Concept:Addition of
HI to an unsymmetrical alkene follows Markovnikov’s rule, giving the more stable carbocation intermediate.
Explanation:The given alkene is 3-methylhex-3-ene.
Protonation of the double bond can occur at two positions.
If
H+ adds at C-4, a tertiary carbocation forms at C-3.
If
H+ adds at C-3, a secondary carbocation forms at C-4.
The tertiary carbocation is more stable due to greater hyperconjugation and the inductive effect of alkyl groups.
Therefore, the reaction proceeds through the tertiary carbocation at C-3.
The iodide ion
I− then attacks this carbocation, placing iodine at C-3.
Thus, the major product is 3-iodo-3-methylhexane.
Answer:Option A: 3-iodo-3-methylhexane.
