Final answer:
Chlorine gives a larger amount of the minor product (CH₃)₂CHCH₂X during the halogenation of 2-methylpropane compared to bromination because chlorine is more reactive than bromine. Bromine is less reactive and therefore reacts more selectively with the weaker, more stable 3° C-H bond, leading to a higher proportion of the major product (CH₃)₃CX.
Step-by-step explanation:
Chlorination of 2-methylpropane results in two products: the minor product (CH₃)₂CHCH₂X and the major product (CH₃)₃CX. When comparing chlorination versus bromination, chlorine results in a higher proportion of the minor product relative to the major product. The reason for this is related to the different reactivities and selectivities of chlorine and bromine as halogens.
The correct answer is: Bromine atoms are less reactive than chlorine, and preferentially attack the weaker 3° C-H bond. This means that chlorine is more reactive and therefore more likely to attack the more substituted carbon center, leading to the formation of the major product. This is because although the 3° C-H bond is weaker and easier to break due to its partial stabilization by the surrounding alkyl groups, the higher reactivity of chlorine can overcome the weaker 1° and 2° C-H bonds easier than bromine, leading to a significant amount of the minor product in chlorination. Bromine, being less reactive, reacts more selectively with the weaker, more stable 3° C-H bond, leading to a higher proportion of the major product when bromination occurs.