I've got a bunch of Calcium Hydroxide on hand. Both Sodium and Calcium Hydroxide will deprotonate my compound.
I am planning to do a simple substitution reaction. Will the fact that a single mole of Calcium Hydroxide can deprotonate two moles of my molecules
have any negative effect on the reaction? Will the Calcium Hydroxide be able to enter a +1 oxidation state as an intermediate during the substituion
reaction? Or do both of its deprotonated ligands have to leave simultaneously?blogfast25 - 7-7-2014 at 04:45
Whether or not calcium hydroxide can be used instead of sodium hydroxide will depend on specific context.
But as a base, oxidation state does not come into anything here. The oxidation state of all atoms in the calcium hydroxide remains unaltered when it
is used as a base.Zyklon-A - 7-7-2014 at 07:59
Yeah, Calcium can never enter the +1 oxidation state.
It needs to lose two electrons to become stable as it has two in it's outer shell. Losing one electron would make it far less stable then in elemental
form (+0), which is not all that stable in it self. The Volatile Chemist - 7-7-2014 at 08:06
What's the context? What're you preparing?JefferyH - 7-7-2014 at 09:02
Do formed enolates/enols/phenolate disassociate from the base base in solution? So hypothetically could both enolate molecules be on one side of the
flask and the cation on the other side, and both enolates react react at different times? Is the cation complete unchanging in this process?
There's no specific experiment but I was planning on doing a simple SN2 of Phenol with 1-bromopropane.
Is this concept applicable for all substitution reactions? Or is there a time where the charge of the base will play some influence in restricting the
reaction (NaOH/KOH opposed to CaOH2 for example). blogfast25 - 8-7-2014 at 04:43
Do formed enolates/enols/phenolate disassociate from the base base in solution? So hypothetically could both enolate molecules be on one side of the
flask and the cation on the other side, and both enolates react react at different times?
You've lost me. And others I think...JefferyH - 8-7-2014 at 07:59
I'm just trying to picture the charge transfer on the molecules.
If I react Calcium Hydroxide with my acidic organic, do they not form an enolate complex? Does this complex separate in solution as ions or is it held
together? Since the Calcium will be attached to two molecules, what happens when only one of those anions reacts? Won't the Calcium cation be left
attached to a single anion, at least until it can receive the substituted anion from the other reactant in solution, or are these complete detached
from eachother when dissolved in solution?
I should probably be asking how this enolate exists in solution. Phenol-enolate can be isolated as Sodium Phenolate, or Calcium Phenolate, (or
whatever other base you use to deprotonate it). Does the fact that Calcium Hydroxide holds maintains two charges, where as NaOH only maintains on,
effect its behavior at all in substitution reactions? Or is it entirely irrelevent? blogfast25 - 8-7-2014 at 09:06
Irrelevant, I think. Ca(OH)2 should probably work.mnick12 - 8-7-2014 at 09:26
Solubility might be an issue, a lot calcium compounds have terrible solubility. Also phenols are pretty acidic pKa ~10, you don't really need a
hydroxide for deprotonation. Additionally as blogfast25 said I don't think the enolate form is relevant for for ether synthesis.
If you really wanted to find out I would try on a small scale and use a lot of solvent, it is basically insoluble in most organics.
