Birch reduction of 1,2-Cyclopentanedione and Diketones This answer given to this problem was that only one of the double bonds is reduced into an alcohol. I
think i understand how the double bond is reduced into an alcohol. An electron is transferred to the carbonyl carbon and the double bond cleaves homolytically, forming a carbanion (stabilised by resonance ) and the oxygen with an extra electron which then will gain another electron and both get hydrogens to form the alcohol.
First of all is my method right? I don't think so as both the double bonds would have been reduced into alcohols. Also i was unable to see any sources for the birch reduction of di-ketones, I was only able to find it for unsaturated ketones.
The course of the reaction depends on whether or not a proton source other than ammonia is present. In the absence of an alcohol (ethanol, tert-butanol) the α-dione 1 adds an electron from sodium to give a resonance stabilized radical anion 2. A second electron is added to afford enediolate 3. [This is the same species formed with sodium in the acyloin condensation on diethyl glutarate.]  Water workup provides the α-hydroxyketone 4. In the presence of alcohol, enolate 3 is protonated to form 5. [The reason I am reluctant to protonate the alkoxide of 5 is to avoid reductive elimination of hydroxide from 4 to leave the enolate of cyclopentanone. Further protonation/reduction may lead to the alkoxide of cyclopentanol.]  One-electron reduction of 5 yields ketyl 6, which via protonation and a second reduction leads to dialkoxide 8. Workup affords the diol 9. 
The bottom line is if you want to form 4 by metal/ammonia reduction, avoid alcohol.