At first glance, this looks like some kind of reductive amination reaction. However, on closer inspection, you can see that there is one less carbon in the product than the starting material. Furthermore, there are no reducing agents present, only acid (and presumably water). Of course the obvious starting point is to react the secondary amine with the aldehyde to form a cyclic imminium structure. Once generated, this is nicely set up to undergo a [3,3]-sigmatropic rearrangement to transfer an allyl group to the imminium carbon. The resulting formaldehyde imminium product is then hydrolyzed in the presence of water to afford the product plus an equivalent of formaldehyde. The full mechanism is shown below. Notice I am a stickler for showing every proton transfer step! No shortcuts here.
Tynchtyk says this problem appeared in a science olypiad for High School Students in Moscow. I wish our high school education here in the states was up to this kind of challenge.Thanks, Tynchtyk, nice problem! In the spirit of problem solving, let me pose a new challenge. This is one of my favorite transformations.
11 comments:
Bravo
good hint to label the post "sigmatropic rearrangement" ;)
btw, good luck with the blog and please keep posting interesting transformations!
I wish they covered organic chemistry in high school in the US. Then again, that would probably lead to more people getting into med school. Wouldn't want that, now would we?
Is there a 4-membered cycle intermediate?
Ψ*Ψ: more chemistry in high school -> more people flee to med school?
But what's the yield? ;)
Nope, no 4-membered rings. The yield is generally good. If I recall it is in the 80-90% range.
Actually, I was one of the lucky kids who did have a year of organic chemistry in high school. That is the sole reason I am a chemist today. General chemistry bored me to tears.
We also had organic chemistry in school.
Things like this won't eliminate on their own. The leaving group, the nitrogen is quarternary, with acyl on it, and the is gonna stay on it.
If you have not isolated a 4-membered ring intermediate it does not mean that it is not in the reaction path. A better chemist of Chuck Norris would catch him.
zwitter-ion reacted acetic anhydride to form a mixed anhydride and acetate anion(base), retro-Michael and amidation gave the product.
Congrats Anonymous! I just posted the answer and I see you have solved it.
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