What is the major product formed by HI on reaction with

Step 1: Identify the reaction

Alkene + HI → Hydrohalogenation (addition of HI across double bond)

Starting alkene: 2-Methylbut-2-ene ($\ce{(CH3)2C=CH-CH3}$)

Step 2: Apply Markovnikov's rule

Markovnikov's rule: In addition of HX to alkene, H adds to carbon with more H atoms, X adds to carbon with fewer H atoms.

Alternatively: Carbocation stability determines product.

Mechanism:

1. Protonation:

   • H⁺ can add to either C of double bond
   • Forms carbocation

2. Two possibilities:

   Path A: H⁺ adds to C-3 $\ce{(CH3)2C=CH-CH3 + H+ -> (CH3)2C+-CH2-CH3}$

   • Tertiary carbocation (3°)
   • Very stable ✓

   Path B: H⁺ adds to C-2 $\ce{(CH3)2C=CH-CH3 + H+ -> (CH3)2CH-CH+-CH3}$

   • Secondary carbocation (2°)
   • Less stable ✗

3. Nucleophilic attack: $\ce{(CH3)2C+-CH2-CH3 + I- -> (CH3)2C(I)-CH2-CH3}$

Product: 2-Iodo-2-methylbutane ($\ce{(CH3)2C(I)-CH2-CH3}$)

Answer: (b)

Key Concepts:

Markovnikov's rule:

• "Rich get richer" - more substituted carbon gets X
• Based on carbocation stability: 3° > 2° > 1°

Carbocation stability:

• Hyperconjugation: More alkyl groups = more stable
• Inductive effect: Alkyl groups donate electrons
• Order: 3° > 2° > 1° > methyl

Anti-Markovnikov addition:

• Peroxide effect (HBr only)
• Free radical mechanism
• H adds to less substituted carbon