[Markovnikov]

I was supposed to do a Corey-Chaykowsky epoxidation of alpha-tetralone, but due to the extensive heating I did (170^oC), the product turned into 1-napthaldehyde. All the NMR-data strengthens the formation of this. I'm just wondering how?

The epoxidation is probably opened forming a terminal alcohol and a benzyllic and tertiary carbocation. I'm guessing that what happens next is a deprotonation of the hydrogen at 2-position forming a conjugated double bond. But what happens next?

[azmanam]

NaH OXIDATION!!! WOO HOO!!!

http://totallysynthetic.com/blog/?p=1903

j/k

forming a terminal alcohol and a benzyllic and tertiary carbocation

Not under strongly basic conditions, you won't form a carbocation.  Perhaps epoxide forms, then deprotonation at the methylene next to the epoxide gives you the first double bond of the ring on the right. Off hand, I don't know what happens after that.  Perhaps advantageous oxygen adsorbed on to your sodium oxidized some things. 

Also, your resonance structure is incorrect.  The ylide is fine, the S=C double bond (with the S=O double bond as well) is no good.  You'd need to kick those electrons up on to oxygen to form O-.  Or just leave it as the ylide, that's probably good enough

[Markovnikov]

Ah! Thanks for pointing out the mistake I made in the resonance and about the carbocation.

I'm not familiar with the different things that could occur during high temperature. But could the heat trigger the deprotonation of one of the hydrogens to form the last conjugation to form a fully aromatic system? Or is there a way for I^- to substitute (somehow?) and then eliminate in that sense?

From reading most of the comments on that blogpost it seems that NaH as an oxidant is farfetched, and they somehow suggest the formation of Na₂O₂ as oxidant ().

[azmanam]

From reading most of the comments on that blogpost it seems that NaH as an oxidant is farfetched, and they somehow suggest the formation of Na2O2 as oxidant

yes, that part was a joke.  I really don't think (and neither do they) that NaH is the active oxidant.  Rather, in that particular case, it's likely autoxidation of benzylic alcohols to benzaldehydes and phenones by oxygen adsorbed onto sodium.  That pathway is known.  I don't know if it's active in your case, because you don't have a benzyllic alcohol, but you may still have adsorbed oxygen in your reaction...

I'm not familiar with the different things that could occur during high temperature. But could the heat trigger the deprotonation of one of the hydrogens to form the last conjugation to form a fully aromatic system? Or is there a way for I- to substitute (somehow?) and then eliminate in that sense?

Think of possible mechanisms as traversing a mountain range.  to get from one side of the range (SMs) to the other side of the range (pdts), you could theoretically go any number of ways.  You could form a radical in the first step, and do a bunch of radical reactions.  That's not likely, because those are high energy steps.  That's like saying you could go across the mountain range by walking over each and every mountain peak along the way.  It'd take a lot of energy to reach the summit of every mountain, but you could do it.  Or, you could walk along the valley the whole way and take the lowest energy pathway.  This is what normal reactions do.  The mechanisms shown in class are the lowest energy possiblities.  At normal reaction conditions (normal operating temperatures, no heat or light), the reactions walk along the valley of the possible reaction pathways.  The reason the more exotic, peak traveling pathways don't occur is that there's simply not enough energy under normal reaction conditions to get over those high energy barriers.

However, when you start heating the reactions to high temperatures (and 170 counts as very high), you start having enough energy available to get over those more exotic peaks.  Plus, you have the pleasure of making a rock stable naphthalene system, so your products are energy sinks.  They're so low in energy - they're so stable - that their formation is not surprising.

That said, I still don't know what the answer is, but those are my thoughts.

For more reading, you might try here:
http://www.chemistry-blog.com/2009/05/13/survivor-mechanisms/

[milo2112]

It has been my experience that 1) epoxide readily open at high temperatures to give the corresponding aldehyde and 2) air oxidation of benzylic systems is pretty facile at high temperatures. Oxygen is actually a pretty good oxidant.

[Markovnikov]

I was wondering if there's some additional stability for these spiro-structures compared to regular epoxides?

The article I was following specifically stated that the reaction mixture was heated under reflux...

[jinclean]

Wow! That is amazing!the NaH can react as a oxdant?!?!but what's the mechanism to this reaction?

[orgopete]

While not stated, the high temperature, since DMSO is a by-product, I am presuming the reaction is being heated in DMSO.  I think the DMSO results in a Kornblum oxidation, then a dehydration of the alcohol, and then aromatization of the dihydronaphthalene would give naphthaldehyde. While I haven't seen a DMSO oxidation of an epoxide, I also haven't looked for one either.

[Markovnikov]

Hmm, orgopete, I'm not at all familiar with the Kornblum oxidation, but the information of it I found, doesn't allow a proton to be deprotonated. If I've misunderstood the mechanism completely, could you perhaps explain in more detail how you imagine it being?

[orgopete]

This is how I imagined the formation of naphthaldehyde from the epoxide. If you need a mechanism also, you will need to ask.

[Markovnikov]

Thanks for the mechanism orgopete.

According to your mechanism, a side-product formed is DMS. While evaporating (our rotary evaporators aren't under fumehoods ) the smell of the room started smelling like old beets. Would that be DMS or does DMSO smell like that? Would it be a plausible indicator of this oxidation taking place?

[Heory]

Very good idea, orgopete !