[Starksays]

Would RNH2 react the same way with cyclohexenone as it would with cyclohexanone? For example, the reaction of cyclohexanone with RNH2 would result in a six-membered ring with a double-bonded nitrogen attached which is attached to the R (ring=N-R).

Would the the double-bound between C2 and C3 make any difference? Or would the result be the same but with a double-bond at C2-C3?

[discodermolide]

I don't think it will make much difference but the cyclohexenone should react faster. You are expecting this product to  form?



Why would this product form?

[Starksays]

Well, I believe that the nitrogen would attack the carbon and the electrons from the double-bond would move to oxygen. Then there would be deprotonation at nitrogen and protonation at oxygen. Lastly water would be heated off.

What I'm unsure of is the problem just had the cyclohexenone reacting with R-NH2 and there was no mention of any other reagents. However, I never saw the reaction like this, I only saw it with cyclohexanone so I was worried there was a trick like what if NHR ends up on one the carbons of the alkene and the ketone portion is left alone?

[discodermolide]

No trick it's just the formation of an imine.

[orgopete]

Unfortuneately, i missed the class this topic was discussed, but I think the product should be the Michael or conjugate addition. Depending on the problem and how it is written, I would expect it to add two times with a primary amine. An imine could form, but it would be reversible. Conjugate addition is also reversible, but protonation of the enolate will make greatly favor the ketone and make the conjugate addition irreversible in this instance. I consider this principle helpful to predict whether a 1,2 or 1,4-addition should occur. 1,2-additions, especially to aldehydes, can occur very rapidly, but they may not be kinetically stable. If so, they may reverse and result in the formation of a slower thermodynamic product.