The correct option is C Reaction proceeds via carbocation intermediates
E1 reaction is a two step process.
In step 1, leaving group leaves and form a carbocation. Thus, carbocation intermediate is formed.
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.
Thus, the rate of the reaction depends on the rate of formation of carbocation.
Therefore, good leaving group will favour the reaction by ease of formation of carbocation.
Polar protic solvent solvates the halide ion formed due to C-X bond breaking. Hence, polar protic solvent increases reactivity.
Since, proton abstraction is not rate determining step we can use any base. However, a stronger base may lead to side reactions, so weak & moderate bases are preferred.