Even though sodium ethoxide is a strong nucleophile and a strong base, only one major product is observed in the reaction below. Explain why.
The formation of only one major product in this reaction is a direct consequence of the stereochemical requirements of the E2 (bimolecular elimination) mechanism within a cyclohexane ring system.
Sodium ethoxide is a strong, sterically unhindered base, which strongly favors the E2 mechanism over SN2 for secondary halides like the one shown. The E2 mechanism requires a specific spatial arrangement: the hydrogen atom being removed (the β-hydrogen) and the leaving group (the chloride) must be in an anti-periplanar conformation. In the context of a cyclohexane chair conformation, this translates to a strict requirement that both the β-hydrogen and the leaving group must be in axial positions.
Let’s analyze the chair conformations of the starting material, (1R,2R)-1-chloro-2-isopropylcyclohexane.
- The More Stable Conformation: The bulky isopropyl group prefers the equatorial position to minimize steric strain. Since the isopropyl group and the chlorine atom are trans to each other (one is up, one is down), placing the isopropyl group in the equatorial position forces the chlorine atom to also be in an equatorial position. In this stable conformation, the chlorine leaving group is not axial, so E2 elimination cannot occur.
- The Less Stable Conformation: Through a ring flip, the molecule can adopt a less stable conformation where the isopropyl group is axial and the chlorine atom is also axial. Although this conformation is less populated due to the steric strain of the axial isopropyl group, it is the only conformation from which E2 elimination is possible because the chlorine is now in the required axial position.
Once in this reactive conformation, the base (ethoxide) must remove an axial β-hydrogen. There are two β-carbons, C2 and C6.
- At C2: The isopropyl group is in the axial position. Therefore, the hydrogen on C2 is in the equatorial position. Since this hydrogen is not axial, it is not anti-periplanar to the axial chlorine, and elimination cannot occur across the C1-C2 bond. This would have formed the more substituted (Zaitsev) product, but it is stereochemically forbidden.
- At C6: This carbon has two hydrogens. One is equatorial, and one is axial. The axial hydrogen on C6 is perfectly positioned anti-periplanar to the axial chlorine on C1. The ethoxide base can readily abstract this proton, leading to the formation of a double bond between C1 and C6.
Because only the hydrogen on C6 meets the strict axial, anti-periplanar requirement for the E2 mechanism, elimination occurs exclusively in one direction. This results in the formation of 3-isopropylcyclohex-1-ene as the single major product
