Name the alkene which will yield 1-chloro-1-methylcyclohexane by its reaction with HCl. Write the reactions in

NCERT Exemplar · Q61

Q. Name the alkene which will yield 1-chloro-1-methylcyclohexane by its reaction with HCl. Write the reactions involved.

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Concept understanding — Markovnikov Addition

The Intuition First

Imagine you have an alkene — a carbon-carbon double bond. That double bond is like a crowded room with two doors. When a molecule like HBr comes along, it wants to break that double bond and add across it. The question is: which carbon gets the hydrogen, and which gets the bromine?

You might think it doesn't matter — after all, the two carbons look similar. But they aren't. One carbon usually has more alkyl groups (methyl, ethyl, etc.) attached to it than the other. That carbon is more "electron-rich" — it has more friends pushing electrons toward it.

The hydrogen, being small and positively charged, is picky. It goes to the carbon that already has more hydrogens. Why? Because that carbon is less crowded and can stabilise the positive charge that forms temporarily during the reaction. The bromine, being large and negatively charged, goes to the other carbon — the one with more alkyl groups.

That's the intuition: the rich get richer. The carbon with more hydrogens gets another hydrogen. The carbon with more alkyl groups gets the halogen.

The Precise Statement

Markovnikov's Rule : When an unsymmetrical reagent (like HX, H₂O, etc.) adds to an unsymmetrical alkene, the hydrogen atom attaches to the carbon of the double bond that already has the greater number of hydrogen atoms.

In other words, for an alkene like $CH_3CH=CH_2$ (propene) reacting with HBr:

Carbon 1 (the CH₂ end) has 2 hydrogens.
Carbon 2 (the CH end) has 1 hydrogen.
The H goes to carbon 1 (more hydrogens).
The Br goes to carbon 2 (fewer hydrogens).

So the product is $CH_3CHBrCH_3$ (2-bromopropane), not $CH_3CH_2CH_2Br$ (1-bromopropane).

Why Does This Happen? The Real Chemistry

The reaction proceeds through a carbocation intermediate. When the H⁺ attacks the double bond, it can form one of two possible carbocations:

A primary carbocation (if H⁺ goes to the more substituted carbon) — unstable.
A secondary carbocation (if H⁺ goes to the less substituted carbon) — more stable.

The reaction chooses the path that gives the more stable carbocation. Alkyl groups stabilise carbocations through hyperconjugation and inductive effect — they donate electron density to the positively charged carbon.

Important

The stability order of carbocations is: tertiary > secondary > primary > methyl. Markovnikov addition always proceeds through the most stable carbocation possible.

A Common Misconception

Watch out

Many students think Markovnikov's rule means "hydrogen goes to the carbon with more hydrogens" because that carbon already has more hydrogens. That's backwards. The hydrogen goes there because that path leads to a more stable carbocation — the number of hydrogens is just a convenient way to predict the outcome, not the cause.

The One Big Exception …

Why this formula?

Markovnikov Addition: Why the Rule Holds

Markovnikov's rule is not a formula in the algebraic sense — it's a predictive principle for electrophilic addition to unsymmetrical alkenes. The "why" comes from carbocation stability and reaction mechanism.

The Rule in Words

When H–X adds to an unsymmetrical alkene, the hydrogen attaches to the carbon with more hydrogen atoms already attached, and the halogen (or X group) attaches to the carbon with fewer hydrogen atoms.

Example:

Propene ( $\ce{CH3-CH=CH2}$ ) + HBr → 2-bromopropane (major product), not 1-bromopropane.

Why This Happens: The Step-by-Step Reasoning

1. The Mechanism (Electrophilic Addition)

The reaction proceeds in two steps:

Slow step (rate-determining): The alkene's $\pi$ bond attacks the electrophilic $\ce{H+}$ from H–X, forming a carbocation intermediate.
Fast step: The carbocation is attacked by the nucleophilic $\ce{X-}$ .

2. The Key: Carbocation Stability

The more stable carbocation intermediate forms faster and determines the major product.

Carbocation Type	Stability Order	Reason
Tertiary ( $3^\circ$ )	Most stable	+3 alkyl groups donate electron density via hyperconjugation and inductive effect
Secondary ( $2^\circ$ )	Intermediate	+2 alkyl groups
Primary ( $1^\circ$ )	Least stable	+1 alkyl group
Methyl ( $\ce{CH3+}$ )	Unstable	No alkyl stabilization

3. Applying to Propene + HBr

Propene: $\ce{CH3-CH=CH2}$

Two possible protonation sites:

Path A (Markovnikov): $\ce{H+}$ adds to $\ce{CH2}$ (terminal carbon) → forms secondary carbocation:

\ce{CH3-CH+-CH3} \quad (2^\circ)

Path B (Anti-Markovnikov): $\ce{H+}$ adds to $\ce{CH}$ (middle carbon) → forms primary carbocation:

\ce{CH3-CH2-CH2+} \quad (1^\circ)

Result: The secondary carbocation is more stable (by ~25–30 kJ/mol), so Path A is faster. The $\ce{Br-}$ then attacks the positively charged carbon, giving 2-bromopropane.

The "Formula" — A Stability-Based Prediction

There is no algebraic formula, but a decision rule:

\text{Major product} = \text{Product from the more stable carbocation}

For alkenes with alkyl substituents, the stability order is:

$\text{Tertiary} > \text{Secondary} > \text{Primary} > \text{Methyl}$ …

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The key is Markovnikov’s rule: the hydrogen of HCl adds to the less substituted carbon of the double bond, placing chlorine on the more substituted carbon. The alkene that gives 1-chloro-1-methylcyclohexane is methylenecyclohexane.

Why Markovnikov’s rule decides the product

When HCl adds to an unsymmetrical alkene, the hydrogen atom attaches to the carbon that already has more hydrogens (the less substituted carbon), and the chlorine attaches to the carbon with fewer hydrogens (the more substituted carbon). This is because the intermediate carbocation is more stable when the positive charge is on a more substituted carbon — tertiary > secondary > primary.

Our target product, 1-chloro-1-methylcyclohexane, has the chlorine and the methyl group both on the same carbon (carbon-1 of the ring). That carbon is tertiary. So the double bond in the starting alkene must be positioned so that after H⁺ adds, the positive charge ends up exactly at that tertiary carbon.

Step-by-step reasoning

Identify the target structure.

1-chloro-1-methylcyclohexane is a cyclohexane ring with a methyl group and a chlorine atom both attached to carbon-1. That carbon is tertiary (connected to three other carbons: two from the ring and one from the methyl group).
Work backwards: what carbocation forms?

For HCl to add and place Cl at carbon-1, the intermediate carbocation must have the positive charge at carbon-1. That carbocation is tertiary — very stable. So the alkene must be such that when H⁺ adds, it creates this tertiary carbocation.
Where must the double bond be?

Two different alkenes both lead to a tertiary carbocation at ring-C1 via Markovnikov addition:
- 1-methylcyclohexene (double bond between ring C1 and C2, methyl on C1): H⁺ adds to C2, leaving the positive charge on C1 — a tertiary carbocation.
- Methylenecyclohexane (an exocyclic double bond between ring-C1 and an outside $CH_2$ group): H⁺ adds to the exocyclic $CH_2$ , placing the positive charge on ring-C1 — the SAME tertiary carbocation. Both routes converge on the identical carbocation intermediate, so both genuinely give 1-chloro-1-methylcyclohexane on chloride attack. There is no real chemistry reason to prefer one over the other — the standard, expected answer for this question is methylenecyclohexane. …

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Here's a breakdown of the common mistakes students make on this specific organic chemistry problem, along with how to avoid each.

The Core Concept: Markovnikov's Rule & Carbocation Stability

The reaction is an electrophilic addition of HCl to an alkene. The key to getting the correct product (1-chloro-1-methylcyclohexane) is identifying an alkene where straightforward Markovnikov addition (H⁺ to the carbon with more hydrogens) places the resulting cation -- and therefore the Cl -- directly on the carbon that ends up bearing both the methyl group and the chlorine.

Common Mistake #1: Assuming methylenecyclohexane needs a rearrangement

The Mistake: Students assume that protonating methylenecyclohexane's exocyclic =CH2 gives a SECONDARY carbocation on the ring, requiring a 1,2-hydride shift to reach the tertiary centre needed for the product.

Why it's wrong: Count the ring carbon's bonds carefully. Methylenecyclohexane's double bond is between the ring carbon (C1) and the exocyclic CH2. C1 is already bonded to its two ring neighbours (C2 and C6) BEFORE the reaction even starts -- when H⁺ adds to the exocyclic CH2 (which has 2 H's, more than the ring carbon's zero H's, so Markovnikov's rule picks it), the cation forms on C1, which is now bonded to C2, C6, AND the new CH3 group -- three carbons. That's already a tertiary carbocation, formed directly, with no rearrangement needed at all.

How to Avoid: Always count ALL of a carbon's bonds -- including the ring bonds it already had before the double bond reacts -- not just the new bond being formed. A ring carbon that's part of an exocyclic double bond is already attached to both of its ring neighbours, so protonating the exocyclic carbon immediately gives a tertiary (not secondary) cation.

Common Mistake #2: Skipping the mechanism in a mechanism question

The Mistake: Students write only the starting alkene and the final product, skipping the carbocation intermediate.

Why it's wrong: This is effectively a mechanism question -- you need to show why the product forms, not just state it.

How to Avoid: Always write the full mechanism in two steps:

Protonation: H⁺ from HCl adds to the exocyclic CH2 carbon (more hydrogens), forming the tertiary carbocation directly on the ring carbon.
Nucleophilic attack: Cl⁻ (from the dissociated HCl) attacks that tertiary carbocation.

Common Mistake #3: Mislabeling the Product (Incorrect IUPAC Name)

The Mistake: Students write the correct structure but misplace the locants when naming it.

Why it's wrong: The name must be precise -- both the chlorine and the methyl group are on the SAME ring carbon (C1). …

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Q.Name the alkene which will yield 1-chloro-1-methylcyclohexane by its reaction with HCl. Write the reactions involved.

Concept understanding — Markovnikov Addition

The Intuition First

The Precise Statement

Why Does This Happen? The Real Chemistry

A Common Misconception

The One Big Exception …

Why this formula?

Markovnikov Addition: Why the Rule Holds

The Rule in Words

Why This Happens: The Step-by-Step Reasoning

1. The Mechanism (Electrophilic Addition)

2. The Key: Carbocation Stability

3. Applying to Propene + HBr

The "Formula" — A Stability-Based Prediction

Why Markovnikov’s rule decides the product

Step-by-step reasoning

Reaction

The Core Concept: Markovnikov's Rule & Carbocation Stability

Common Mistake #1: Assuming methylenecyclohexane needs a rearrangement

Common Mistake #2: Skipping the mechanism in a mechanism question

Common Mistake #3: Mislabeling the Product (Incorrect IUPAC Name)

Q.Name the alkene which will yield 1-chloro-1-methylcyclohexane by its reaction with HCl. Write the reactions involved.

Why Markovnikov’s rule decides the product

Step-by-step reasoning

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