Does bicyclo[3.2.0]heptane have a plane of symmetry?

Most stable form of cyclobutane is the "envelope form" and most stable conformation of cyclopentane is either the "envelope form" or the "twist" form.

1. In the above compound do the individual rings exist in their most stable conformation? Or is the actual situation different?
2. What is the most stable conformation of the above compound? And does it have a plane of symmetry?
• I figure that the question "why is the cyclobutane ring planar" is a good question in itself for main site. But you'll probably have to specify that you are looking for an org chem answer and not simply some calculations that say "because it is the most stable conformation". – orthocresol Oct 16 '16 at 23:26

It turns out that bicyclo[3.2.0]heptane does have a plane of symmetry (courtesy PubChem):

The link explains the experimental procedure in deduction of the actual structure.There are two possible stable structures/forms for the molecule namely endo and exo.The endo form is supposed to be more favourable than the exo form.

From here, we can conclude Cs symmetry.

• I'm sorry to say.But you have given a very vague answer with no reason at all.Why doesnt the cyclopentane take up an envelope or twist form here?Why doesnt the cyclobutane take up an envelope form?Really vague answer :-( – user14857 Oct 16 '16 at 6:56
• @ZOZ Satisfied? – DHMO Oct 16 '16 at 7:07
• No :-(!You gave the bond angles.Ok good.You gave the predicted structure.Good.But you still did not give the reason for the structure's occurence and stability..and you did'nt answer my previous questions..in the comments as well as the original question.. – user14857 Oct 16 '16 at 7:11
• @ZOZ Reason? This is experimentally determined. You could solve the Schroedinger's equation if that counts for a reason. There's no reason. Everything is just rationalization. – DHMO Oct 16 '16 at 7:12
• If the solution of the Schrodinger equation is the only admissible "reason", then chemistry is in a lot of trouble. @ZOZ The cyclopentane ring is already in an envelope conformation, as pH13 notes. Remember that any distortion of geometry in the cyclobutane ring is transferred to the cyclopentane ring via the bridging carbons. Presumably, distortion of the cyclobutane ring towards a puckered geometry simply creates more strain (bond/angle) in the rest of the molecule than is relieved in the cyclobutane ring. – orthocresol Oct 16 '16 at 23:02

The most stable conformation is the endo-form which obeys $C_\mathrm s$ symmetry. The cyclobutane-part is in envelope form and the cyclopentane-part is also in envelope form. The endo version with the methylene group bent towards the cyclobutane moiety is the preferred one.

The next most stable form would then be the exo form, where neither the cyclobutane- nor the cyclopentane-part is in its favorite form, both are slightly distorted. At least after my optimization, this molecule does not have a symmetry, except $C_1$. Though, I did not try a symmetric version.

With a quite high energy gap of ~90 kJ/mol the most unstable form is where both parts are in their twisted forms. Though, this molecule is $C_2$ symmetric.

The structures were optimized with RI-TPSS-D3BJ/def2-TZVP and a following frequency calculation showed that the results are local minima. Those geometries where then used for a single point energy calculation, using the Tight-PNO DLPNO-CCSD(T)/def2-TZVP scheme. All calculations have been made with ORCA 3.0.3.

• I see my answer here not really as an answer but more as an additional help for some qualified people to discuss. – pH13 - Yet another Philipp Oct 18 '16 at 7:32