The correct option is
C 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.
Slowest step in the reaction is the rate determining step.
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. Stable carbocation will be formed easily so rate of E1 reaction depends on the stability of the carbocation formed.
Compound (a), forms primary carbocation with 2
α− hydrogen.
Compound (b), forms primary carbocation with 0
α− hydrogen.
Compound (c), forms tertiary carbocation with 9
α− hydrogen.
Compound (d), forms secondary carbocation with 5
α− hydrogen.
Tertiary carbocation is more stable due to +I effect and hyperconjugation effect (9
α− hydrogen).
Therefore, compound (c) t-butyl bromide is the most reactive alkyl halide towards
E1 reaction.