Question

Problem IV The Elimination reaction allows the removal from a molecule of a Leaving group and a neighboring Hydrogen to generate a double bond. When that reaction is performed on a cyclohexane, the leaving group (Br in the example below, and the H must be Axial as shown below. When the reaction is performed on the two molecule below (A and B), it is found that the reaction works very well on molecule A but very slowly on molecule B. Propose an explanation for this observation.

Answer 1

First see attached diagram.

In molecule 'A" the eliminating group i.e. Hydrogen and Bromine are anti (both axial, means one up and each other. So their elimination is easy. Morever there is steric hindrance between bromine and tertiary butyl group. So bromine is easily eliminated.

I IL

In structure 'B" the eliminating groups are not anti. also the steric hindrance is not present as the Br and tertiary butyl group are far away and hence there is no steric hindrance and the structure is quite stable and doesnt favor elimination.

So the anti position of eliminating group and steric hindrance is present in str 'A" hence it undergo elimination faster while 'B" doesnt have anti (eliminating group) so and there is no steric hindrance in "B" so it undergo elimination slower.