Question

Write the mechanism of the bimolecular nucleophilic substitution reaction (S_N2) of an alkyl halide.

Hint:

Think one concerted step: the nucleophile attacks the carbon from the side opposite the halogen, giving a five-coordinate transition state and inversion of configuration.

Answer 1

Step 1: Order and rate. The S_N2 pathway follows second order kinetics because two molecules meet in the rate-determining step. Doubling either the alkyl halide or the nucleophile doubles the rate: \(\text{Rate} = k[R\text{-}X][Nu^-]\).
Step 2: Energy picture. As the nucleophile pushes in from the rear and the halide slides out the front, the system climbs to a single energy maximum, the transition state, and then falls to products. No stable intermediate is formed, unlike S_N1.
Step 3: Geometry of the transition state. At the top of the energy hill the carbon is partly bonded to five atoms: the entering nucleophile and the leaving halide are on a straight line (\(Nu \cdots C \cdots X\)), while the other three substituents flatten into a plane like the ribs of an umbrella about to flip.
Step 4: Stereochemistry. Once the halide departs, the three flattened groups swing through to the far side, so an optically active substrate gives a product of opposite configuration (Walden inversion).
Step 5: Reactivity trend. Because the nucleophile must reach the carbon, crowding slows it down. So the ease of S_N2 is: methyl > primary > secondary > tertiary halide.
Illustration: \(CN^- + CH_3CH_2Br \rightarrow CH_3CH_2CN + Br^-\), where cyanide attacks the rear of the C-Br carbon in one concerted move.
\(\boxed{\text{Concerted, 2nd order, backside attack, inversion of configuration}}\)