Which one of the following compounds is more reactive towards SN2 reaction and why ? CH₃CH(Cl)CH₂CH₃ or CH₃CH₂

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S_N2

Karnataka PUCCBSE Class XII Board 2018Subjective· 2mImportance★★★★★

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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.

SN2 mechanism in wedge-dash notation: hydroxide attacks the carbon of CH3Cl from the back side, passing through a five-coordinate transition state (partial HO and Cl bonds, both delta-minus) before chloride leaves

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.

Backside attack in the SN2 reaction (NCERT Fig 6.2): the incoming hydroxide (red) approaches the carbon opposite the outgoing halide (green), and the three hydrogens flip through like an umbrella inverting

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)

Watch out

SN2 is impossible on tertiary carbons — the three bulky groups block the backside completely. The nucleophile simply cannot get close enough.

Steric effects in the SN2 reaction (NCERT Fig 6.3): relative rates fall from methyl (30) to ethyl (1) to isopropyl (0.02) to tert-butyl (0) as alkyl groups crowd the backside approach of the nucleophile

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 two steps of the SN1 mechanism for 2-bromo-2-methylpropane: slow reversible ionisation to the planar tert-butyl carbocation, then fast attack by hydroxide to give 2-methylpropan-2-ol

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.

SN1 racemisation of 2-bromobutane: the planar carbocation is attacked by hydroxide on either face, giving equal amounts of (+)- and (-)-butan-2-ol — an optically inactive racemic mixture

Note

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)

Watch out

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 …

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$CH_3CH_2CH_2Cl$ (primary) beats $CH_3CH(Cl)CH_2CH_3$ (secondary) in $S_N2$ because a primary carbon has less steric hindrance to backside nucleophilic attack.

Concept. The $S_N2$ mechanism (CBSE Class-12 haloalkanes-and-haloarenes) is a one-step, concerted attack of the nucleophile from the side opposite the leaving group.

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

KCET 2026Set D31 markMCQ
$_N$
›Reveal solutionSolution
$S_N1$ hydrolysis proceeds via a planar carbocation attacked equally from both faces, but racemisation is only observable when the resulting alcohol has a stereocentre — true only for 2-bromobutane among the four options.

Step 1 — How racemisation arises in $S_N1$

In an $S_N1$ reaction, the alkyl halide first ionises to form a planar, $sp^2$ -hybridised carbocation, with the leaving group ( $Br^-$ ) departing completely before the nucleophile (water, in hydrolysis) attacks. Because the carbocation is planar, water can attack from either face of the plane with nearly equal probability, generating both possible spatial arrangements at that carbon in roughly equal amounts. If — and only if — that carbon becomes a genuine stereocentre in the product, this equal attack from both faces produces a racemic mixture (equal amounts of the two enantiomers).

Step 2 — Checking each option for a resulting stereocentre

(A) Tertiary butyl bromide ionises to a carbocation with three identical methyl groups attached; the product, tert-butanol, has no stereocentre at all (the carbon bears three identical $CH_3$ groups), so there is nothing to racemise — no chirality is possible either way.

(B) 2-bromobutane ionises at $C_2$ , which bears four different groups in the product ( $CH_3$ , $OH$ , $H$ , $C_2H_5$ ) — a genuine stereocentre. Water attacking the planar carbocation from either face gives both the (R)- and (S)-2-butanol in equal amounts: a true racemic mixture. …
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Q.Which one of the following compounds is more reactive towards SN2S_N2SN​2 reaction and why ? CH3CH(Cl)CH2CH3CH_3CH(Cl)CH_2CH_3CH3​CH(Cl)CH2​CH3​ or CH3CH2CH2ClCH_3CH_2CH_2ClCH3​CH2​CH2​Cl

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Q.Which one of the following compounds is more reactive towards $S_N2$ reaction and why ?
$CH_3CH(Cl)CH_2CH_3$ or $CH_3CH_2CH_2Cl$

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