# Product of the reaction between 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?

• $\ce{-OCH3}$ will attach on $\beta$ carbon, then keto-enol tautomerization would occur. Finally bromination at $\alpha$ carbon will be there. – Zenix Apr 3 '20 at 9:27

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