Q.Which one of the following compounds is more reactive towards SN2 reaction and why ?
CH3CH(Cl)CH2CH3 or CH3CH2CH2Cl
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — SN1 and SN2 Mechanism
The Core Idea: Two Ways to Swap a Group
Imagine you have a molecule with a leaving group (like a halogen) attached to a carbon. You want to replace that leaving group with a nucleophile (something that loves positive charge). There are two fundamentally different ways this can happen — like two different ways to replace a lightbulb.
SN2 is like unscrewing the old bulb and screwing in the new one in one smooth motion. SN1 is like first pulling the old bulb out completely, leaving an empty socket, and then putting the new bulb in.
That empty socket — the carbocation — is the key difference.
SN2: One Step, Backside Attack
The name says it all: Substitution, Nucleophilic, Bimolecular. "Bimolecular" means two molecules (the nucleophile and the substrate) are involved in the rate-determining step.
The Mechanism
The nucleophile attacks the carbon from the backside — directly opposite the leaving group. As the nucleophile approaches, the leaving group starts to leave. At the transition state, the nucleophile is partially bonded and the leaving group is partially detached. Then the leaving group departs completely, and the nucleophile is fully bonded.
All of this happens in one step — no intermediate.
The Stereochemistry: Inversion
Because the nucleophile attacks from the back, the configuration at the carbon inverts — like an umbrella turning inside out in a strong wind. If you start with an R configuration, you get S (and vice versa). This is called Walden inversion.
What Favours SN2?
- Primary carbon (least steric hindrance — the backside is wide open)
- Strong nucleophile (needs to push its way in)
- Good leaving group (but not too good — it needs to wait for the nucleophile)
- Polar aprotic solvent (doesn't solvate the nucleophile too tightly)
SN2 is impossible on tertiary carbons — the three bulky groups block the backside completely. The nucleophile simply cannot get close enough.
SN1: Two Steps, Carbocation Intermediate
Substitution, Nucleophilic, Unimolecular. "Unimolecular" means only one molecule (the substrate) is involved in the rate-determining step.
The Mechanism
Step 1 (slow, rate-determining): The leaving group leaves on its own, forming a carbocation (a carbon with only six electrons — positively charged and very unstable).
Step 2 (fast): The nucleophile attacks the carbocation. Since the carbocation is flat (trigonal planar), the nucleophile can attack from either side with equal probability.
The Stereochemistry: Racemisation
Because the nucleophile can attack from either face of the flat carbocation, you get a racemic mixture — equal amounts of R and S. If the starting material is optically pure, the product will be optically inactive.
In practice, you often get slightly more inversion than retention (about 60:40) because the leaving group can partially block one face as it departs. But the key idea is loss of stereochemistry.
What Favours SN1?
- Tertiary carbon (the carbocation is stabilised by three alkyl groups — hyperconjugation and inductive effect)
- Weak nucleophile (doesn't need to force its way in — the carbocation is desperate for electrons)
- Excellent leaving group (must be able to leave on its own)
- Polar protic solvent (stabilises the carbocation and the leaving group)
SN1 is impossible on primary carbons — a primary carbocation is so unstable it effectively doesn't exist. The leaving group would never leave on its own.
The Big Comparison Table …
SN2 rate is governed by steric crowding at the carbon bearing the halogen: a primary (less hindered) carbon reacts faster than a secondary one. …
CH3CH2CH2Cl (primary) beats CH3CH(Cl)CH2CH3 (secondary) in SN2 because a primary carbon has less steric hindrance to backside nucleophilic attack.
Concept. The SN2 mechanism (CBSE Class-12 haloalkanes-and-haloarenes) is a one-step, concerted attack of the nucleophile from the side opposite the leaving group.
…
- KEAM 2025Set eng-2025-04234 marksMCQQ.The order of reactivity of the following compounds towards SN2 displacement reaction is(i) 2-Bromo-2-methylbutane(ii) 1-Bromopentane(iii) 2-Bromopentane (A)(ii) >(i) >(iii) (B)(iii) >(i) >(ii) (C)(ii) >(iii) >(i) (D)(i) >(ii) >(iii) (E)(iii) >(ii) > (i)
›Reveal solutionSolution
SN2 reactivity order is (ii) 1-bromopentane > (iii) 2-bromopentane > (i) 2-bromo-2-methylbutane.
Concept and Intuition
SN2 proceeds by backside attack, so it is fastest with the least steric hindrance at the carbon bearing the halogen. Reactivity decreases as: primary > secondary > tertiary.
Step-by-Step Solution
- (ii) 1-Bromopentane = primary → fastest SN2.
- (iii) 2-Bromopentane = secondary → intermediate. …
- KEAM 2025Set eng-2025-04254 marksMCQQ.The order of reactivity of the following compounds towards SN2 displacement reaction is(i) C6H5CH(CH3)Br(ii) C6H5CH(C6H5)Br(iii) C6H5C(CH3)(C6H5)Br(iv) C6H5CH2Br (A)(ii) >(i) >(iii) >(iv) (B)(iv) >(ii) >(i) >(iii) (C)(ii) >(iii) >(i) >(iv) (D)(i) >(ii) >(iii) >(iv) (E)(iv) >(i) >(ii) > (iii)
›Reveal solutionSolution
SN2 requires backside attack, so reactivity decreases as steric crowding at the carbon bearing the halogen increases: primary > secondary > tertiary, with bulkier substituents slowing it further. Order: (iv) > (i) > (ii) > (iii).
Classifying each carbon bearing Br:
- (iv) C6H5CH2Br: primary benzylic — least hindered, fastest.
- (i) C6H5CH(CH3)Br: secondary, with one small methyl. …
- KEAM 2025Set eng-2025-04284 marksMCQQ.The order of reactivity towards SN2 reaction among the following is(a) CH3Cl(b) CH3CH(Cl)CH3(c) (CH3)3CCl(d) CH3CH2Cl (A) a > d > b > c (B) a > b > c > d (C) a > c > d > b (D) b > a > c > d (E) c > b > d > a
›Reveal solutionSolution
SN2 needs backside attack, so less-hindered carbons react faster: CH3>1∘>2∘>3∘.
Ranking the substrates by steric hindrance at the carbon bearing Cl:
- (a) CH3Cl — methyl, least hindered, fastest
- (d) CH3CH2Cl — primary …
- KEAM 2024Set eng-2024-06074 marksMCQQ.Which of the following statement is incorrect? (A) (-)-2-bromooctane reacts with NaOH gives (+)-octan-2-ol by SN2 reaction. (B) 2-Bromobutane reacts with NaOH gives racemic mixture by SN1 reaction. (C) β-elimination of 2-bromopentane gives pent-1-ene as major product. (D) The hybridization of the carbon in the intermediate formed in SN1 reaction is sp2. (E) Primary alkyl halide undergoes SN2 faster than secondary alkyl halide.
›Reveal solutionSolution
Statement (C) is incorrect: β-elimination of 2-bromopentane follows Zaitsev's rule and gives the more substituted alkene, pent-2-ene, as the major product — not pent-1-ene.
Checking each: (A) SN2 on optically active (-)-2-bromooctane inverts configuration to give (+)-octan-2-ol — correct (a classic Walden inversion). (B) The secondary substrate 2-bromobutane can react by SN1 via a planar carbocation, giving a racemic alcohol — correct. (D) The SN1 intermediate is a carbocation, sp2 hybridised and planar — correct. (E) Primary halides …
- KEAM 2024Set pha-2024-06104 marksMCQQ.The decreasing order of reactivity of butyl bromides in $S_N2$ reaction is (A) $(CH_33CBr>CH3CH2CH2CH2Br>CH3CHCH_3CH2Br>CH3CH2CHBrCH3 (B) $CH_3CH_2CH_2CH_2Br > CH_3CH(CH_32Br>CH_33CBr>CH3CH2CHBrCH3 (C) $(CH_33CBr>CH3CHCH_3CH2Br>CH3CH2CH2CH2Br>CH3CH2CHBrCH3 (D) $CH_3CH_2CH_2CH_2Br > (CH_33CBr>CH3CHCH_3CHBr>CH3CHCH_3CH2Br (E) $CH_3CH_2CH_2CH_2Br > CH_3CH(CH_32Br>CH3CH2CHBrCH3>CH_33CBr
›Reveal solutionSolution
SN2 rate falls with increasing steric hindrance at the carbon bearing Br.
SN2 reactivity decreases with steric crowding: primary > secondary > tertiary, and among primaries the less branched one reacts faster. So:
- n-butyl bromide (CH3CH2CH2CH2Br, 1°, least hindered) — fastest
- isobutyl bromide ((CH3)2CHCH2Br, 1° but β-branched)
- sec-butyl bromide (CH3CH2CHBrCH3, 2°) …
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