Q.Three compounds are given, all benzyl-type alcohols:
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🔒 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
SN1 Reactivity Order (for alkyl halides):
Allylic≈Benzyl>3∘>2∘≫1∘≈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 (CHX3X+) 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.
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 π electrons of the double bond can delocalize the positive charge onto the second carbon:
CHX2=CH−CHX2X+↔+CHX2−CH=CHX2
In a benzylic carbocation, the positive charge is adjacent to a benzene ring. The π system of the ring delocalizes the charge across multiple carbons:
CX6HX5−CHX2X+↔(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:
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
| Carbocation Type | Stability | SN1 Reactivity | Example |
|------------------|-----------|----------------|---------| …
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:
Rate=k[RX]
Only the substrate appears in the rate law — the nucleophile does not participate in the slow step. The slow step is:
RXslowR++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:
Allylic>Benzyllic>Tertiary>Secondary>Primary>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:
CH2=CH−CH2+⟷CH2+−CH=CH2
- Benzylic carbocation: The positive charge is delocalized into the aromatic ring:
C6H5−CH2+⟷several resonance forms involving the ring
Those resonance forms look like this:
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) …
These alcohols react with HX by an SN1-type route through a benzylic carbocation. Electron-withdrawing groups destabilise that cation and slow the reaction. -NO2 withdraws more strongly than -Cl, so reactivity increases b < c < a. …
Substitution of -OH by halogen from HX on a benzylic alcohol goes through a benzylic carbocation intermediate. The more stable that cation, the faster the reaction. Electron-withdrawing ring substituents (-NO2 strongly, -Cl moderately) destabilise the cation, so reactivity increases in the order b < c < a.
Concept
Benzylic alcohols react with HBr/HCl largely by an SN1 pathway: protonation of -OH, loss of water to give a resonance-stabilised benzylic carbocation, then capture by the halide. Anything that stabilises the positive charge speeds the reaction; anything that destabilises it slows the reaction.
Analysis of substituent effects
- (b) p-nitrobenzyl alcohol: -NO2 is a strong electron-withdrawing group (-I and -M). It pulls electron density away from the electron-deficient benzylic carbon, strongly destabilising the carbocation -> slowest. …
Method: Carbocation-Stability Method for SN1 Reactivity of Benzylic Alcohols
Core Concept
Benzylic alcohols react with HX (HBr/HCl) through an SN1 pathway: the -OH is protonated and leaves as water, generating a resonance-stabilised benzylic carbocation that is then captured by the halide ion. Whichever alcohol forms the MOST stable benzylic cation reacts FASTEST, so reactivity ranking reduces to ranking carbocation stability.
Steps
- Draw the mechanism skeleton for all the alcohols being compared: protonate -OH -> lose H2O -> benzylic carbocation (delocalised into the ring) -> X- attacks the cation.
- Identify every ring substituent present in each compound and classify it as electron-donating (+M/+I, stabilises the cation, speeds the reaction) or electron-withdrawing (-M/-I, destabilises the cation, slows the reaction).
- Rank the substituents by the STRENGTH of their effect, not just the direction: -NO2 (strong -M and -I) destabilises far more than -Cl (weak net deactivator, -I dominant with a small +M).
- Order the compounds from most destabilised cation (slowest reaction) to least destabilised/no destabilisation (fastest reaction).
- Convert that stability order directly into a reactivity order (more stable cation = faster reaction = higher position in an "increasing reactivity" list). …
- GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQQ.For the following compounds, what is the correct increasing order of reactivity towards SN1 displacement? (I) 2-Bromo-2-methylbutane (II) 1-Bromopentane (III) 2-Bromopentane(a) I < III < II(b) II < III < I(c) III < II < I(d) I < II < III
›Reveal solutionSolution
SN1 reaction rate depends on the stability of the intermediate carbocation: tertiary > secondary > primary.
The SN1 mechanism proceeds via a carbocation intermediate formed by ionisation of the C–X bond in the rate-determining step. The MORE stable this carbocation, the FASTER the SN1 reaction — and carbocation stability increases with alkyl substitution (more +I donation and hyperconjugation stabilise the positive charge): 3° > 2° > 1°.
- (II) 1-Bromopentane — a primary halide, forms a primary carbocation (least stable) → SLOWEST SN1. …
- GUJCET 2024Set 131 markMCQQ.Predict the order of reactivity of the following compounds in SN1 reaction.(i) C6H5⋅CH2Br(ii) C6H5⋅CH⋅(C6H5)Br(iii) C6H5⋅CH(CH3)Br(iv) C6H5⋅C⋅(CH3)(C6H5)Br (A)(ii) >(iii) >(iv) >(i) (B)(ii) >(iv) >(iii) >(i) (C)(iv) >(iii) >(ii) >(i) (D)(iv) >(ii) >(iii) > (i)
›Reveal solutionSolution
More stabilising groups on the cationic carbon → faster SN1. Two phenyls beat one phenyl + methyl.
Concept. SN1 rate depends on the stability of the carbocation formed. Phenyl groups stabilise through resonance more strongly than a methyl stabilises by induction/hyperconjugation.
- (iv) C6H5C+(CH3)(C6H5): two phenyl + methyl — most stable.
- (ii) C6H5C+H(C6H5): two phenyl (benzhydryl). …
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.Predict the order of reactivity of the following compounds in SN1 reaction:(i) CH3CH2CH(Br)CH3(ii) (CH3)2CHCH2Br(iii) (CH3)3CBr(a)(iii) <(ii) <(i)(b)(ii) <(i) <(iii)(c)(i) <(ii) <(iii)(d)(iii) <(i) < (ii)
›Reveal solutionSolution
SN1 reaction rate depends on the stability of the carbocation intermediate formed; more substituted (more alkyl-stabilised) carbocations react faster via SN1.
- CH3CH2CH(Br)CH3 - a secondary alkyl halide (sec-butyl bromide), gives a secondary carbocation.
- (CH3)2CHCH2Br - a primary alkyl halide (isobutyl bromide), gives a primary carbocation (least stable).
- (CH3)3CBr - a tertiary alkyl halide (tert-butyl bromide), gives a tertiary carbocation (most stable, fastest SN1). …
- GUJCET 2021Set 151 markMCQQ.Which would undergo SN1 reaction faster from following? (A) Chloromethane (B) 2-bromo-3-methylbutane (C) 2-chloro-3-methylbutane (D) 2-bromo-2-methylpropane
›Reveal solutionSolution
SN1 rate tracks carbocation stability: tertiary + good leaving group = fastest.
Concept: SN1 is rate-determined by ionisation to a carbocation. Order of stability 3° > 2° > 1° > methyl; and C−Br ionises more easily than C−Cl (weaker bond, better leaving group).
- (A) Chloromethane → methyl cation (impossible) — slowest.
- (B) 2-bromo-3-methylbutane → 2° cation. …
- GSEB Higher Secondary Certificate (HSC) Examination 2020Set ANNUAL1 markMCQQ.Which of the following compound has highest reactivity towards SN1 reaction?(a) C6H5CH(C6H5)Br(b) C6H5CH2Br(c) C6H5C(CH3)(C6H5)Br(d) C6H5CH(CH3)Br
›Reveal solutionSolution
SN1 reactivity tracks carbocation stability: the more substituted and the more resonance-stabilised (benzylic) the resulting cation, the faster the SN1 reaction.
Ranking the carbocations that would form on loss of Br-:
- (b) C6H5CH2+ - primary benzylic cation, stabilised by only one phenyl ring.
- (d) C6H5CH(CH3)+ - secondary benzylic cation, one phenyl + one methyl.
- (a) C6H5CH(C6H5)+ - secondary but doubly-benzylic (two phenyl rings delocalise the charge) - more stable than (d). …
- GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQQ.Which compound will give unimolecular nucleophilic substitution reaction easily with aqueous NaOH?(a) C6H5-CH2-CH2-Cl(b) C6H5-CH(Cl)-CH3(c) C6H5-C(Cl)(C6H5)-CH3(d) C6H5-CH2-Cl
›Reveal solutionSolution
SN1 reactions proceed through a carbocation intermediate, so the rate is fastest when the substrate can form the MOST STABLE carbocation -- tertiary and/or benzylic (resonance-stabilised) cations react fastest.
Comparing the stability of the carbocation each substrate would form on loss of Cl-:
- C6H5-CH2-CH2-Cl: ionisation gives a primary carbocation (not benzylic, since the CH2-Cl carbon is not directly attached to the ring) -- very unstable, SN1 disfavoured, reacts by SN2.
- C6H5-CH(Cl)-CH3: ionisation gives a SECONDARY benzylic carbocation (one phenyl ring for resonance stabilisation) -- reasonably stable, moderate SN1 reactivity. …
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