The addition of HCl to an alkene proceeds in two steps. The first step is the attack of the proton H+ on the c

NCERT Exemplar · Q15

Q. The addition of HCl to an alkene proceeds in two steps. The first step is the attack of the proton H+ on the carbon-carbon double bond >C=C<. Several representations show the proton approaching the double bond together with a curved (electron-pushing) arrow that indicates how the electron pair moves. Which representation correctly shows this first step?

(i) H is drawn approaching the double bond with the curved arrow originating at H (pointing from the hydrogen toward a doubly-bonded carbon), i.e. the electrons are shown flowing out of H into the alkene.

(ii) H+ is drawn near the double bond, but the curved arrow originates at a carbon atom / is directed so that the electron pair does not flow cleanly from the pi bond onto H+.

(iii) H+ is drawn near the double bond with a single curved arrow that starts at the C=C pi bond and points to H+, showing the pi-electron pair attacking the proton.

(iv) All of these are possible.

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Concept understanding — Electrophilic Addition Reactions

Electrophilic Aromatic Substitution – The First Meeting

Imagine you have a benzene ring — that perfect, flat hexagon of six carbons with alternating double bonds. It's stable, almost stubbornly so. You want to attach something new to it, say a bromine atom or a nitro group. But benzene doesn't react like an alkene. It doesn't just add across a double bond. Instead, it does something more elegant: it kicks out a hydrogen and keeps its aromatic ring intact.

That's the heart of Electrophilic Aromatic Substitution (EAS).

The Intuition: Why "Substitution" and Not "Addition"?

Benzene's stability comes from its delocalised $\pi$ electrons — a cloud above and below the ring. This cloud is electron-rich, so it attracts electrophiles (electron-loving species). But if an electrophile simply added to a double bond, the ring would break its aromaticity, losing that huge stabilisation. That would be energetically costly.

So benzene does something smarter: it lets the electrophile attack, temporarily breaks aromaticity to form a high-energy intermediate (the arenium ion), and then loses a proton to restore the aromatic ring. The net result? A hydrogen is replaced by the electrophile. The ring is back to its stable, aromatic self.

Note

The key trade-off: temporary loss of aromaticity is acceptable because the final product regains it. Addition reactions would permanently destroy aromaticity — benzene avoids that.

The Precise Statement

Electrophilic Aromatic Substitution is a reaction in which an electrophile ( $E^+$ ) replaces a hydrogen atom on an aromatic ring, proceeding through a sigma complex (arenium ion) intermediate, and restoring aromaticity after deprotonation.

The general equation:

\ce{Ar-H + E+ -> Ar-E + H+}

where Ar represents an aromatic ring.

The Mechanism in Three Steps

Generation of the electrophile – Many EAS reactions need a catalyst to create a strong enough $E^+$ . For example, bromination uses $\ce{FeBr3}$ to polarise $\ce{Br2}$ into $\ce{Br^+}$ .
Attack by the aromatic ring – The $\pi$ electrons of benzene attack the electrophile, forming a sigma complex (also called the arenium ion or Wheland intermediate). This intermediate is non-aromatic — it has four $\pi$ electrons delocalised over five carbons, and one $sp^3$ carbon bearing the electrophile and a hydrogen.

\text{Benzene} + E^+ \longrightarrow \text{Sigma complex (non-aromatic)}

Deprotonation – A base (often the counterion of the catalyst, like $\ce{FeBr4-}$ ) removes the proton from the $sp^3$ carbon. The pair of electrons from the C–H bond flows back into the ring, restoring the aromatic sextet.

\text{Sigma complex} + \text{Base} \longrightarrow \text{Product} + \ce{H+}

Watch out

The sigma complex is not aromatic. It's a high-energy intermediate. Students often mistakenly think it's still aromatic — it isn't. That's why the step is fast and the complex is short-lived.

Why This Matters for Exams

EAS is the gateway to understanding how to put groups onto benzene rings. The rate-determining step is usually the formation of the sigma complex (step 2). The regiochemistry (where the electrophile goes) depends on whether the ring already has a substituent — that's the topic of activating/deactivating groups and ortho/para vs. meta directors. …

Why this formula?

Electrophilic Addition Reactions: Why the Mechanism Works

Electrophilic addition is a cornerstone of alkene and alkyne chemistry. Instead of memorising the "arrow pushing," let's understand why the reaction proceeds the way it does — driven by electron density, stability, and charge.

1. The Core Idea: Why Alkenes React This Way

Alkenes have a π-bond — a cloud of electrons above and below the plane of the σ-bond. This π-electron cloud is:

Electron-rich (nucleophilic)
Exposed (not shielded by σ-bonds like in alkanes)

An electrophile (electron-lover) is attracted to this high electron density. The reaction is electrophilic addition because the electrophile attacks first.

Key principle: The π-bond acts as a Lewis base (electron donor). The electrophile is a Lewis acid (electron acceptor).

2. The General Mechanism (Two-Step)

Step 1: Formation of a Carbocation (or Bridged Intermediate)

The electrophile (E⁺) attacks the π-bond. The π-electrons form a new σ-bond to E⁺, leaving the other carbon with a positive charge — a carbocation.

\ce{C=C} + \ce{E+} \longrightarrow \ce{C+-C-E}

Why does this happen?

The π-bond is weaker than a σ-bond (~260 kJ/mol vs ~350 kJ/mol). Breaking the π-bond to form a σ-bond is energetically favourable because the new σ-bond is stronger. The carbocation is a high-energy intermediate, but it's stabilised by:

Hyperconjugation (alkyl groups donate electron density)
Inductive effect (alkyl groups push electrons toward the positive carbon)

Step 2: Nucleophilic Attack

A nucleophile (Nu⁻) attacks the carbocation, forming a second σ-bond.

\ce{C+-C-E} + \ce{Nu-} \longrightarrow \ce{C-Nu-C-E}

Why does this happen?

The carbocation is electron-deficient (positive charge). The nucleophile is electron-rich. Opposite charges attract — this is electrostatic and orbital overlap driven.

3. The Key "Formula" — Markovnikov's Rule

Statement: In the addition of HX to an unsymmetrical alkene, the hydrogen attaches to the carbon with more hydrogens already, and the X attaches to the carbon with fewer hydrogens.

Why does this rule hold? (The reasoning)

Consider propene: $\ce{CH3-CH=CH2}$ + HBr.

Possible carbocations:
1. Primary carbocation: $\ce{CH3-CH+-CH2Br}$ (less stable)
2. Secondary carbocation: $\ce{CH3-CHBr-CH2+}$ (more stable)

The more substituted carbocation (secondary > primary) is more stable due to:

Hyperconjugation: More alkyl groups = more C–H σ-bonds that can donate electron density into the empty p-orbital of the carbocation.
Inductive effect: Alkyl groups are electron-donating, stabilising the positive charge.

Result: The reaction proceeds via the more stable carbocation, leading to Markovnikov addition.

Markovnikov's rule is not a law — it's a consequence of carbocation stability.

4. The "Anti-Markovnikov" Exception (Why It Happens)

With HBr in the presence of peroxides (ROOR), the addition is anti-Markovnikov — Br goes to the less substituted carbon.

Why? The mechanism changes from ionic to free-radical.

Peroxide decomposes to radicals: $\ce{ROOR -> 2 RO•}$
RO• abstracts H from HBr: $\ce{RO• + HBr -> ROH + Br•}$
Br• adds to the alkene — at the less substituted carbon (because the radical formed is more stable — tertiary > secondary > primary).
The new radical abstracts H from another HBr, regenerating Br•. …

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Adding HCl to an alkene is electrophilic addition. In the first step the electron-poor proton H+ (the electrophile) is attacked by the electron-rich pi bond of the alkene. A curved arrow therefore starts at the C=C double bond and points to H+, producing a carbocation. Only option (iii) draws the arrow in this correct direction.

Concept: which species is the nucleophile?

A curved (double-barbed) arrow always points from a source of electrons to an electron-poor site.

H+ is a bare proton — it has no lone pair and no bonding electrons to donate, so an arrow can never start on H+.
The C=C double bond has a loosely held pi electron pair, which makes the alkene nucleophilic.

Hence the electron pair flows from the pi bond to H+:

C=C< + H+ gives a carbocation >C(+)—CH<

Why each option is right or wrong

(i): the arrow is shown coming out of H toward the alkene, implying that hydrogen donates electrons. A proton has none — wrong direction. …

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Concept: which species is the nucleophile?

Why each option is right or wrong

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