0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ It's rather called hydrazide. They're stronger bases then respective alkoxides (and OH-). $\endgroup$ – Mithoron Nov 22 '18 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 '18 at 23:11
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.