The number of mono bromo structural isomers formed by free radical bromination of isopentane are

4

The correct answer is Option (1) → 4

Isopentane is 2-methylbutane with structure:

$(CH_3)_2CH–CH_2–CH_3$

In free-radical bromination, substitution can occur at non-equivalent hydrogen atoms. Count the distinct types of hydrogens:

1. Tertiary hydrogen
   – On the carbon attached to three other carbons (C-2) → 1 type
2. Secondary hydrogens
   – On the CH₂ group → 1 type
3. Primary hydrogens (two types)
• Two equivalent CH₃ groups attached to C-2 → 1 type
• One terminal CH₃ group → 1 different type

Total distinct positions =

1 (tertiary) + 1 (secondary) + 2 (primary) = 4

Hence, 4 mono-bromo structural isomers are formed.