In reactions with Grignard reagents, I've always seen the alkyl group being the nucleophilic component, with the R–MgBr bond being broken. This is always rationalised by the fact that the alkyl carbon has a partial negative charge, whereas the electrophile (e.g. a carbonyl group) has a partial positive charge.

However, the halogen atom should also have a partial negative charge. Why doesn't it attack the electrophile instead, breaking the RMg–Br bond?

  • 4
    $\begingroup$ And if bromide did add to a carbonyl group? It is reversible step. $\endgroup$
    – user55119
    Jul 25, 2020 at 18:40
  • $\begingroup$ but isn't addition of iodine the only halogen addition that is unstable? $\endgroup$ Jul 25, 2020 at 19:13
  • $\begingroup$ You may be thinking about addition of iodine to a CC double bond. The species R2C(OH)X where X=halogen. Moreover, the species you propose has a negative charge on oxygen, which is a worse case scenario. Although cyanohydrins are stable when formed, they are reversed by base. $\endgroup$
    – user55119
    Jul 29, 2020 at 20:50

1 Answer 1


In all likelihood the halide ion does attack the substrate. But, the product such a reaction mode would form does not accumulate whereas the alkylated product does.

Recall that in a nucleophilic reaction halide ions are good leaving groups, at least if you stay away from fluorides which are rarely used for Grignard reagents. So while the halide ion may add to the substrate, it may also be displaced if an alkyl-anion moiety from another Grignard molecule comes in, or possibly the halide ion may just launch on its own to attach to a magnesium atom (which, in a Grignard reagent, acts as an electrophilic center, for example by adding to the oxygen end of a carbonyl group).

In contrast, when the alkyl-anion moiety attacks, the carbon-carbon bond it forms is strongly favored thermodynamically and a lot harder to break; here we do not have a good leaving group. So the alkylated product rather than the halogenated product is what accumulates.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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