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.

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  • $\begingroup$ I think this needs a diagram $\endgroup$ – Waylander Jun 26 '20 at 16:06
  • $\begingroup$ @Waylander I have added it now. $\endgroup$ – booma vijay Jun 26 '20 at 16:23

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.

  • $\begingroup$ Just to clarify. If the alkoxide of 5 is protonated, the OR- group will act as a base and eliminate the -OH group as water which leads to the formation of an alpha-beta unsaturated ketone that leads to an enolate which on further reaction forms the alkoxide of cyclopentanol. But the same can also happpen with h2o right?(3 to 4). Hydroxide acts as a base forms the enolate again? $\endgroup$ – booma vijay Jun 27 '20 at 4:40
  • $\begingroup$ could u pls clear the above doubt? Thanks! $\endgroup$ – booma vijay Jun 28 '20 at 0:38
  • $\begingroup$ If 5 is protonated on oxygen by an alcohol to form 4, then adding an electron to the double bond forms a ketyl (like 6 but protonated on oxygen). This species could eliminate hydroxide and after the addition of a second electron one has the enolate of cyclopentanone, which can protonate with ROH and cyclopentanone can reduce to the alkoxide of cyclopentanol. Aqueous workup gives cyclopentanol. Enediolate 3 is stable to Na/NH3 in the absence of ROH. Aqueous workup gives 4. I'm just showing possible outcomes. $\endgroup$ – user55119 Jun 28 '20 at 0:56
  • $\begingroup$ Your second structure with the electron added to oxygen is my enediolate 3. $\endgroup$ – user55119 Jun 28 '20 at 0:59
  • $\begingroup$ thx a lot. got it now $\endgroup$ – booma vijay Jun 28 '20 at 1:43

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