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Unimolecular nucleophilic substitution: the mechanism, step by step

The leaving group goes first, on its own, leaving a flat carbon with a positive charge. Whatever is around then attacks it from either side.

rule-derived3 stepsNot a record of what was observed

The pathway runs in 3 steps. What sits between them is a real structure: it is what the reaction passes through, and on a bad day it is what you isolate instead of the product.

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 leaving group leaves on its own

    The bond to the leaving group breaks by itself. Both electrons go with the leaving group, and the carbon is left flat and positively charged.

    Rate-determining heterolysis of the C-leaving-group bond, giving a planar carbocation. Nothing attacks in this step, which is why the rate does not depend on the nucleophile.

  2. The nucleophile attacks the flat carbon

    The carbon is short of electrons and flat, so the nucleophile can reach it from either side. It shares a pair and the bond forms.

    Capture of the planar cation. Both faces are open, so a carbon that was a single configuration is left as a mixture.

  3. A base takes the proton

    The oxygen is carrying a positive charge and one hydrogen too many. Something basic takes it, and the neutral product is left.

    Proton transfer to any base present, most often another molecule of the solvent.

What this drawing assumes

Read these before you quote the mechanism
  • The first step is drawn as the slow one, which is what unimolecular means: the rate does not depend on the nucleophile.
  • The cation is drawn as a plain flat carbon. Where a neighbouring group can reach round and stabilise it the picture is more complicated than this.

Other pathways this class runs by

  • A less stabilised carbon goes by the single-step route instead, with the nucleophile arriving as the leaving group departs.
  • The cation can also lose a proton rather than be attacked, which gives the alkene by elimination.

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.