Conformational Analysis of Cyclic Compounds | Stereochemistry | Organic Chemistry

Structure of Cyclopropane (C3H6).
The formula for cyclopropane is C3H6. In cyclopropane, all the three carbon atoms are sp3 hybridized. Two of sp3 orbitals of carbon atoms overlap to form C - C sigma bonds. However, the overlap is not exactly coaxial and hence ‘bent’ bonds are formed. These sigma bonds lie mostly outside of the ring and that’s why they show more of pi bond character instead of sigma bond. The tetrahedral bond angle of 109.5-degree changes to 104 degrees in cyclopropane. Due to these strained bonds cyclopropane is highly reactive.

Structure of Cyclobutane (C4H8).
In the cyclobutane (C4H8) ring, which has four carbon atoms. If it has a planar structure, then there will be a large angle strain. Further, the adjacent CH bond will be eclipsed hence there will be strong torsional and steric strain. But the experiments showed that the total strain in cyclobutane is less than the calculated value. This is because
cyclobutane exists in non-planar conformation. It exists in a folded form in which one of the carbon atom is slightly out of the plane. In these conformations of cyclobutane, the torsional strain is slightly relieved due to the staggering of C - H bonds. By thermodynamic calculations, it can be found that Cis isomer is more stable than the Trans isomer. e.g. methyl -3-methyl cyclobutane carboxylate.

Structure of Cyclopentane (C5H10).
Similarly, if we consider the planar structure of cyclopentane, there is practically no angle strain but due to five pairs of eclipsed C - H bonds, there will be the torsional and steric strain. Hence it exists in two non-planar conformations. (i) Envelop form (ii) Half chair form
In envelope form four carbon atoms are in the same plane and one carbon atom is out of the plane by 0.5 A. In half-chair form three carbon atoms are in the plane and two carbon atoms are out of plane (one above the plane and other below the plane).

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