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Chemistry Forums for Students => Organic Chemistry Forum => Organic Chemistry Forum for Graduate Students and Professionals => Topic started by: Dan on April 11, 2012, 11:26:15 AM

Title: Problem of the... month... 11-04-12
Post by: Dan on April 11, 2012, 11:26:15 AM
Sorry I've not had time to post one for a while.

Here we go, predict the product of the following sequence of conditions (I know there's lots of steps but at least half are straightforward so don't worry):

Note: The ABCD ring system is (almost) unaffected in this sequence, so for now we will focus on the transformation of the anisole portion of the starting material.

mCPBA = meta-Chloroperbenzoic acid
DMP = Dess-Martin periodinane
Title: Re: Problem of the... month... 11-04-12
Post by: Honclbrif on April 11, 2012, 04:32:46 PM
So far I'm thinking that the Birch reduction yields the 1-methoxy-1,4-cyclohexadiene (the isomer with the most substituted double bond). The enol ether is cleaved by methanolic HCl to yield the cyclohex-3-en-1-one. LiAlH4 burns that down to 3-ene-1-ol, mCPBA epoxidizes the double bond, and DMP re-oxidizes the alcohol back to a carbonyl. After that the tosyl hydrazine produces the aryl hydrazone and I have no idea what hot acidic hydroxylamine does to that.

Normally I'd look for Beckman rearrangement under those conditions, but I can't find a good place to put the oxime. At least not without displacing the aryl hydrazone which seems unlikely because why would you put it there if you were just going to knock it off in the next step when you could have gone there directly from the ketone?

Aryl hydrazones have very rich chemistry, but all the reactions I'm recalling are usually initiated by strong bases rather than acids.

Am I on track so far?
Title: Re: Problem of the... month... 11-04-12
Post by: Dan on April 11, 2012, 06:31:40 PM
So far I'm thinking that the Birch reduction yields the 1-methoxy-1,4-cyclohexadiene (the isomer with the most substituted double bond). The enol ether is cleaved by methanolic HCl to yield the cyclohex-3-en-1-one. LiAlH4 burns that down to 3-ene-1-ol, mCPBA epoxidizes the double bond, and DMP re-oxidizes the alcohol back to a carbonyl. After that the tosyl hydrazine produces the aryl hydrazone

All good up to here, the hydrazone reacts further before the hydroxylamine step.

Title: Re: Problem of the... month... 11-04-12
Post by: stewie griffin on April 11, 2012, 07:58:48 PM
I think you can get a Wharton reaction with the hydrazone to give an allylic alcohol.
Title: Re: Problem of the... month... 11-04-12
Post by: Dan on April 12, 2012, 02:41:57 AM
Very close with the Wharton! You'd get a Wharton with hydrazine, but you get another related reaction with sulfonylhydrazine.
Title: Re: Problem of the... month... 11-04-12
Post by: stewie griffin on April 12, 2012, 11:36:06 AM
Well the only reaction I know of similar to a Wharton is the Shapiro reaction, but there's no alkyl lithium present. I'll have to think more..
Title: Re: Problem of the... month... 11-04-12
Post by: Honclbrif on April 12, 2012, 01:47:57 PM
There's the Eschenmoser fragmentation, but wouldn't that require an alpha-beta epoxy aryl hydrazone? I'm getting a methylene between the epoxide and the hydrazone.
Title: Re: Problem of the... month... 11-04-12
Post by: Dan on April 12, 2012, 03:09:42 PM
There's the Eschenmoser fragmentation,

Correct, it is an Eschenmoser fragmentation

Quote
but wouldn't that require an alpha-beta epoxy aryl hydrazone? I'm getting a methylene between the epoxide and the hydrazone.

Indeed. I misread your original post in a bit of a hurry:

Quote
The enol ether is cleaved by methanolic HCl to yield the cyclohex-3-en-1-one

You only have to look at cyclohex-3-en-1-ones to isomerise their C=C double bonds into conjugation with the ketone, it will be unstable in the presence of acids or bases.
Title: Re: Problem of the... month... 11-04-12
Post by: Honclbrif on April 12, 2012, 06:15:18 PM
You're absolutely right, I should have seen that. I once got a heck of a headache trying to play with an enone with skipped conjugation (and not just because of how it smelled).

Here's what I've got so far

(http://www.majhost.com/gallery/poorjon/misc/potm_11_04_12a.gif)

Eschenmoser followed by a Beckmann which could lead to A or B, depending on what migrates... I've never been totally clear on the rules about migration in a Beckmann, but being this deep into a synthesis I doubt you would want to throw away half your product so there's got to be some. I suppose B would involve the migration of the more highly substituted branch. Looks like I've got to do more reading on this one.
Title: Re: Problem of the... month... 11-04-12
Post by: stewie griffin on April 12, 2012, 07:55:43 PM
It's been a while since I've looked at the Beckmann but I'm pretty sure that the more substituted carbon migrates to the nitrogen.
Title: Re: Problem of the... month... 11-04-12
Post by: Dan on April 13, 2012, 02:39:00 AM
It's the substituent anti to the N-O bond that migrates in a Beckmann, so you have to predict the geometry of the oxime to determine the product.

However, the reaction is not a Beckmann rearrangement. They require more forcing conditions (example (http://www.organic-chemistry.org/abstracts/literature/153.shtm)) than simply heating with hydroxylamine hydrochloride.

You are half way there with the oxime though, that is formed, but something else happens as well - it is simple but quite surprising...
Title: Re: Problem of the... month... 11-04-12
Post by: Honclbrif on April 18, 2012, 12:38:28 PM
In step v, how many equivalents of NH2OH are added?
Title: Re: Problem of the... month... 11-04-12
Post by: Dan on April 19, 2012, 04:57:12 AM
Two. Sorry, I should have included that information from the start.
Title: Re: Problem of the... month... 11-04-12
Post by: Honclbrif on April 22, 2012, 02:52:56 PM
Sorry I haven't been around in a while.

With 2 equivalents of NH2OH, does it do something like this?

(http://www.majhost.com/gallery/poorjon/misc/potm_11_4_12b.gif)
Title: Re: Problem of the... month... 11-04-12
Post by: Dan on April 23, 2012, 05:39:37 AM
Yes! Nice one. The only thing I'd say is that the oxime closer to the chain terminus will probably be E rather than Z.

Only two steps to go....
Title: Re: Problem of the... month... 11-04-12
Post by: camptzak on July 21, 2012, 02:24:02 AM
The sodium borohydride reduces the imines to sodium amides

The sodium amides take a proton from methanol?
Title: Re: Problem of the... month... 11-04-12
Post by: Dan on July 21, 2012, 05:35:55 AM
There is no amide functionality present, we have two oximes. Can you draw what you mean?
Title: Re: Problem of the... month... 11-04-12
Post by: camptzak on July 23, 2012, 01:20:19 AM
I am guessing that the oxime reacts with the sodium borohydride to form a primary amine. I cant seem to make the arrow pushing work though.

I have the sodium borohydride attacking the ketone, then the negative charge ends up on the nitrogen. It stands to reason (at least in my head it does ;D) that the negative charge would move over to the oxygen to form sodium hydroxide.

Does the boron accept the electron pair from nitrogen, charging the boron?
Then perhaps the nitrogens lumo could be attacked by a hydride again..


heres a really bad drawing to accompany my idea
Title: Re: Problem of the... month... 11-04-12
Post by: Dan on July 23, 2012, 02:46:37 AM
I am guessing that the oxime reacts with the sodium borohydride to form a primary amine. I cant seem to make the arrow pushing work though.

So you are suggesting: oxime  :rarrow: hydroxylamine  :rarrow: amine by the action of sodium borohydride?

No, as far as I know boron hydrides do not generally reduce hydroxylamines to amines, and do not in this case either.

Quote
I have the sodium borohydride attacking the ketone, then the negative charge ends up on the nitrogen.

Not ketone, oxime. You need to use correct terminology, nomenclature exists to help chemists communicate.

Right idea to start with. Now you have generated a nucleophile, what might that do? (also don't ignore the ZrCl4).
Title: Re: Problem of the... month... 11-04-12 [Still unsolved!]
Post by: yesway on August 27, 2012, 06:57:36 PM
Hello,

after nucleophile formation due to hydride attack, I would suggest formation of a N-hydroxyl piperidine moiety by loss of hydroxylamine (which is followed by another hydride attack on activated cyclic "oxime" that formed after expelling the hydroxylamine). I only can guess what ZrCl4 does: 1. Lewis acid? 2. Chelating Oximes via their hydroxyl group? 3. Both?  When there is activation by ZrCl4 I suppose overreduction to primary amines could be possible as a side product. Can that be? I also wonder if there is any stereochemistry involved in the first/second hydride attack?

For the second step, I'm guessing Zn/AcOH reduces the hydroxylamine to an amine (although I have never seen this reduction but I believe so simply because I dont see any other unprot. fct. group).

Best
Title: Re: Problem of the... month... 11-04-12 [Still unsolved!]
Post by: Dan on August 30, 2012, 07:09:34 AM
after nucleophile formation due to hydride attack, I would suggest formation of a N-hydroxyl piperidine moiety by loss of hydroxylamine (which is followed by another hydride attack on activated cyclic "oxime" that formed after expelling the hydroxylamine).

Well done, that is correct.

Quote
I only can guess what ZrCl4 does: 1. Lewis acid? 2. Chelating Oximes via their hydroxyl group? 3. Both? 

Yes. Following reduction of one of the oximes to a hydroxylamine, the presence of Lewis acidic Zr(IV) species facilitates condensation of the hydroxylamine with the remaining oxime, forming a nitrone, which is reduced by hydride to the corresponding N-hydroxypiperidine.

Quote
For the second step, I'm guessing Zn/AcOH reduces the hydroxylamine to an amine (although I have never seen this reduction but I believe so simply because I dont see any other unprot. fct. group).

Yes, Zn/AcOH reduces the N-O bond.

Quote
I also wonder if there is any stereochemistry involved in the first/second hydride attack?

Me too. The yield for this sequence (borohydride/ZrCl4, then Zn/AcOH, followed by [not shown here] TFAA protection) is 32% for the desired stereoisomer. It is claimed that this is the major product, but product ratios are not given in the paper.

Chemistry is from Mander's synthesis of Galbulimima alkaloid GB13: J. Am. Chem. Soc. 2003, 125, 2400. DOI: 10.1021/ja029725o (http://dx.doi.org/10.1021/ja029725o)