Question

Consider the elimination reactions shown below and explain the observed selectivity NaOCH ToRxn 1 NaOCH d o momenten - Rxn 2 A. The E2 elimination reaction requires that the leaving group and B-hydrogen be in the equatorial positions for a cyclohexanering KOB. The E2 elimination reaction requires anti periplanar geometry for the leaving group and the B-hydrogen. C. The starting material for Rxn 1 is able to adopt a more stable conformation D. The starting material for Rxn 2 is unable to yield a stable carbocation for the E1 elimination reaction

Answer 1

In E2 Reaction hydrogen must be coplanar with leaving group should it should be anti direction to the leaving group.

In E1 reaction the hydrogen does not have to be antiperiplanar to the leaving group, because in the E1 reaction the deprotonation happen in the separate step.

Stereochemistry prepared this because it places the C-H bonding orbital anti to C-LG (leaving group) antibonding orbital.

In the transition state those are the orbital that create the C-C Pi bond.

So the option second is correct.