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Chalcone and sodium methoxide in methanol

I'm a bit lost on how to do this, for Product C I can see two things happening:

1) The $\ce{-OCH3}$ attacks the alpha hydrogen, but that doesn't really make sense as then you would get two double bonds next to one another

2) The $\ce{-OCH3}$ attacks the electrophilic carbon, but then we know the negatively charged oxygen cannot collapse back because then the $\ce{-OCH3}$ would leave. So, would it just get printed by $\ce{HOCH3}?$ And if it does that how would it react with the second set of reactions?

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  • $\begingroup$ $\ce{-OCH3}$ will attach on $\beta$ carbon, then keto-enol tautomerization would occur. Finally bromination at $\alpha$ carbon will be there. $\endgroup$ – Zenix Apr 3 '20 at 9:27
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The first step is called Michael addition. Have a look at this wikipedia page.

The Michael reaction or Michael addition is the nucleophilic addition of a carbanion or another nucleophile to an α,β-unsaturated carbonyl compound. A newer definition, proposed by Kohler, is the 1,4-addition of a doubly stabilized carbon nucleophile to an α,β-unsaturated carbonyl compound.

Of course, in your case, the nucleophile is $\ce{OCH3-}$. Perform 1,4-addition by adding it to the double bond at the beta position (with respect to the carbonyl), and you'll expect an enolate ion, which looks like this -

enter image description here

Finally protonate the $\ce{O-}$ and tautomerise to get your product as @Zenix already stated.

For the second step, check out this link.

enter image description here

References: Michael reaction, wikipedia || Ketone halogenation, wikipedia || Enol, wikipedia

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  • $\begingroup$ So for question 6 would I just be getting the same product again? $\endgroup$ – user690808 Apr 3 '20 at 21:51
  • $\begingroup$ @user690808 Nope, I'm not sure if you got the idea. For question 6, the answer would be 3-methoxy-1,3-phenyl-propan-1-one. $\endgroup$ – Nikhil Anand Apr 4 '20 at 3:04
  • $\begingroup$ Im still a bit confused could you draw a mechanism of the first reaction for me. $\endgroup$ – user690808 Apr 4 '20 at 3:08

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