SN2 Reaction Mechanism | JEE Notes

S_{N} 2

means substitution, nucleophilic, bimolecular. Bimolecular means two species are in the slowest step.

The mechanism

The carbon is slightly positive, and the nucleophile approaches from the side opposite the halogen, because that is where the empty $σ^{*}$ orbital of the C-X bond has its large lobe. As the new bond forms the old one breaks, and there is no intermediate.

$S_{N} 2$ of (R)-2-bromobutane

One pair arrives as the other leaves, so there are two curly arrows in a single step. The nucleophile aims at the back lobe of the C-Br $σ^{*}$ orbital.

At the top of the energy hill, both bonds are partial and the three other groups lie in one plane. This is a transition state, so it cannot be isolated.

The three groups have flipped. The product here is (S)-butan-2-ol. OH takes the rank that Br held, so the letter changes.

Watch the animation: SN2: attack from the back

Walden inversion: the three groups flip like an umbrella in the wind.

Explore in 3D: Inversion, R to S

S_{N} 2

: one hump, no intermediate.

The transition state has five groups around carbon, with the three remaining groups in a plane. It is an energy maximum, so it cannot be isolated.

Rate

rate = k [R X] [N u^{-}]

Double either concentration and the rate doubles. Doubling both makes it four times faster.

Try the simulator: Rate law: SN1 or SN2

Crowding

The nucleophile has to reach a small target behind the carbon. Each alkyl group blocks more of it.

C H_{3} X > 1^{\circ} > 2^{\circ} ≫ 3^{\circ}

Primary does not always mean fast. Branching next door slows the reaction too. Neopentyl bromide, $(C H_{3})_{3} C C H_{2} B r$ , is primary, yet it is very slow by $S_{N} 2$ . Primary is only a label. What matters is the space behind the carbon.

Watch the animation: Why crowding stops SN2

Faster than expected

Allylic and benzylic halides. The p orbital at the reacting carbon in the transition state overlaps with the neighbouring $π$ system. This was tested in April 2024.
$α$ -Halo ketones such as $C_{6} H_{5} C O C H_{2} C l$ . The carbonyl group lowers the transition state strongly, so they are among the fastest $S_{N} 2$ substrates.

Never by $S_{N} 2$

Tertiary halides are too crowded. Vinylic and aryl halides have no open path, because the ring or the double bond sits behind the carbon.

Worked example: rank for $S_{N} 2$

Rank by $S_{N} 2$ rate with $I^{-}$ in acetone: (a) 1-bromobutane, (b) 2-bromobutane, (c) 1-bromo-2,2-dimethylpropane, (d) 3-bromopropene.

Show the solution

All four are bromides, so only the carbon matters. (d) is allylic, so the transition state is helped by the $π$ bond and it is fastest. (a) is primary and open. (b) is secondary. (c) is neopentyl, and the bulky group next door nearly stops the reaction.

So (d) > (a) > (b) > (c).

JEE Advanced: inversion in a ring

In a ring, inversion shows up as cis turning into trans. cis-1-Bromo-4-methylcyclohexane with $O H^{-}$ by $S_{N} 2$ gives trans-4-methylcyclohexanol. If a question draws a ring with wedges, look for the group to change face. Inside one molecule the same reaction closes rings: with base, 4-bromobutan-1-ol gives tetrahydrofuran, which is an intramolecular Williamson reaction.

$S_{N} 2$ : one step, backside attack

In 60 seconds

The mechanism

$S_{N} 2$ of (R)-2-bromobutane

Rate

Crowding

Faster than expected

Never by $S_{N} 2$

Worked example: rank for $S_{N} 2$

JEE Advanced: inversion in a ring

SN2: one step, backside attack

In 60 seconds

The mechanism

SN2 of (R)-2-bromobutane

Rate

Crowding

Faster than expected

Never by SN2

Worked example: rank for SN2

JEE Advanced: inversion in a ring

$S_{N} 2$ : one step, backside attack

$S_{N} 2$ of (R)-2-bromobutane

Never by $S_{N} 2$

Worked example: rank for $S_{N} 2$