SN2 Reaction: Mechanism, Walden Inversion, Reactivity Order and Exam Traps

ChemistryClass 12, Haloalkanes and HaloarenesFor JEE and NEET aspirants

The SN2 reaction ( $S_{N} 2$ : substitution, nucleophilic, bimolecular) replaces a leaving group on an sp $^{3}$ carbon in one step. The nucleophile attacks the carbon from the side opposite the leaving group while the old bond breaks, so the carbon's configuration is inverted. Both the haloalkane and the nucleophile take part in that single step, so the rate depends on both concentrations. SN2 is fastest at methyl and primary carbons and does not happen at tertiary ones. It is in the Class 12 NCERT chapter on haloalkanes and haloarenes and is tested in JEE Main, JEE Advanced and NEET, often against SN1.

At a glance

Example:
$C H_{3} Br + O H^{-} C H_{3} OH + B r^{-}$
Rate law: rate $= k [C H_{3} Br] [O H^{-}]$ : second order overall, first order in each.
Mechanism: one step, no intermediate. In the transition state the carbon is half bonded to both the nucleophile and the leaving group.
Stereochemistry: inversion of configuration (Walden inversion) at the carbon attacked.
Substrate order: $C H_{3} X$ $>$ 1° $>$ 2° $≫$ 3° (3° does not react by SN2).
Leaving group: $I^{-}$ $>$ $B r^{-}$ $>$ $C l^{-}$ $≫$ $F^{-}$ .
Best solvent: polar aprotic (acetone, DMF, DMSO) for an anionic nucleophile.

1. Mechanism

The nucleophile brings a lone pair to the back of the C–X bond, into the C–X antibonding ( $σ^{*}$ ) orbital, whose larger lobe on carbon points away from X. As the new bond forms, the C–X bond breaks. There is no intermediate, only a transition state.

Figure 1: SN2 of (R)-2-bromobutane with

O H^{-}

. Bond making and bond breaking happen together, and the nucleophile enters opposite the leaving group, so the carbon's configuration is inverted (Walden inversion).

Approach. $O H^{-}$ comes in at 180° to the C–Br bond, where that $σ^{*}$ lobe is and away from the electron-rich bromine.
Transition state. Carbon has five groups around it: two partial bonds (to O and to Br) on one axis and three groups flat in the plane between them. The negative charge is shared between the entering O and the leaving Br.
Inversion. The C–Br bond breaks completely and the three groups flip to the other side, like an umbrella turned inside out in the wind. $B r^{-}$ leaves, and the product has the opposite configuration.

Exam Trick

"One step, back door, inside out." SN2 = one step (rate depends on substrate and nucleophile), attack from the back (so crowding at the carbon kills it) and an inverted product. If a question says the product is optically active with the opposite configuration, it is SN2.

2. What makes SN2 fast

Everything that crowds the back of the carbon slows SN2, and everything that makes a better nucleophile or a better leaving group speeds it up.

Factor	Fast SN2	Slow or no SN2
Substrate	$C H_{3} X$ , 1° halides, allyl and benzyl halides	2° slower; 3° none; neopentyl, $(C H_{3})_{3} CC H_{2} X$ , very slow
Nucleophile	Strong: $R O^{-}$ , $H O^{-}$ , $C N^{-}$ , $I^{-}$ , $R S^{-}$ ; also neutral $N H_{3}$ , amines	Weak, neutral O nucleophiles: $H_{2} O$ , $ROH$
Leaving group	Weak bases: $I^{-}$ , $B r^{-}$ , tosylate	$F^{-}$ ; $O H^{-}$ (an alcohol must first be protonated)
Solvent	Polar aprotic: acetone, DMF, DMSO	Polar protic ( $H_{2} O$ , $C H_{3} OH$ ) cages anions by H-bonding

Why 3° fails. Three alkyl groups block the back of the carbon, so the nucleophile cannot reach it. Neopentyl bromide is primary, but its bulky $(C H_{3})_{3} C$ group sits right beside the reacting carbon and blocks it almost as well: it reacts about 100,000 times more slowly than bromoethane.

Why aryl and vinyl halides fail. Their C–X bond has partial double-bond character (the halogen's lone pair is delocalised into the ring or the C=C), the carbon is sp $^{2}$ , and in an aryl halide the ring itself sits where the nucleophile would have to come in. This is why aniline cannot be made by the Gabriel synthesis.

Nucleophiles in a protic solvent. Small anions are held tightly by hydrogen bonds, so in water or methanol the order is $I^{-}$ $>$ $B r^{-}$ $>$ $C l^{-}$ $>$ $F^{-}$ , the reverse of basicity. Among oxygen anions, the less stable (more basic) one is the better nucleophile: $H O^{-}$ $>$ $C_{6} H_{5} O^{-}$ $>$ $C H_{3} CO O^{-}$ . Large, polarisable nucleophiles beat small oxygen anions in protic solvents: $I^{-}$ $>$ $R O^{-}$ and $R S^{-}$ $>$ $R O^{-}$ . In a polar aprotic solvent the anions are not caged, and the halide order follows basicity: $F^{-}$ $>$ $C l^{-}$ $>$ $B r^{-}$ $>$ $I^{-}$ .

3. SN2 reactions you meet by name

Reaction	SN2 step	Watch for
Finkelstein	$R - Cl$ + NaI in dry acetone gives $R - I$	NaCl precipitates, which drives it forward
Williamson ether synthesis	$R O^{-}$ + $R^{'} X$ gives $RO R^{'}$	R′X must be methyl or 1°; a 3° halide gives an alkene (E2)
Gabriel phthalimide synthesis	Phthalimide anion + $R - X$ , then hydrolysis gives $R - N H_{2}$	Only 1° alkyl amines; no aryl amines
Nitrile from KCN	$C N^{-}$ + $R - X$ gives $R - CN$ (one carbon more)	AgCN gives the isocyanide $R - NC$ instead
Nitrite or nitro compound	$KN O_{2}$ gives the alkyl nitrite $R - O - N = O$	$AgN O_{2}$ gives the nitroalkane $R - N O_{2}$
Ether cleavage by HI	$I^{-}$ attacks the less crowded alkyl carbon of the protonated ether	The aryl–O bond never breaks: anisole gives phenol + $C H_{3} I$ . If one group is 3°, the reaction goes by SN1 and the 3° iodide forms: $(C H_{3})_{3} COC H_{3}$ gives $(C H_{3})_{3} CI$ + $C H_{3} OH$

The same reagent can substitute or eliminate. Aqueous KOH with a 1° or 2° halide mainly substitutes (alcohol); alcoholic KOH, heat or a bulky base such as $(C H_{3})_{3} C O^{-}$ favour elimination (alkene). A 3° halide never reacts by SN2: with a strong base it eliminates, and in water it gives the 3° alcohol by SN1.

JEE Advanced

Solvent and charge. When an anion attacks a neutral halide ( $H O^{-}$ + $RCl$ ), the charge is spread out in the transition state, so a more polar solvent stabilises the reactants more than the transition state and slows the reaction. When two neutral molecules react ( $R_{3} N$ + $RCl$ ), charge builds up in the transition state, and a more polar solvent speeds it up. Polar aprotic solvents still help an anion, because they cannot hydrogen-bond to it and leave it free; the polarity effect on the anion case is small, while for $R_{3} N$ + $RCl$ it is large.

A stereocentre next door. SN2 inverts only the carbon that is attacked. In 1-bromo-2-methylbutane the stereocentre is C-2, not the $C H_{2} Br$ carbon, so it keeps its configuration.

R/S is a label, inversion is geometry. The 3D arrangement always inverts. If the four groups keep the same priority order (the new group takes the old group's rank, as OH takes Br's), R becomes S. If the order changes, assign R or S again from the drawing: the letter may stay the same.

Proof that every SN2 step inverts. When an optically active alkyl iodide exchanges with radioactive iodide ion, it loses its optical activity twice as fast as it takes up the radioactive iodine: each exchange inverts one molecule, which cancels the rotation of another.

4. Worked examples

Solved Example 1

(R)-2-Bromobutane is warmed with aqueous NaOH under SN2 conditions. Name the product and its configuration.

Solution:

$O H^{-}$ attacks from the back, so the configuration inverts. OH takes bromine's place as the highest priority group and the other three groups keep their order, so the label changes too: the product is (S)-butan-2-ol.

Solved Example 2

Arrange in order of decreasing SN2 rate with NaI in acetone:

(C H_{3})_{3} CBr

C H_{3} C H_{2} Br

(C H_{3})_{2} CHBr

C H_{3} Br

Solution:

Less crowding at the carbon means faster back-side attack: $C H_{3} Br$ $>$ $C H_{3} C H_{2} Br$ $>$ $(C H_{3})_{2} CHBr$ $>$ $(C H_{3})_{3} CBr$ . The tertiary bromide gives practically no SN2 product.

Solved Example 3

Which pair gives tert-butyl methyl ether:

(C H_{3})_{3} CBr

with

C H_{3} ONa

, or

(C H_{3})_{3} CONa

with

C H_{3} Br

Solution:

The second. Williamson synthesis is an SN2 reaction, so the halide must be easy to attack: $C H_{3} Br$ is ideal. With $(C H_{3})_{3} CBr$ , methoxide cannot reach the crowded carbon and acts as a base instead, giving 2-methylpropene by elimination.

5. Exam traps

Watch out

SN2 always inverts the 3D arrangement, but R does not always become S: re-rank all four groups.
SN2 has no carbocation, so there is no rearrangement and no racemisation (contrast SN1).
Inversion and R/S do not predict the sign of rotation: (−)-2-bromooctane with NaOH gives (+)-octan-2-ol.
A tertiary halide does not react by SN2. Neopentyl halides are primary but also very slow.
Aryl and vinyl halides do not undergo SN2, so aniline cannot be made by the Gabriel synthesis.
The rate depends on both concentrations: doubling $[O H^{-}]$ doubles the rate (unlike SN1).
In water or alcohol, $I^{-}$ is the best halide nucleophile and $F^{-}$ the worst, the reverse of basicity.
A stereocentre that is not the reacting carbon keeps its configuration in SN2.
Aqueous KOH substitutes; alcoholic KOH eliminates. Do not mix them up in a reaction chain.
Williamson ethers: put the bulky group on the alkoxide and use a methyl or 1° halide, never a 3° halide.

Past-paper questions on SN2 reaction

Our past papers: JEE Main 2025-2026, JEE Advanced 2022-2026, NEET 2018-2019, 2022-2026. We found 26 questions in them that test this:

JEE Main 13 (2025-2026)
JEE Advanced 8 (2022-2023, 2025-2026)
NEET 5 (2022, 2024-2025)

Each question was read and checked by hand (list checked 2026-10-14). A question that shows its structures only as pictures can be missed by our search. Each opens in a new tab, with the question to try and our step-by-step solution.

Learn it with the chapter

Related named reactions

SN1 reactionUnimolecular substitution: the leaving group departs first to give a carbocation, which the nucleophile captures; rate = k[RX].

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