Which of the following haloalkanes reacts with aqueous KOH most easily?

The correct option is 2-Bromo-2-methyl propane.

Step-by-step Explanation:

The reaction of haloalkanes with aqueous KOH is a nucleophilic substitution reaction. Aqueous KOH provides hydroxide ions (OH^-) which act as a nucleophile to replace the halogen atom (leaving group) in the haloalkane.

Nucleophilic substitution reactions of tertiary haloalkanes primarily proceed via the S_N1 (Substitution Nucleophilic Unimolecular) mechanism. This is because tertiary carbocations are highly stable due to inductive effect (+I effect) and hyperconjugation from the three neighboring alkyl groups.

Let us analyze the structures and classification of the given options:

1. 1-Bromobutane is a primary (1°) haloalkane: CH₃-CH₂-CH₂-CH₂-Br
2. 2-Bromobutane is a secondary (2°) haloalkane: CH₃-CH(Br)-CH₂-CH₃
3. 2-Bromo-2-methylpropane is a tertiary (3°) haloalkane: (CH₃)₃C-Br
4. 2-Chlorobutane is a secondary (2°) haloalkane: CH₃-CH(Cl)-CH₂-CH₃

For S_N1 reactions, the rate of reaction depends on the stability of the carbocation intermediate formed in the slow rate-determining step. The stability order of carbocations is:
Tertiary (3°) > Secondary (2°) > Primary (1°)

Since 2-bromo-2-methylpropane is a tertiary haloalkane, it undergoes ionization rapidly to form a highly stable tertiary carbocation, (CH₃)₃C⁺. Additionally, bromine is a better leaving group than chlorine due to its weaker carbon-halogen bond (C-Br bond is weaker than C-Cl bond because of larger size of bromine). Therefore, 2-bromo-2-methylpropane reacts with aqueous KOH most easily among the given choices.