In the Hofmann Rearrangement Reaction (in basic medium), when $\ce{RNH-}$ ion is produced, why does it not attack $\ce{Br2}$ or take $\ce{H+}$from the amide itself? Also, why does it take $\ce{H+}$ from water only?
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$\begingroup$ How much water is there and how much of everything else is there? $\endgroup$– jerepierreJul 17, 2017 at 16:32
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$\begingroup$ @jerepierre, we have taken bromine in excess and 1 equivalent of water. $\endgroup$– AyushmaanJul 17, 2017 at 16:45
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1$\begingroup$ It's very unlikely that there's only one equivalent of water... $\endgroup$– jerepierreJul 17, 2017 at 16:51
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3$\begingroup$ @Ayushmaan Wait, wait, wait. There are 3 equivalents of sodium hydroxide and no water? Is it just a solid sitting at the bottom of the flask? $\endgroup$– ZheJul 17, 2017 at 17:45
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2$\begingroup$ Most of the steps in the mechanism are reversible, especially the proton transfers. So, the exact source of a proton doesn't matter, as that proton might have been exchanged with other protons many times before. Except for the actual rearrangement (i.e. the isocyanate forming step), most are reversible steps. $\endgroup$– Eashaan GodboleJul 17, 2017 at 17:50
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1 Answer
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Water is present in large excess when compared to the amide, so it is statistically not likely that the proton would be removed from the amide rather then water. Even if it happens, proton exchange is reversible so it would be formed/unformed many times that eventually the removal of the proton by water would happen and proceed with the mechanism.
Regarding the attack to $\ce{Br2}$, I believe the reason it doesn't happen is that the ion would be surrounded by many many water molecules because: 1) the amide forms H-bonds with water, 2) the ion is solvated efficiently by water, so more molecules would be around to have their protons removed.
