[physstudent1]

Sorry for all the questions today guys but they decided to put homework on webassign the day it was due without any warning so I only have had today to work on it, compounded with my thermodynamics homework I've been strained for time today...anyway



Draw the skeletal structure of the major organic product A which is the compound of interest. (There is another product that can be considered the by-product which is of no synthetic interest.)
There is one equivalent of base used in step 1.
Assume no alcohol proton exchange with solvent during the acid workup.

I'm thinking that the NaOPh is going to remove a hydrogen from the cyclohexanone and cause an enolate? then the double bond will attack the carbonyl carbon and form a ketone and a O- structure(the adol before the H is on the O- not sure what thats called) I have a feeling this will be like one of the other one's i've done where a good leaving group is formed though But I'm not sure what to do

[azmanam]

remove a hydrogen

it removes a proton.

It's a Claisen condensation reaction.

A better question:  why the weird base (NaOPh)?

[physstudent1]

I see, I don't know why the weird base or why they gave me two questions about this kind of reaction in my homework when it doesn't appear in the book for another 200+ pages

[russellm72]

Dont forget your quenching with deuterated water so where will the D go in your product...

[azmanam]

You will get an undesired side reaction if you use, say, NaOH as the base.  do you see what it might be?

[aldoxime_amine]

I have a doubt also...the poster says 1 equivalent of the base, meaning one equivalent of enolate (without complicating things too much), this means that after the attack of the enolate, the PhO^- (phenoxide) passes in the solution, being a very good L.G., giving us a ketone. So why use the deuterated water....to get deuterated phenol?...seems unlikely...

And suppose i use 2 equivalents of the base, will another enolate attack the above ketone?...everything seems to be too sterically hindered here...the base, the ketone (both of them), the leaving group...

[azmanam]

the PhO- (phenoxide) passes in the solution, being a very good L.G., giving us a ketone. So why use the deuterated water

Not just a ketone... a (very acidic) beta-keto aldehyde.  The reaction mechanism doesn't stop after enoate attack and acyl substitution.  Deuterium quench helps prove that.

[aldoxime_amine]

Oh thanks, i understand the role of the deuterium now...