[mir]

Attached is a reaction with a strong base.
Why is the base attacking the allylic position shown, and not allylic to the endocyclic double bond?
Has it something to do with the steric repulsion from the long chained base?

[arvind1990]

I am not sure about my answer.
But I think the base cannot attach itself to the endocyclic  double bond and can attach itself only
to allylic position as the former is not resonance stabilised but the latter is.
Resonance structures are quite less in the former when compared to latter.
If I am wrong please provide the correct answer.

[Dan]

Both anions would be resonance stabilised, but consider inductive effects in a secondary carbanion relative to a primary one.

[mir]

Both anions would be resonance stabilised, but consider inductive effects in a secondary carbanion relative to a primary one.

So a secondary carbanion is destabilized by induction?
But I got the impression from my textbook that this model is a bit out of date in explaining carbanion stabilities.

After a bit studying I guess I ound the answer,
It might also be the bulky molecule that interfere with the solvation shell. And thereby make the protons at the endocyclic allylic sites less available.

[Dan]

So a secondary carbanion is destabilized by induction?
But I got the impression from my textbook that this model is a bit out of date in explaining carbanion stabilities.

I would definately argue induction, but that's just me. I wasn't aware that the model is considered out of date, what's the text, I'd like to read it? does it also question rationalising the stability of tertiary carbocations by inductive effects? If not, I would have thought that the door swings both ways.

I like your solvation idea though.

[Mitch]

Doesn't n-BuLi form the kinetic carbanion not the thermodynamic carbanion.

[Dan]

Doesn't n-BuLi form the kinetic carbanion not the thermodynamic carbanion.

This did occur to me when I replied. Wouldn't a secondary carbanion produced deprotonate another molecule to form a primary carbanion though, if the only reagent is BuLi and limonene starting material. It will always result in loss of the most acidic proton.

[mir]

I would definately argue induction, but that's just me. I wasn't aware that the model is considered out of date, what's the text, I'd like to read it? does it also question rationalising the stability of tertiary carbocations by inductive effects? If not, I would have thought that the door swings both ways.
I like your solvation idea though.

My reference:  Carrol, Felix A. 1998, "Perspectives on structure and mechanism in organic chemistry", Brooks/Cole Publishing Company....
Here is a short review of the discussion in the book (in somewhat bad english due to a hurry):

The stabilities of carbocations from tertiary to primary, might be explained in terms of the model of hyperconjugation. The same reasoning is used for explaining radical center stabilities. Of course, this is one of many models.

Maybe (my thought) the hyperconjugation model might be valid for carbanions too. Because increasing the number of alkylsubstituents on an alcohol, is actually increasing its acidity in gas phase. So, alkylgroups is perhaps able to delocalize or polarize negative charge away from the center. Just like they stabilize carbocations through hyperconjugation. The effect of the alkyl substituents is defended by theoretical calculations.

Funny enough, in solution the acidity is opposite. Maybe because the solvation of the anion is overcoming the electronic effects because of an favorable entropy term in the solvation process. A large bulky molecule is demanding in the ionization process, more solvent molecules around the molecule surface: an unfavorable increase in order.

Just to add to the confusion:
The acidity in both gas and in solution for halomethanes is: CH4 < CH3F < CH3Cl
So it seems that we cannot use simple electronegivity values to predict the acidity.
Acidity, so it seems, is a better explained by polarization.

Doesn't n-BuLi form the kinetic carbanion not the thermodynamic carbanion.

But is the actually kinetic carboanion formed?
You have also some primary proton on the endocyclic allylic position.

[movies]

I think the answer to this question can be found by considering the strain involved in the two possible structures.  Since the allyl carbanion is planar (in order to maximize the delocalization of charge) that forces the two C-C bond that make up the rest of the ring in the endocyclic isomer to be coplanar as well.  This would result in significant allylic strain (similar to a 1,3 diaxial interaction in cyclohexanes).  In the exocyclic isomer, this isn't a problem since the substituents are all Hs at the two termini of the allyl carbanion.  As for the third possibility, deprotonation at the CH₃ next to the endocyclic alkene (which has been alluded to but not directly mentioned yet), that allyl carbanion would suffer from similar strain to that observed in the completely endocyclic allyl carbanion, but the strain would be between the 1 and 2 positions of the allylic system instead of the 1 and 3 positions.

[Dan]

Another outrageously sensible suggestion from movies!