[cupid.callin]

I was studying aldehydes and ketones and came across this aldol rxn i just couldn't understand the reason . (please look at the first pic before reading ahead)

So in the pic,, the second structure shows that H+ is taken from C3 ... i think it should be from C1 ... (next pic)

I searched everywhere i can but still got same reaction where H+ is taken from C3,,, and no reason ...

Please help me,,, i have my exam next week and i know aldol would be there ... please help me!!!!!!!!!!

[Dan]

The reaction you suggest can happen, but notice that these intermediates are all in equilibrium. This means that you will get the thermodynamic product from the reaction. Draw the structure of the product you'd get from your suggested mode of cyclisation - will this be more or less thermodynamically favourable than the cyclopentene product? Why?

[discodermolide]

The reaction you suggest can happen, but notice that these intermediates are all in equilibrium. This means that you will get the thermodynamic product from the reaction. Draw the structure of the product you'd get from your suggested mode of cyclisation - will this be more or less thermodynamically favourable than the cyclopentene product? Why?

Look up Baldwin's rules for ring closure and I think you will find the 5 membered ring favoured

[Dan]

Look up Baldwin's rules for ring closure and I think you will find the 5 membered ring favoured

Both 7-enolendo-exo-trig and 5-enolexo-exo-trig are Baldwin favoured, so it is difficult to predict on those grounds. 6-Enolendo-exo-trig is usually dominant over other favoured modes, but not possible in this case.

[orgopete]

Just to give a slightly different interpretation on this reaction, let's just think of the result as data. Our job is to interpret what that tells us about the chemistry in operation. I agree that the pKa of a CH3-group should be lower than a CH2-group, therefore enolization from the methyl should be faster. But that doesn't explain the product!

From other reactions, we know that five and six-membered rings are favored over all other sizes. We must conclude that deprotonation is not rate limiting, right? If it isn't rate limiting, then cyclization must be. If that is the case, then the five membered ring can form faster. It doesn't change the mechanism, it just changes the kinetics or control of the reaction. You could just think of it like a Fischer esterification and acid catalyzed ester hydrolysis. The mechanisms really are the same, but in opposite directions. It isn't that protonation is favored in one direction as opposed to another. It is just convention to write mechanism designed to arrive at the product. Another example is a Claisen condensation and why the alcohol and ester match to avoid ester exchange from addition to the carbonyl, even though formally that is not part of the mechanism. (If you haven't gotten to either of these reactions, I just couldn't think of an example that I knew you would know.)

[saden99]

Maybe this has to do with the stability of the enol tautomer?

If you draw the enolates from the attack of the Base electrons onto the hydrogens for each of the disputed cases you will get the keto form of the enolate. Draw the enol form and you have one with an internal double-bond and another with a terminal double-bond. Terminal double-bonds are less stable than internal...you can see this on hydrogenation of a molecule. Not MUCH more stable, but still more stable.

[cupid.callin]

But guys dont you think the deprotonation will completely depend on the stability of the anion formed ... and as 1 degree anion is more stable than 2 degree so C1 should give proton!!!

Even if we consider -I effect from Oxygen atom, which is at C7, then we must also consider +I effect from all carbons

So -I of Oxygen C7 and +I of carbon from C3 to C6 ... I guess +I will be more.

So it will form 7 member ring ...
THEN WHY IS THIS WRONG!!!

@ orgopete
Please tell me why 6 and 5 member rings are favoured?
i thought even member rings are favoured over odd member rings due to symmetric structures . due to tetrahedral shape of sp3 orbital!!!
Is this funda wrong?

[orgopete]

What would you conclude if I told you that if the reaction were done in D2O/NaOD and the reaction were stopped when 10% of starting material remained. If that were analyzed by MS, it contained 10xD atoms. What would that tell you about the reaction? Did you know that the hydrogens of acetone can be exchanged with D2O? Every seems to be thinking that is not possible.

Re: ring sizes in cyclization reactions
I believe the preference for 5 and 6-membered rings is a combination of entropic and structural effects. That is, intramolecular collisions occur faster then intermolecular ones. If you did something like putting the atoms with a cis or trans-cyclohexane or a trans double bond, this will also affect the rates. A trans double bond cannot exist in a five or six membered ring because they will be constrained in a way that they will not collide. In that case, intermolecular reactions or larger ring sizes would be preferred. Similarly, geminal substitution (on the same atom) will increase the rate of ring formation.

[cupid.callin]

So you mean in intra molecular aldol, no matter what other product can be formed ... If 5 member or 6 member ring can form, it will be the major product ... ?

And of course case in which both 5 & 6 member ring can form is impossible in aldol . Right???

[orgopete]

So you mean in intra molecular aldol, no matter what other product can be formed ... If 5 member or 6 member ring can form, it will be the major product ... ?

And of course case in which both 5 & 6 member ring can form is impossible in aldol . Right???

Let's be certain we are comparing that which should be compared, we have the choice of forming a 7-membered ring from the kinetic enolate or a 5-membered ring from the thermodynamic enolate. I am saying that based on this data, the kinetic enolate (by definition) must be forming but not leading to the product (check the product, it is a cyclopentane). I am not a genius, there is a preference for the 5-membered ring. The poster was correct, a methyl group should form the enolate the fastest, but it did not lead to the product. The incorrect conclusion would be to think that in this instance and this instance only, the enolization does not favor the methyl ketone to form. I think it forms, but that is not the rate determining step in the product formation. 

"And of course…Right?" Come again?

[cupid.callin]

I meant that like in this case a 5 member and a 7 member ring is possible ... and acc. to you 5 member and 6 member rings is major product in any aldol.
i said that no reactant has a possibility of forming both 5 member and 6 member ring ... so we dont need top bother out of 5 & 6 member rings which will form major product.

I am saying that based on this data, the kinetic enolate (by definition) must be forming but not leading to the product (check the product, it is a cyclopentane).

Kinetic enolate donot give product, what does that mean? what happens to the carbanion after it is formed???

[orgopete]

In an aldol reaction, a common procedure is to use hydroxide or alkoxide as base in water or alcohol. In either case, the pK_a of acetone, for example, is about 20, so the equilibrium favors the ketone. After formation of an enolate, it can become protonated again. This would appear to be no reaction because the starting material is recovered. Because the pK_a of a CH₂- and CH₃-group are close, even though the CH₃ will form an enolate faster, it can become protonated by the solvent to reform ketone and hydroxide or alkoxide. Therefore it is possible for the reaction to form many enolates which return to ketone before any aldol reaction occurs.

For most reactions, the kinetic (the one that forms the fastest) and the thermodynamic product (the one that is the most stable), are the same. In this instance, the kinetic intermediate does not become converted to the product. The overall product of this reaction comes from the thermodynamic enolate. That was the original question I was trying to answer.

[cupid.callin]

Thanks for your help orgopete!!!

Thanks All!!!!