Consider the following reaction (species labelled (a)–(e) as printed in the Exemplar): $\mathrm{\underset{(a)}{HO^-} + \underset{(b)}{CH_3Cl} \rightarrow \underset{(c)}{[HO\cdots CH_3\cdots Cl]^-} \rightarrow \underset{(d)}{CH_3OH} + \underset{(e)}{Cl^-}}$ In the printed diagram (b) is drawn with its three H atoms arranged tetrahedrally (one on a wedge, one on a dash, one in plane); (c) is the trigonal-bipyramidal transition state shown in square brackets with partial (dashed) bonds to the incoming $\mathrm{HO}$ and the leaving $\mathrm{Cl}$; in the product (d) the umbrella of H atoms is drawn inverted. Which of the following statements are correct about the reaction intermediate? (Two or more than two options may be correct.) (i) Intermediate (c) is unstable because in this carbon is attached to 5 atoms. (ii) Intermediate (c) is unstable because carbon atom is $sp^2$ hybridised. (iii) Intermediate (c) is stable because carbon atom is $sp^2$ hybridised. (iv) Intermediate (c) is less stable than the reactant (b).

> [!TLDR] > The reaction is an $\mathrm{S_N2}$ mechanism. The intermediate (c) is actually the **transition state**, not a stable intermediate. Carbon is bonded to five atoms and is **$sp^2$ hybridised** in this trigonal-bipyramidal geometry. It is **unstable** because carbon cannot accommodate five full bonds, and it is **less stable** than the reactant. Therefore, statements **(i)** and **(iv)** are correct. ![The SN2 mechanism: backside attack and inversion](/figures/haloalkanes-and-haloarenes/mechanism-incoming-hydroxide-red-outgoing-halide-green-fig-6-2.svg) 1. **Identify the reaction type.** The species shown — $\mathrm{HO^-}$ attacking $\mathrm{CH_3Cl}$ with inversion of configuration — is the classic **bimolecular nucleophilic substitution ($\mathrm{S_N2}$)**. In an $\mathrm{S_N2}$ reaction, bond formation and bond breaking occur **simultaneously** in a single step. There is no discrete intermediate that can be isolated; the species in square brackets (c) is the **transition state**, not a stable intermediate. 2. **What is the geometry and hybridisation at carbon in (c)?** In the transition state, the carbon is partially bonded to five atoms: the three hydrogens (still in their original positions), the incoming $\mathrm{HO}$ group, and the leaving $\mathrm{Cl}$ atom. This arrangement is **trigonal bipyramidal**. For a trigonal-bipyramidal geometry, the central atom uses **$sp^2$ hybridisation** for the three equatorial bonds (here, the C–H bonds) and the two axial positions (the incoming and leaving groups) involve **$p$ orbitals**. So carbon is **$sp^2$ hybridised** in this transition state. 3. **Evaluate each statement.** - **(i) Intermediate (c) is unstable because in this carbon is attached to 5 atoms.** This is **correct**. Carbon normally forms only four covalent bonds (octet rule). In the transition state, carbon has five partial bonds — it is **pentacoordinate**. This is a high-energy, unstable arrangement. The transitio

Consider the following reaction (species labelled (a)–(e) as printed in the Exemplar): mathrmunderset(a)HO^- +

NCERT Exemplar · Q34

\mathrm{\underset{(a)}{HO^-} + \underset{(b)}{CH_3Cl} \rightarrow \underset{(c)}{[HO\cdots CH_3\cdots Cl]^-} \rightarrow \underset{(d)}{CH_3OH} + \underset{(e)}{Cl^-}}

(b) is drawn with its three H atoms arranged tetrahedrally (one on a wedge, one on a dash, one in plane);

\mathrm{HO}

and the leaving

\mathrm{Cl}

; in the product

(d) the umbrella of H atoms is drawn inverted.
Which of the following statements are correct about the reaction intermediate? (Two or more than two options may be correct.)

(i) Intermediate

(ii) Intermediate

sp^2

hybridised.

(iii) Intermediate

sp^2

hybridised.

(iv) Intermediate

Andhra Pradesh BieapMCQ· 1mImportance★★★★★

50% · 74/147 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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) …

Start your 14-day free trial to unlock the full solution →

The reaction is an $\mathrm{S_N2}$ mechanism. The intermediate (c) is actually the transition state, not a stable intermediate. Carbon is bonded to five atoms and is $sp^2$ hybridised in this trigonal-bipyramidal geometry. It is unstable because carbon cannot accommodate five full bonds, and it is less stable than the reactant. Therefore, statements (i) and (iv) are correct.

The SN2 mechanism: backside attack and inversion

Identify the reaction type.

The species shown — $\mathrm{HO^-}$ attacking $\mathrm{CH_3Cl}$ with inversion of configuration — is the classic bimolecular nucleophilic substitution ( $\mathrm{S_N2}$ ). In an $\mathrm{S_N2}$ reaction, bond formation and bond breaking occur simultaneously in a single step. There is no discrete intermediate that can be isolated; the species in square brackets (c) is the transition state, not a stable intermediate.
What is the geometry and hybridisation at carbon in (c)?

In the transition state, the carbon is partially bonded to five atoms: the three hydrogens (still in their original positions), the incoming $\mathrm{HO}$ group, and the leaving $\mathrm{Cl}$ atom. This arrangement is trigonal bipyramidal.

For a trigonal-bipyramidal geometry, the central atom uses $sp^2$ hybridisation for the three equatorial bonds (here, the C–H bonds) and the two axial positions (the incoming and leaving groups) involve $p$ orbitals. So carbon is $sp^2$ hybridised in this transition state.
Evaluate each statement.
- (i) Intermediate (c) is unstable because in this carbon is attached to 5 atoms.
  
  This is correct. Carbon normally forms only four covalent bonds (octet rule). In the transition state, carbon has five partial bonds — it is pentacoordinate. This is a high-energy, unstable arrangement. The transition state cannot be isolated; it exists only fleetingly at the energy maximum of the reaction coordinate.
- (ii) Intermediate (c) is unstable because carbon atom is $sp^2$ hybridised.
  
  This is incorrect in reasoning. $sp^2$ hybridisation itself is not inherently unstable (e.g., in ethene, carbon is $sp^2$ and perfectly stable). The instability here comes from pentacoordination, not from hybridisation. So the cause given is wrong.
- (iii) Intermediate (c) is stable because carbon atom is $sp^2$ hybridised. …

Start your 14-day free trial to unlock the full solution →

Common Mistakes & How to Avoid Them

This question tests your understanding of SN2 reaction mechanism and transition state theory. Here are the most frequent errors students make:

Mistake 1: Calling (c) an "intermediate" instead of a "transition state"

The error: Students treat (c) as a stable, isolable species with a finite lifetime.

Why it's wrong: The species (c) is shown in square brackets with partial bonds — this is the transition state, not an intermediate. Intermediates have finite lifetimes and appear in energy wells; transition states are energy maxima.

How to avoid: Always check the notation:

Square brackets [ ] with dashed/partial bonds → transition state
No brackets, full bonds, often with a charge → intermediate

Mistake 2: Assuming "5 groups around carbon" automatically means $sp^3d$

The error: Students see the trigonal-bipyramidal-LOOKING geometry (5 groups around carbon) and reach for $sp^3d$ -- the hybridisation used for period-3+ elements like phosphorus or sulfur in a genuine trigonal bipyramidal molecule (e.g. $PCl_5$ ).

Why it's wrong: Carbon is a period-2 element and has no accessible d-orbitals -- $sp^3d$ is not a chemically meaningful hybridisation for carbon at all. The correct picture: the three H atoms stay on $sp^2$ orbitals (becoming coplanar as the "umbrella" flattens mid-inversion), while the incoming $\mathrm{HO}$ and leaving $\mathrm{Cl}$ share the two lobes of carbon's remaining, unhybridised p-orbital -- giving an electronically "5-coordinate-looking" transition state without invoking any d-orbital.

How to avoid: The $sp^3d$ /trigonal-bipyramidal model is for elements that genuinely have accessible d-orbitals (period 3 and below). For carbon specifically, always default to $sp^2$ + p-orbital for this 5-group transition-state picture, never $sp^3d$ .

Mistake 3: Thinking the transition state is "stable"

The error: Selecting option (iii) — "Intermediate (c) is stable because carbon atom is $sp^2$ hybridised."

Why it's wrong: Even if the hybridisation were $sp^2$ (which it isn't), a pentacoordinate carbon is always unstable due to:

Steric crowding — 5 atoms around a small carbon
Partial bonds — bonds are neither fully formed nor fully broken
High energy — it's the peak of the reaction energy profile

How to avoid: Remember: Transition states are always unstable — they represent the highest energy point along the reaction coordinate. Stability is not about hybridisation alone; it's about bond count, geometry, and energy.

Mistake 4: Misreading the question — "reaction intermediate" …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 17 on this concept.

AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
$X$
›Reveal solutionSolution
X is a secondary bromide, Y a primary bromide, Z a tertiary bromide. Since SN1 rate depends on carbocation stability (3° > 2° > 1°), the reactivity order is Z > X > Y.

Concept and Intuition

$S_N1$ reactions proceed through a carbocation intermediate in the rate-determining step, so their rate is governed almost entirely by how stable that carbocation is — the more stable the carbocation (more substituted = more hyperconjugation/inductive electron donation from alkyl groups), the faster the $S_N1$ reaction. This immediately tells us the reactivity order among alkyl halides towards $S_N1$ : tertiary > secondary >> primary (primary halides essentially never go by $S_N1$ due to the very unstable primary carbocation).

So the actual chemistry of the reactions is secondary to correctly identifying the type of halide (1°, 2°, or 3°) each one produces.

Step-by-Step Solution

Reaction I: $\mathrm{CH_2=CH-CH_2-CH_3}$ (but-1-ene) + HBr, normal ionic addition → Markovnikov's rule: $\mathrm{H^+}$ adds to the terminal (less substituted) carbon, generating the more stable secondary carbocation at C2, which $\mathrm{Br^-}$ then attacks → X = 2-bromobutane (a secondary halide).

Reaction II: same alkene + HBr in presence of a peroxide, $(\mathrm{C_6H_5CO})_2\mathrm{O_2}$ → this triggers the free-radical (peroxide/Kharasch) mechanism, which is anti-Markovnikov: $\mathrm{Br}^{\bullet}$ adds first to the terminal carbon (forming the more stable secondary radical at C2), and the halogen ends up on the terminal carbon → Y = 1-bromobutane (a primary halide). …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
$C_4H_9Br$
›Reveal solutionSolution
A is a primary $C_4H_9Br$ isomer: its Wurtz coupling product B is a fairly symmetric $C_8H_{18}$ alkane giving only a small number of monochloro products on photochlorination. Being primary, A reacts with a moderate base/nucleophile like sodium ethoxide mainly by substitution ( $S_N2$ ), not elimination. Answer: predominantly substitution with $NaOC_2H_5$ .

Concept and Intuition

The Wurtz reaction couples two identical alkyl halide molecules via loss of the halide (as NaX) to form a symmetric alkane with double the carbon count. Photochemical (free-radical) chlorination of that alkane substitutes different types of hydrogen atoms, and the number of distinct monochloro products equals the number of chemically distinct hydrogen environments in the molecule — fewer distinct environments (a more symmetric alkane) means fewer monochloro products, which is a strong clue about how symmetric/branched the parent alkyl halide is.

Among the four possible reactivity/behaviour statements, we test each candidate structure:

A primary halide (e.g. n-butyl or isobutyl bromide) is not chiral, does not favor $S_N1$ (primary carbocations are unstable), reacts with alkoxide bases predominantly via $S_N2$ substitution (only mildly reduced by beta-branching), and cannot be made by HX addition to an internal alkene like but-2-ene (which would give a secondary halide instead).

A secondary halide (2-bromobutane) is chiral AND can be made from but-2-ene + HBr — giving two simultaneously 'true' statements, which is inconsistent with a single-answer MCQ, ruling it out as the intended compound.

A tertiary halide (tert-butyl bromide) reacts via $S_N1$ /E1 and gives (famously) only ONE type of monochloro product from its highly symmetric coupling product — its defining reactivity statement is the $S_N1$ one, not the substitution-with-ethoxide one.

Given the clue of only a small number (few, not many) monochloro products from B, and that this must correspond to an answer choice that is uniquely and unambiguously true, the primary-halide identification (giving statement C) is the internally consistent choice.

Step-by-Step Solution

A is a primary $C_4H_9Br$ isomer; Wurtz coupling with Na/dry ether gives a symmetric $C_8H_{18}$ alkane B.

Photochemical chlorination of B substitutes at its few distinct types of C–H bonds, consistent with a small number of monochloro products for a primary-derived, moderately symmetric alkane. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
$C_6H_5Cl$
›Reveal solutionSolution
This tests the distinction between allylic/benzylic positions (which give resonance-stabilized carbocations on ionization) and aryl/vinylic positions (which do not); only benzyl chloride (II) and allyl bromide (IV) qualify.

Concept and Intuition

Resonance stabilization of a carbocation requires the empty p-orbital on the cationic carbon to overlap with an adjacent $\pi$ system so the positive charge can delocalize. This overlap geometry exists when the leaving group sits on an $sp^3$ carbon next to a ring (benzylic) or a double bond (allylic) — ionization then produces a cation whose empty orbital is perfectly aligned with the neighbouring $\pi$ system. It does not exist when the leaving group is itself on the $sp^2$ carbon of the ring or double bond (aryl/vinylic), because then there is no adjacent, differently-oriented $\pi$ system to delocalize into — the resulting aryl or vinyl cation is high-energy and not resonance stabilized, and in practice these C–X bonds barely ionize at all due to partial double-bond character.

Step-by-Step Solution

I. Chlorobenzene: Cl is bonded directly to an aromatic ring carbon. Loss of $Cl^-$ would need to break a bond with significant double-bond character (from ring conjugation with the halogen lone pair) and would give a phenyl cation in the plane of the ring, orthogonal to the ring's $\pi$ system — not stabilized by resonance. Excluded.

II. Benzyl chloride: Cl is on the $sp^3$ $CH_2$ carbon attached to the ring. Loss of $Cl^-$ gives the benzylic cation $C_6H_5-\overset{+}{C}H_2$ , whose empty p-orbital aligns with the ring $\pi$ system, delocalizing the charge onto the ortho/para ring carbons. Resonance stabilized. Included. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
$S_N1$
(1) a methylcyclohexene ring reacting with HI to give A;
(2) a vinylcyclohexane ring reacting with HBr in the presence of $(C_6H_5CO)_2O_2$ peroxide to give B;
(3) cyclohexanol, a cyclohexane ring bearing an OH group, reacting with $SOCl_2$ to give C;
(4) a benzene ring bearing a diazonium salt group $N_2^+Cl^-$ reacting with KI to give D) (A) A>B>C>D (B) D>B>A>C (C) A>C>B>D (D) B>C>A>D
›Reveal solutionSolution
Identify each product's carbon type (3°, 1°, 2°, aryl respectively for A, B, C, D) and rank by $S_N1$ -relevant carbocation stability: A (3°) > C (2°) > B (1°) > D (aryl, unreactive).

Concept and Intuition

$S_N1$ reaction rate depends entirely on how stable the intermediate carbocation is, since the rate-determining step is ionisation of the C–X bond. More substituted carbons (tertiary > secondary > primary) stabilise positive charge better through hyperconjugation and inductive electron donation. Aryl (and vinyl) halides essentially never undergo $S_N1$ : the C–X bond has partial double-bond character from ring conjugation, and the resulting aryl cation would be extremely unstable (sp² cation in the ring plane, no adjacent orbital stabilisation) — so aryl halides always rank last regardless of the specific reagent used to make them.

Step-by-Step Solution

Product A: A methyl-substituted cyclohexene reacts with HI. Markovnikov addition places $I^-$ on the more substituted (tertiary) carbon of the double bond (bearing the methyl group) — giving a tertiary alkyl iodide.

Product B: Vinylcyclohexane + HBr in the presence of a peroxide follows the peroxide (anti-Markovnikov) effect — a free-radical mechanism places Br on the terminal, less-hindered carbon of the vinyl group — giving a primary alkyl bromide.

Product C: Cyclohexanol (a secondary alcohol) + $SOCl_2$ replaces –OH with –Cl (via the chlorosulfite intermediate, $S_Ni$ -type) — giving a secondary alkyl chloride, with no change in the carbon's substitution level. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
$S_N1$
›Reveal solutionSolution
Order I correctly reflects the $S_N1$ trend driven by carbocation stability (1° < 2° < 3°), and order II correctly reflects the $S_N2$ trend driven by steric hindrance (unbranched primary > branched primary > secondary) — both are correct.

Concept and Intuition

$S_N1$ reactions proceed through a carbocation intermediate, so their rate is controlled almost entirely by how stable that carbocation is: tertiary carbocations (stabilised by three alkyl/hyperconjugative groups) form fastest, followed by secondary, then primary (least stable, slowest $S_N1$ ).

$S_N2$ reactions, in contrast, proceed through a single concerted backside-attack transition state, so their rate is controlled by steric accessibility of the carbon under attack: more crowding around (or even beta to) the reacting carbon slows the reaction down. Straight-chain primary halides react fastest, branched (e.g., isobutyl, which has a beta-branch) primary halides are slower due to steric hindrance from the nearby branch, and secondary halides are slower still because the reacting carbon itself bears an extra alkyl group.

Step-by-Step Solution

Order I ( $S_N1$ ): Isobutyl iodide is a primary halide (least stable carbocation if it ionised, so slowest $S_N1$ ); sec-butyl iodide is secondary (more stable carbocation, faster than primary); t-butyl bromide is tertiary (most stable carbocation, fastest $S_N1$ despite bromide being a slightly poorer leaving group than iodide, the carbocation-stability effect dominates). This gives isobutyl iodide < sec-butyl iodide < t-butyl bromide — matches the trend 1° < 2° < 3°, so order I is correct. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
$\xrightarrow{\mathrm{HBr}} X$
›Reveal solutionSolution
Tests $S_N1$ reactivity trends based on carbocation stability and leaving-group ability; the answer is (C) y > x > z.

Concept and Intuition

$S_N1$ reactions proceed through a carbocation intermediate, so their rate is governed chiefly by:

Carbocation stability: 3° > 2° > 1° (more alkyl substitution stabilises the positive charge by hyperconjugation/induction).

Leaving group ability: for the same carbon skeleton type, a better leaving group (weaker conjugate base) ionises faster — Br⁻ is a better leaving group than Cl⁻ since HBr is a stronger acid than HCl, so C-Br bonds ionise more readily than C-Cl bonds.

First identify what X, Y, Z actually are:

X: Cyclohexene + HBr → Markovnikov addition of HBr across the double bond gives bromocyclohexane, a secondary alkyl bromide.

Y: 1-Methylcyclohex-1-ene + HCl → Markovnikov addition puts H on the carbon that gives the more stable (more substituted) carbocation; since C1 already bears the methyl group and two ring bonds, the carbocation forms at C1 (tertiary), so Cl ends up at C1 too, giving 1-chloro-1-methylcyclohexane, a tertiary alkyl chloride.

Z: Cyclohexanol + SOCl2 → gives chlorocyclohexane, a secondary alkyl chloride (via the chlorosulfite intermediate).

Step-by-Step Solution

Y is a tertiary halide → forms the most stable (tertiary) carbocation → fastest $S_N1$ reactant of the three. Y is highest. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
$S_N1$
›Reveal solutionSolution
SN1 reactivity tracks carbocation stability, and stability here is set by how many phenyl rings (resonance donors) and alkyl groups (hyperconjugation) stabilize the intermediate cation: III > II > I > IV.

Concept and Intuition

The rate-determining step of an SN1 reaction is the formation of a carbocation. Any substituent that stabilizes this positive charge — through resonance (as aryl/phenyl groups do, by delocalizing the charge into the ring) or through hyperconjugation/induction (as alkyl groups do) — speeds up SN1. Aromatic (benzylic-type) stabilization is generally very strong, and stacking multiple phenyl groups (plus alkyl groups) compounds this stabilization.

Step-by-Step Solution

I, $C_6H_5CH_2Br$ : ionisation gives the benzyl cation, stabilized by resonance delocalization into one phenyl ring — moderately stabilized (primary carbon, one aryl group).

II, $(C_6H_5)_2CHBr$ : ionisation gives the benzhydryl (diphenylmethyl) cation, stabilized by resonance into two phenyl rings simultaneously — more stable than I.

III, $(C_6H_5)_2C(CH_3)Br$ : ionisation gives a tertiary cation stabilized by two phenyl rings and hyperconjugation/induction from the extra methyl group — the most stabilized cation of the four. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
$S_N1$
›Reveal solutionSolution
This tests the link between carbocation stability and $S_N1$ reactivity; ethyl chloride, forming an unstabilised primary carbocation, is the least reactive.

Concept and Intuition

$S_N1$ reactions proceed through a carbocation intermediate in the rate-determining step, so anything that stabilises that carbocation speeds up the reaction. Stability order: tertiary > allylic/benzylic (resonance-stabilised) > secondary > primary. A primary carbocation, lacking both hyperconjugative stabilisation (from multiple alkyl groups) and resonance stabilisation, is highly unstable, so alkyl halides that would have to form one (like ethyl chloride) react very slowly, if at all, by the $S_N1$ pathway — they instead prefer $S_N2$ .

Step-by-Step Solution

Tertiary butyl chloride → forms a tertiary carbocation (most stable due to hyperconjugation/inductive donation from three methyl groups) → highly reactive in $S_N1$ .

Allyl chloride → forms an allylic carbocation stabilised by resonance with the adjacent double bond → quite reactive in $S_N1$ , despite being formally 'primary' in connectivity.

Isopropyl chloride → forms a secondary carbocation → moderately reactive in $S_N1$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
$\mathrm{C_6H_5CH_2Cl}$
›Reveal solutionSolution
This tests reactivity toward nucleophilic substitution for benzylic, allylic, and differently-substituted aryl halides. The answer is (C): the meta-nitro chlorobenzene, since -NO $_2$ only activates the ring toward nucleophilic aromatic substitution from the ortho/para position, not meta.

Concept and Intuition

Nucleophilic substitution reactivity depends heavily on the type of C–X bond and any stabilizing/activating groups. Benzylic and allylic halides are highly reactive because the developing positive charge (SN1) or the transition state (SN2) is stabilized by resonance with the adjacent pi system. Plain aryl (vinyl-type) C–Cl bonds are normally very unreactive to nucleophilic substitution because of partial double-bond character (resonance donation from the ring) and the difficulty of forming an aryl cation or backside attack on an sp2 carbon. This inertness can be overcome ('activated') by a strong electron-withdrawing group like $-\mathrm{NO_2}$ , but ONLY if it is positioned ortho or para to the leaving group, because only then can its resonance structures place negative charge directly on the carbon bearing the leaving group (stabilizing the Meisenheimer-type intermediate in nucleophilic aromatic substitution).

Step-by-Step Solution

(A) Benzyl chloride: the $-\mathrm{CH_2Cl}$ carbon is benzylic; ionization gives a resonance-stabilized benzylic carbocation, making it highly reactive via SN1 (also reactive via SN2). Reactive.

(D) Allyl chloride: the C–Cl carbon is allylic; ionization gives a resonance-stabilized allylic carbocation, also highly reactive. Reactive.

(B) 1-chloro-4-nitrobenzene: the $-\mathrm{NO_2}$ is para to Cl, so its resonance structures put negative charge directly on the ring carbon bearing Cl, strongly stabilizing the anionic intermediate of nucleophilic aromatic substitution — this activates the ring, making it noticeably more reactive than plain chlorobenzene. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
$S_N1$
(ii) }H_2O_2/OH^-} $B.$ \mathrm{C_6H_5-CH_2-CH=CH-CH_3}\xrightarrow{HBr} $C.$ \mathrm{C_6H_5-C(CH_3)=CH_2}\xrightarrow{HBr} $D.$ \mathrm{C_6H_5-CH_2-CH=CH-CH_3}\xrightarrow{HBr,\ (C_6H_5CO)_2O_2}$ (A) C, D only (B) A, B, C only (C) B, C only (D) A, D only
›Reveal solutionSolution
This tests whether you can connect how an alkyl halide is made (Markovnikov vs anti-Markovnikov addition) to how it will react later (SN1 needs a stabilised — tertiary/benzylic/allylic — carbocation).

Concept and Intuition

SN1 hydrolysis proceeds through a carbocation intermediate, so it is fast only when that carbocation is well stabilised — tertiary, allylic, or benzylic (ring directly or nearly conjugated). A plain primary or an "ordinary" secondary halide, with no such stabilisation, hydrolyses by SN2 instead. So for each reaction you must first work out the structure of the halide product, then judge how stable its carbocation would be.

Step-by-Step Solution

A: $(BH_3)_2$ then $H_2O_2/OH^-$ is hydroboration–oxidation — strictly anti-Markovnikov, so the terminal alkene $C_6H_5-CH_2-CH=CH_2$ gives the terminal (primary) alcohol $C_6H_5CH_2CH_2CH_2OH$ . $SOCl_2$ then swaps $-OH$ for $-Cl$ with retention of position, giving a primary chloride three carbons from the ring. A primary carbocation is never favourable, so this halide hydrolyses by SN2 only — excluded.

C: plain $HBr$ (no peroxide) on $C_6H_5-C(CH_3)=CH_2$ (α-methylstyrene) adds Markovnikov-wise: $H^+$ goes to the terminal $=CH_2$ , generating the carbocation on the ring-attached carbon, $C_6H_5-\overset{+}{C}(CH_3)-CH_3$ . This cation is tertiary and directly benzylic (fully conjugated with the ring) — about as stable as a carbocation gets. $Br^-$ then attacks there, giving $C_6H_5C(CH_3)(Br)CH_3$ , a textbook SN1 substrate (structurally like tert-cumyl bromide). Definitely included.

B vs D both start from the same internal alkene $C_6H_5-CH_2-CH=CH-CH_3$ , so the only difference is where $Br$ ends up.

B (plain $HBr$ , ionic/Markovnikov): the reaction proceeds through whichever secondary carbocation is more stable, which is the one adjacent to the benzylic $CH_2$ (weak but real stabilisation from the nearby aromatic ring through hyperconjugation). $Br^-$ ends up there, giving a secondary bromide close to the ring — favourable enough for SN1 in this comparison. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
$\mathrm{C_5H_{11}Br}$
›Reveal solutionSolution
First-order kinetics for the hydrolysis of a C5H11Br isomer identifies it as the tertiary bromide (S_N1 requires the more stable, more substituted carbocation). Grignard formation + D2O quench installs D exactly where the Br (and hence MgBr) was, with no skeletal rearrangement.

Concept and Intuition

The rate of $\mathrm{S_N1}$ solvolysis depends on carbocation stability: tertiary > secondary > primary. Only a tertiary alkyl halide among the $\mathrm{C_5H_{11}Br}$ isomers would show clean first-order (unimolecular, carbocation-mediated) hydrolysis kinetics at an appreciable rate under mild conditions — this pins down $X$ as the tertiary isomer, 2-bromo-2-methylbutane, $\mathrm{CH_3CH_2C(CH_3)_2Br}$ .

A Grignard reagent, $R\text{–}MgBr$ , is formed by inserting Mg directly into the C–Br bond — the carbon skeleton is completely untouched; only the halogen is replaced by the $-\mathrm{MgBr}$ group. Quenching a Grignard reagent with $\mathrm{D_2O}$ (heavy water) protonates (deuterates) the carbanion-like carbon, replacing $-\mathrm{MgBr}$ with $-\mathrm{D}$ — again with no skeletal rearrangement. This is a standard method for site-specific deuterium labeling.

Step-by-Step Solution

$\mathrm{C_5H_{11}Br}$ hydrolyzing by first-order kinetics ⟹ $\mathrm{S_N1}$ mechanism ⟹ the carbon bearing Br must be tertiary (to form the stabilized 3° carbocation). So $X = \mathrm{CH_3CH_2C(CH_3)_2Br}$ (2-bromo-2-methylbutane).

$X + \mathrm{Mg}$ (dry ether) → Grignard reagent $\mathrm{CH_3CH_2C(CH_3)_2MgBr}$ , formed at the same tertiary carbon (no rearrangement in Grignard formation).

Grignard reagent $+ \mathrm{D_2O}$ → the $-\mathrm{MgBr}$ is replaced by $-\mathrm{D}$ : $Y = \mathrm{CH_3CH_2C(CH_3)_2D}$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
$S_N1$
›Reveal solutionSolution
Benzyl chloride's benzylic cation is resonance-stabilized, favouring $S_N1$ ; chlorobenzene resists substitution because resonance gives its C–Cl bond partial double-bond character — and this resonance argument is exactly why chlorobenzene doesn't react, making R the correct explanation of A.

Concept and Intuition

In benzyl chloride ( $C_6H_5-CH_2-Cl$ ), the C–Cl bond is on an $sp^3$ carbon attached to the aromatic ring, not directly on the ring itself. When this bond ionizes, it generates a benzylic carbocation, which is strongly stabilized by resonance delocalization of the positive charge into the aromatic ring — making $S_N1$ ionization very favourable.

In chlorobenzene ( $C_6H_5-Cl$ ), the chlorine is directly bonded to an $sp^2$ ring carbon. Chlorine's lone pair can delocalize into the ring's $\pi$ system through resonance, giving the C–Cl bond significant partial double-bond character. This makes the bond shorter and stronger than a typical C–Cl single bond, and also means the ring carbon bearing Cl is not electrophilic in the way needed for nucleophilic attack — so chlorobenzene resists both $S_N1$ (no easy ionization, since breaking a partial-double bond and losing aromatic stabilization is costly) and $S_N2$ (steric and electronic shielding from the ring).

Step-by-Step Solution

Assess the assertion: benzyl chloride is $S_N1$ -reactive (true, benzylic cation is resonance-stabilized), chlorobenzene is not (true, due to bond strengthening). Assertion is correct.

Assess the reason: chlorobenzene's C–Cl bond has partial double-bond character from resonance with the ring lone pair on Cl. Reason is correct (this is textbook resonance structure reasoning for haloarenes). …
Start your 14-day free trial to see the full solution →

←Previous questionQ33 — Consider the following reaction (species labelled (a)–(e) as printed i…→Next questionQ35 — Answer on the basis of the following reaction (species labelled (a)–(d…

🎓Unlock everything free for 14 days

Unlock everything free for 14 days