Which of the following haloalkanes reacts with aqueous KOH most easily? Explain giving reason. (i) 1-Bromobuta

NCERT Exemplar · Q62

Q. Which of the following haloalkanes reacts with aqueous KOH most easily? Explain giving reason.

(i) 1-Bromobutane

(ii) 2-Bromobutane

(iii) 2-Bromo-2-methylpropane

(iv) 2-Chlorobutane

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Concept understanding — SN1 Reactivity Order

The Core Intuition: Who Wants to Leave, and Who Can Wait?

Imagine you're at a party where the host (the leaving group) is about to leave. The party (the reaction) happens in two stages. First, the host walks out the door — that's the slow, painful step. Then, a new guest (the nucleophile) rushes in to take the empty spot.

The SN1 reaction works exactly like this: the leaving group leaves first, forming a carbocation intermediate. The nucleophile attacks after the leaving group is gone. This means the rate of the reaction depends only on how easily the leaving group can leave — it does not depend on the nucleophile at all.

So the question becomes: What makes a carbocation form easily? The answer is stability. A carbocation that is more stable will form faster and last longer, making the SN1 reaction faster.

The Precise Statement of SN1 Reactivity Order

\text{Allylic} \approx \text{Benzyl} > 3^\circ > 2^\circ \gg 1^\circ \approx \text{Methyl}

This is the order of how fast the SN1 reaction proceeds. Let's unpack why.

Why This Order? The Stability Ladder

A carbocation is a carbon with only six electrons in its valence shell — it's electron-deficient and positively charged. The more you can spread out (delocalize) that positive charge, the more stable the carbocation becomes.

1. Methyl and 1° Carbocations: The Unstable Ones

A methyl carbocation ( $\ce{CH3+}$ ) has no alkyl groups attached to the positive carbon. There is zero electron-donating effect to stabilize the charge. It is so unstable that it practically never forms in an SN1 reaction — the reaction simply doesn't happen.

A primary (1°) carbocation has one alkyl group attached. Alkyl groups are weakly electron-donating (through hyperconjugation and inductive effect), so it's slightly more stable than methyl — but still far too unstable to form under normal SN1 conditions.

Watch out

Never say "SN1 happens on a primary carbon" in an exam. It is essentially impossible under standard conditions because the carbocation is too unstable.

2. Secondary (2°) Carbocations: The Borderline Case

A secondary carbocation has two alkyl groups donating electron density. It is moderately stable — stable enough to form, but only under certain conditions (like a good leaving group and a polar protic solvent). SN1 reactions on secondary carbons are possible, but they are slower than on tertiary carbons.

3. Tertiary (3°) Carbocations: The Sweet Spot

Three alkyl groups donate electron density to the positive carbon. This makes the carbocation very stable. Tertiary alkyl halides undergo SN1 reactions readily — they are the classic example.

4. Allylic and Benzylic: The Champions

These are special cases. In an allylic carbocation, the positive charge is adjacent to a carbon-carbon double bond. The $\pi$ electrons of the double bond can delocalize the positive charge onto the second carbon:

\ce{CH2=CH-CH2+} \leftrightarrow \ce{+CH2-CH=CH2}

Two resonance structures of the allyl carbocation, CH2=CH-CH2+, with the positive charge delocalised between the two terminal carbons (Structure I and Structure II)

In a benzylic carbocation, the positive charge is adjacent to a benzene ring. The $\pi$ system of the ring delocalizes the charge across multiple carbons:

\ce{C6H5-CH2+} \leftrightarrow \text{(several resonance structures)}

Here are those resonance structures — the positive charge cycles from the CH₂ carbon onto the ortho and para positions of the ring:

Four resonance structures of the benzyl carbocation, C6H5-CH2+, with curved electron-pushing arrows showing the positive charge delocalising from the exocyclic CH2 carbon onto the ortho and para ring carbons

This resonance stabilization makes allylic and benzylic carbocations even more stable than tertiary ones. They form the fastest in SN1 reactions.

The Complete Picture in a Table

|------------------|-----------|----------------|---------| …

Why this formula?

SN1 Reactivity Order: Why It Holds

The SN1 reaction (Substitution Nucleophilic Unimolecular) proceeds via a carbocation intermediate. The reactivity order is determined entirely by the stability of this carbocation — because the rate-determining step is its formation.

The Core Principle

The rate law for SN1 is:

\text{Rate} = k[\text{RX}]

Only the substrate appears in the rate law — the nucleophile does not participate in the slow step. The slow step is:

\text{RX} \xrightarrow{\text{slow}} \text{R}^+ + \text{X}^-

Thus, anything that stabilizes the carbocation (R⁺) lowers the activation energy and increases the reaction rate.

The Reactivity Order

For alkyl halides (RX), the SN1 reactivity order is:

\text{Allylic} > \text{Benzyllic} > \text{Tertiary} > \text{Secondary} > \text{Primary} > \text{Methyl}

Let's break down why each step holds.

1. Why Tertiary > Secondary > Primary > Methyl?

This is purely about hyperconjugation and inductive effect.

Tertiary carbocation: Three alkyl groups donate electron density via hyperconjugation (C–H σ bonds overlap with empty p orbital) and +I effect. This spreads the positive charge over more atoms → most stable.
Secondary: Two alkyl groups → less stabilization.
Primary: Only one alkyl group → very little stabilization.
Methyl: No alkyl groups → least stable (only inductive effect from H atoms, which is negligible).

Key formula: The number of α-hydrogens (H on carbons adjacent to the positive carbon) determines hyperconjugation. More α-H → more resonance structures → more stable.

2. Why Allylic and Benzylic Are Even Faster

These carbocations are resonance-stabilized.

Allylic carbocation: The positive charge is delocalized over two carbon atoms via π-bond conjugation:

\text{CH}_2 = \text{CH} - \text{CH}_2^+ \longleftrightarrow \text{CH}_2^+ - \text{CH} = \text{CH}_2

Two resonance structures of the allyl carbocation with the positive charge shared between the two terminal CH2 carbons

Benzylic carbocation: The positive charge is delocalized into the aromatic ring:

\text{C}_6\text{H}_5 - \text{CH}_2^+ \longleftrightarrow \text{several resonance forms involving the ring}

Those resonance forms look like this:

Resonance structures of the benzylic carbocation C6H5-CH2+ showing the positive charge delocalised onto the ortho and para carbons of the benzene ring

This resonance stabilization is so powerful that even a primary allylic or benzylic carbocation is more stable than a tertiary alkyl carbocation.

3. The Complete Order (with reasoning) …

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With aqueous KOH the reaction is hydrolysis (substitution), and in the strongly ionising aqueous medium it proceeds by the SN1 mechanism — so the rate is governed by carbocation stability, not steric hindrance. 2-Bromo-2-methylpropane (iii) forms the stable tertiary carbocation on ionisation, so it reacts most easily. The answer is (iii).

The question asks which haloalkane reacts most easily with aqueous KOH. Two things about "aqueous" matter here:

Aqueous KOH means substitution. OH⁻ in water replaces the halogen to give an alcohol (contrast with alcoholic KOH, which favours elimination to an alkene).
Water is a polar, protic, strongly ionising solvent. It stabilises ions superbly — both the carbocation and the departing halide — which is exactly the environment in which a haloalkane ionises. Hydrolysis under these conditions is governed by the SN1 pathway: the slow, rate-determining step is ionisation of the C–X bond, and its ease depends on the stability of the carbocation formed.

Carbocation stability is the deciding factor. The SN1 reactivity order is:

Tertiary (most stable carbocation) > Secondary > Primary

Now let's classify each compound:

1-Bromobutane — bromine on a primary carbon. Ionisation would give a highly unstable primary carbocation, so ionisation is very difficult.
2-Bromobutane — bromine on a secondary carbon. A secondary carbocation is more stable than primary, but far less than tertiary.
2-Bromo-2-methylpropane — bromine on a tertiary carbon. Ionisation gives $(\text{CH}_3)_3\text{C}^+$ , stabilised by hyperconjugation and the +I effect of three methyl groups. This is by far the easiest ionisation of the four, so this compound reacts most easily.
2-Chlorobutane — secondary like (ii), but with the poorer leaving group Cl⁻ (the C–Cl bond is stronger than C–Br), so it reacts more slowly than 2-bromobutane.

Compound	Type	Leaving group	Carbocation on ionisation	Ease of reaction with aq. KOH
2-Bromo-2-methylpropane (iii)	Tertiary	Br⁻ (good)	Tertiary (most stable)	Easiest
2-Bromobutane (ii)	Secondary	Br⁻ (good)	Secondary	Moderate
2-Chlorobutane (iv)	Secondary	Cl⁻ (poorer)	Secondary	Slower
1-Bromobutane (i)	Primary	Br⁻ (good)	Primary (least stable)	Slowest by this pathway

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Method: Carbocation-Stability Analysis for Hydrolysis with Aqueous KOH

This problem tests whether you pick the mechanism from the medium. Aqueous KOH means hydrolysis (substitution) in a polar, protic, strongly ionising solvent — conditions under which the reaction proceeds by SN1, so the rate is governed by carbocation stability.

Step 1 — Decide the mechanism from the medium

Aqueous KOH → substitution (hydrolysis). Water is a strongly ionising solvent: it stabilises the carbocation and the leaving halide, so the C–X bond ionises first (slow step) — SN1.
(Alcoholic KOH would instead favour elimination to an alkene — a different question.)

SN1 reactivity order: Tertiary > Secondary > Primary (the carbocation-stability order).

Step 2 — Classify each haloalkane by the carbon bearing the halogen

Haloalkane	Structure	Carbon type
1-Bromobutane (i)	CH₃–CH₂–CH₂–CH₂–Br	Primary
2-Bromobutane (ii)	CH₃–CH(Br)–CH₂–CH₃	Secondary
2-Bromo-2-methylpropane (iii)	(CH₃)₃C–Br	Tertiary
2-Chlorobutane (iv)	CH₃–CH(Cl)–CH₂–CH₃	Secondary

Step 3 — Rank by carbocation stability

Only (iii) ionises to a tertiary carbocation, $(\text{CH}_3)_3\text{C}^+$ , stabilised by hyperconjugation and the +I effect of three methyl groups — the easiest ionisation by far.
(ii) and (iv) would give secondary carbocations; (i) a very unstable primary one. …

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Common Mistakes & How to Avoid Them

Mistake 1: Applying the SN2 steric argument and picking the primary halide

The error: Students reason "aqueous KOH supplies OH⁻, a good nucleophile → SN2 → least hindered wins" and pick (i) 1-Bromobutane.

Why it's wrong: Water is a polar, protic, strongly ionising medium — it stabilises a carbocation and the departing halide ion superbly, so hydrolysis with aqueous KOH proceeds through ionisation of the C–X bond (SN1). Under SN1, the rate is set by carbocation stability (3° > 2° > 1°), so the tertiary halide reacts most easily — the exact opposite of the steric order. The NCERT Exemplar's own key confirms this: (iii); the tertiary carbocation formed in the reaction is stable.

How to avoid: Decide the mechanism from the medium first. Aqueous (ionising, protic) medium → SN1 → rank by carbocation stability. Only when conditions genuinely favour a one-step bimolecular attack does the steric (primary-fastest) order apply.

Mistake 2: Confusing aqueous KOH with alcoholic KOH

The error: Students see "KOH" and write an elimination product (an alkene).

Why it's wrong: Aqueous KOH gives substitution (hydrolysis to the alcohol); it is alcoholic KOH that favours elimination. This pairing is standard NCERT doctrine and the question hinges on it.

How to avoid: Memorise the pair: aq. KOH → substitution; alc. KOH → elimination.

Mistake 3: Forgetting the Leaving Group Effect

The error: Treating (ii) 2-Bromobutane and (iv) 2-Chlorobutane as equally reactive because both are secondary.

Why it's wrong: The C–Br bond is weaker than the C–Cl bond, so Br⁻ is a better leaving group than Cl⁻. Ionisation (the SN1 slow step) is easier for the bromide, so 2-bromobutane reacts faster than 2-chlorobutane.

How to avoid: Always factor in leaving-group ability: I⁻ > Br⁻ > Cl⁻ > F⁻.

Mistake 4: Stopping at "tertiary is most stable" without saying why

The error: Writing "(iii), because tertiary" with no reason. …

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Q.Which of the following haloalkanes reacts with aqueous KOH most easily? Explain giving reason.

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