Compound ‘A’ with molecular formula $\mathrm{C_4H_9Br}$ is treated with aq. KOH solution. The rate of this reaction depends upon the concentration of the compound ‘A’ only. When another optically active isomer ‘B’ of this compound was treated with aq. KOH solution, the rate of reaction was found to be dependent on concentration of compound and KOH both. (i) Write down the structural formula of both compounds ‘A’ and ‘B’. (ii) Out of these two compounds, which one will be converted to the product with inverted configuration.

Compound ‘A’ with molecular formula mathrmC₄H₉Br is treated with aq. KOH solution. The rate of this reaction d

NCERT Exemplar · Q55

\mathrm{C_4H_9Br}

(i) Write down the structural formula of both compounds ‘A’ and ‘B’.

(ii) Out of these two compounds, which one will be converted to the product with inverted configuration.

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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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The key is that a reaction whose rate depends only on the substrate concentration follows SN1 (unimolecular) kinetics, while a reaction whose rate depends on both substrate and base concentration follows SN2 (bimolecular) kinetics. Compound A is tert-butyl bromide (2-bromo-2-methylpropane), and compound B is 2-bromobutane (optically active). Only the SN2 reaction (compound B) gives inversion of configuration.

1. Understanding the rate dependence

The problem gives two different kinetic behaviours for two isomers of $\mathrm{C_4H_9Br}$ .

For compound A: rate depends only on [A]. That means the rate law is:

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

This is characteristic of an SN1 reaction — the slow step is the unimolecular dissociation of the C–Br bond to form a carbocation. The nucleophile (here, OH⁻ from aq. KOH) does not appear in the rate law because it attacks in a fast second step.

For compound B: rate depends on both [B] and [KOH]. That means:

\text{Rate} = k[\text{B}][\text{OH}^-]

This is characteristic of an SN2 reaction — the slow step is bimolecular, involving both the substrate and the nucleophile simultaneously.

Important

The molecular formula $\mathrm{C_4H_9Br}$ has four structural isomers. Only one of them is chiral (optically active): 2-bromobutane. The other three are achiral.

2. Identifying compound A (SN1 substrate)

For an SN1 reaction to be favourable, the substrate must form a relatively stable carbocation. Among $\mathrm{C_4H_9Br}$ isomers, the one that gives the most stable carbocation is tert-butyl bromide (2-bromo-2-methylpropane): $(CH_3)_3C{-}Br$

Why? Because the carbocation formed after Br⁻ leaves is a tertiary carbocation — stabilised by three alkyl groups via hyperconjugation and inductive effects. Tertiary carbocations are much more stable than secondary or primary ones, so SN1 is fast.

Also, tert-butyl bromide is achiral (no chiral centre), so it cannot have an optically active isomer. That matches the fact that compound A is not described as optically active.

\text{Allylic/Benzylic} > 3^\circ > 2^\circ > 1^\circ > \text{Methyl}

So compound A is:

\boxed{\text{CH}_3\text{--C(Br)(CH}_3\text{)--CH}_3 \quad (\text{2-bromo-2-methylpropane})}

3. Identifying compound B (SN2 substrate)

Compound B is an optically active isomer of $\mathrm{C_4H_9Br}$ . Optical activity requires a chiral carbon — a carbon with four different substituents.

The only $\mathrm{C_4H_9Br}$ isomer that has a chiral centre is 2-bromobutane:

      CH₃
      |
H—C—Br
      |
      CH₂CH₃

The chiral carbon (C2) is attached to: H, Br, CH₃, and CH₂CH₃ — all four different. This molecule exists as a pair of enantiomers, and a pure enantiomer is optically active.

For SN2 to be the dominant mechanism, the substrate should be sterically unhindered so that the nucleophile can approach from the back side. 2-Bromobutane is a secondary alkyl halide — it can react by either SN1 or SN2 depending on conditions. But here, the rate law clearly shows bimolecular kinetics, so under these conditions (aq. KOH, a strong nucleophile in a polar protic solvent), SN2 is favoured. …

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Method: SN1 vs SN2 Reactivity Analysis

The key is to identify the substitution mechanism from the rate law, then deduce the structure.

Step 1 — Identify the mechanism from rate dependence

Compound A: Rate depends only on [A] → SN1 (unimolecular, first-order).
Compound B: Rate depends on [B] and [KOH] → SN2 (bimolecular, second-order).

Step 2 — Relate mechanism to alkyl halide structure

For $\mathrm{C_4H_9Br}$ , the possible isomers are:

Isomer	Structure	Type
1-bromobutane	$\mathrm{CH_3CH_2CH_2CH_2Br}$	Primary
2-bromobutane	$\mathrm{CH_3CH_2CHBrCH_3}$	Secondary (chiral)
1-bromo-2-methylpropane	$\mathrm{(CH_3)_2CHCH_2Br}$	Primary
2-bromo-2-methylpropane	$\mathrm{(CH_3)_3CBr}$	Tertiary

SN1 (A) requires a stable carbocation → tertiary halide.
SN2 (B) requires an optically active isomer → chiral secondary halide.

Step 3 — Assign structures

(i) Structural formulas

Compound A (SN1, rate depends only on [A]):

2-bromo-2-methylpropane (tertiary)

$\mathrm{(CH_3)_3CBr}$
Compound B (SN2, rate depends on both, optically active):

2-bromobutane (secondary, chiral)

$\mathrm{CH_3CH_2CHBrCH_3}$ …

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Concept understanding — SN1 Reactivity Order

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

The Precise Statement of SN1 Reactivity Order

Why This Order? The Stability Ladder

1. Methyl and 1° Carbocations: The Unstable Ones

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

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

4. Allylic and Benzylic: The Champions

The Complete Picture in a Table

Why this formula?

SN1 Reactivity Order: Why It Holds

The Core Principle

The Reactivity Order

1. Why Tertiary > Secondary > Primary > Methyl?

2. Why Allylic and Benzylic Are Even Faster

3. The Complete Order (with reasoning) …

1. Understanding the rate dependence

2. Identifying compound A (SN1 substrate)

3. Identifying compound B (SN2 substrate)

Method: SN1 vs SN2 Reactivity Analysis

Step 1 — Identify the mechanism from rate dependence

Step 2 — Relate mechanism to alkyl halide structure

Step 3 — Assign structures

Mistake 1: Confusing the Rate Law with the Mechanism

Mistake 2: Forgetting Optical Activity in SN2

Mistake 3: Mixing Up Inversion vs. Retention

1. Understanding the rate dependence

2. Identifying compound A (SN1 substrate)

3. Identifying compound B (SN2 substrate)

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