# Why doesn't the N-Br bond dissociate quickly in NBS?

NBS is used to provide a constant, small amount of $\ce{Br2}$ which then dissociates to give $\ce{Br^.}$ free radical. Why doesn't the $\ce{N-Br}$ bond simply dissociate to give $\ce{Br}$? This dissociation should be just as fast or faster because the free radical formed would be stabilized by delocalization over the two oxygen atoms.

• It does, and that's why it is used in the first place. Aug 22 '17 at 8:54

Actually, the sequence of reactions leading to a low but constant concentration of $\ce{Br^.}$ radicals is very simple:

\begin{align}\ce{R2N-Br &<<=> R2N^. + Br^.}\tag{1}\\[0.4em] \ce{Br^. + Br-NR2 &<=> Br2 + R2N^.}\tag{2}\\[0.4em] \ce{Br2 &<<=> 2 Br^.}\tag{3}\end{align}

The $\ce{N-Br}$ bond is very weak and can easily dissociate homolytically to give both bromine radicals and succinimide radicals. Bromine radicals can combine with other NBS molecules to form free bromine and another succinimide radical and bromine molecules can dissociate to give bromine radicals.

All of these steps are reversible and the equilibrium favours the non-radicals side. Whether your reaction needs a constant supply of bromine or a constant supply of radicals doesn’t matter as NBS can supply them.

Only if there is already some bromide in solution can a fourth reaction take place:

$$\ce{R2N-Br + Br- -> R2N- + Br2}\tag{4}$$

This is the only reaction that has a dedicated forward-direction. If bromide is present, bromine molecules can be generated much more rapidly than radicals, so the generation of radicals will have to come from bromine dissociation.

While it is true that the NBS radical is somewhat well stabilised, that does not make it that much easier for the $\ce{N-Br}$ bond to dissociate: radicals are still radicals and two-electron-two-centre bonds are still energetically favoured. The $\mathrm{S_N2}$ displacement is much better since an anion is equally well stabilised but has a lower general energy.