The major product of the following chemical reaction is :

The correct answer is (CH₃)₂CH-CH₂-CH₂-Br (1-bromo-3-methylbutane).

Reaction Analysis:
The given reaction is the addition of hydrogen bromide (HBr) to an alkene, specifically 3-methylbut-1-ene, in the presence of benzoyl peroxide, $(C_6{H}_5\mathrm{CO}{)}_2{O}_2$.
When HBr reacts with an unsymmetrical alkene in the presence of a peroxide, the addition follows an anti-Markovnikov pathway (also known as the peroxide effect or Kharasch effect), which proceeds via a free-radical mechanism.

Step-by-step Mechanism:

1. Chain Initiation:
The peroxide initiator decomposes homolytically to produce radicals, which then react with HBr to generate a bromine radical (Br•):
$(C_6{H}_5\mathrm{COO}{)}_2\to 2C_6{H}_5{\mathrm{COO}}^{•}\to 2C_6{H_5}^{•}+2{\mathrm{CO}}_2$
$C_6{H_5}^{•}+\mathrm{H-Br}\to C_6{H}_6+{\mathrm{Br}}^{•}$

2. Chain Propagation:
The bromine radical attacks the double bond of 3-methylbut-1-ene. The addition occurs selectively at the terminal carbon atom to form a more stable secondary carbon radical instead of a less stable primary carbon radical:
$({\mathrm{CH}}_3{)}_2\mathrm{CH-CH}={\mathrm{CH}}_2+{\mathrm{Br}}^{•}\to ({\mathrm{CH}}_3{)}_2\mathrm{CH-}{\mathrm{CH}}^{•}-{\mathrm{CH}}_2\mathrm{Br}$
This secondary radical then abstracts a hydrogen atom from another HBr molecule to form the final product and regenerate the bromine radical:
$({\mathrm{CH}}_3{)}_2\mathrm{CH-}{\mathrm{CH}}^{•}-{\mathrm{CH}}_2\mathrm{Br}+\mathrm{H-Br}\to ({\mathrm{CH}}_3{)}_2{\mathrm{CH-CH}}_2-{\mathrm{CH}}_2\mathrm{Br}+{\mathrm{Br}}^{•}$

Unlike carbocation intermediates, free-radical intermediates do not undergo rearrangement under these standard conditions. Therefore, the major product is the non-rearranged anti-Markovnikov addition product: 1-bromo-3-methylbutane.