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

Question

Explain the SN2 mechanism for the reaction of methyl bromide with NaOH. Why does backside attack occur? What is Walden inversion, and why is SN2 a single-step process?

(JEE Main and NEET both test this mechanism regularly)

Solution — Step by Step

In SN2, the nucleophile attacks at the same time as the leaving group departs. There is no intermediate — just a single transition state.

CH_3Br + OH^- \xrightarrow{\text{one step}} CH_3OH + Br^-

The rate depends on both reactant concentrations: $\text{Rate} = k[CH_3Br][OH^-]$ . This is why it is called “substitution, nucleophilic, bimolecular.”

The nucleophile approaches from the side opposite to the leaving group. Why? Because the leaving group’s electron cloud creates repulsion on its side. The antibonding orbital ( $\sigma^*$ ) of the C–Br bond is accessible from the back — that is where the nucleophile’s electrons can overlap.

Think of it like an umbrella flipping in the wind — the nucleophile pushes from behind while the leaving group pops off the front.

At the transition state, the central carbon is sp² hybridised (temporarily) with five groups partially bonded:

[HO \cdots CH_3 \cdots Br]^{\ddagger}

Both the incoming nucleophile and the departing leaving group are partially bonded. The three remaining groups (H atoms in methyl bromide) lie in a plane perpendicular to the Nu–C–LG axis.

Because of backside attack, the configuration at the carbon is inverted — like an umbrella turning inside out. If the starting material is R, the product is S (and vice versa). This is called Walden inversion and gives 100% inversion of configuration.

This is how we experimentally distinguish SN2 from SN1 (which gives racemisation).

flowchart LR
    A["OH⁻ approaches<br/>from backside"] --> B["Transition State<br/>[HO---C---Br]‡<br/>sp² carbon"]
    B --> C["Br⁻ departs<br/>from front side"]
    C --> D["Product: CH₃OH<br/>Configuration INVERTED"]
    style B fill:#ffeb3b,stroke:#333

Why This Works

SN2 is favoured when the substrate is unhindered (methyl or primary) because the nucleophile needs clear access to the backside of the carbon. In a tertiary substrate, three bulky groups block the backside — the nucleophile simply cannot reach. That is why 3° substrates go SN1 instead.

The bimolecular rate law confirms that both species are involved in the rate-determining step. If you increase the nucleophile concentration, the reaction speeds up — unlike SN1 where nucleophile concentration does not matter.

Alternative Method

Quick identification for MCQs: 1° substrate + strong nucleophile + polar aprotic solvent = SN2. Polar aprotic solvents (like DMSO, acetone, DMF) do not solvate the nucleophile, keeping it “naked” and highly reactive — perfect for SN2.

Common Mistake

Students often draw the nucleophile attacking from the same side as the leaving group (front-side attack). This does not happen in SN2 because of electron-electron repulsion between the nucleophile and the leaving group. The backside attack is not optional — it is the only geometrically and electronically feasible pathway. Front-side attack would require the nucleophile’s electrons to overlap with the filled bonding orbital, which is energetically unfavourable.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next