Q.What is ‘A’ in the following reaction?

Concept: Electrophilic addition of HX to an alkene follows Markovnikov’s rule — the hydrogen adds to the carbon with more hydrogens, and the halogen adds to the more substituted carbon.

Reasoning:

1. The given alkene is $\mathrm{C_6H_5-CH_2-CH=CH_2}$. The double bond is between the terminal and the second carbon. …

This is a classic electrophilic addition of HX to an unsymmetrical alkene. The molecule is 3-phenylprop-1-ene — a terminal alkene with a benzene ring three carbons away. The double bond is between C1 and C2 (counting from the phenyl end). HCl adds across the double bond, and the regiochemistry is governed by Markovnikov’s rule.

The rule says: the hydrogen atom of HX attaches to the carbon of the double bond that already has more hydrogens. That means H goes to the terminal carbon (C1, which has two hydrogens), and Cl goes to the internal carbon (C2). Why? Because the carbocation that forms on the more substituted carbon (secondary vs. primary) is more stable.

Let’s walk through it.

1. Identify the alkene structure.

   The compound is $\mathrm{C_6H_5-CH_2-CH=CH_2}$. The double bond is between the last two carbons:

   $\mathrm{C_6H_5-CH_2-CH=CH_2}$

   Carbon numbering: C1 (terminal, $\mathrm{=CH_2}$), C2 ($\mathrm{-CH=}$), C3 (benzylic, $\mathrm{-CH_2-}$), and then the ring.

2. Protonation step — which carbon gets H?

   The $\pi$ electrons attack the electrophilic proton of HCl. Markovnikov’s rule says the proton adds to the less substituted carbon (the one with more hydrogens already). That’s C1 (terminal, $\mathrm{=CH_2}$).

   This gives a carbocation at C2:

   $\mathrm{C_6H_5-CH_2-\overset{+}{C}H-CH_3}$

   This is a secondary carbocation — but not a benzylic one. The positive charge sits on C2, which is one carbon away from the ring (C3, the carbon actually attached to the ring, is the benzylic position). Still, a plain secondary carbocation is stable enough for Markovnikov addition to proceed cleanly here.

   Watch out

   A common mistake is to think the carbocation is benzylic. It is not — the benzene ring is attached to C3, not C2. The positive charge is on C2, so it’s a simple secondary carbocation, not resonance-stabilised by the ring. That’s fine; it’s still more stable than a primary one.

3. Nucleophilic attack by chloride. …

The reaction shown is not a carboxylic acid halogenation — it is an electrophilic addition of HCl to an alkene.

The concept tested here is Markovnikov’s rule for addition of HX to unsymmetrical alkenes.

Method: Markovnikov Addition of HX

Step 1 – Identify the alkene and its symmetry

The substrate is 3-phenylprop-1-ene:

$\mathrm{C_6H_5-CH_2-CH=CH_2}$

The double bond is between C1 and C2 (terminal alkene). It is unsymmetrical.

Step 2 - Determine the more stable carbocation intermediate

Protonation of the double bond can give two carbocations:

• Proton at C1 (terminal carbon) -> secondary carbocation:

  $\mathrm{C_6H_5-CH_2-\overset{+}{C}H-CH_3}$

  (a plain secondary carbocation -- NOT benzylic: a saturated CH2 sits between this carbon and the ring, so there is no resonance overlap with the phenyl group. It is stabilised only by the usual two alkyl groups a secondary carbocation has.)

• Proton at C2 -> primary carbocation: …

Let’s break this down step-by-step. The reaction given is:

This is 3-phenylprop-1-ene (an alkene with a phenyl ring two carbons away from the double bond) reacting with HCl.

🧠 Core Concept: Electrophilic Addition to Alkenes

The reaction follows Markovnikov’s rule:

In the addition of HX to an alkene, the hydrogen attaches to the carbon with more hydrogens, and the halogen attaches to the carbon with fewer hydrogens.

Here, numbering per the IUPAC name 3-phenylprop-1-ene (C1 = the terminal $\mathrm{=CH_2}$ carbon; the phenyl sits on C3), the double bond is between C1 and C2:

• C2 has 1 H
• C1 has 2 H

So, H⁺ adds to C1 (more H’s), and Cl⁻ adds to C2.

Product:

That matches option (iii).

✗ Common Mistakes & How to Avoid Them

1. ✗ Mistake: Adding Cl to the benzene ring

Why it happens: Students see the phenyl group and think of electrophilic aromatic substitution (chlorination of benzene).

How to avoid:

• Recognize that HCl does not chlorinate an aromatic ring under normal conditions — that requires $\mathrm{Cl_2}$ and a Lewis acid catalyst ($\mathrm{FeCl_3}$).
• Here, HCl adds across the double bond, not to the ring.

2. ✗ Mistake: Applying anti-Markovnikov addition

Why it happens: Confusing with radical addition (requires peroxides) or with other reagents like HBr/peroxide.

How to avoid:

• HCl does not show peroxide effect — it always follows Markovnikov’s rule.
• Remember: Only HBr shows anti-Markovnikov addition in the presence of peroxides.

3. ✗ Mistake: Adding Cl to the terminal carbon (option (ii))

Why it happens: Students think “Cl goes to the less substituted carbon” (anti-Markovnikov).

How to avoid:

• Always check substitution:
  • C2 (internal) = more substituted → gets Cl
  • C1 (terminal) = less substituted → gets H
• Draw the carbocation intermediate: …