# Reaction mechanism of aqueous ammonia with 3,4-dibromopyridine

I am struggling on how this mechanism would occur. I am aware that the bromine will be removed by for some reason I cannot figure out the $\ce{NH3}$ attaches onto the ring, can anyone prove me with any help?

I have also had a feeling that deprotonation occurs and a benzyne is formed, in which a bromine gets kicked off and one of the $\ce{:NH3}$ attaches onto the ring. Is that correct?

The reaction between ammonia and 3,4-dibromopyridine is a nucleophilic aromatic substition reaction (SNAr). This is possible because the ring is rather deactivated due to the presence of the electron withdrawing nitrogen atom of the pyridine drawing electron density towards it. The ammonia attacks carbon at the 4-position preferably, rather than the carbon at the 3-position as this position is more deactivated than the latter. This attack results in the breakage of the double bond, resulting in a loss of aromaticity. This results in the formation of a carbanionic Meisenheimer intermediate. The bromine at the 4-position then leaves and the double bond reforms, resulting in the restoration of aromaticity in the ring. This results in a net substitution of $\ce {-Br}$ with an $\ce {-NH2}$ group.