enter image description here

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?


2 Answers 2


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.

grignard reagent fails to remain in synthetically useful form-reaction mech

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.

  • 1
    $\begingroup$ Your reaction scheme isn't perfect, but, oh well... BTW there's direct reduction of amine hydrogen may be more important reaction then formation and destruction of Grignard reagent. $\endgroup$
    – Mithoron
    Commented Feb 19, 2020 at 16:31
  • $\begingroup$ Well, it is possible that there are lapses in my knowledge (freshman year chemist here), but I'm not sure if piperidine can be reduced any further.. Even a Google search didn't yield anything useful. Could you elaborate? $\endgroup$ Commented Feb 19, 2020 at 16:40
  • 1
    $\begingroup$ It's just like reaction of alcohol with sodium - you get alcoholate and hydrogen. $\endgroup$
    – Mithoron
    Commented Feb 19, 2020 at 16:42
  • $\begingroup$ @Mithoron Ahh.. makes sense. But is the H that acidic (N is additionally in a ring, not a simple amine)? $\endgroup$ Commented Feb 19, 2020 at 16:48
  • $\begingroup$ @WilliamR.Ebenezer For what it's worth,the H of a aliphatic alcohol isn't that acidic either(considering the high basicity of the RO- anion). So qualitatively speaking,I think I would give the H in the reagent here a pass as well for this reaction $\endgroup$ Commented Feb 19, 2020 at 17:02

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.

enter image description here


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:

enter image description here

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

  • $\begingroup$ I found the first point quite satisfying. Good answer! $\endgroup$ Commented Feb 19, 2020 at 14:17
  • 2
    $\begingroup$ @WilliamR.Ebenezer And I didn't. There's no atomic Mg in solution and no radical flying around to react. $\endgroup$
    – Mithoron
    Commented Feb 19, 2020 at 16:15
  • $\begingroup$ @Mithoron I based this answer on the observations given in the quoted article,and the possible side reactions that can occur. From what I saw,I didn't really talk about atomic Mg,but the R• generated does seem like an independent species $\endgroup$ Commented Feb 19, 2020 at 16:35
  • 3
    $\begingroup$ Reaction is heterogeneous - reagent adsorbs on Mg surface. Br, or in this case probably mostly N, coordinates with surface atoms and reduction occurs in bound state. Also @WilliamR.Ebenezer $\endgroup$
    – Mithoron
    Commented Feb 19, 2020 at 16:51
  • 1
    $\begingroup$ It's simplified and schematic, that's unsurprising, because mechanisms of heterogeneous reactions are poorly known. I think nobody knows what exactly happens on the surface there. I personally doubt it's SET. $\endgroup$
    – Mithoron
    Commented Feb 19, 2020 at 17:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.