Q.CH3-CH=CH2 + HBr (No peroxide) -> CH3-CH2-CH2Br, is the major product. (True/False)
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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 CH3CH=CH2 (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 CH3CHBrCH3 (2-bromopropane), not CH3CH2CH2Br (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.
The stability order of carbocations is: tertiary > secondary > primary > methyl. Markovnikov addition always proceeds through the most stable carbocation possible.
A Common Misconception
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 (CHX3−CH=CHX2) + 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 π bond attacks the electrophilic HX+ from H–X, forming a carbocation intermediate.
- Fast step: The carbocation is attacked by the nucleophilic XX−.
2. The Key: Carbocation Stability
The more stable carbocation intermediate forms faster and determines the major product.
| Carbocation Type | Stability Order | Reason |
|---|---|---|
| Tertiary (3∘) | Most stable | +3 alkyl groups donate electron density via hyperconjugation and inductive effect |
| Secondary (2∘) | Intermediate | +2 alkyl groups |
| Primary (1∘) | Least stable | +1 alkyl group |
| Methyl (CHX3X+) | Unstable | No alkyl stabilization |
3. Applying to Propene + HBr
Propene: CHX3−CH=CHX2
Two possible protonation sites:
- Path A (Markovnikov): HX+ adds to CHX2 (terminal carbon) → forms secondary carbocation:
CHX3−CHX+−CHX3(2∘)
- Path B (Anti-Markovnikov): HX+ adds to CH (middle carbon) → forms primary carbocation:
CHX3−CHX2−CHX2X+(1∘)
Result: The secondary carbocation is more stable (by ~25–30 kJ/mol), so Path A is faster. The BrX− then attacks the positively charged carbon, giving 2-bromopropane.
The "Formula" — A Stability-Based Prediction
There is no algebraic formula, but a decision rule:
Major product=Product from the more stable carbocation
For alkenes with alkyl substituents, the stability order is:
Tertiary>Secondary>Primary>Methyl …
Whether HBr adds to an alkene in a Markovnikov or anti-Markovnikov sense depends entirely on whether peroxide is present, since the two conditions proceed through completely different mechanisms. …
The statement is FALSE: without peroxide, Markovnikov addition gives 2-bromopropane as the major product, not the 1-bromo isomer shown.
When HBr adds across the double bond of propene (CH3-CH=CH2), which product dominates depends on whether peroxide is present:
- In the ABSENCE of peroxide (normal ionic/electrophilic mechanism): the reaction follows Markovnikov's rule — H adds to the carbon of the double bond that already carries more hydrogens (the terminal CH2), and Br adds to the carbon with fewer hydrogens, because the reaction proceeds through the more stable secondary carbocation. This gives CH3-CHBr-CH3 (2-bromopropane) as the major product. …
- CBSE 2025Set ANNUAL1 markQ.Draw the structure of the major monohalo product: 1-methylcyclohexene (a cyclohexene ring with a CH3 substituent on one of the double-bond carbons) +HI→?
›Reveal solutionSolution
Markovnikov addition of HI to 1-methylcyclohexene puts I on the more substituted carbon (C-1, which already bears –CH3), via the more stable tertiary carbocation, giving 1-iodo-1-methylcyclohexane.
In 1-methylcyclohexene the double bond is between ring carbons C-1 and C-2, with a –CH3 group on C-1. Electrophilic addition of HI proceeds in two steps:
Step 1 (protonation): H+ adds to the alkene carbon that gives the more stable carbocation. Adding H+ to C-2 places the positive charge on C-1, which is a tertiary carbocation (bonded to the ring's C-2 and C-6, plus the –CH3 group). Adding H+ instead to C-1 would place the charge on C-2, only a secondary carbocation. Since 3° carbocations are more stable (greater hyperconjugation/inductive stabilisation from three alkyl groups) than 2°, the reaction proceeds through the C-1 cation — this is Markovnikov's rule (the H goes to the carbon already bearing more hydrogens, i.e. C-2).
1-methylcyclohexene+H+→1-methylcyclohexan-1-yl cation (3°, at C-1)
…
- CBSE 2024Set ANNUAL1 markQ.Complete the following reaction: 1-methylcyclohexene (a cyclohexene ring with a CH3 substituent on one of the double-bond carbons) +HI→?
›Reveal solutionSolution
Markovnikov's rule: with unsymmetrical alkenes, H+ from HX adds to the carbon that generates the more stable (here, tertiary) carbocation, and the halide ion then bonds to that carbon.
1-Methylcyclohexene has its ring double bond between C1 (bearing the CH3 substituent) and C2 (bearing only H). Protonation of the alkene can occur in two ways:
- H+ adds to C1 ⇒ carbocation forms at C2, a secondary carbocation (flanked by C1 and C3, both ring carbons).
- H+ adds to C2 ⇒ carbocation forms at C1, a tertiary carbocation (bonded to CH3, C2 and C6 — three carbon substituents). …
- CBSE 2020Set NC1 markQ.Write the reaction with conditions for conversion of 2-methylpropene into 1-bromo-2-methylpropane.
›Reveal solutionSolution
Ordinary Markovnikov addition of HBr to 2-methylpropene would put Br on the more-substituted carbon; getting the anti-Markovnikov product (Br on the terminal carbon) needs the peroxide-initiated free-radical mechanism (Kharasch/peroxide effect).
Target: 2-methylpropene, (CH3)2C=CH2 (isobutylene), converted into 1-bromo-2-methylpropane, (CH3)2CH–CH2Br — i.e. Br ends up on the terminal (less substituted) carbon, which is the opposite regiochemistry to normal Markovnikov addition (which would instead place Br on the more substituted carbon, giving 2-bromo-2-methylpropane).
…
- CBSE 2019Set ANNUAL1 markQ.Identify the products A and B formed in the following reaction: CH3–CH2–CH=CH–CH3+HCl→A+B
›Reveal solutionSolution
CH3–CH2–CH=CH–CH3 is pent-2-ene; electrophilic addition of HCl proceeds via protonation to give whichever secondary carbocation is more stabilised, so the products are a mixture of 2-chloropentane (major) and 3-chloropentane (minor).
The alkene CH3–CH2–CH=CH–CH3 is pent-2-ene, numbered C1H3–C2H2–C3H=C4H–C5H3, with the double bond between C3 and C4 (equivalently C2–C3 counting from the other end). Since this alkene is unsymmetrically substituted but both carbons of the double bond are internal, protonation of either carbon gives a secondary carbocation, so a mixture of two constitutional isomers is obtained:
- H+ adds to C3: leaves the cation at C4, giving CH3–CH2–CH2–CH+–CH3 — this secondary cation is flanked by a −CH2CH2CH3 (propyl) group on one side and a −CH3 on the other, i.e. more hyperconjugating α-hydrogens overall, making it the somewhat more stabilised carbocation. Cl− then attacks here, giving CH3–CH2–CH2–CHCl–CH3, i.e. 2-chloropentane (major). …
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