Step 1: Read the structure:
The skeletal drawing has a methyl branch on the carbon at one end of the double bond. Counting the zigzag chain gives six carbons with the double bond at C3=C4 and the methyl at C3. So the alkene is 3-methylhex-3-ene.
Step 2: Apply Markovnikov's rule:
Hydrogen goes to the carbon that already has more hydrogens, here C4, and iodine goes to the more substituted carbon, C3.
The reason is cation stability: $\text{C}_3^+$ is tertiary (stabilised by hyperconjugation and the inductive effect of three alkyl groups), while $\text{C}_4^+$ would be secondary.
Step 3: Name the product:
C3 now carries I, a methyl, an ethyl and a propyl chain. The longest chain is still six carbons, so the product is 3-iodo-3-methylhexane.
Option (B) 4-iodo-4-methylhexane is not possible because there is no methyl on C4. Options (C) and (D) are minor secondary-cation type products or wrongly named.
Final Answer:
The major product is 3-iodo-3-methylhexane, option (A).
\[ \boxed{\text{3-iodo-3-methylhexane (A)}} \]
