Skip to the content
Browse the library

Bimolecular nucleophilic substitution: the mechanism, step by step

One step. The nucleophile comes in on the opposite side from the leaving group, and the carbon turns inside out as the exchange happens.

rule-derivedOne concerted step, no intermediateNot a record of what was observed

This class has one step and no intermediate. Everything happens at once, and that is not a simplification — it is what the class means. Bonds break and form in the same motion, through a single transition state that is never isolated.

Below is the sequence as it is taught. On the platform it is drawn on whatever structures you supply, with the curly arrows resolved to the actual atoms involved rather than to a general case with R groups on it.

The steps

  1. The nucleophile pushes the leaving group out

    The nucleophile comes in on the far side from the leaving group and pushes it off. The carbon turns through, like an umbrella in the wind — though on a carbon with two hydrogens there is nothing to notice about it.

    Backside attack: the nucleophile's lone pair enters the σ* of the C–leaving-group bond while that bond breaks, through a single transition state with both partly made. The carbon passes through a flat arrangement, so a stereogenic centre comes out inverted; where the carbon is not stereogenic the same thing happens and leaves no trace.

What this drawing assumes

Read these before you quote the mechanism
  • Drawn as a single step with no intermediate, which is what bimolecular means here: bond making and bond breaking happen together.
  • The nucleophile can only arrive from behind the leaving group, and that is true of every reaction of this kind. Whether it shows is another matter: a carbon carrying four different groups is turned inside out and the configuration inverts, while a CH2 is attacked from behind just the same and there is nothing to see.

Other pathways this class runs by

  • A tertiary or strongly stabilised carbon goes by the stepwise route instead, through a planar cation, and then the configuration is lost rather than inverted.

A class is not one pathway. Which of these runs depends on the substrate, the solvent and what else is in the flask, and the drawing above does not decide that for you.

Run it on your own structures

Enter the reactants and the product you expect. If the transformation is one the engine can perform and this pathway reaches that product from those structures, it is drawn on them — with the arrows on the right atoms and the intermediates you would actually pass through. If it does not reach the product, you are told that instead of being shown a drawing that does not apply.