Making Enolate Anions: Substitution Question

[justadream]

---

Members don't see this ad.

Why is deprotonating the more substituted alpha-carbon (to the carbonyl) the "thermodynamic product" while deprotonating the less-substituted alpha carbon (to the same carbonyl) the "kinetic product"?

Are the judgements about thermodnyamic/kinetic based on the stability (and thus, substitution) of the alkene form of the enolate anion?

 

[Concrete Jungle]

kinetics are determined by reaction rate and thermodynamics deal with comparing differences in energy between products and reactants

something that is the fastest reaction doesn't have to be the most stable....

 

[Czarcasm]

Why is deprotonating the more substituted alpha-carbon (to the carbonyl) the "thermodynamic product" while deprotonating the less-substituted alpha carbon (to the same carbonyl) the "kinetic product"?

Are the judgements about thermodnyamic/kinetic based on the stability (and thus, substitution) of the alkene form of the enolate anion?

And just to add on to what was explained above, the thermodynamic product is the result of stability. A more substituted alkene is more stable. Enolates are resonance stabilized (has partial alkene character), and thus the more substitued alpha carbon forms the more stable alkene product. The kinetics has to do with speed rather than stability. A less substituted alpha carbon is more acidic and therefore easier to deprotonate, so protonation can occur more quickly.

 

[justadream]

@Czarcasm
@Concrete Jungle

Right, I understand that the thermodynamic product results when the alkene product is more substituted.

However, what if you don't let the reaction go to the alkene product? An example from TBR Ochem book II page 58 #4:

When you treat acetone with a small base, the thermodynamic intermediate is preferred. What is the major organic product when 2-methylcyclohexanone is first treated with potassium hydride at 40C followed by treatment with ethyl iodide?

The answer is not an alkene (2-ethyl-2-methylcyclohexanone).

Is my original explanation (using the stability (and thus, substitution) of the alkene form of the enolate anion to make judgements about theromodynamic v. kinetic) wrong?

 

[Concrete Jungle]

Took me a while to find the actual example you were talking about. The passage gives you some idea about the function of temperature on the ratio of product distributions between kinetic and thermo products. If you read it carefully it says "the researcher methylated..." why are you thinking an alkene is being formed? The question says 40 degrees C. Look at the table and use that to get an idea of how it fits in to product distribution. What does the first step do? It is a base, so find the location(s) for the most acidic hydrogens. This is the kinetic and thermo part, are you sure you understand this? What is your intermediate now? The reason why it is the most acidic is explained in your intermediate structure. Now follow the workup, your intermediate plus an alkyl halide... some basic mechanism should be jumping out at you.

 

[justadream]

@Concrete Jungle

I understand the reaction. The base can deprotonate at either of the 2 alpha carbons. Here, it deprotonates at the more substituted alpha carbon.

Deprotonating at the more substitued alpha-carbon is the "thermodynamic product".

I was trying to understand (conceptually) why deprotonating at the more substitued alpha-carbon is the thermodynamic product.

With aldol condensations (where , after deprotonating, attack, and elimination, you form an alkene - you can then make judgements about stability based on the alkene's substitution), I can understand why deprotonating at the more substituted alpha-carbon is thermodynamic (because it results in the MORE substituted alkene product).

This problem is similar to aldol condensations (in that you also have to decide which alpha-carbon to deprotonate) but the difference is that the final product isn't an alkene. Instead, the two possible products are simply cyclic ketones (which, unlike alkenes, I cannot judge the stability of).

Thus, I cannot understand (conceptually) why deprotonating at the more substituted carbon is "thermodynamic" for this particular reaction.

 

[Concrete Jungle]

Okay I see why I wasn't answering your question. You can't always just look at structure and figure this out. It is based on experimental data or some sophisticated modelling. The aldol is the "textbook" example because we can easily understand the stability differences. The more substituted enolate is NOT always the most thermodynamically stable intermediate. The passage tells you in this case it is.