[Dan]

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

[Honclbrif]

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?

[Dan]

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.

[stewie griffin]

I think you can get a Wharton reaction with the hydrazone to give an allylic alcohol.

[Dan]

Very close with the Wharton! You'd get a Wharton with hydrazine, but you get another related reaction with sulfonylhydrazine.

[stewie griffin]

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..

[Honclbrif]

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.

[Dan]

There's the Eschenmoser fragmentation,

Correct, it is an Eschenmoser fragmentation

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:

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.

[Honclbrif]

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

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.

[stewie griffin]

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.

[Dan]

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) 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...

[Honclbrif]

In step v, how many equivalents of NH2OH are added?

[Dan]

Two. Sorry, I should have included that information from the start.

[Honclbrif]

Sorry I haven't been around in a while.

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

[Dan]

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....