When sec-butylcyclohexane reacts with bromine in the presence of sunlight, the major product is:

Step 1: Identify the structure

The sec-butyl group is: $\ce{-CH(CH3)-CH2-CH3}$ attached to cyclohexane.

Step 2: Understand the reaction

Bromination in sunlight = free radical halogenation

Reactivity order for free radical bromination: 3° > 2° >> 1°

Bromine is highly selective and preferentially attacks the most stable radical (tertiary).

Step 3: Identify carbon types

In sec-butylcyclohexane:

• Tertiary (3°): The carbon where sec-butyl attaches to cyclohexane ring AND the $\ce{CH}$ carbon in $\ce{-CH(CH3)-}$ of the side chain
• Secondary (2°): Carbons in the cyclohexane ring, $\ce{-CH2-}$ in the side chain
• Primary (1°): $\ce{CH3}$ groups

Wait, let me reconsider. The sec-butyl group is $\ce{-CH(CH3)-CH2-CH3}$:

• The $\ce{CH}$ carbon (where it branches) is tertiary when attached to the ring
• The $\ce{CH2}$ is secondary
• The $\ce{CH3}$ groups are primary

Step 4: Determine major product

Bromination will occur at the most substituted (tertiary) carbon because:

• Tertiary radical is most stable
• Bromine radical is selective (less reactive than Cl•)

The tertiary carbon is the one in the sec-butyl group: $\ce{-CH(CH3)-CH2-CH3}$

Major product: Br attached to the tertiary carbon of the side chain.

Answer: (b)

Key Points:

• Free radical bromination: 3° >> 2° > 1° (highly selective)
• Free radical chlorination: 3° > 2° > 1° (less selective)
• Br• is more selective than Cl• due to lower reactivity