Q.Which of the products will be major product in the reaction given below? Explain.
$\mathrm{CH_3CH{=}CH_2 + HI \rightarrow \underset{(A)}{CH_3CH_2CH_2I} + \underset{(B)}{CH_3CHICH_3}}$

The key idea is Markovnikov’s rule: in the addition of HX to an unsymmetrical alkene, the hydrogen attaches to the carbon with more hydrogens, and the halogen attaches to the carbon with fewer hydrogens.

Reasoning:

1. Propene ($\mathrm{CH_3CH{=}CH_2}$) is unsymmetrical. The double bond carbons are $\mathrm{CH_2}$ (two H’s) and $\mathrm{CH}$ (one H).
2. According to Markovnikov’s rule, the H of HI adds to the $\mathrm{CH_2}$ end, and the I adds to the $\mathrm{CH}$ end. …

Why this reaction follows Markovnikov’s rule

The addition of HX to an unsymmetrical alkene is governed by the stability of the carbocation intermediate. Propene ($\mathrm{CH_3CH{=}CH_2}$) is unsymmetrical — one carbon of the double bond is terminal (primary) and the other is internal (secondary). When HI adds, the proton can attach to either carbon, creating two possible carbocations:

• If H⁺ adds to the terminal carbon (C-1), the positive charge ends up on the secondary carbon (C-2) — a secondary carbocation.
• If H⁺ adds to the internal carbon (C-2), the positive charge ends up on the primary carbon (C-1) — a primary carbocation.

Secondary carbocations are significantly more stable than primary ones (by about 25–30 kJ/mol in the gas phase) because alkyl groups are electron-donating via hyperconjugation and inductive effects. The reaction pathway that goes through the more stable intermediate is faster, so it dominates.

Step-by-step reasoning

1. Identify the alkene structure

   Propene: $\mathrm{CH_3CH{=}CH_2}$. The double bond is between C-2 (internal, attached to a methyl and a hydrogen) and C-1 (terminal, attached to two hydrogens).

2. Protonation step — two possible pathways

   • Pathway 1: H⁺ adds to C-1 (terminal). The positive charge forms on C-2: $\mathrm{CH_3CH{=}CH_2 + H^+ \rightarrow CH_3\overset{+}{C}HCH_3}$ (secondary carbocation).
   • Pathway 2: H⁺ adds to C-2 (internal). The positive charge forms on C-1: $\mathrm{CH_3CH{=}CH_2 + H^+ \rightarrow CH_3CH_2\overset{+}{C}H_2}$ (primary carbocation).

3. Compare carbocation stability

   The secondary carbocation (pathway 1) is more stable than the primary one (pathway 2). Therefore, pathway 1 is much faster and essentially all product comes from it.

4. Nucleophilic attack by iodide …

Concept: Markovnikov’s Rule for Electrophilic Addition to Alkenes

When an unsymmetrical reagent like HI adds to an unsymmetrical alkene, the hydrogen atom attaches to the carbon that already has more hydrogen atoms (the less substituted carbon). The halide (I⁻) then attaches to the more substituted carbon.

Method: Markovnikov Addition

Steps:

1. Identify the alkene structure

   $\mathrm{CH_3CH{=}CH_2}$ is propene. The double bond is between C1 and C2:

   • C1 has 2 H atoms (terminal carbon)
   • C2 has 1 H atom (internal carbon)

2. Apply Markovnikov’s rule

   The H⁺ from HI adds to the carbon with more hydrogens — that is C1.

   The I⁻ adds to the other carbon — that is C2.

3. Write the product

   This gives: …

Here are the common mistakes students make when solving this Markovnikov addition problem, along with how to avoid each.

1. Forgetting the Carbocation Stability Rule (Markovnikov’s Rule)

• The Mistake: Students often guess product (A) (the primary iodide) because they think the iodine simply adds to the end of the double bond, or they confuse the rule with anti-Markovnikov addition (which requires peroxides for HBr, not HI).
• Why it’s wrong: The reaction proceeds via a carbocation intermediate. The proton ($H^+$) from HI attacks the double bond. The more stable carbocation forms faster.
  • Attack at C-1 gives a secondary carbocation ($CH_3CH^+CH_3$).
  • Attack at C-2 gives a primary carbocation ($CH_3CH_2CH_2^+$).
  • The secondary carbocation is more stable (due to hyperconjugation and inductive effects), so it forms preferentially.
• How to avoid: Always draw the carbocation intermediate. Ask: “Which carbon will hold the positive charge more stably?” The answer determines where the negative iodide ($I^-$) attacks.

2. Misidentifying the “Rich” and “Poor” Carbon

• The Mistake: Students try to apply the “rich gets richer” version of Markovnikov’s rule but misidentify which carbon of the double bond already has more hydrogens.
• Why it’s wrong: In propene ($CH_3CH=CH_2$), the terminal carbon ($=CH_2$) has two hydrogens, while the middle carbon ($CH=$) has one. The rule says the hydrogen (from H–I) adds to the carbon with more hydrogens (the terminal one). This leaves the positive charge on the more substituted (middle) carbon.
• How to avoid: Memorize the rule precisely: “The hydrogen adds to the carbon of the double bond that already has the greater number of hydrogens.” Then, the halogen adds to the other carbon.

3. Ignoring the “No Peroxide” Condition for HI

• The Mistake: Students apply the anti-Markovnikov rule (which works for HBr with peroxides) to HI, assuming it works the same way.
• Why it’s wrong: Anti-Markovnikov addition requires a radical mechanism (peroxides). For HI, the H–I bond is too weak to undergo the radical chain reaction efficiently. Even with peroxides, HI gives Markovnikov addition.
• How to avoid: Remember the special cases:
  • HBr + peroxide → anti-Markovnikov.
  • HCl, HI → always Markovnikov (no anti-Markovnikov version).
  • If the problem doesn’t mention peroxide, assume Markovnikov for all hydrogen halides.

4. Confusing the Major Product with the Kinetic Product

• The Mistake: Students think the reaction is reversible and choose the more stable (thermodynamic) product, which might be the primary iodide if they consider sterics. …