[sharbeldam]

I really feel bad for asking a lot, this is a synthesis problem for one of my student, my solution was long and I used O-chem 1 reaction (hydroboration), is there a shorter/better way? if not, is my solution correct? I feel like it's very messy.

Question and solution attached

Thanks!

[rolnor]

I think its a strange way to make this compound. I see many, very bad steps in your solution but also this is not so easy.

[sharbeldam]

Can you elaborate? this is a synthesis for college not lab.

[rolnor]

The E2-elimination will be difficult with an acid bromide, I can not see how this should work. The step where you react with hydrazine you will get Michael addition to the unsaturated ketone.
I am thinking you could try to protect the ketone starting material as 1,3-dithiane, use LDA on the ester and methylate, then make a weinreb amide and react that with phenyllithium. Then you make the reduction of the non-protected carbonyl with hydrazine followed by deprotection of the dithiane with mercuric chloride or NBS/2,6-lutidine.

[sharbeldam]

Thank you rolnor, I appreciate it, some reactions that you mentioned are not in syllabus and can not be used unfortunately like the usage of weinreb amide but very good way!
also , are you saying that acid bromide is senstive to heat like carboxylic acid? it's not possible to do an e2 with acid bromides/chlorides for what reason?
And Which enolate will attack the alpha beta unsaturated ketone in a michael reaction? i have no alpha hydrogens
And thank you again for your correction.

[rolnor]

The acid bromide will be forming a ketene with base.
You will get attack by hydrazine on the double-bond in a Michael-type reaction.

[sharbeldam]

Thanks, can you direct me to this reaction? (ketene forming with a base?)

[rolnor]

Thanks, can you direct me to this reaction? (ketene forming with a base?)

Here is a page;

https://en.m.wikipedia.org/wiki/Ketene

[sharbeldam]

I get it! Thank you.
If you let me ask an unrelated question (part b)

There is this synthesis problem done by Prof Stuart, I understood that the hemiacetal is with equilibrium with the aldehyde, so the aldehyde can undergo wittig with the pph3, but my question is how did the CH2Ph group on the second carbon turned into Obn, how did that Oxygen stuck itself there?

Everyday i think i know it all, but every day i realize im just more ignorant, so many reactions

[rolnor]

I get it! Thank you.
If you let me ask an unrelated question (part b)

There is this synthesis problem done by Prof Stuart, I understood that the hemiacetal is with equilibrium with the aldehyde, so the aldehyde can undergo wittig with the pph3, but my question is how did the CH2Ph group on the second carbon turned into Obn, how did that Oxygen stuck itself there?

Everyday i think i know it all, but every day i realize im just more ignorant, so many reactions

Are you sure this is not just a miss-type? Its very difficult to do this.

[spirochete]

Learn acetoacetic ester synthesis (and its related cousin, malonic ester synthesis). This is a very "by the book" type synthesis problem. There is one clear best way to do it. You would need to be very clever to discover this on your own. Just read about acetoacetic ester synthesis.

[sharbeldam]

Holy...
like this?

[rolnor]

I don understand at all, the question was how to convert a benzyl to a benzyloxy?

[sharbeldam]

I think he answered the original synthesis, about that second question, as you said, hopefully its a mistake, cause i cant figure it out.