0
$\begingroup$

Cyclobutadiene is like benzene in that it has alternating single and double bonds in a ring. However, it's bonds are not the same length, the double bonds being shorter than single bonds. The molecule is rectangular, not square.

How would this be explained? I have one plausible explanation that I tried to give...that is as follows

Cyclobutadiene is anti aromatic with 4n = 4 pi electrons, moreover it has huge strain in the molecule. To avoid such strain the molecule goes out of plane to compensate, but thus loses its ability to resonate.

Thus resonance structures like ones in benzene cannot be drawn for cyclobutadiene.

Is this correct? Please correct me if I am wrong..and some more data about this would be appreciated

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ @MathewMahindaratne no, I specifically want to know why we cannot draw resonance structures like we do in benzene, for cyclobutadiene $\endgroup$ – Techie5879 Feb 16 '20 at 4:26
  • 1
    $\begingroup$ I think this is a question from Solomons and Fryhle; Its answer key states a weird reason - the position of atoms differ in the two resonating structures which can't be the case in resonance phenomenon. And thus no resonance structures are possible. Even I had this doubt and noticed the question linked by Mathew. But it didn't help here. $\endgroup$ – user14250 Feb 16 '20 at 5:11
  • $\begingroup$ @GuruVishnu yes it is a question from Solomon's and fryhle. Where did you get the answer key from? $\endgroup$ – Techie5879 Feb 16 '20 at 5:13
  • $\begingroup$ @GuruVishnu The adapted version $\endgroup$ – Techie5879 Feb 16 '20 at 5:18
  • $\begingroup$ Ok. You must login to see the key here - wileyindia.com/customer/account/login; Although it contains some errors (like the one I mentioned in my first comment) it's quite useful. $\endgroup$ – user14250 Feb 16 '20 at 5:19
-1
$\begingroup$

Strictly speaking cyclobutadiene is not antiaromatic but non aromatic, ie the same energy 4A + 4B is attained by applying the Hückel LCAO, as for two isolated double bonds, 2 times 2A + 2B.

Given that cyclobutadiene can only be isolated as non delocalized system when in matrixes and a very low T points to strain as a reason for instability of both forms, the delocalized one remaining hypothetical.

Edit for the readers not familiar with the terms: A system obeying all the requisites (cyclic, pi orbitals on each atoms) is antiaromatic when the hypothetical delocalization destabilizes it as compared to the localized system. If no energy difference arises, the system is called non aromatic.

Beware there are plenty of pages discussing non aromaticity in a trivial way, e.g. aliphatic or linear compounds. Cyclobutadiene is non aromatic within the Hückel picture. Also it seems worth to point out again that antiaromatic refers anyway to an hypothetical structure, not to an actual delocalization. A system which cannot avoid it by distorting would be unstable tout-court. Cyclobutadiene, albeit being non aromatic, is on the same line. Indeed you can find antithetical assertions about it, such as it being "the archetype of antiaromatics“ along with “it reacts as diradical“ as if the two things would be straight forward correlated. Normally a system escape antiaromaticity. Being cyclobutadiene non aromatic results in a distorted rectangular molecule, but only when it cannot escape this anyway unstable situation. See also Resonance structure of cyclobutadiene?. Although the situation of cyclobutadiene is again erroneously described as an antiaromatic case, it gives the answer. Also worth reading is the answer by Georg.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Downvoter should comment though the first version was not particularly clear. $\endgroup$ – Alchimista Feb 16 '20 at 9:38
  • $\begingroup$ Why when I prepare sketches I am not able to upload the photo? This one was about the Frost diagram and just of 70 kb... $\endgroup$ – Alchimista Feb 16 '20 at 10:14

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