Final answer:
The alkyl bromide least likely to form a carbocation in a polar protic solvent is (CH3)3CBr, which is a tertiary bromide, because tertiary carbocations are the most stable and there is no driving force for it to lose a bromide to form a carbocation.
Step-by-step explanation:
In a polar protic solvent, the alkyl bromide least likely to form a carbocation is (CH3)3CBr, which is option c). The ability to form a carbocation depends on the stability of the carbocation that would be created during the reaction process. Carbocations are stabilized by the presence of alkyl groups through hyperconjugation and the inductive effect.
Tertiary carbocations, such as the one that would form if (CH3)3CBr underwent heterolytic cleavage, are more stable than secondary, which are more stable than primary carbocations. Therefore, a methyl carbocation (CH3+) would be the least stable, followed by a primary carbocation (CH3CH2+), and then a secondary carbocation (CH3CHCH3+). The most stable would be a tertiary carbocation ((CH3)3C+).
Since a compound's likelihood to form a carbocation correlates with the stability of the potential carbocation, (CH3)3CBr, being a tertiary bromide, is least likely to form a carbocation because it would already be at a relatively stable tertiary state, and there is no driving force for it to lose a bromide to form a carbocation.