Question

Consider the following reaction and predict the number of possible product:

• A. 2
• B. 4
• C. 6
• D. 3

Answer 1

The correct option is D 3

Reaction proceeds by E1 mechanism
E1 reaction is a two step process.
In step 1, leaving group leaves and form a carbocation.
In step 2, the base will attack the proton and proton abstraction takes place.
In E1, formation of carbocation is the slowest step. Hence, it is the rate determining step.

Formed carbocation will undergo deprotonation to give alkene product.
In given reaction, the carbocation (I) formed is tertiary and it undergoes 1, 2- H shift to form another stable tertiary carbocation (II). Both (I) and (II) are in equilibrium and both carbocation will give respective alkene products.


Carbocation (I) undergo deprotonation to form two alkene products (a) and (b). (a) is less substituted alkene and (b) is more substituted alkene.

Carbocation (II) undergo deprotonation to form two alkene products (d) and (b). (d) is less substituted alkene and (b) is more substituted alkene.
Product (b) is common in both carbocation (I) and (II). This is due to symmetry of molecule.
Thus, there are three possible products of the given reaction.