Question

Consider the following elimination reaction.

When reaction is carried out with $C_{2}H{}_{5}ONa$ in ethanol, 2-menthene is the major product while 3-menthene is the major product if the substrate is heated in ethanol only. It is due to

• A. with strong base, less substituted alkene is always the major product.
• B. with strong base, reaction proceeds by E2 mechanism while with weak base, E1 reaction takes place.
• C. methyl group gives steric hindrance to strong base in E2 reaction.
• D. isopropyl group gives steric hindrance to strong base in E2 reaction.

Answer 1

The correct option is B with strong base, reaction proceeds by E2 mechanism while with weak base, E1 reaction takes place.

With strong base $C_{2}H{}_{5}ONa$, reaction proceeds by E2 mechanism. Since, there is no H at anti position of tertiary $\beta -$ carbon, deprotonation occur from secondary $\beta -$ carbon giving 2-menthene as the major product.
On the other hand, with weak base, reaction proceeds by E1 mechanism and the most stable teritary carbocation formed, gives the major product 3-methene (more substituted).