# Which one is the stronger base?

I'm studying Wolff-Kishner reduction and I don't understand why after this step the leaving group is $$\ce{OH-}$$ instead of $$\ce{^-HN-NH2}$$. I guess it's because hydroxide ion is less strong than the other although it would be useful to know if i can calculate pkb of diazanide (namely $$\ce{^-NH-NH2}$$) starting from data on hydrazine. Also I'm considering pkb of $$\ce{OH-}$$ equal to $$0$$ because it's acting as a base.

• It's rather called hydrazide. They're stronger bases then respective alkoxides (and OH-). – Mithoron Nov 22 '18 at 22:31
• A hydrazide is a derivative of a carboxylic acid [RC(=O)NHNH2]. A hydrazone is a derivative of a ketone [R2C=NNH2] or aldehyde [RHC=NNH2]. ;) – user55119 Nov 22 '18 at 23:11

## 1 Answer

Mirko: The mechanism you present for the early stages of the Wolff-Kishner Reduction includes hydroxide necessary for the decomposition of the yet to be formed hydrazone 6 of the ketone 1 (or aldehyde). One may preform the hydrazone and then do the actual reduction in the presence of hydroxide in a high boiling hydroxylic solvent (Huang-Minlon variation1).

The mechanism for hydrazone formation is a reversible reaction and is often conducted by removal of water to shift the equilibrium to the right.2 Realizing that most commercial hydrazine contains 15% water, it can act as the agent for proton exchange in adduct 3. One need only use the electron pair in structure 4 to eject hydroxide contrary to the elimination proposed in the last step of your mechanism. There is no need to be concerned about the anion of hydrazine as a leaving group. Hydroxide only has to deprotonate the acid 5.

1) Huang-Minlon, J. Am. Chem. Soc., 1946, 68, 2487.
2) Hydrazones and oximes are not as susceptible to hydrolysis as are imines owing to the resonance stabilization provided by the second heteroatom.