Question

Consider the hydrocarbons whose structures are listed below. Which of these molecules would be planar, meaning that all of the atoms must always lie in the same plane? Explain your answer in terms of orbital hybridizations.

Given hydrocarbons: Cyclohexane, Benzene, Methlbenzene

Answer 1

Benzene molecule is a planar. All carbon and hydrogen atoms lie in the same plane. It has a regular hexagon structure with all six carbon atoms lying at the corners, each carbon atom is bonded to three other atoms. All carbon-carbon bond lengths are equal at 139 pm. All CH angles are equal at 120⁰. It means that each carbon atom in benzene molecule is sp2 hybridized. All sp2 hybrid orbitals lie in the same plane and are oriented towards the corners of an equilateral triangle. Each carbon in benzene has three sp2 hybrid orbitals lying in the same plane and one –unhybridized p orbital.

So, each bond in benzene has a character intermediate between a single and a double bond

All the carbons in cyclohexane are sp3 hybridized, so the ideal bond angle would be 109.5 degrees. However, the planarity of the ring would force the carbons to have bond angles of 120 degrees, which makes it unstable. To gain more stability, cyclohexane adopts the chair conformation. The chair conformation is a six-membered ring in which atoms 2, 3, 5, and 6 lie in the same plane, atom 1 lies above the plane, and atom 4 lies below the plane. 6 carbons of the benzene ring of toluene are sp2 hybridized, planar molecule. The methyl group on the other hand, has all single bonds attached to the central carbon atom. So with 4 bonds, the carbon of the methyl group is sp3 hybridized, and it has a tetrahedral geometry.