Which of the following is halogen exchange reaction? (i) mathrmRX + NaI arrow RI + NaX (ii) mathrm>C=C< + HX a

NCERT Exemplar · Q5

Q. Which of the following is halogen exchange reaction?

(i)

\mathrm{RX + NaI \rightarrow RI + NaX}

(ii)

\mathrm{>C{=}C< + HX \rightarrow >\!\underset{\underset{\displaystyle H}{|}}{C}-\underset{\underset{\displaystyle X}{|}}{C}\!<}

(iii)

\mathrm{R-OH + HX \xrightarrow{ZnCl_2} R-X + H_2O}

(iv)

\mathrm{C_6H_5CH_3 + X_2 \xrightarrow[dark]{Fe} \textit{o-}X\text{-}C_6H_4CH_3 + \textit{p-}X\text{-}C_6H_4CH_3}

, shown below:

Toluene + X2/Fe (dark) giving ortho- and para-halotoluene, drawn as real benzene rings matching the NCERT Exemplar page (Cl used as the concrete halogen; X = halogen generically) — Figure

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Concept understanding — Nucleophilic Addition

Nucleophilic Addition – From Intuition to Precision

Imagine you have a molecule with a carbon–oxygen double bond — a carbonyl group ( $\ce{C=O}$ ). That oxygen is greedy for electrons; it pulls them away from carbon, leaving the carbon slightly positive ( $\delta^+$ ) and the oxygen slightly negative ( $\delta^-$ ). Now, if you bring a species that is rich in electrons (a nucleophile, meaning "nucleus-loving"), it will naturally be attracted to that electron-deficient carbon. The nucleophile attacks the carbon, the $\pi$ bond breaks, and the oxygen picks up a proton (or some other electrophile) to become stable. That, in a nutshell, is nucleophilic addition.

Note

The key idea: a nucleophile adds across a polar multiple bond (usually $\ce{C=O}$ or $\ce{C≡N}$ ), breaking the $\pi$ bond and forming two new sigma bonds.

The Precise Statement

Nucleophilic addition is a reaction in which a nucleophile (an electron-rich species) forms a sigma bond with an electrophilic carbon atom of a polar multiple bond (typically a carbonyl group, $\ce{C=O}$ , or a nitrile, $\ce{C≡N}$ ), while the $\pi$ bond breaks. The resulting intermediate then captures a proton (or another electrophile) to give a neutral product.

In general form:

\ce{R2C=O + Nu^- ->[H^+] R2C(OH)Nu}

The nucleophile ( $\ce{Nu^-}$ ) attacks the carbonyl carbon; the oxygen becomes negatively charged; then a proton ( $\ce{H^+}$ ) from the medium attaches to the oxygen, yielding an alcohol.

Why It Happens – The Driving Force

The carbonyl carbon is electrophilic because:

Oxygen is more electronegative than carbon, so the $\ce{C=O}$ bond is polarised: $\ce{C^{\delta+}=O^{\delta-}}$ .
The $\pi$ bond is weaker than a $\sigma$ bond, so it can break relatively easily.

A nucleophile (like $\ce{OH^-}$ , $\ce{CN^-}$ , or $\ce{NH3}$ ) has a lone pair or a negative charge. It seeks positive centres. The attack forms a new $\sigma$ bond, and the $\pi$ electrons move entirely to oxygen, creating an alkoxide ion ( $\ce{R2C-O^-}$ ). This intermediate is then quenched by a proton.

Watch out

A common mistake: thinking the nucleophile attacks the oxygen. No — oxygen is already electron-rich; the nucleophile goes to the carbon because it is electron-deficient.

A Concrete Example – Addition of HCN to a Ketone

Take acetone ( $\ce{CH3COCH3}$ ) and hydrogen cyanide ( $\ce{HCN}$ ). In the presence of a base, $\ce{CN^-}$ (the nucleophile) attacks the carbonyl carbon:

\ce{CH3COCH3 + CN^- -> CH3C(O^-)(CN)CH3}

The alkoxide intermediate then picks up a proton from $\ce{HCN}$ (or from water) to give a cyanohydrin:

\ce{CH3C(O^-)(CN)CH3 + H^+ -> CH3C(OH)(CN)CH3}

The product is acetone cyanohydrin. Notice: two new sigma bonds formed ( $\ce{C-CN}$ and $\ce{O-H}$ ), and the $\pi$ bond is gone.

What Makes a Good Nucleophile?

Strong nucleophiles are usually negatively charged or have lone pairs:

$\ce{OH^-}$ , $\ce{CN^-}$ , $\ce{NH2^-}$ , $\ce{CH3O^-}$ , $\ce{H^-}$ (from hydride reagents like $\ce{NaBH4}$ or $\ce{LiAlH4}$ )
Neutral but polarisable: $\ce{NH3}$ , $\ce{H2O}$ (weaker, but can add under acidic conditions) …

Why this formula?

Nucleophilic Addition: Why the Mechanism Works the Way It Does

Let's build this from first principles — understanding why nucleophilic addition happens, not just memorising the steps.

1. The Core Problem: Why Does Addition Happen at All?

A carbonyl group ( $\ce{C=O}$ ) has a polarised double bond:

Oxygen is more electronegative than carbon → it pulls electron density toward itself.
This creates a partial positive charge on carbon ( $\delta^+$ ) and a partial negative charge on oxygen ( $\delta^-$ ).

\ce{C^{\delta+}=O^{\delta-}}

Key insight: The carbon is electron-deficient — it wants electrons. A nucleophile (Nu⁻) is electron-rich — it wants to give electrons. This is a natural match.

2. The Two-Step Mechanism (Why Two Steps?)

Step 1: Nucleophilic Attack (Slow, Rate-Determining)

The nucleophile donates its lone pair to the electrophilic carbonyl carbon.

\ce{Nu^- + C=O -> [Nu-C-O]^-}

Why this happens:

The $\pi$ bond between C and O breaks — the electrons move entirely to oxygen.
Oxygen now has a full negative charge (alkoxide ion).
The carbon changes from $sp^2$ (trigonal planar) to $sp^3$ (tetrahedral).

This step is slow because the $\pi$ bond must break — it requires energy.

Step 2: Protonation (Fast)

The negatively charged oxygen picks up a proton ( $\ce{H^+}$ ) from the solvent or acid.

\ce{[Nu-C-O]^- + H^+ -> Nu-C-OH}

Why this happens:

The alkoxide ion is a strong base — it wants to neutralise its charge.
Protonation gives a stable neutral alcohol product.

3. The Key Formula: Rate Law Derivation

For a general nucleophilic addition:

\ce{Nu^- + R2C=O ->[k] products}

The rate law comes from the slow step (Step 1):

\text{Rate} = k[\text{Nu}^-][\ce{R2C=O}]

Why this form?

The reaction is bimolecular — two species must collide with correct orientation.
Doubling either concentration doubles the rate (first order in each).
This is second order overall.

Exam tip: This is why nucleophilic addition is often called addition-elimination when followed by loss of a leaving group (like in acyl substitution), but here it's just addition.

4. Why the Tetrahedral Intermediate Forms (And Why It's Unstable)

The intermediate is tetrahedral ( $sp^3$ hybridised carbon):

Bond angles: ~109.5°
Four groups around carbon: Nu, R, R', O⁻

Why it's unstable:

The negative charge on oxygen is high-energy.
The tetrahedral geometry is sterically crowded (especially with bulky R groups).
The intermediate collapses quickly — either back to starting materials or forward to product. …

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A halogen exchange reaction is a nucleophilic substitution where one halogen in an organic halide is replaced by another halogen from an inorganic salt. The correct option is (i), because only it shows this direct displacement: $\mathrm{RX + NaI \rightarrow RI + NaX}$ .

The key is to recognise what "halogen exchange" actually means. It is a specific type of nucleophilic substitution reaction where the leaving group is a halide ion and the attacking nucleophile is also a halide ion. The net effect is that one halogen atom in an organic molecule is swapped for another. This is most famously seen in the Finkelstein reaction, where an alkyl chloride or bromide is converted to an alkyl iodide using sodium iodide in acetone.

Why does this work so well? Iodide is a much better nucleophile than chloride or bromide (it is larger, more polarisable, and less solvated in acetone), and sodium chloride or bromide precipitates out of the acetone solution, driving the equilibrium forward. The reaction is essentially: $\mathrm{R-X + I^- \rightarrow R-I + X^-}$ .

Now, let's examine each option carefully.

Option (i): $\mathrm{RX + NaI \rightarrow RI + NaX}$

This is the textbook definition. An alkyl halide ( $\mathrm{RX}$ ) reacts with sodium iodide. The iodide ion ( $\mathrm{I^-}$ ) acts as a nucleophile and displaces the halide ion ( $\mathrm{X^-}$ ) from the carbon. The sodium salt of the displaced halide ( $\mathrm{NaX}$ ) is formed. This is a direct, one-step halogen exchange. This fits perfectly.
Option (ii): $\mathrm{>C{=}C< + HX \rightarrow >\!\underset{\underset{\displaystyle H}{|}}{C}-\underset{\underset{\displaystyle X}{|}}{C}\!<}$

This is the electrophilic addition of hydrogen halide to an alkene. The $\pi$ bond acts as a nucleophile, attacking the hydrogen of HX. The halide ion then adds to the carbocation. No halogen is being exchanged; a new carbon-halogen bond is being formed from an alkene. The starting material had no halogen at all. This is not a substitution, let alone a halogen exchange.
Option (iii): $\mathrm{R-OH + HX \xrightarrow{ZnCl_2} R-X + H_2O}$

This is the conversion of an alcohol to an alkyl halide using a hydrogen halide, often catalysed by zinc chloride (the Lucas reagent for secondary/tertiary alcohols). The hydroxyl group ( $\mathrm{-OH}$ ) is a poor leaving group; it gets protonated to become $\mathrm{-OH_2^+}$ , which then leaves as water. The halide ion from HX then attacks. Again, the starting material had no halogen — it had an alcohol. This is a substitution reaction, but it is not a halogen exchange because no halogen was present initially to be exchanged. …

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Concept: Halogen Exchange Reaction

A halogen exchange reaction is a nucleophilic substitution where one halogen atom in an organic compound is replaced by another halogen atom. The key is that both the reactant and product are alkyl/aryl halides — the halogen changes, but the carbon skeleton remains the same.

Method: Identify the Halogen Exchange

Step 1: Check if the reactant is an organic halide (R–X).

Step 2: Check if the product is also an organic halide (R–Y), where Y is a different halogen.

Step 3: If both conditions are true, it is a halogen exchange reaction.

Applying the Method

Option	Reactant	Product	Halogen exchange?
(i)	$\mathrm{RX}$ (alkyl halide)	$\mathrm{RI}$ (alkyl iodide)	✓ Yes — X replaced by I
(ii)	Alkene (no halogen)	Alkyl halide	✗ No — addition, not exchange

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Common Mistakes on Halogen Exchange Reactions

1. Confusing Halogen Exchange with Halogenation

Mistake: Students often pick (iv) (halogenation of toluene) or (iii) (conversion of alcohol to alkyl halide) thinking any reaction that introduces a halogen is a "halogen exchange."

Why it's wrong:

Halogen exchange specifically means one halogen atom replaces another on the same carbon.
In (iii), $\mathrm{-OH}$ is replaced by $\mathrm{-X}$ — that's substitution, not exchange.
In (iv), a hydrogen on the benzene ring is replaced by halogen — that's electrophilic aromatic substitution, not exchange of halogens.

How to avoid:

Remember the keyword: exchange = halogen for halogen.
Look for a reaction where both reactant and product have a halogen on the same carbon, but the halogen atom changes.

2. Picking Addition Reactions (Option (ii))

Mistake: Thinking addition of HX to an alkene involves "exchange" because a halogen appears in the product.

Why it's wrong:

Addition reactions add atoms across a double bond — no halogen is being replaced.
The starting alkene has no halogen to exchange.

How to avoid:

Check the reactant: does it already contain a halogen? If not, it cannot be a halogen exchange.
Halogen exchange requires a C–X bond in the starting material.

3. Misidentifying the Finkelstein Reaction (Option (i))

Mistake: Some students reject (i) because they think $\mathrm{NaI}$ is just a reagent, not part of an "exchange."

Why it's correct:

$\mathrm{RX + NaI \rightarrow RI + NaX}$ is the classic Finkelstein reaction — a textbook example of halogen exchange.
The halogen $\mathrm{X}$ (often Cl or Br) is replaced by $\mathrm{I}$ .

How to avoid:

Memorise the Finkelstein reaction as the prime example of halogen exchange.
Look for the pattern: alkyl halide + metal halide → different alkyl halide + metal halide.

4. Overlooking the Role of the Reagent …

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Q.Which of the following is halogen exchange reaction?

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