What happens when (i) n-butyl chloride is treated with alcoholic KOH, (ii) bromobenzene is treated with Mg in

Exercises · 6.22

Q. What happens when

(i) n-butyl chloride is treated with alcoholic KOH,

(ii) bromobenzene is treated with Mg in the presence of dry ether,

(iii) chlorobenzene is subjected to hydrolysis,

(iv) ethyl chloride is treated with aqueous KOH,

(v) methyl bromide is treated with sodium in the presence of dry ether,

(vi) methyl chloride is treated with KCN?

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Concept understanding — Ambident Nucleophile Reactivity

Ambident Nucleophile Reactivity

Most nucleophiles attack through a single, obvious atom — a single lone pair, a single reactive site. An ambident nucleophile is unusual: it has TWO different atoms that each carry enough electron density to act as the attacking site, so it can bond to an electrophile through either one, giving two structurally different products from the same reagent.

Why This Happens: Resonance Delocalisation

An ambident nucleophile's negative charge (or lone pair) is delocalised by resonance across more than one atom, so more than one atom is genuinely nucleophilic.

Cyanide ion, $CN^-$ : $^-C{\equiv}N: \;\leftrightarrow\; :C{=}N^-$ . Both the carbon and the nitrogen carry real electron density and can attack an electrophile.

Attack through carbon gives an alkyl cyanide (nitrile), $R{-}C{\equiv}N$ .
Attack through nitrogen gives an alkyl isocyanide (isonitrile), $R{-}N{\equiv}C$ .

Nitrite ion, $NO_2^-$ : the negative charge is shared between nitrogen and the oxygens.

Attack through oxygen gives an alkyl nitrite, $R{-}O{-}N{=}O$ .
Attack through nitrogen gives a nitroalkane, $R{-}NO_2$ .

What Decides Which End Attacks: The Counter-Ion Matters

For cyanide specifically, the identity of the metal counter-ion changes which end of $CN^-$ ends up bonded to the electrophile — this is the classic KCN-vs-AgCN contrast:

KCN is genuinely ionic: it dissociates fully to give a FREE $CN^-$ ion. The carbon end is intrinsically the more nucleophilic site (more polarisable, and it forms the stronger C–C bond with the alkyl carbon), so KCN reacts through carbon, giving the nitrile as the major product.
AgCN is covalent, with silver bonded to the CARBON of the cyanide group ( $Ag{-}C{\equiv}N$ ). With the carbon end already occupied by silver, it is the NITROGEN lone pair that is left free to attack the alkyl halide — so AgCN gives the isocyanide as the major product.

Watch out

A common mistake is to assume silver coordinates to nitrogen (since nitrogen is "more electronegative" or "harder"). It is the opposite: silver bonds to carbon, and it is precisely THAT occupation of the carbon end that forces attack to happen through nitrogen instead. …

Why this formula?

Ambident Nucleophile Reactivity: Why the Rules Hold

Ambident nucleophiles are nucleophiles that have two (or more) different atoms capable of donating a lone pair to form a bond with an electrophile. Classic examples include:

Cyanide ion ( $\ce{CN-}$ ): can attack via carbon or nitrogen
Nitrite ion ( $\ce{NO2-}$ ): can attack via oxygen or nitrogen
Enolate ions: can attack via carbon or oxygen

The key question: Why does one atom react preferentially over the other?

The Core Principle: Hard-Soft Acid-Base (HSAB) Theory

The reactivity of ambident nucleophiles is governed by HSAB theory, which states:

Hard acids prefer hard bases; soft acids prefer soft bases.

Why this holds — the reasoning:

Hard species are small, highly charged, and non-polarizable. Their interactions are dominated by ionic (electrostatic) forces.
Soft species are large, polarizable, and have diffuse electron clouds. Their interactions are dominated by covalent (orbital overlap) forces.

For an ambident nucleophile, the two attacking atoms differ in hardness/softness:

Ambident Nucleophile	Harder Atom	Softer Atom
$\ce{CN-}$	N (hard)	C (soft)
$\ce{NO2-}$	O (hard)	N (soft)
Enolate ( $\ce{CH2=CH-O-}$ )	O (hard)	C (soft)

The Key Formula(e) and Their Derivation

1. Charge Density Rule (for hard-hard interactions)

For a hard electrophile (e.g., $\ce{H+}$ , $\ce{CH3+}$ , $\ce{AlCl3}$ ):

The nucleophile attacks via the atom with higher charge density (more negative charge).

Why?

Hard-hard interactions are electrostatic. The force between charges is:

F = \frac{k \cdot q_1 \cdot q_2}{r^2}

$q_1$ , $q_2$ = charges on the species
$r$ = distance between them

A hard electrophile has a localized positive charge. The nucleophile's atom with greater negative charge density (more concentrated charge) exerts a stronger electrostatic attraction. This atom is typically the more electronegative one (e.g., O in enolate, N in cyanide).

Example:

Enolate with $\ce{CH3I}$ (hard electrophile) → O-alkylation (harder O attacks)

2. Polarizability Rule (for soft-soft interactions)

For a soft electrophile (e.g., $\ce{CH3CH2I}$ , $\ce{Hg^{2+}}$ , $\ce{Br2}$ ):

The nucleophile attacks via the atom with higher polarizability (softer atom).

Why?

Soft-soft interactions are covalent and depend on orbital overlap. The softer atom has:

Larger, more diffuse orbitals (e.g., 3p vs 2p)
Lower electronegativity
Greater polarizability — its electron cloud can distort easily to form a bond

The energy of orbital overlap is approximated by:

\Delta E \propto \frac{(\text{overlap integral})^2}{\text{energy gap}}

A softer atom has a higher-energy HOMO (closer to the electrophile's LUMO), giving a smaller energy gap and stronger interaction.

Example: …

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This question tests your understanding of how the reagent and reaction conditions (alcoholic vs. aqueous, presence of Mg/Na, type of halide) dictate the product. Each subpart maps to a classic named reaction: dehydrohalogenation, Grignard reagent formation, nucleophilic aromatic substitution, simple hydrolysis, Wurtz reaction, and nucleophilic substitution with an ambident nucleophile.

1. n-Butyl chloride + alcoholic KOH → Elimination (Dehydrohalogenation)

Alcoholic KOH is a strong base in a poorly solvating solvent (ethanol). It favours E2 elimination over substitution. The base abstracts a $\beta$ -hydrogen, the C–Cl bond breaks, and a double bond forms.

The only $\beta$ -hydrogen available is on C2 (since n-butyl is CH $_3$ –CH $_2$ –CH $_2$ –CH $_2$ Cl, and C1 is the only carbon bonded to Cl). Removing a hydrogen from C2 forms the C1=C2 double bond, giving but-1-ene. There is no other $\beta$ -hydrogen available to eliminate — reaching but-2-ene would require removing a hydrogen from C3, which is the $\gamma$ -carbon relative to the chlorine on C1, not accessible in a single E2 step — and E2 has no carbocation intermediate that could rearrange to reach it either. But-1-ene is therefore the only alkene this elimination can give.

Watch out

Do not assume a rearranged or more-substituted alkene can always form. Elimination is limited to the actual $\beta$ -hydrogens available on the specific carbon adjacent to the leaving group — here there is only one such hydrogen-bearing carbon, so only one alkene is possible.

Product: CH $_3$ –CH $_2$ –CH=CH $_2$ (but-1-ene)

2. Bromobenzene + Mg / dry ether → Grignard reagent formation

Mg inserts into the C–Br bond in anhydrous ether. The ether solvates the Mg, stabilising the highly polar C–Mg bond. This is the classic preparation of a Grignard reagent.

The product is phenylmagnesium bromide, C $_6$ H $_5$ –Mg–Br. This is a powerful nucleophile and base, used in countless organic syntheses.

Important

The ether must be absolutely dry. Even traces of water will protonate the Grignard reagent, giving benzene and Mg(OH)Br.

Product: C $_6$ H $_5$ MgBr (phenylmagnesium bromide)

3. Chlorobenzene + hydrolysis → Phenol (via the Dow process, forcing conditions)

Chlorobenzene is an aryl halide. The C–Cl bond has partial double-bond character due to resonance with the ring. The lone pair on Cl is delocalised into the ring, making the C–Cl bond stronger and the carbon less electrophilic. Ordinary, mild nucleophilic substitution (S $_N$ 1 or S $_N$ 2, as would work on an alkyl halide) does not occur.

However, this is the standard NCERT "hydrolysis of chlorobenzene" question, taught in this very chapter's own phenol-preparation section: under the specific, drastic Dow-process conditions — fused with NaOH at 623 K (350°C) and 300 atm pressure — chlorobenzene DOES undergo nucleophilic aromatic substitution (via a benzyne-type elimination–addition pathway) to give phenol.

Watch out

Do not simply write "no reaction." Ordinary, mild aqueous hydrolysis genuinely does nothing to chlorobenzene -- but this question is testing the forcing Dow-process conditions specifically taught in this chapter, under which phenol IS formed. State both halves: hydrolysis fails under normal conditions, but succeeds under the Dow process.

Product: Phenol, $C_6H_5OH$ (formed only under Dow-process conditions: fused NaOH, 623 K, 300 atm)

C_6H_5Cl + 2NaOH \xrightarrow[300\,\text{atm}]{623\,K} C_6H_5ONa \xrightarrow{H_3O^+} C_6H_5OH

4. Ethyl chloride + aqueous KOH → Nucleophilic substitution (S $_N$ 2)

Aqueous KOH provides OH $^-$ ions in a protic, highly solvating solvent (water). This favours S $_N$ 2 substitution on a primary alkyl halide. The OH $^-$ attacks the carbon bearing Cl, displacing Cl $^-$ .

The product is ethanol (CH $_3$ CH $_2$ OH). No elimination occurs because water is a poor base for E2, and the primary halide favours S $_N$ 2. …

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Here is the clear solution method for each reaction, with the concept and steps named and explained.

Method: Reagent-Substrate Matching (for Ambident Nucleophile & Reaction Type)

Concept: The product depends on the nature of the nucleophile (ambident vs. simple), the solvent (protic vs. aprotic), and the substrate (alkyl halide vs. aryl halide). For ambident nucleophiles like CN⁻, the site of attack (C vs. N) is controlled by the reaction conditions.

(i) n-butyl chloride + alcoholic KOH

Method: Elimination (E2) in strong base / polar protic solvent

Step 1: Identify the reagent. Alcoholic KOH is a strong base and a poor nucleophile (due to solvation by alcohol).
Step 2: Recognize the substrate. n-butyl chloride is a primary alkyl halide.
Step 3: Apply the rule: in the alcoholic medium, OH⁻/ethoxide is poorly solvated and strongly basic, so E2 elimination is favoured over substitution. (The base is not bulky — it is its strength in this poorly-solvating medium that matters.)
Step 4: Remove β-hydrogen and Cl⁻ to form an alkene.

Result: But-1-ene (major) + KCl + H₂O

\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl} + \text{KOH (alc.)} \xrightarrow{\Delta} \text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2 + \text{KCl} + \text{H}_2\text{O}

(ii) bromobenzene + Mg / dry ether

Method: Grignard Reagent Formation (Oxidative Addition)

Step 1: Ensure dry ether (anhydrous) — water destroys the reagent.
Step 2: Mg inserts into the C–Br bond via a single-electron transfer mechanism.
Step 3: The product is phenylmagnesium bromide, a strong nucleophile and base.

Result: C₆H₅MgBr (Grignard reagent)

\text{C}_6\text{H}_5\text{Br} + \text{Mg} \xrightarrow{\text{dry ether}} \text{C}_6\text{H}_5\text{MgBr}

(iii) chlorobenzene + hydrolysis (aqueous NaOH, high pressure/high temperature)

Method: Nucleophilic Aromatic Substitution (Elimination–Addition via Benzyne)

Step 1: Chlorobenzene is an aryl halide — the C–Cl bond is strong due to resonance; direct Sₙ2 is impossible.
Step 2: Under drastic conditions (high T, high P, strong base), elimination of HCl occurs to form a benzyne intermediate.
Step 3: Water (or OH⁻) attacks the benzyne, followed by tautomerization.

Result: Phenol (C₆H₅OH)

\text{C}_6\text{H}_5\text{Cl} + \text{H}_2\text{O} \xrightarrow[623\,\text{K}, 300\,\text{atm}]{\text{aq. NaOH}} \text{C}_6\text{H}_5\text{OH} + \text{HCl}

(iv) ethyl chloride + aqueous KOH

Method: Nucleophilic Substitution (Sₙ2) in aqueous medium

Step 1: Aqueous KOH provides OH⁻ as a strong nucleophile in a protic solvent (water).
Step 2: Ethyl chloride is a primary alkyl halide — Sₙ2 is favoured.
Step 3: OH⁻ attacks the carbon bearing Cl, displacing Cl⁻ via backside attack.

Result: Ethanol (CH₃CH₂OH)

\text{CH}_3\text{CH}_2\text{Cl} + \text{KOH (aq.)} \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{KCl}

(v) methyl bromide + sodium / dry ether

Method: Wurtz Reaction (Coupling)

Step 1: Sodium metal in dry ether acts as a reducing agent and coupling agent. …

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This set of reactions is a classic test of your understanding of reaction mechanisms, reagent specificity, and solvent effects. Students often lose marks here by memorizing products without understanding the why.

Here is a breakdown of the common mistakes for each reaction and how to avoid them.

(i) n-butyl chloride + alcoholic KOH

The Concept: Alcoholic KOH is a strong base in ethanol — a less-polar protic solvent that solvates OH⁻ poorly, leaving it strongly basic. This favors Elimination (E2) over Substitution (SN2).

Common Mistake: Writing the product as n-butyl alcohol (from SN2).
Why it's wrong: Aqueous KOH ( $KOH_{aq}$ ) gives substitution (SN2). Alcoholic KOH ( $KOH_{alc}$ ) gives elimination.
How to Avoid: Associate the solvent with the mechanism.
- Aqueous (water) → Substitution (SN1/SN2).
- Alcoholic (ethanol) → Elimination (E2).
Correct Product: But-1-ene ( $CH_3CH_2CH=CH_2$ ). (But-1-ene is the only alkene possible here — the only β-hydrogens are on C2, and E2 has no carbocation intermediate that could rearrange.)

(ii) Bromobenzene + Mg in dry ether

The Concept: This is the Grignard reaction. The key is the formation of a Grignard reagent ( $C_6H_5MgBr$ ).

Common Mistake: Writing the product as biphenyl ( $C_6H_5-C_6H_5$ ) or benzene ( $C_6H_6$ ).
Why it's wrong: Biphenyl forms if moisture is present (Wurtz-type coupling). Benzene forms if the Grignard reagent is hydrolyzed by water. The question asks for the initial reaction product.
How to Avoid: Remember the purpose of dry ether. It is an inert, aprotic solvent that stabilizes the highly reactive Grignard reagent by coordinating to the Mg atom. The reaction stops at the Grignard reagent unless you add another reagent (like $CO_2$ , $H_2O$ , or a carbonyl compound).
Correct Product: Phenylmagnesium bromide ( $C_6H_5MgBr$ ).

(iii) Chlorobenzene + hydrolysis

The Concept: Aryl halides (like chlorobenzene) are extremely resistant to nucleophilic substitution under normal conditions due to resonance stabilization and the $sp^2$ hybridized carbon.

Common Mistake: Writing the product as phenol via a simple, unqualified SN2/SN1 hydrolysis (as if chlorobenzene reacted like an ordinary alkyl halide).
Why it's wrong: SN2 requires a backside attack, which is impossible on the planar, hindered aromatic ring. SN1 would require a very unstable phenyl carbocation. Simple hydrolysis (e.g. boiling with water, or even dilute NaOH at room temperature) genuinely does nothing to chlorobenzene.
How to Avoid: Recognize the "Aryl Halide Inertness" rule -- but also recognize the DRASTIC-conditions exception this question is testing. This is a standard NCERT exercise question about the industrial (Dow) process for phenol from chlorobenzene, taught in this very chapter's preparation section: fused with NaOH at 623 K and 300 atm pressure, chlorobenzene DOES hydrolyse to phenol (via a benzyne-type mechanism), precisely because those forcing conditions overcome the inertness that blocks ordinary hydrolysis.
Correct Product: Phenol, formed only under the drastic NaOH/623K/300atm conditions -- under any milder/ordinary hydrolysis attempt, there is no reaction at all. Both halves matter: state clearly that ordinary hydrolysis fails, AND that the specific forcing conditions taught in this chapter do succeed.

(iv) Ethyl chloride + aqueous KOH

The Concept: Aqueous KOH provides a polar, protic solvent (water) and a strong nucleophile ( $OH^-$ ). This favors Substitution (SN2) .

Common Mistake: Writing the product as ethene (from elimination).
Why it's wrong: Aqueous conditions favor substitution. The $OH^-$ ion is a strong nucleophile and readily attacks the primary carbon.
How to Avoid: Contrast with (i). Same reagent (KOH), different solvent (aqueous vs. alcoholic). The solvent dictates the pathway.
Correct Product: Ethanol ( $CH_3CH_2OH$ ).

(v) Methyl bromide + Na in dry ether

The Concept: This is the Wurtz reaction. Two alkyl halides couple in the presence of sodium metal.

Common Mistake: Writing the product as methane ( $CH_4$ ) or thinking it's a Grignard reaction.
Why it's wrong: Sodium metal is a reducing agent, but it doesn't insert into the C-Br bond like Mg does. Instead, it forms a radical intermediate that couples with another radical.
How to Avoid: Identify the metal.
- Mg in dry ether → Grignard reagent.
- Na in dry ether → Wurtz reaction (coupling).
Correct Product: Ethane ( $CH_3-CH_3$ ).

(vi) Methyl chloride + KCN …

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Q.What happens when

1. n-Butyl chloride + alcoholic KOH → Elimination (Dehydrohalogenation)

2. Bromobenzene + Mg / dry ether → Grignard reagent formation

3. Chlorobenzene + hydrolysis → Phenol (via the Dow process, forcing conditions)

4. Ethyl chloride + aqueous KOH → Nucleophilic substitution (SN_NN​2)

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4. Ethyl chloride + aqueous KOH → Nucleophilic substitution (S $_N$ 2)

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