Q.Which of the following does not follow Markovnikov's rule?
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The Intuition First
Imagine you have a double bond between two carbon atoms. That double bond is like a crowded room — it has a lot of electron density, and it's looking for something to react with. When you add H–X (like HCl, HBr, or water in acid), the molecule splits into H⁺ and X⁻. The question is: which carbon gets the H⁺, and which gets the X⁻?
If the alkene is symmetrical — say, ethene — it doesn't matter. Both carbons are identical. But if the alkene is unsymmetrical — like propene, where one carbon has two hydrogens and the other has one hydrogen plus a methyl group — then the two carbons are different. Which one gets the hydrogen?
The answer comes from stability. The reaction goes through a carbocation intermediate — a carbon with a positive charge. That carbocation is unstable and wants to be as stable as possible. So the H⁺ will add to the carbon that leads to the more stable carbocation.
Carbocation stability order: tertiary > secondary > primary > methyl. More alkyl groups attached to the positive carbon stabilise it by hyperconjugation and inductive effect.
The Precise Statement
Markovnikov's Rule (formally, Vladimir Markovnikov, 1870) states:
In the addition of a protic acid HX to an unsymmetrical alkene, the hydrogen atom adds to the carbon of the double bond that already has the greater number of hydrogen atoms.
That is: "the rich get richer" — the carbon with more hydrogens gets one more hydrogen.
Why It Works — The Mechanism
Take propene (CHX3−CH=CHX2) and add HBr.
The double bond has two carbons:
- Carbon 1 (terminal): has 2 hydrogens
- Carbon 2 (internal): has 1 hydrogen
The H⁺ can add to either carbon. If it adds to carbon 1, you get a secondary carbocation (the positive charge is on carbon 2, which is attached to one methyl and one hydrogen). If it adds to carbon 2, you get a primary carbocation (positive charge on carbon 1, attached to two hydrogens and one methyl).
The secondary carbocation is more stable. So the H⁺ adds to carbon 1 — the one with more hydrogens — giving the secondary carbocation. Then Br⁻ attacks the positive carbon, giving 2-bromopropane as the major product.
The rule is a consequence of carbocation stability, not a separate law. If you ever forget the rule, just ask: "Which carbocation is more stable?"
The Product
For propene + HBr:
- Major product: CHX3−CHBr−CHX3 (2-bromopropane)
- Minor product: CHX3−CHX2−CHX2Br (1-bromopropane) …
Markovnikov's rule distinguishes between two different possible carbons for H/X addition, so it has no distinguishing effect on a symmetrical alkene where both alkene carbons carry the same substitution. …
CH3CH=CHCH3 (2-butene) is a symmetrical alkene, so Markovnikov's rule (which predicts a preference between two different possible addition products) does not apply in the usual sense.
Markovnikov's rule predicts that, in the addition of HX to an unsymmetrical alkene, the H atom adds to the carbon already bearing more hydrogen atoms (equivalently, X adds to the more substituted carbon, forming the more stable carbocation intermediate). This rule is meaningful only when the two alkene carbons are different, since only then are there two distinguishable possible products.
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- CBSE 2026Set ANNUAL1 markQ.Write Markownikoff's rule.
›Reveal solutionSolution
Markovnikov's rule: the negative part of an unsymmetrical reagent adds to the carbon atom of the double bond that has fewer hydrogen atoms.
When an unsymmetrical reagent such as HX (X = halogen, or H-OH, etc.) adds across an unsymmetrical carbon-carbon double bond (i.e. a double bond where the two carbons are not equivalently substituted), two different products are, in principle, possible. Markovnikov's rule (given by Vladimir Markovnikov in 1869) predicts which one predominates: the hydrogen atom of the reagent goes to the carbon of the double bond already carrying more hydrogen atoms, and the other (negative/electronegative) part of the reagent goes to the carbon with fewer hydrogen atoms (the more substituted carbon).
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- CBSE 2025Set ANNUAL1 markMCQQ.Which of the following does not follow Markovnikov's rule?(a) CH3CH=CH2(b) CH3CH=CHCH3(c) (CH3)2CH-CH=CH2(d) CH3CH2CH=CH2
›Reveal solutionSolution
CH3CH=CHCH3 (2-butene) is a symmetrical alkene, so Markovnikov's rule (which predicts a preference between two different possible addition products) does not apply in the usual sense.
Markovnikov's rule predicts that, in the addition of HX to an unsymmetrical alkene, the H atom adds to the carbon already bearing more hydrogen atoms (equivalently, X adds to the more substituted carbon, forming the more stable carbocation intermediate). This rule is meaningful only when the two alkene carbons are different, since only then are there two distinguishable possible products.
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- CBSE 2025Set ANNUAL1 markMCQQ.CH3-CH=CH2 + HBr -> A, A is(a) 2-Bromopropane(b) 1-Bromopropane(c) 3-Bromopropane(d) Propane
›Reveal solutionSolution
Markovnikov's rule: when HX adds across an unsymmetrical alkene, H goes to the carbon that already has MORE hydrogens, and X (here Br) goes to the more substituted carbon — this gives 2-bromopropane, not 1-bromopropane.
Reaction: CH3-CH=CH2 + HBr -> A
The double bond is between C2 and C3 of propene. This is an UNSYMMETRICAL alkene, since C2 (=CH-) has 1 H and 1 alkyl group (CH3) attached, while C3 (=CH2) has 2 H atoms attached — the two double-bond carbons are chemically different.
Markovnikov's rule (the classic 'rich get richer' rule for electrophilic addition to alkenes) states that in the addition of HX to an unsymmetrical alkene, the hydrogen atom attaches to the carbon of the double bond that already bears the greater number of hydrogen atoms, while the halogen attaches to the carbon with fewer hydrogens (the more substituted carbon).
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- CBSE 2021Set annual1 markQ.Complete the following equation:(e) H3C-C(CH3)=CH2 + H2O -> A [condition: H+, Markovnikov addition].
›Reveal solutionSolution
Isobutylene + H2O/H+ gives tert-butyl alcohol by Markovnikov addition.
The starting alkene, H3C-C(CH3)=CH2, is isobutylene (2-methylpropene), (CH3)2C=CH2. Acid-catalysed hydration (H+ catalyst, water as nucleophile) of an unsymmetrical alkene follows Markovnikov's rule: the reaction proceeds via protonation of the double bond to form the more stable carbocation, and then water attacks that carbocation.
Mechanism:
- H+ adds to the terminal =CH2 carbon (which has more H's already), generating the tertiary carbocation (CH3)3C+ (much more stable than the alternative primary carbocation).
- Water then attacks this tertiary carbocation, and loss of a proton gives the alcohol: (CH3)3C+ + H2O -> (CH3)3C-OH2+ -> (CH3)3C-OH + H+
Overall: (CH3)2C=CH2 + H2O --(H+)--> (CH3)3C-OH …
- CBSE 2018Set annual1 markMCQQ.Addition of HBr to 3-methyl-1-pentyne follows:(a) Markownikov's rule(b) Anti-Markownikov's rule(c) Saytzeff rule(d) None of these
›Reveal solutionSolution
Simple (non-peroxide) HBr addition to an unsymmetrical alkyne follows Markovnikov's rule: H goes to the carbon with more hydrogens, Br to the more substituted carbon, via the more stable carbocation intermediate.
Markovnikov's rule
When a protic acid HX (such as HBr) adds across an unsymmetrical carbon-carbon multiple bond (double or triple bond), the hydrogen atom of HX attaches to the carbon of the multiple bond that already carries the greater number of hydrogen atoms, while the halogen (X) attaches to the carbon bearing fewer hydrogens (the more substituted carbon). Mechanistically, this happens because the reaction proceeds through the more stable carbocation intermediate - protonation occurs in the way that generates the more substituted (more stable) carbocation, which is then attacked by the bromide ion.
3-methyl-1-pentyne: HC triple-bond C-CH(CH3)-CH2-CH3. The terminal alkyne carbon bears no alkyl substituents, while the internal alkyne carbon is attached to the rest of the chain (more substituted). Following Markovnikov's rule, H adds to the terminal carbon and Br adds to the internal (more substituted) carbon, via the more stable carbocation.
Why not anti-Markovnikov?
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- CBSE 2018Set ANNUAL1 markQ.State Markonikoff's rule.
›Reveal solutionSolution
Markovnikov's rule: 'rich gets richer' -- H adds to the carbon of the C=C that already has more hydrogens.
Step 1 -- statement: When an unsymmetrical reagent such as HX (X = halogen) is added across the double bond of an unsymmetrical alkene, the negative part (X) of the reagent attaches to the carbon atom bearing the fewer number of hydrogen atoms, while the positive part (H) attaches to the carbon atom already bearing the greater number of hydrogen atoms. …
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