[sam79]

Hi

I was reading about elimination reactions of Alcohols
http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/alcohol1.htm
If i understand correctly, an alcohol like ethanol is a good solvent for strong bases. Are alcohols used as solvents for strong bases in labs/industry?
What are the potential pitfalls?

It would even make solvent removal easier, compared to dissolving the strong base in water.

[discodermolide]

Hi

I was reading about elimination reactions of Alcohols
http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/alcohol1.htm
If i understand correctly, an alcohol like ethanol is a good solvent for strong bases. Are alcohols used as solvents for strong bases in labs/industry?
What are the potential pitfalls?

It would even make solvent removal easier, compared to dissolving the strong base in water.

Alcohols are only used in industry if you require the corresponding alkoxide.
If we wish to use a strong base we usually try and get away with hexyllithium as a solution in hexane or THF, other strong bases (LDA, KHMDS) are usually dissolved in toluene or THF.
Pitfalls of alcohols are potential transesterification or substitution.
Absolute (100%) alcohols are usually good solvents for recrystallisation of the API (unless it is methanol, which is toxic).

[sam79]

So if i understand correctly, dissolving NaOH or KOH in say ethanol, will yield an alkoxide.
But what will be the equilibirum between the forward and backward reactions? Will it be heavily in favor of the forward?

[discodermolide]

So if i understand correctly, dissolving NaOH or KOH in say ethanol, will yield an alkoxide.
But what will be the equilibirum between the forward and backward reactions? Will it be heavily in favor of the forward?

Dissolution of NaOH or KOH does not give the alkoxide, where did that from from? You generate the alkoxide by adding a base (stronger than (NaOH) to the alcohol, this gives a solution of the alkoxide in the alcohol.

[orgopete]

So if i understand correctly, dissolving NaOH or KOH in say ethanol, will yield an alkoxide.
But what will be the equilibirum between the forward and backward reactions? Will it be heavily in favor of the forward?

If you added NaOH to methanol, you should get a considerable amount of sodium methoxide due to the greater concentration of methanol than water and the pKa values being quite similar. You should be able to calculate the amounts from the pKa values. The pKa is the negative log of the equilibrium. If you know the amount of alcohol, you should get an approximate value for the amount of alcohol and water present. Obviously, the disadvantage of making an alkoxide with NaOH or KOH is water will accompany it formation. If you were catalyzing a transesterification reaction, you would sacrifice some of your ester with the formation of carboxylate from the water present.

[sam79]

I am confused. The last 2 posts seem to contradict each other. No offence intended from a Chemistry newbie
@discodermolide, suggests dissolving NaOH or KOH in ethanol will not yield an alkoxide.
@orgopete, seems to suggest otherwise.
Confused.

Let's say the aim is to dissolve .1-1.0 Moles of KOH or NaOH in 1 liter of ethanol

[discodermolide]

I am confused. The last 2 posts seem to contradict each other. No offence intended from a Chemistry newbie
@discodermolide, suggests dissolving NaOH or KOH in ethanol will not yield an alkoxide.
@orgopete, seems to suggest otherwise.
Confused.

Let's say the aim is to dissolve .1-1.0 Moles of KOH or NaOH in 1 liter of ethanol

Is the NaOH/KOH in water or as a solid?

[sam79]

I was thinking solid. How do the results differ if the NaOH/KOH were dissolved in water?

[discodermolide]

I was thinking solid. How do the results differ if the NaOH/KOH were dissolved in water?

Water will increase the solubility.
If NaOH and/or KOH are dissolved in alcohols, you will get a solution. The pka's are similar, around 14-15, I don't think much alkoxide will form at all. You need a stronger base such as sodium hydride.

[fledarmus]

Water is more acidic than alkoxides - if you have water present, you will not be forming alkoxides, you will form hydroxide and alcohol.

This is an equilibrium, though, so if you start with very dry alcohol and solid hydroxide, you can form some alkoxide. You can form more alkoxide is you also have a way to remove water as it is formed. I have seen some reports of this being done on an industrial scale, but on a lab scale it is much easier to form the alkoxide using either sodium metal or sodium hydride in alcohol. You do have to be very careful to keep the evolved hydrogen from igniting and lighting off the whole pot, however. The metal or hydride should be added in very small portions under an inert gas.

As a solvent for "strong bases", however, this is limited to bases no stronger than the alkoxide itself, and many things that laboratory organic chemists consider useful bases are considerably stronger than alkoxides. A base stronger than alkoxide (for example sodium amide, methyllithium, grignard reagents, lithium hexamethyldisilazide, all very useful synthetic bases) will simply deprotonate the alcohol and neutralize the base.

Industrially, alkoxides are about the strongest bases you can work with on large scale without some fairly elaborate precautions; in a lab setting, alkoxides are only medium strong and are usually prepared by reacting the alcohol with an even stronger base rather than by trying to react it with the weaker base (hydroxide) and removing water to force the reaction.

[discodermolide]

Industrially, alkoxides are about the strongest bases you can work with on large scale without some fairly elaborate precautions;
[/quote]


It is actually easier to work with "strong bases" such as LDA, Grignards, KHMDS, Hexyllithium, RedAl, DIBAL-H, etc because they come in cylinders to your specification usually around 100Kgs of reagent solution. The only problem is their concentration, around 0.5 - 1.0M, more concentrated and they are pyrophoric and can't be transported without tremendous effort and paperwork. So the reactions are quite dilute. You pump them into the reactor with nitrogen. It is much harder to use sodium, sodium hydride on an industrial scale because they are hydroscopic solids and very difficult to handle effectively. There is no really good way of adding solids to a reactor.
Most industrial plants have low temperature reactors, that will get you down to -80°C but that is about the limit. So using LDA etc is not a problem. I've done it many times.

[sam79]

I don't want to form an alkoxide. My intentions are more about preventing it's formation.
I just want a solution of NaOH or KOH in ethanol.
So i guess i could dissolve NaOH and KOH in a small amount of water, then add ethanol to make up the volume.

[discodermolide]

I don't want to form an alkoxide. My intentions are more about preventing it's formation.
I just want a solution of NaOH or KOH in ethanol.
So i guess i could dissolve NaOH and KOH in a small amount of water, then add ethanol to make up the volume.

As Fledarmaus said "Water is more acidic than alkoxides - if you have water present, you will not be forming alkoxides, you will form hydroxide and alcohol."

So that's the way to go.

[orgopete]

I wasn't trying to answer a specific question, just a more general description. In Evan's pka table, the pka of methanol is 15.5 and water is 15.7. If you add solid NaOH to methanol, you will get a lot of methoxide forming. I'm sure someone could calculate the actual amount for a given amount of NaOH and CH3OH. Even without calculating, the pKa values and the law of mass action predicts there will be more methoxide than hydroxide present, but both will be present.

If one were saponifying an ester, alcohols are commonly used as a co-solvent. Arguably some alkoxide does form, but the equilibrium product will be the carboxylate from hydroxide attack. Even an unfavorable kinetic attack of an alkoxide would be overcome by the greater equilibrium formation of the carboxylate.