enter image description here While reading of birch reduction I was confused as of why 1,4-cyclohexadiene is prefered over 1,3-cyclohexadiene, the latter which ought to have been the major product due to conjugation. One of the reason I felt could be of this outcome was "symmetry". The symmetry produced by the product could cancel the dipole forces acting on the cis- bonds, making the molecule stable. But is this sole reason sufficient to overpower the conjugation? Please explain. Edit: another possible reason I feel could be is the stereochemistry of the product. The product could facilitate more efficient structure in space about it's double bond (again symmetry), where as 1,3-cyclohexadiene formation could face issues of rotational barriers and might increase torsional strain on it.

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    $\begingroup$ Symmetry does not inherently lead to any stability. The product is not thermodynamically favoured, but is kinetically favoured: en.wikipedia.org/wiki/Birch_reduction#Reaction_regioselectivity see the last paragraph in this section. $\endgroup$ Mar 7, 2019 at 20:41
  • $\begingroup$ Orthocresol thank you. I got it. So as it is kinetically favoured, the reaction was too quick to occur. Instead of stabilising it spontaneously formed the kinetically favoured product, 1,4-cyclohexadiene. I hope I got the right track. $\endgroup$ Mar 7, 2019 at 20:51
  • $\begingroup$ So would there be a notable amount of 1,3-cyclohexadiene at low and controlled temperature and pressure $\endgroup$ Mar 7, 2019 at 20:54
  • $\begingroup$ The temperature isn't the key thing here. The only requirement for a kinetically controlled reaction is that the reaction is not reversible. The last step in the Birch reduction, protonation of a highly basic anion, is for all intents and purposes completely irreversible. $\endgroup$ Mar 7, 2019 at 21:04
  • $\begingroup$ Thank you. Yeah I didn't consider the protonation step here. I got it 😁 $\endgroup$ Mar 8, 2019 at 5:23


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