[Altered State]

I'll explain it with pictures

I need to identify L-O compounds:



First is ester in basic media, so I guess it will undergo a Claisen Autocondensation, and I find two posible products to be L:



But now is where I'm stucked. Next step is protecting carbonyl groups with ethylenglicol. Thats ok for both L posibilities, but then I use LiAlH4 in ether, but I do not think anything would happen. LiAlH4 can't reduce either -OH, ether or alkene. What did I do wrong in the first steps? Or is it correct and LiAlH4/Et2O does something in this situation?

[clarkstill]

Your two isomeric "L's" are correct, and will rapidly interconvert in acid or base.  You need to draw out what you think the product of the ethylene glycol protection will be (hint: esters are much less prone to acetal formation than ketones).  You're unlikely to form an ortho-ester (the ester version of an acetal) under these conditions...

Once you've identified the protected molecule M, the role of the LiAlH4 will become more clear.

[AlphaScent]

The products of the condensation are tautomers.  Once you protect with ketone with the glycol, what reacts readily with LAH? 

What common synthetic technique involves a base, alcohol then a primary alkyl halide?

[Dan]

Next step is protecting carbonyl groups with ethylenglicol. Thats ok

Can you show your structure for M please?

[Altered State]

Next step is protecting carbonyl groups with ethylenglicol. Thats ok

Can you show your structure for M please?

At first I thought about a double acetal formation, and about a ester acetal formation, but now I'm almost sure that M is this guy:



But I found another problem.
I treat compound M with LiAlH4/Et2O, and I get the alcohol. As far as I'm concerned, the acetal doesn't break in this step, so N should be:



But then I treat it with NaH/MeI. These conditions are appropriated for enolate formation and Methylation in alpha to carbonyl position. But it is supposed to be no carbonyl in this stage, so that wont occur.
Let's suppose that acetal does break and we got back the ketone in one of these steps (guess strong base NaH might have done, but I think ketone acetal won't break with no acid media), and that alpha position (most substituted) to carbonyl is methylated, so we have as Ñ:



That might be acceptable. But last step is H+ and heat. And we jus thave a ketone and an alcohol. I don't think that would react anyway...
So I suppose I missed something in some step between L->M to Ñ->Ñ

What do you think?

[AlphaScent]

The ketal is not going to react under the basic conditions.  What else there would react with sodium hydride then react with the PRIMARY alkyl halide with an ALCOHOL present under basic conditions?? 

No enol due to no resonance stabilization..

[discodermolide]

If the structure of N is correct then treatment with NaH/MeI will do a Williamson's ether synthesis. and methylate the alcohol.

[Altered State]

D'oh, I see now, so dumb of me.
Thanks