Question

How many monochloro derivatives are possible when 3-methylpentane is subjected to free radical chlorination (including isomers)?

• A. 7
• B. 5
• C. 6
• D. 4

Hint:

Whenever secondary C becomes substituted, check for \textbf{optical isomerism}.

Answer 1

The Correct Option is C

Step 1: Write the structure of 3-methylpentane: \[ CH_3–CH_2–CH(CH_3)–CH_2–CH_3 \] Carbon types: - C$_1$ and C$_5$ – equivalent primary (2 sites) - C$_2$ and C$_4$ – equivalent secondary (2 sites) - C$_3$ – tertiary (1 site) - branch methyl – primary (1 site)
Step 2: Non-equivalent substitution positions: 1. on terminal primary → 1 product 2. on secondary → 1 product 3. on tertiary → 1 product 4. on branch methyl → 1 product Total constitutionally = 4.
Step 3: Secondary carbon substitution creates a chiral centre. Each such gives two enantiomers. There are 2 secondary positions (equivalent) → 2×2 = 4 stereoisomers.
Step 4: Add non-chiral others (primary + tertiary + branch = 2).
Step 5: Overall monochloro derivatives = 4 + 2 = 6. Hence → (C).