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This reaction is mentioned as incorrect in my exercise book. Why is this so? My guess was that the Magnesium, instead of forming a grignard reagent with the compound and bromine, would instead form a complex with the N atom. However, wouldn't the complex formed be quite unstable due to the size of the ligand and the steric hindrance associated with it?
Is this reaction incorrect? If so, what reaction would take place here?
Product would not form as intended, not proceeding in the path of synthesis usually seen with such a sequence of reagents.
This is due to the (quite) acidic hydrogen atom attached to the heteroatom nitrogen in the ring, which would be readily abstracted by the dense electron cloud on the initial Grignard complex that would form.
The species drawn are highly unstable and would have a really short half-life.
The nitrogen could be protonated subsequently. The desired product simply will not form because a reaction much faster than it will consume the formed reagent instantly.
The Grignard reagent fails to remain in a synthetically useful form.
I would just like to throw in some more perspective here, although your main question has been answered.
First, let's see a postulated mechanism of formation of Grignard reagent in the presence of Mg and diethyl ether.
The tentative problems that would may arise by the presence of the N-H bond would be:
1)The R radical made in the first step might begin a homolysis reaction with the H of the N-H, and thereby forming a kind of amino radical radical. This may lead to further erroneous coupling in the synthesis.
2) Like the answer and comments above state, the major reason for the failure of this synthesis is the reaction of the highly basic Grignard reagent with the acidic H after formation.
3) Non-polar solvents like diethyl ether stabilize the Grignard reagent formed like so:
The N-H bond will have significant polarity due to difference in electronegativities of N and H, so a lot of ether will be expended in stabilizing this bond, so again, it might hinder the formation of the Grignard reagent
Note: This answer is more like food-for-thought on this issue as far as point 1 and 3 are considered, the clear-cut answer is in point 2.
P.S. As pointed out in the commnts by Mithoron, the reaction happens in a heterogenous state, so the mechanism cited above and all these arguments are highly simplified ways of looking at a picture perhaps not completely understood. This reaction probably bears more similarity with that of sodium and an alcohol
