Match the reactions given in Column I with the reaction types in Column II. Column I (i) CH2=CH2 + H2O --H+-->

NCERT Exemplar · Q43

Q. Match the reactions given in Column I with the reaction types in Column II. Column I

(i) CH2=CH2 + H2O --H+--> CH3CH2OH

(ii) CH2=CH2 + H2 --Pd--> CH3-CH3

(iii) CH2=CH2 + Cl2 --> Cl-CH2-CH2-Cl

(iv) 3 CH≡CH --Cu tube/Heat--> C6H6
Column II

(a) Hydrogenation

(b) Halogenation

(d) Hydration

(e) Condensation

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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 idea is to match each reaction of ethene or ethyne with its characteristic type: hydration adds water, hydrogenation adds hydrogen, halogenation adds halogen, and cyclotrimerisation polymerises three alkynes into benzene.

Reaction (i): $CH_2=CH_2 + H_2O \xrightarrow{H^+} CH_3CH_2OH$

Water adds across the double bond in the presence of an acid catalyst. This is the classic hydration of an alkene to form an alcohol. The hydrogen from water goes to one carbon, and the hydroxyl group goes to the other, following Markovnikov’s rule (though ethene is symmetric, so the product is unique).

Match: (i) → (d) Hydration.
Reaction (ii): $CH_2=CH_2 + H_2 \xrightarrow{Pd} CH_3-CH_3$

Molecular hydrogen adds across the double bond on a palladium catalyst. This reduces the alkene to an alkane. The process is called hydrogenation.

Match: (ii) → (a) Hydrogenation.
Reaction (iii): $CH_2=CH_2 + Cl_2 \longrightarrow Cl-CH_2-CH_2-Cl$

Chlorine adds directly to the double bond, giving a vicinal dichloride. This is an addition of halogen — specifically, halogenation.

Match: (iii) → (b) Halogenation.
Reaction (iv): $3\, CH \equiv CH \xrightarrow{Cu\, tube/Heat} C_6H_6$

Three molecules of ethyne (acetylene) join together to form benzene. This is a cyclotrimerisation, which is a type of polymerisation (specifically, addition polymerisation where monomers link to form a cyclic product). …

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Q.Match the reactions given in Column I with the reaction types in Column II.
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