The number of mono bromo structural isomers formed by free radical bromination of isopentane are
Answer & explanation
Correct answer: option 1
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:
- Tertiary hydrogen
– On the carbon attached to three other carbons (C-2) → 1 type - Secondary hydrogens
– On the CH₂ group → 1 type - 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.