I'm studying Wolff-Kishner reduction and I don't understand why after this step enter image description here 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.

  • $\begingroup$ It's rather called hydrazide. They're stronger bases then respective alkoxides (and OH-). $\endgroup$
    – Mithoron
    Nov 22, 2018 at 22:31
  • 1
    $\begingroup$ 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]. ;) $\endgroup$
    – user55119
    Nov 22, 2018 at 23:11

1 Answer 1


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).

![enter image description here

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.


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