[davidenarb]

Hi all,

My textbook provides this thermodynamic explanation about why does the loss of N2 occur:

"The step of N2 loss warrants special attention, because formation of a carbanion in a solution of aqueous hydroxide is thermodynamically unfavorable (significantly uphill in energy). Why, then, does this step occur? It is true that the equilibrium for this step greatly disfavors formation of the carbanion, and therefore, only a very small number of molecules will initially lose N2 to form the carbanion. However, the resulting N2 gas then bubbles out of the reaction mixture, and the equilibrium is adjusted to form more nitrogen gas, which again leaves the reaction mixture. The evolution of nitrogen gas ultimately renders this step irreversible and forces the reaction to completion. As a result, the yields for this process are generally very good."

As knowing the basics of thermodynamics, I have two questions :

1.formation of a carbanion in a solution of aqueous hydroxide is thermodynamically unfavorable (significantly uphill in energy). Why? (please I would like to have a short explanation as I am a beginner in thermodynamics )

2. I didn't understand the explation of the textbook about why the step does occur: "Why, then, does this step occur? It is true that the equilibrium for this step greatly disfavors formation of the carbanion, and therefore, only a very small number of molecules will initially lose N2 to form the carbanion. However, the resulting N2 gas then bubbles out of the reaction mixture, and the equilibrium is adjusted to form more nitrogen gas, which again leaves the reaction mixture. The evolution of nitrogen gas ultimately renders this step irreversible and forces the reaction to completion. As a result, the yields for this process are generally very good."

Thank you for making this clear and short.

[orgopete]

Let's start with an easy question. If you take the equilibrium between propane, for example, and hydroxide, how much carbanion would you expect? What does the equilibrium favor?

The W-K reaction generates a carbanion with hydroxide. Does this seem to be favored or disfavored thermodynamically (giving the lowest energy product)?

[davidenarb]

Let's start with an easy question. If you take the equilibrium between propane, for example, and hydroxide, how much carbanion would you expect? What does the equilibrium favor?

The W-K reaction generates a carbanion with hydroxide. Does this seem to be favored or disfavored thermodynamically (giving the lowest energy product)?

I don't know. how do I know the direction of the equilibirum in your question ?

[rwiew]

Think about the pKa's of the hydroxide and for example propane, as orgopete suggested. Why is the difference between those telling you?

[zsinger]

Also, the N₂ is the driving force of this reaction as it escapes as a gas!
         -Zack

[davidenarb]

I know that a great difference between pka values suggest that the reaction is not reversible, but I don't know why?

Also, I know that OH is extreemly basic ! but, what about carbanion !

[rwiew]

Ok, so pKa's tell you he equilibrium constants for the deprotonation reaction of a species (pKa = -logKa), the higher the pKa, the lower the Ka, the less acidic a species is - it becomes thermodynamically less favourable to remove that proton as pKa's increase. You're familiar with all this, right?

Let's use our example reaction with propane now, the equilibrium would be:

OH^- + C₃H₈  H₂O + C₃H₇^-

Check the pKa's of water and propane - from that can you tell me which of them would rather keep the proton? The answer will of course be propane, it's pKa is much higher than the one of water - hence the equilibrium is heavily shifter towards the left side.

About what you wrote above: a great difference in pKa's doesn't suggest the reaction is not reversible, but that the deprotonation equilibrium will he heavily shifter to one side (every reaction is technically speaking reversible, people use "not reversible" to talk about reactions with heavily shifter equilibria sometimes, but it's not a good term - also in this case we're just not getting much deprotonation, so it's more like "not happening"). Also, OH^- is not extremely basic in the big scheme of things, see hydrides, lithium amides, organometallic carbanionic reagents for much higher basicity. (Though OH- is the strongest possible base in an aqueous solution, everything more basic will rip protons off water to make it).