SN2 mechanism step by step — backside attack, Walden inversion, transition state

Question

Describe the SN2 mechanism. Why does backside attack cause Walden inversion, and which substrates favour SN2?

Solution — Step by Step

In SN2, the nucleophile attacks the substrate carbon from the back side (180 degrees opposite to the leaving group) at the same time as the leaving group departs. There is no intermediate — only a transition state.

Nu^- + R\text{-}X \xrightarrow{\text{one step}} Nu\text{-}R + X^-

The rate depends on both the nucleophile and substrate concentrations:

\text{Rate} = k[Nu^-][R\text{-}X]

That is why it is called SN2 — bimolecular.

At the transition state, the carbon is bonded to five groups simultaneously (partially bonded to both the nucleophile and the leaving group). The geometry is trigonal bipyramidal.

[Nu \cdots C \cdots X]^\ddagger

The three non-reacting groups are in a plane perpendicular to the Nu-C-X axis. As the nucleophile pushes in, these groups flip to the other side — like an umbrella inverting in wind.

Because the nucleophile attacks from behind, all three substituents on the carbon flip to the opposite side. If the starting material was R-configuration, the product becomes S-configuration (and vice versa).

This 100% inversion of configuration is called Walden inversion. Unlike SN1, there is no racemisation — the stereochemistry is completely inverted.

graph LR
    A["Nu- approaches from back"] --> B["Transition state: Nu...C...X"]
    B --> C["X- leaves from front"]
    C --> D["Product with inverted configuration"]
    style B fill:#fff3cd

Why This Works

SN2 favours methyl and primary substrates because backside attack requires an unhindered pathway to the carbon. With tertiary substrates, three bulky alkyl groups block the back side — the nucleophile cannot reach the carbon.

Steric hindrance order: methyl < 1-degree < 2-degree < 3-degree

So SN2 reactivity follows: methyl > 1-degree > 2-degree >> 3-degree (essentially zero)

Strong nucleophiles (like $OH^-$ , $CN^-$ , $I^-$ ) and polar aprotic solvents (like DMSO, DMF, acetone) favour SN2. Polar aprotic solvents do not solvate the nucleophile, keeping it “naked” and reactive.

Alternative Method

A useful shortcut for exams: check the substrate + nucleophile combination.

Substrate	Nucleophile	Likely Mechanism
Methyl, 1-degree	Strong ( $CN^-$ , $OH^-$ )	SN2
3-degree	Weak (water, ROH)	SN1
2-degree	Strong, polar aprotic	SN2
2-degree	Weak, polar protic	SN1

Secondary substrates are the borderline case — the nucleophile strength and solvent tip the balance.

For NEET: if a question shows an optically active halide reacting with $OH^-$ and asks about the product’s optical activity, it is testing SN2. The answer is always “optically active with inverted configuration.”

Common Mistake

Students often confuse inversion with loss of optical activity. SN2 gives complete inversion — the product IS optically active, just with opposite rotation. It is SN1 (racemisation) that causes loss of optical activity. If the question says “the product is optically inactive,” think SN1 (racemic mixture), not SN2.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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