I am facing trouble in the following question

What is the stability order of the following compounds? enter image description here

In my view cis is less stable than trans. So if prior groups are on the same side of the double bond then it must have less stability. Also conjugated double bond in short non aromatic cyclic system is highly unstable. So 1,2,3,4,5,6,7,8-octahydrobiphenylene (X) should be the least stable. However I could not guess the correct order.

  • $\begingroup$ Google is your friend. What is the strain energy in cyclobutane and cyclobutene; what is the strain energy in cyclohexane and cyclohexene. $\endgroup$ – ron Dec 26 '16 at 23:00
  • $\begingroup$ Also, cyclobutadiene has other properties to consider. $\endgroup$ – Zhe Dec 26 '16 at 23:44
  • 3
    $\begingroup$ I don’t understand a single thought of your reasoning. $\endgroup$ – Jan Dec 27 '16 at 0:24
  • $\begingroup$ Whoever voted to close this as primarily opinion based should take another good look at the original (homework) question and realise that it is very clear and obviously answerable; no opinions required. $\endgroup$ – Jan Dec 27 '16 at 15:33

While solving other related questions, I came up with a idea that

More is the number of $\alpha$ hydrogen present(hydrogen attached to carbon which is adjacent to a carbon containing double bond ) more stable is the compound.

In the question $Z$ has maximum 10 $\alpha$ hydrogen so it is the most stable.$X$ and $Y$ both have 8 $\alpha$ hydrogen.But $Y$ is more stable since conjugated double bond in short non aromatic cyclic system is highly disfavoured.

Hence the correct stability order is $Z>Y>X$

Side note- I am still not satisfied with my second argument.Any other explanations or thought will be highly appreciated.

| improve this answer | |
  • $\begingroup$ Forcing 3 or 4 carbon atoms to form a ring is energetically difficult due to the high strain energy in such rings (try building a model). Putting double bonds in these small rings increases the strain energy even further. The correct order of stabilities (from most stable to least stable) is Y (4-membered ring) > Z (4-membered ring with one double bond) > X (4-membered ring with two double bonds). $\endgroup$ – ron Dec 28 '16 at 2:32
  • $\begingroup$ @ron should not the presence of $\alpha$ hydrogen play a role here cause there would be more number of resonance structures possible. $\endgroup$ – Pink Dec 28 '16 at 2:50
  • $\begingroup$ Such resonance structures are referred to as hyperconjugative resonance structures. I think the ring strain effect I mentioned outweighs such resonance effects. Post back if your professor says otherwise. $\endgroup$ – ron Dec 28 '16 at 3:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.