# Why is in the reaction of cyclic unsaturated aldehydes with hydrazine and hydrogen peroxide the aldehyde preserved?

Question What is the product of this reaction?

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?

• 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: $$\ce{H2N2 + H2O2 -> N2 + 2H2O}$$ May 28 '15 at 10:22
• en.wikipedia.org/wiki/Reductions_with_diimide However, peroxide and hydrazine is not listed as a common method of preparing diimide in situ.
– bon
Jun 3 '15 at 20:13
• Sanofi have demonstrated this reduction method at large scale on artemisinic acid and it's not without its calorimetric challenges. Jun 4 '15 at 21:15

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.
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.