Match the reagent from Column I which on reaction with CH3-CH=CH2 gives some product given in Column II as per

NCERT Exemplar · Q40

Q. Match the reagent from Column I which on reaction with CH3-CH=CH2 gives some product given in Column II as per the codes given below: Column I

(i) O3/Zn + H2O

(ii) KMnO4/H+

(iii) KMnO4/OH-

(iv) H2O/H+

(v) B2H6/NaOH and H2O2
Column II

(a) Acetic acid and CO2

(b) Propan-1-ol

(d) Acetaldehyde and formaldehyde

(e) Propane-1,2-diol

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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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Propene undergoes different reactions depending on the reagent: ozonolysis cleaves the double bond to give aldehydes or ketones; acidic KMnO₄ cleaves it to carboxylic acids and CO₂; basic KMnO₄ gives a diol; acid-catalysed hydration gives the more stable carbocation product (propan-2-ol); hydroboration-oxidation gives the anti-Markovnikov alcohol (propan-1-ol).

The key to matching these reagents is understanding the mechanism each one follows. Propene ( $CH_3-CH=CH_2$ ) is an unsymmetrical alkene, so regiochemistry matters in some reactions.

Let’s go through each reagent step by step.

O₃/Zn + H₂O (Ozonolysis)

Ozone adds across the double bond to form an ozonide, which is then reductively cleaved by Zn/H₂O. The double bond is broken completely, and each carbon gets a carbonyl group.

For propene:

$CH_3-CH=CH_2 \xrightarrow{O_3/Zn, H_2O} CH_3CHO + HCHO$

The products are acetaldehyde and formaldehyde. This matches (d).
KMnO₄/H⁺ (Hot acidic permanganate)

Under hot acidic conditions, KMnO₄ is a strong oxidising agent. It cleaves the double bond completely, oxidising the alkene carbons to their highest oxidation states.

For propene:

$CH_3-CH=CH_2 \xrightarrow{KMnO_4/H^+, \Delta} CH_3COOH + CO_2$

The terminal carbon ( $=CH_2$ ) becomes $CO_2$ , and the internal carbon ( $CH_3-CH=$ ) becomes acetic acid. This matches (a).
KMnO₄/OH⁻ (Cold basic permanganate)

In cold, dilute, alkaline conditions, KMnO₄ acts as a hydroxylating agent. It adds two —OH groups across the double bond syn (from the same side), forming a vicinal diol. No bond cleavage occurs.

For propene:

$CH_3-CH=CH_2 \xrightarrow{KMnO_4/OH^-, cold} CH_3-CH(OH)-CH_2OH$

The product is propane-1,2-diol. This matches (e).
H₂O/H⁺ (Acid-catalysed hydration)

Water adds across the double bond in the presence of an acid catalyst. The reaction follows Markovnikov’s rule: the hydrogen adds to the less substituted carbon (the $=CH_2$ end), and the —OH adds to the more substituted carbon (the $CH_3-CH=$ end). This happens because the intermediate carbocation is more stable at the secondary carbon.

$CH_3-CH=CH_2 \xrightarrow{H_2O/H^+} CH_3-CH(OH)-CH_3$

The product is propan-2-ol. This matches (c).
B₂H₆/NaOH and H₂O₂ (Hydroboration-oxidation) …

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Q.Match the reagent from Column I which on reaction with CH3-CH=CH2 gives some product given in Column II as per the codes given below: Column I

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Q.Match the reagent from Column I which on reaction with CH3-CH=CH2 gives some product given in Column II as per the codes given below:
Column I

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