# Give the action of the Clemmensen reduction on 2-butenal?

Mar 9, 2016

CLEMMENSEN REDUCTION

The Clemmensen Reduction involves adding $\text{Zn(Hg)}$ dissolved in heated $\text{HCl}$ to something reducible. Note though that this process can accidentally chlorinate a double bond also present on the reactant.

The principal action of the Clemmensen Reduction is to reduce a -stackrel("O")stackrel(||)"C"- to a $- {\text{CH}}_{2} -$ or -stackrel("O")stackrel(||)"C"-"H" to a $- {\text{CH}}_{3}$.

So, using it on 2-butenal is supposed to turn it into 2-butene.

But when you have $\text{HCl}$, a strong acid, in the presence of a double bond, there's a good chance it'll protonate the double bond. At that point a hydrochlorination would happen across the double bond, forming an alkyl chloride.

That's why if you wanted to remove the carbonyl oxygen, you should do a similar reduction in basic media, and the Wolff-Kischner reduction qualifies.

WOLFF-KISHNER REDUCTION

The basic counterpart to the acidic Clemmensen reduction is the Wolff-Kishner reduction, which is run in base. It aims to accomplish the same thing, but in different, sometimes more favorable conditions.

The general reactants are hydrazine (${\text{H"_2"N"-"NH}}_{2}$) and a strong base, like $\text{NaOH}$.

WOLFF-KISHNER VS. CLEMMENSEN

The main benefit of the Wolff-Kishner reduction is that because it's in base, it wouldn't accidentally protonate a hydroxyl group or double bond in the reactant and allow side reactions to occur for acid-sensitive reagents.

Confer with a reaction involving a hydroxyaldehyde reactant (Organic Chemistry, Bruice, Ch. 15.18), and you would see the following conundrum is resolved by choosing one reduction method over the other: