unique135

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The reaction is yielding 90%.

I believe the hydrogen on C1 is more acidic compared to other hydrogens on C 2/3/5. Am I right? Can anyone explain why so (or is this too complex)?





If above is the case then enolate would prefer ethyl in methyl ethyl ketone. Would it contradict with carbanion stability (since they prefer CH3>1>2>3) ?

Probably, I am thinking too much and confusing myself. Any help appreciated. :)
 

Samchik

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I think the major product forms because of its thermodynamical stability, in other words it produces 5-membered ring which does not have too much ring strain. Compared to other options of intramolecular aldol cyclization only alpha hydrogen at C1 will produce a stable 5-membered ring product.
 
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I agree with the ring strain explanation, but it would only account for the preference of 1 over 2 and 3.

5 is still a legitimate candidate that could yield an the other cyclopentane configuration, but it faces the problem of being a primary carbanion, which is more high-energy and less thermodynamically stable intermediate than the secondary carbanion at 1.
 

unique135

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Sorry, I kinda misguided the question with yield. True, the product formation with C2 and C3 won't be preferred because of ring strain. However, enolates might form at that these carbons (based on how acidic hydrogen are at these position). I wanted to focus on enolates formation and stability.

For C5, I think C5 is not preferred for the same reason - ring strain (as the strain increases after cyclohexane). docelh, isn't primary carbanion more stable than secondary one?

I could say that the "product formation" at C1 is preferred over C2,C3,C5 because of ring strain. However, I am confused about the "enolate formation" and its stability.

In short, should we think of carbanion stability for enolate formation?
 
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Few things:

C5 bonds to the carbonyl C b/n 1 and 2 to yield an equivalent cyclopentane, so ring strain shouldn't be valid comparing 1 to 5.

For related reasons that a secondary carbocation is more stable than a primary, I think the same applies to carbanions. The extra electron can freely move to delocalize it.

With regards to enolate stability, I don't see any obvious advantage that one position has over the others, excluding 4.
 

boaz

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I agree with the ring strain explanation, but it would only account for the preference of 1 over 2 and 3.

5 is still a legitimate candidate that could yield an the other cyclopentane configuration, but it faces the problem of being a primary carbanion, which is more high-energy and less thermodynamically stable intermediate than the secondary carbanion at 1.
Another reason for the preference of 1 over 5 is that the product is more stable. With 1 it's a more highly substituted double bond.

...
In short, should we think of carbanion stability for enolate formation?
Only if the reaction is under kinetic control.

The more the reaction is under therm control, the more we are concerned about the stability of the products, and the less we're concerned about stability of intermediates.
 

loveoforganic

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Using hydroxide, I wouldn't have too much concern about sterics of any protons.
 

MegaSpectacular

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You mean sterically?
You said unless we are under kinetic control, which would mean a change of conditions. Obviously we can't say "we are now under kinetic control" without changing the conditions. Kinetic control is favored with a hindered base/colder temperature.

If we keep the same conditions here, we would get the same 90% yield...
 

boaz

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You said unless we are under kinetic control, which would mean a change of conditions. Obviously we can't say "we are now under kinetic control" without changing the conditions. Kinetic control is favored with a hindered base/colder temperature.

If we keep the same conditions here, we would get the same 90% yield...
Kinetic control does not necessarily have to include hindered base. My use of the term kinetic control was limited to the use of identical reagents at low temperature. I was addressing his question about stability of carbanions. Under low temperature the stability of the transition state (carbanion-like in this case) needs to be considered.
 

MegaSpectacular

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Kinetic control does not necessarily have to include hindered base. My use of the term kinetic control was limited to the use of identical reagents at low temperature.
You should be more specific then. You didn't talk about changing any of the conditions, temperature or base. I can't assume you drop temperature but keep the base.

btw,

Kinetic control is favored with a hindered base/colder temperature.
obviously implies that it does not necessarily have to include a hindered base, hence the word "favored" is present.
 

boaz

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You should be more specific then. You didn't talk about changing any of the conditions, temperature or base. I can't assume you drop temperature but keep the base.
Like I said, I was addressing his/her question about carbanion stability, and hindered base has nothing to do with that. So I think it's fair to assume that kinetic control refers to the conditions that affect carbanion stability, i.e. temperature.
 

MegaSpectacular

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Like I said, I was addressing his/her question about carbanion stability, and hindered base has nothing to do with that. So I think it's fair to assume that kinetic control refers to the conditions that affect carbanion stability, i.e. temperature.
Then it would have been better to say lower temperature.

As the term "kinetic control", which I think is fair to assume, would refer to the conditions that are most likely to give the kinetic enolate (hindered base/colder tempture).