Q.Predict the major product formed when HCl is added to isobutylene. Explain the mechanism involved.

Concept: Markovnikov Addition of HX to an Alkene — The question asks for HCl addition to isobutylene (2-methylpropene). The key idea is that the proton adds to the less substituted carbon of the double bond, forming the more stable tertiary carbocation.

Reasoning:

1. Isobutylene is $(CH_3)_2C=CH_2$. The double bond is asymmetric. …

Why This Reaction Works the Way It Does

Isobutylene (2-methylpropene) is an unsymmetrical alkene. When you add HCl, the question is: which carbon gets the chlorine and which gets the hydrogen? The answer comes from the stability of the intermediate carbocation that forms during the reaction.

The key insight is that the reaction doesn’t happen in one step. Instead, the π bond acts as a nucleophile and attacks the hydrogen of HCl, leaving a positively charged carbon (a carbocation) and a chloride ion. The chloride then attacks the carbocation to form the final product. The major product is determined entirely by which carbocation is more stable.

Step-by-Step Mechanism

1. Protonation of the double bond

The π electrons of isobutylene attack the partially positive hydrogen of HCl. This breaks the H–Cl bond, giving a carbocation and a chloride ion.

There are two possible sites for protonation:

• Protonation at C1 (the terminal carbon): This puts the positive charge on the more substituted carbon (C2), forming a tertiary carbocation.
• Protonation at C2 (the internal carbon): This puts the positive charge on the less substituted carbon (C1), forming a primary carbocation.

The tertiary carbocation is far more stable than the primary one — roughly 25–30 kcal/mol more stable — because the three alkyl groups donate electron density through hyperconjugation and inductive effects. So the reaction overwhelmingly takes the first path.

2. Nucleophilic attack by chloride

The chloride ion (a good nucleophile) attacks the positively charged carbon of the tertiary carbocation. This forms a new C–Cl bond and gives the neutral product.

The overall reaction is:

$\text{(CH}_3\text{)}_2\text{C=CH}_2 + \text{HCl} \rightarrow \text{(CH}_3\text{)}_3\text{CCl}$ …

Concept: Electrophilic Addition to Alkenes (Markovnikov’s Rule)

When a protic acid (like HCl) adds to an unsymmetrical alkene, the hydrogen attaches to the carbon with more hydrogens already, and the halide attaches to the more substituted carbon. This is Markovnikov’s rule.

Method: Markovnikov Addition via Carbocation Intermediate

Step 1 — Identify the alkene

Isobutylene is:

It is an unsymmetrical alkene — one carbon has 2 H’s, the other has 0 H’s.

Step 2 — Protonation of the double bond

The $\text{H}^+$ from HCl attacks the less substituted carbon (the one with more H’s). This gives the more stable carbocation.

• Attack on $\text{CH}_2$ (primary carbon) → forms a tertiary carbocation (most stable):

Why? The tertiary carbocation is stabilised by +I effect of three methyl groups.

Step 3 — Nucleophilic attack by chloride

The $\text{Cl}^-$ ion attacks the positively charged carbon:

Step 4 — Major product

The product is 2-chloro-2-methylpropane (tert-butyl chloride):

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Here is a breakdown of the common mistakes students make when tackling the addition of HCl to isobutylene, along with the correct reasoning and exam-ready explanations.

The Core Concept: Markovnikov’s Rule & Carbocation Stability

The reaction is an electrophilic addition of a protic acid (H⁺ from HCl) to an alkene. The key principle is Markovnikov’s Rule: the hydrogen atom adds to the carbon of the double bond that already has more hydrogen atoms (the less substituted carbon), while the halogen adds to the more substituted carbon.

For isobutylene ($\text{CH}_2=\text{C}(\text{CH}_3)_2$), the correct product is tert-butyl chloride ($(\text{CH}_3)_3\text{CCl}$).

Common Mistake #1: Getting the Product Wrong (Anti-Markovnikov Product)

The Mistake: Students often predict the product as isobutyl chloride ($(\text{CH}_3)_2\text{CHCH}_2\text{Cl}$), where the Cl ends up on the primary carbon. This is the anti-Markovnikov product.

Why it happens: Students either forget Markovnikov’s rule or apply it mechanically without understanding the reason behind it.

How to Avoid It (The Why): The rule is a consequence of carbocation stability.

1. Protonation step: The H⁺ from HCl attacks the double bond. It can attack either carbon.
   • Attack on the less substituted carbon (the $\text{CH}_2$ group) gives a tertiary carbocation ($(\text{CH}_3)_3\text{C}^+$). This is very stable.
   • Attack on the more substituted carbon (the $\text{C}(\text{CH}_3)_2$ group) gives a primary carbocation ($(\text{CH}_3)_2\text{CHCH}_2^+$). This is very unstable.
2. The stable path wins: The reaction proceeds exclusively through the more stable tertiary carbocation.
3. Nucleophilic attack: The $\text{Cl}^-$ ion then attacks this tertiary carbocation, placing the chlorine on the tertiary carbon.

Exam Tip: Always draw the two possible carbocations. The product comes from the most stable one.

Common Mistake #2: Ignoring Carbocation Rearrangement (When It Doesn’t Happen)

The Mistake: Students sometimes assume a rearrangement (a hydride or methyl shift) must occur, even when the initially formed carbocation is already the most stable possible.

Why it happens: Students over-generalize from examples where a less stable carbocation rearranges to a more stable one (e.g., in the addition of HBr to 3-methyl-1-butene).

How to Avoid It: Check the stability of the carbocation formed directly after protonation.

• In isobutylene, the direct protonation on the terminal carbon gives a tertiary carbocation — the most stable type possible.
• No rearrangement is possible or needed. A rearrangement would only make the carbocation less stable or identical.
• The mechanism is a simple two-step process: protonation → nucleophilic attack.

Common Mistake #3: Writing an Incorrect or Incomplete Mechanism …