Reacting Grignard reagent with amides

Question

Propene in the presence of $\ce{HBr}$ and $\ce{H2O2}$ gives A.
A on reaction with $\ce{Mg}$ in dry ether gives B.
B on reaction with acetamide, $\ce{CH3C(O)NH2}$, gives C.

In the above question, I got A as 1-bromopropane and B as propylmagnesium bromide. As for C, I got

$$\ce { CH3-CH2-CH3 + CH3C(=O)NHMgBr}$$

The ratio of moles of the Grignard reagent vs. acetamide is not specified. Should I add an additional propyl Grignard and arrive at $\ce{CH3-CH2-CH3}$, $\ce{CH4}$, and $\ce{ CH3C(OH)N(MgBr)2}$ ?

I need help with figuring out the product in the reaction of B and acetamide.

Answer 1

In the reaction that you have mentioned above, the reaction essentially stops after the reaction has taken place in a 1:1 ratio. That is because the electrophilicity of the polar bond of the acid derivative has reduced sharply so it cannot undergo further reaction.

However, had the nucleophile been $\ce{RLi}$ instead of $\ce{RMgX}$ ($\ce{RLi}$ is more ionic and hence a better nucleophile), further reaction would've been possible, although it provides the product in low yield. R attacks C=O resulting in the formation of $\ce{R-C-O- Li+}$ (The other groups remain as they were- the only difference being that it is $\ce{NHLi}$ instead of $\ce{NHMgX}$)