chemrox
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alcoholic KOH & experiment suggestions
McElvain made nicotinic acid from pyridine via 3-cyanopyridine which he hydrolized with KOH in alcohol as follows: 'A solution of 3.6 g. of
3-cyanopyridine and 4 g. of sodium hydroxide in 40 ml. of 70% alcohol was refluxed for three hours after which the *solvent* was removed ...'
(emphasis added). My question is what base species is involved here? Is it hydroxide or alkoxide? or both? In general, when alcoholic KOH is cited
as a reagent can one generalize as to which species is present and/or particpating? Can we sub in the alkoxide to advantage in the above synthesis?
It says 70% alcohol and I assume the rest is water so does the 30% water preclude alkoxide being there?
FYI- A couple of months ago I posted frustration with rusty technique and said I was going to do a bunch of distillations and other simple procedures
to get my feeling for the lab back. This reaction is the next step in my re-familiarizing myself with wet chemistry techniques. (I'm reading up on
mechanisms at the same time.) I have the materials, they're relatively cheap and the procedure is uncomplicated but has a few nuances, especially
toward the end. More of the kind will be sought so any suggestions for future exercises are welcome.
Cheers,
CRX
[Edited on 13-11-2007 by chemrox]
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Antwain
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My guess would be water, since I believe that NaOH is rather insoluble in EtOH, unlike KOH. To a good approximation, there will be no alkoxide in a
solution containing water. That includes KOH in ethanol I would assume. The equalibrium EtO- + H2O <==> EtOH + OH- is just so far to the right.
The only way to make significant alkoxide is with alkali metal + alcohol.
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The_Davster
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| Quote: | Originally posted by Antwain
The only way to make significant alkoxide is with alkali metal + alcohol. |
https://sciencemadness.org/talk/viewthread.php?tid=2656
Nifty eh? It is so counterintuitive
The NaOH/ethoxide is just catalytic, Antwain is correct in saying that the water present is what is reacting.
With the ammount of water present, you are correct, substitution of ethoxide for hydroxide in this case would have no effect as it would hydrolyse
back to alcohol and OH-.
In the procedure you gave a part of, an acidic workup follows at some point, yes?
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Nicodem
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| Quote: | Originally posted by Antwain
My guess would be water, since I believe that NaOH is rather insoluble in EtOH, unlike KOH. To a good approximation, there will be no alkoxide in a
solution containing water. That includes KOH in ethanol I would assume. The equalibrium EtO- + H2O <==> EtOH + OH- is just so far to the right.
The only way to make significant alkoxide is with alkali metal + alcohol. |
NaOH is relatively well soluble in EtOH. At room temperature you can make an approximately 12% NaOH/EtOH solution. Such a solution contains more
ethoxide than hydroxide anions. Check the pKa's of EtOH and H2O and calculate yourself if you do not believe me. The equilibrium of EtO- + H2O
<==> EtOH + OH- is only somewhat to the right since ethoxide is only a 1.8-times stronger base than hydroxide (in water at least). Thus, in
ethanol as solvent the [EtOH] component of the equilibrium equation is high enough to force the equilibrium to the left.
| Quote: | Originally posted by The_Davster
The NaOH/ethoxide is just catalytic, Antwain is correct in saying that the water present is what is reacting. |
Not really. Since NaOH gets consumed in the reaction, it can not be called a catalyst but a reagent required to yield the solution of sodium
nicotinate.
Whether the attacking nucleophile in the first step is hydroxide or ethoxide is completely irrelevant. All the end products are the same for both
pathways. But the attack of the nitrile by H2O is not feasible at such conditions. Water is a many, many magnitudes weaker nucleophile than either
hydroxide or ethoxide.
The role of ethanol is only as a cosolvent to make the starting material more soluble and thus speed up the hydrolysis. Lower alcohols, THF, dioxane
and other such cosolvents are practically always used for hydrolyses with alkali hydroxides. In some cases this might cause a biphasic system
(isopropanol and THF generally form a separate layer with aq. NaOH) but this does not prevent the hydrolysis.
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The_Davster
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| Quote: | Originally posted by Nicodem
Not really. Since NaOH gets consumed in the reaction, it can not be called a catalyst but a reagent required to yield the solution of sodium
nicotinate.
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But once the nucleophile attacks the -CN what about the negative charge left on the nitrogen which reacts with water reforming the base?
Wait, hmm yes I was mistaken, the base formed will deprotonate the formed nicotinate, I just forgot about the further steps, and was only thinking of
the initial part of the mechanism.
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