I expect there to be more steric repulsions between the two methyl groups if they are on the same face as each other. Why is this not the case?

  • 1
    $\begingroup$ Hint: the four carbon atoms are not coplanar. It is not a flat square. $\endgroup$
    – getafix
    Sep 22 '16 at 13:52
  • $\begingroup$ You should find a realistic stick drawing of the cyclobutane ring, you should then be able to answer your own question. $\endgroup$
    – porphyrin
    Sep 22 '16 at 15:31

The four carbon atoms in cyclobutane are not coplanar, as that would lead to a large degree of unfavourable eclipsing interactions. So, it is not a flat square; it adopts a puckered conformation. One of the carbon atoms makes a $25^\circ$ angle with the plane formed by the other three carbons, and this mitigates some of the eclipsing interactions, at the cost of a small increase in angle strain. The conformation is commonly called the "butterfly", and two equivalent puckered conformations interconvert rapidly. Here's a visual reference to help you


Pictured above is an unsubstituted cyclobutane.

Take a closer look at the figure on the left, and notice how there would be some transannular interaction between the two $\ce{H'}$ hydrogens at $\ce{C-1}$ and $\ce{C-3}$, marked in $\color{red}{\text{red}}$. This is bad. In the figure on the right, the $\ce{H'}$ hydrogens are no longer axial, but equatorial, and this is better.

In short, placing substituents in an equatorial position is better than placing them in an axial position.

Now, imagine what the cis and trans isomers would look like for 1,3-dimethylcyclobutane.

In the cis isomer both $\ce{H'}$ on $\ce{C-1}$ and $\ce{C-3}$ would be replaced by $\ce{-CH3}$ groups (since they are on the same side). Of course there will be considerable repulsions if they are arranged in the conformation on the left, which is why ring flip takes place and we will predominantly get the conformation on the right with both the methyl groups in equatorial positions. This is ideal.

cis dimethyl

The trans isomer would be like replacing one $\color{red}{\text{red}}$ $\ce{H'}$ and one $\color{blue}{\text{blue}}$ $\ce{H}$ at $\ce{C-1}$ and $\ce{C-3}$ with methyl groups. No matter what you do, you can only get one of the methyl groups at the equatorial position, but never both.

trans dimethyl



It is because cyclobutane looks like a square piece of paper with a fold along the diagonal. If you add the methyl groups to the corners not on the fold, they can be both axial or both equatorial. The axial is the less stable conformation than diequatorial. If it is the trans compound, then it will have one axial and one equatorial. Just as two equatorial groups are more stable, only one equatorial group will be less stable.

You need to understand that cyclobutane is a bent structure to reduce angle strain and torsional strain.Read more here http://www.masterorganicchemistry.com/2014/04/03/cycloalkanes-ring-strain-in-cyclopropane-and-cyclobutane/


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