Q.Write the products of the following reactions:

Concept: Markovnikov Addition — the hydrogen of HX adds to the carbon with more hydrogens already, giving the more stable carbocation intermediate. Peroxide reverses the regioselectivity (anti-Markovnikov) for HBr only.

Step 1: For (i) $C_6H_5CH=CH_2$, the double bond is between the benzylic carbon and the terminal carbon. Protonation at the terminal carbon gives a resonance-stabilized benzylic carbocation.

Step 2: Bromide attacks the benzylic carbon. Product: $C_6H_5CHBrCH_3$.

Step 3: For (ii) $CH_3CH_2CH=CH_2$, protonation at the terminal carbon gives a secondary carbocation (more stable than primary). Chloride attacks the secondary carbon. Product: $CH_3CH_2CHClCH_3$. …

(i) $C_6H_5CH=CH_2 + HBr$

Styrene. Protonating the terminal carbon puts the positive charge on the ring-attached carbon, giving a benzylic carbocation stabilised by resonance into the ring — far more stable than the alternative primary cation. Markovnikov addition:

(ii) $CH_3CH_2CH=CH_2 + HCl$

But-1-ene. Protonation at the terminal carbon gives a secondary carbocation, more stable than the primary cation from protonating the internal carbon. Markovnikov addition:

(iii) $C_6H_5CH_2CH=CH_2 + HBr \xrightarrow{\text{Peroxide}}$

Allylbenzene: $C_6H_5-CH_2-CH{=}CH_2$. Peroxide reverses which carbon gets the halogen — via a free-radical chain, Br adds to the LESS substituted (terminal) carbon, leaving the radical on the more substituted (internal) carbon, because a more substituted radical is more stable regardless of what else is nearby. …

Markovnikov Addition — Solution Method

Method: Markovnikov's Rule

Rule statement: When an unsymmetrical alkene adds a protic acid (H–X), the hydrogen atom attaches to the carbon of the double bond that already has more hydrogen atoms, and the halogen attaches to the carbon with fewer hydrogen atoms.

Steps to apply the method

1. Identify the double bond and the two sp² carbons.
2. Count the number of H atoms directly bonded to each sp² carbon.
3. Apply Markovnikov's rule:
   • H goes to the carbon with more H's.
   • X (halogen) goes to the carbon with fewer H's.
4. Write the product — the halogen adds to the more substituted carbon.

Solutions

(i) $C_6H_5CH=CH_2 + HBr \rightarrow$

• Double bond: $C_6H_5CH=CH_2$
  • Left sp² carbon: 1 H (attached to phenyl ring)
  • Right sp² carbon: 2 H's
• H adds to the right carbon (more H's) → Br adds to the left carbon

Product:

(ii) $CH_3CH_2CH=CH_2 + HCl \rightarrow$

• Double bond: $CH_3CH_2CH=CH_2$
  • Left sp² carbon: 1 H (attached to $CH_3CH_2-$)
  • Right sp² carbon: 2 H's
• H adds to the right carbon → Cl adds to the left carbon

Product:

(iii) $C_6H_5CH_2CH=CH_2 + HBr \xrightarrow{\text{Peroxide}}$

⚠️ Important: Peroxide changes the mechanism to free radical addition — this follows anti-Markovnikov addition. …

Here are the most common mistakes students make with Markovnikov addition problems, along with how to avoid each.

Mistake 1: Forgetting the Peroxide Effect (Anti-Markovnikov Addition)

The Mistake: For reaction (iii), students apply the normal Markovnikov rule and place the bromine on the more substituted carbon, ignoring the "Peroxide" condition.

• Wrong product: $C_6H_5CH_2CHBrCH_3$ (Br on the internal carbon)

Why it happens: Students memorize "Markovnikov = H goes to the carbon with more H's" but fail to recognize that peroxides ($R-O-O-R$) completely reverse the regioselectivity for HBr addition via a free-radical mechanism.

How to Avoid:

• Spot the trigger: Whenever you see "Peroxide" or "$h\nu$" (light) with HBr, immediately switch to Anti-Markovnikov thinking.
• Mechanism check: Peroxides initiate a free-radical chain. The radical intermediate is more stable on the more substituted carbon, so the bromine atom ends up there, and hydrogen goes to the less substituted carbon.
• Correct product for (iii): $C_6H_5CH_2CH_2CH_2Br$ (Br on the terminal carbon, the anti-Markovnikov position).

Key rule: Peroxide + HBr → Br adds to the less substituted carbon (Anti-Markovnikov).

Mistake 2: Misidentifying the "More Substituted" Carbon

The Mistake: In reaction (i), students incorrectly identify which carbon of the double bond is more substituted.

• Wrong reasoning: "The bulky benzene ring crowds the benzylic carbon, so the bromine must go to the free terminal carbon."
• Wrong product: $C_6H_5CH_2CH_2Br$ (Br on the terminal carbon)

Why it happens: Students either treat the phenyl group as a blocking group, or forget that a benzene ring counts as a carbon substituent. The benzylic carbon of $C_6H_5CH=CH_2$ has one carbon attached (the ring) while the terminal $=CH_2$ carbon has none — so the benzylic carbon is the more substituted one, and the resonance-stabilised benzylic carbocation reinforces the same outcome.

How to Avoid:

• Count carbons directly attached to the double bond:
  • Left carbon ($C_6H_5CH=$): attached to 1 carbon (the benzene ring) + 1 H.
  • Right carbon ($=CH_2$): attached to 0 carbons + 2 H's.
• Apply Markovnikov: The H from HBr adds to the carbon with more H's (the $=CH_2$ group). The Br adds to the carbon with fewer H's (the benzylic carbon).
• Correct product for (i): $C_6H_5CHBrCH_3$

Key rule: Count the number of carbon atoms directly bonded to each alkene carbon. More C's = more substituted.

Mistake 3: Ignoring Carbocation Stability in Asymmetric Alkenes

The Mistake: For reaction (ii), students apply Markovnikov blindly without checking if the intermediate carbocation is stable.

• Wrong reasoning: "Markovnikov says H goes to the carbon with more H's, so product is $CH_3CH_2CHClCH_3$." (This is actually correct here, but the reasoning is incomplete.)

Why it happens: Students treat Markovnikov as a memorized rule rather than understanding it's a consequence of carbocation stability.

How to Avoid:

• Draw the two possible carbocations:
  1. H⁺ adds to terminal carbon → secondary carbocation: $CH_3CH_2\overset{+}{C}HCH_3$ (stable)
  2. H⁺ adds to internal carbon → primary carbocation: $CH_3CH_2CH_2\overset{+}{C}H_2$ (less stable)
• Choose the more stable path: The reaction proceeds via the more stable secondary carbocation.
• Correct product for (ii): $CH_3CH_2CHClCH_3$ (2-chlorobutane) …