand this, we need to discuss the reactivity and stability of a leaving group.
A leaving group is a molecular fragment that departs from a molecule, and it is usually associated with a negative charge. The leaving group must be able to stabilize the negative charge upon departure, which affects the overall reaction.
Now, consider a stronger base. A strong base has a greater affinity for protons, meaning it readily accepts a proton to form a new covalent bond. Generally, a strong base is a good nucleophile, meaning it can attack electrophilic centers and form new bonds.
However, if a molecule acts as a strong base, it tends to hold on to its electrons tightly, which makes it less willing to let go of them. This means that the molecule is less likely to function effectively as a leaving group because it is not prone to leaving the molecule.
Conversely, a weaker base has a lesser affinity for protons and electrons. It is less likely to form a new covalent bond by accepting a proton or reacting at electrophilic centers. Due to its weaker basicity, a weaker base is more likely to leave the molecule, making it a better leaving group.
In summary, a weaker base is a better leaving group because it is less likely to hold on to its electrons tightly and is more willing to leave the molecule, thereby facilitating the reaction.