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Topic: Methylation and Reduction of a conjugated enol  (Read 11068 times)

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Offline lemonoman

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Methylation and Reduction of a conjugated enol
« on: February 03, 2006, 04:57:38 PM »
Hello everyone!

I'd just like to ask a question; I have to assign the proper reagent(s) in the proper sequence to get from reactants to products.  The reaction is below in the picture.

I don't want to sway any votes on how to do this, but here's what I think:

1. Addition of HCl or HBr, etc, to add to the double bond
2. Two alpha-hydrogen replacements with methyl groups (MeOTf should work)
3. removal of HCl, this time the hydrogen has to come from the other cyclohexane ring, to give a double bond back.

I am only worried that the methylations will not be specific to the two alpha-hydrogens on the same alpha-carbon.

Thanks again for any input/criticism! :)

Offline sundberg

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Re:Methylation and Reduction of a conjugated enol
« Reply #1 on: February 04, 2006, 05:24:45 AM »
My (very short) advice is to take a look at enolate alkylations! :)
« Last Edit: February 04, 2006, 05:25:12 AM by sundberg »

Nucleophile

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Re:Methylation and Reduction of a conjugated enol
« Reply #2 on: February 06, 2006, 06:31:39 AM »
[q]
1. Addition of HCl or HBr, etc, to add to the double bond
2. Two alpha-hydrogen replacements with methyl groups (MeOTf should work)
3. removal of HCl, this time the hydrogen has to come from the other cyclohexane ring, to give a double bond back.
[/q]

ain't going to happen. after the addition of HCl on the double bond, you're left with two secondary carbons in the alpha positions of the carbonyl group.

I'd go at it another way.

I'd place a Bromine at the allylic end of the conjugate system with NBS. I am having this vibe that you'll have to protect the ketone with CH2(OH)CH2(OH) first. I am not sure.

Now you'll have a bromine that will let you easily create the needed double bond in the end with a zaytseff elimination.

Next, we have to place methyls in the anti-markovnikov postion. First, Hydrogen peroxide and BH3 will place a -BH2 group just where you want it. Then, knock it down, and place the Methyl instead with some Methyl Iodide.

Now, you to place a second methyl there - easy! First, NaH, at around 0C, will give you the thermodynamic enolate ( the one which is more substituted ). Then, Methylation with some more MeI.

Last but not least, you eliminate some HBr, and you're all set.  ;D
« Last Edit: February 06, 2006, 06:34:05 AM by Nucleophile »

Offline movies

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Re:Methylation and Reduction of a conjugated enol
« Reply #3 on: February 06, 2006, 08:46:38 PM »
My (very short) advice is to take a look at enolate alkylations! :)


That's what I'd do!  You can get there in one step!

Nucleophile

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Re:Methylation and Reduction of a conjugated enol
« Reply #4 on: February 07, 2006, 09:48:49 AM »
1 step? I am know some enolate alkylations but how do do this in one step? Come on, tell the secret!

Guess: MeLi
« Last Edit: February 07, 2006, 09:55:31 AM by Nucleophile »

Offline movies

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Re:Methylation and Reduction of a conjugated enol
« Reply #5 on: February 07, 2006, 11:56:45 PM »
MeLi is too nucleophilic and would just add to the carbonyl group.

In this case, you want to make a thermodynamic enolate which will actually be generated by deprotonation at the gamma position of the alpha-beta unsaturated system.  This extended enolate is more thermodynamically favorable because it is a longer conjugated system than the cross-conjugated system you would get from deprotonation at the other acidic site.  Extended enolates like this still react at the alpha position of the enolate (for molecular orbital reasons).

So, if you use a thermodynamic base like KO-t-Bu and a methyl electrophile like MeI, then you can form the extended enolate, alkylate to give you the first methyl group, and then for the thermodynamic enolate at the more substituted side of the carbonyl (again generating an extended enolate) and then alkylate again at the alpha position.  This can be done all in one pot!!

Offline lemonoman

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Re:Methylation and Reduction of a conjugated enol
« Reply #6 on: February 08, 2006, 03:53:40 AM »
Wow, what a turn of events.

This was for an organic chemistry assignment I was working on, and it looked like there was very little response.  Actually on Monday, the professor asked if we had questions - I volunteered this one, and he asked me to draw it on the board.  Long story short, the entire 50-minute class was me, working through the synthesis...trying different bases for the enol, making different suggestions...he didn't tell me if I was going on the wrong path, but I suppose that enhanced my understanding.

Anyways long story short, the final thing we agreed on was movies' suggestion EXACTLY.  We agreed on using pottasium t-butoxide as a base (after thinking about NaOEt and NaO-t-but).  We also agreed on the fact that the gamma-hydrogen acidity is key, and that the methylations will be SN2 displacements of a methyl halide.

The SN2 stuff and gamma-hydrogen stuff seems common enough...but I find it a coincidence that both movies and my prof suggested KOt-Butoxide.  Do great minds think alike THAT much?

Nucleophile

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Re:Methylation and Reduction of a conjugated enol
« Reply #7 on: February 08, 2006, 04:16:42 AM »
It's quite possible.  :)

Quote
Extended enolates like this still react at the alpha position of the enolate (for molecular orbital reasons).

ahh... this was the piece of info I was missing. Good to know. Could you explain the MO reasoning, perhaps?

Offline Mitch

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Re:Methylation and Reduction of a conjugated enol
« Reply #8 on: February 08, 2006, 04:53:03 AM »
Which is the gamma position? Also, where is alpha and beta?
« Last Edit: February 08, 2006, 04:56:52 AM by Mitch »
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Offline plu

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Re:Methylation and Reduction of a conjugated enol
« Reply #9 on: February 08, 2006, 05:02:23 PM »
The alpha, beta, and gamma carbons refer to the carbons one position, two positions, and three positions away from the carbonyl carbon respectively.  By the way, what would be the difference between movies's "potassium butoxide" and lemonoman's "NaO-t-but" (aside from the fact that one contains potassium and the other sodium  :P) ?

Offline Borek

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Re:Methylation and Reduction of a conjugated enol
« Reply #10 on: February 08, 2006, 05:32:41 PM »
what would be the difference between movies's "potassium butoxide" and lemonoman's "NaO-t-but" (aside from the fact that one contains potassium and the other sodium  :P) ?

lemonoman's compound is more precisely defined - its t. movies was not as precise :)
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Offline movies

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Re:Methylation and Reduction of a conjugated enol
« Reply #11 on: February 09, 2006, 02:42:00 PM »
lemonoman's compound is more precisely defined - its t. movies was not as precise :)

I said t-butoxide too!  Check the post above!

As for why we came to the same set of reagents, I think it's just that those are standard thermodynamic alkylation conditions.  t-Butoxides are the classic way to generate thermodynamic enolates for reasons I'm sure your professor discussed (high basicity, low nucleophilicity, etc.).  As for MeI, that's pretty much one of the very best alkylating agents there is.  It's cheap, easy to work with, and gives almost exclusively C-alkylated products (as opposed to O-alkylated products, again for MO reasons).  So, there is no magic to two organic chemists coming up with nearly identical conditions, it's just knowing some fundamental organic chemistry!

Here is a brief MO rationale for the alpha-alkylation:
Alkylating agents like MeI are strongly influenced by overlap with the molecular orbital of the nucleophile.  With "harder" alkylating agents such as MeOTf, the electrostatic attraction becomes more important, so the electrophile is attracted to the most electron rich part of the nucleophile; in this case that would be oxygen.  However, as I said, MeI reacts preferentially with the highest energy orbital, i.e. the HOMO of the nucleophile.  If you draw out the HOMO of an enolate, you would find that you have a small lobe on oxygen and a large lobe on carbon, so when orbital overlap is important, you expect reaction at carbon because the carbon centered lobe contributes more to the HOMO.  To explain the O-alkylation products you need to look at the next lowest MO, or the sHOMO, which contains a large lobe on oxygen and a small lobe on carbon.  The sHOMO is lower in energy because it has a lot of electron density on the more electronegative atom, oxygen.

When you get to extended enolates, you can draw similar HOMO diagrams, but you will find that if you do the calculations, the size of the orbitals decreases the further out that you go on the extended enolate, so the largest lobe of the HOMO of an extended enolate is still on the alpha carbon.  This can be explained by the principle of least motion, that is, you don't have to move the electrons as far from the oxygen to the alpha carbon as you would need to move them from the oxygen to the gamma carbon (think resonance structures!)

I hope this helps.  I can draw some pictures later, if need be.

Offline Borek

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Re:Methylation and Reduction of a conjugated enol
« Reply #12 on: February 09, 2006, 04:08:59 PM »
I said t-butoxide too!  Check the post above!

That's what happens when you are too lazy to check original source ;)
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Nucleophile

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Re:Methylation and Reduction of a conjugated enol
« Reply #13 on: February 11, 2006, 05:36:58 AM »
Quote
When you get to extended enolates, you can draw similar HOMO diagrams, but you will find that if you do the calculations, the size of the orbitals decreases the further out that you go on the extended enolate, so the largest lobe of the HOMO of an extended enolate is still on the alpha carbon.  This can be explained by the principle of least motion, that is, you don't have to move the electrons as far from the oxygen to the alpha carbon as you would need to move them from the oxygen to the gamma carbon (think resonance structures!)

I hope this helps.  I can draw some pictures later, if need be.

No need to - I understood you fully.  :)

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