Question

${\mathbf{FADH}}_{\mathbf{2}}$reduces $\mathbf{\alpha }\mathbf{,}\mathbf{\beta }\mathbf{-}$unsaturated thioesters to saturated thioesters. The reaction is thought to take place by a mechanism that involves radicals. Propose a mechanism for this reaction.

Short answer

Here, unsaturated thioester is reduced to saturated thioester using ${\mathrm{FADH}}_2$, i.e. dihydroflavine-adenine-dinucleotide.

Step-by-step solution

Reaction of FADH2 with unsaturated thioester

Initially, electron shifting occurs in ${\mathrm{FADH}}_2$and forms positive and negative charge on nitrogen and oxygen respectively.Similarly, electron shifting occurs in thioester and forms positive and negative charges on carbon and oxygen respectively. Both of them reacts together to form FADH radical and thioester radical by losing an electron and protonfrom ${\mathrm{FADH}}_2$and gaining electron and proton from thioester respectively. This is represented below:

Step 2: Tautomerization of thioester radical

Protons are transferred from one position to another through tautomerization. Through tautomerization, thioester radical in enol forms a carbonyl group. While FADH radical loses an electron and a proton to form FAD.Then the thioester radical is converted to saturated thioesters by gaining an electron and proton.

Thus the final products are FAD and saturated thioester.

The whole mechanism is represented below: