# Isn't the reaction of chloral with sodium hydroxide a disproportionation reaction?

Chloral undergoes a haloform reaction with sodium hydroxide. Now the carbon atoms in chloral are in the $\mathrm{+III}$ and $\mathrm{+I}$ oxidation state. Chloroform’s carbon is in the $\mathrm{+II}$ oxidation state, as is sodium formiate’s. Now, carbon has both been reduced and oxidised. Then why is it not a disproportionation reaction ?

• Who said it wasn't? o.o Anyway, it's not always useful to classify organic reactions into classes like disproportionation etc. But if you really want to, yes you could probably still call it a disproportionation. – orthocresol May 1 '16 at 16:22
• OK! It's given in my text that it isn't a disproportionation reaction. – TESLA____ May 1 '16 at 16:28
• I’m not sure if that was intended or a slip of the pen: If anything, the reaction would be called a comproportionation. – Jan May 1 '16 at 17:00
• Oh, true true, it's a comproportionation not a dis... My bad – orthocresol May 1 '16 at 17:42
• – Mithoron May 2 '16 at 16:30

Disproportionations are defined such that you start at two species of the same type and oxidation state (e.g. two formal $\ce{Cr^4+}$ species) and end up with two different species (e.g. a formal $\ce{Cr^3+}$ and a formal $\ce{Cr^6+}$ species). This is not the case here, the oxidation states start out differently.
Comproportionations are defined oppositely: You start at two different species and create two of the same. But in no way can chloroform and sodium formiate be called ‘the same’, even though their oxidation states are. Carbon monoxide (also $\mathrm{+II}$) is not the same as either, either.