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Why isn't benzene reduced all the way down to cyclohexane when performing a Birch reaction? (Why is it not reduced further?)

Similarly for the reaction with sodium dissolved in liquid ammonia, why isn't an alkyne reduced fully to an alkane?

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  • $\begingroup$ If you look at the mechanism for the Birch Reduction, you will see that the mechanism wouldn't work with only 2 double bonds. As for the metal reactions, they do go fully to alkanes, unless it is a poisoned catalyst like in lindlars. $\endgroup$ – CognisMantis Aug 28 '15 at 16:54
  • $\begingroup$ Without going in a lot of detail a first order simple answer is that 1,4-cyclohexadiene doesn't give the Birch reaction and so the initial reaction from benzene stops there. Also what is "disolv. metal reaction" other than a Birch reaction? $\endgroup$ – K_P Aug 28 '15 at 20:20
  • $\begingroup$ @K_P Sodium metal in liquid ammonia reduces alkynes to trans alkenes $\endgroup$ – orthocresol Aug 29 '15 at 6:14
  • $\begingroup$ @CognisMantis yes it would. You could add an electron into any of the remaining double bonds (and if they were next to each other, they could even stabilise by resonance). $\endgroup$ – Jan Sep 28 '15 at 22:50
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The mechanism of the Birch reaction includes single-electron transfers from the sodium dissolved in ammonia to the aromatic compound. These transferred electrons will attempt to occupy the lowest possible orbitals.

In the case of benzene, we have a delocalised $\unicode[Times]{x3C0}$-system. Delocalisation can be understood here to mean higher occupied orbitals and lower unoccupied orbitals. Therefore, the electron will find an orbital at a lower energy and it will gladly hop in to create the radical-anion intermediat.

Once this is protonated, the remaining radical system can still be seen as a delocalised pentadienyl radical. Thus, the energy level a new electron will occupy is still rather low and another anion can be created. Further protonation yields the product of the Birch reaction, a cyclohexa-1,4-diene — most importantly, a non-conjugated system. This can be regarded as a pair of isolated double bonds.

The energy of the unoccupied $\unicode[Times]{x3C0}^*$ orbitals in an alkene seems to be just too unfavourable for a free electron to occupy them. I don’t have the facilities to do calculations (and neither am I firm enough to be able to do them) so I can’t provide you with values to compare, but it has to be the case otherwise the reaction would continue. If it were not so, there would be no reason for the reaction to stop.

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The radical anion after the first step of the Birch reduction needs to be conjugated with (=stabilised by) another pi-system, otherwise it simply will not be formed from a pair of sp2 carbon atoms.

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