Compound I is obviously the Z isomer. I thought it was obviously the less stable one too, given that there's a phenyl and t-butyl on the same side. The answer says that since the phenyl is planar, it actually exhibits less steric hindrance than a methyl group.

Is this because the planar phenyl ring can basically rotate its bulkiness away from the t-butyl, where as the methyl, while also free to rotate, always ends up with a hydrogen in the t-butyl's face?



1 Answer 1


The molecule contains conjugated system of double bonds, being essentially a substituted styrene. The phenyl cannot therefore freely rotate and its steric requirements are difficult to estimate just by hand-waving arguments.

Correct answer needs some quantitative argument, either e.g. heats of formation or calculation.

From the second, the energies come out using MMFF94 (using Avogadro):

  • Z: 177.6 kJ/mol
  • E: 175.6 kJ/mol

Also, both the isomers rotated the styrene substantially from planarity. The difference in results is quite small and can easily change when using other method. According to the numbers I'd agree with your answer, Z is less stable.

I would not consider this a valid question in test, the answer is not straightforward.


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