Question What is the product of this reaction?
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The answer given:
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At first, I thought this was a typical Wolff-Kishner reduction, but then I saw the reagents didn't match up (no $\ce{OH-}$).

How is this getting formed? Shouldn't the $\ce{CHO}$ group be reduced? What am I missing here?

  • 3
    $\begingroup$ This reaction worries me: mostly because hydrazine is known to be explosive (via oxidation to $\ce{N2(g)}$) in the presence of strong oxidants and hydrogen peroxide is not the most stable molecule either: $$\ce{H2N2 + H2O2 -> N2 + 2H2O}$$ $\endgroup$
    – Ben Norris
    May 28, 2015 at 10:22
  • $\begingroup$ en.wikipedia.org/wiki/Reductions_with_diimide However, peroxide and hydrazine is not listed as a common method of preparing diimide in situ. $\endgroup$
    – bon
    Jun 3, 2015 at 20:13
  • $\begingroup$ Sanofi have demonstrated this reduction method at large scale on artemisinic acid and it's not without its calorimetric challenges. $\endgroup$
    – Beerhunter
    Jun 4, 2015 at 21:15

2 Answers 2


The reaction of $\ce{H2O2}$ with hydrazine forms diazene, $\ce{HN=NH}$. This reacts in a syn fashion to deliver both hydrogens to the same side of the alkene.

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    $\begingroup$ Thanks, I can relate this question to our course now - diimide indeed selectively reduces C-C double bonds in preference to polar bonds. $\endgroup$ Sep 16, 2015 at 14:41

Actually it's a di-imide which itself gets oxidised to give up nascent hydrogen and reduce the other reactant. Thus, when you react it with an alkene , hydrogen add on the same side i.e. syn addition. Unlike $\ce{N2H4/KOH}$, it is not strong enough to reduce carbonyl groups. Therefore you get a different product.


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