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

2-Bromo-2-methyl propane

The correct answer is option 3. 2-Bromo-2-methyl propane.

The reactivity of haloalkanes with nucleophiles, such as aqueous \(KOH\) (potassium hydroxide), depends on the stability of the carbocation intermediate formed during the reaction.

When a haloalkane reacts with aqueous KOH, it undergoes nucleophilic substitution, where the halogen atom is replaced by the hydroxide ion (\(OH^-\)). This process involves the formation of a carbocation intermediate, followed by the attack of the nucleophile (hydroxide ion) on the carbocation to form the alcohol product.

The stability of the carbocation intermediate depends on the degree of substitution at the carbon bearing the leaving halogen group. Generally, the stability of carbocations follows the order: tertiary > secondary > primary.

In the given options:

1. 1-Bromobutane is a primary haloalkane.

2. 2-Bromobutane is a secondary haloalkane.

3. 2-Bromo-2-methylpropane is a tertiary haloalkane.

4. 2-Chlorobutane is a secondary haloalkane.

Tertiary haloalkanes are the most reactive towards nucleophilic substitution reactions because the carbocation intermediate formed during the reaction is stabilized by adjacent alkyl groups. This stabilization is due to hyperconjugation and inductive effects, which disperse the positive charge of the carbocation over multiple carbon atoms, making it more stable.

Therefore, 2-Bromo-2-methylpropane (option 3) is expected to react most easily with aqueous \(KOH\) compared to the other options due to the increased stability of the carbocation intermediate formed during the reaction.