10
$\begingroup$

Question What is the product of this reaction?
enter image description here

The answer given:
enter image description here

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?

$\endgroup$
3
  • 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 '15 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 '15 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 '15 at 21:15
8
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ 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 '15 at 14:41
2
$\begingroup$

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.

$\endgroup$

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.