Question

Write two differences between $S_{N}1$ and $S_{N}2$ reactions. Out of chlorobenzene and 2,4,6-trinitrochlorobenzene, which one is more reactive towards nucleophilic substitution reactions?

Hint:

Electron-withdrawing groups (like $-NO_2$) strongly activate the benzene ring for nucleophilic substitution.

Answer 1

Step 1: Describe the $S_N1$ mechanism.
$S_N1$ (unimolecular) proceeds in two steps: first, the leaving group departs heterolytically to form a carbocation intermediate; then the nucleophile attacks. The rate law is Rate $= k[R$-$X]$, involving only the substrate. The flat carbocation intermediate leads to racemisation of the product.
Step 2: Describe the $S_N2$ mechanism.
$S_N2$ (bimolecular) is a single concerted step: the nucleophile attacks the carbon from the back face while the leaving group departs from the front, passing through a trigonal bipyramidal transition state. Rate $= k[R$-$X][Nu^-]$, and the product shows complete inversion of configuration (Walden inversion).
Step 3: Summarise two key differences.
(1) $S_N1$ is a stepwise, two-step process forming a carbocation intermediate, whereas $S_N2$ is a one-step concerted process with no intermediate. (2) $S_N1$ gives racemisation; $S_N2$ gives inversion of configuration.
Step 4: Compare reactivity toward nucleophilic substitution.
In chlorobenzene, C-Cl bond has partial double-bond character (lone pair delocalisation into ring), making nucleophilic attack very difficult. In 2,4,6-trinitrochlorobenzene, three strong $-NO_2$ groups stabilise the Meisenheimer complex intermediate by dispersing the negative charge, making it far more reactive toward nucleophilic aromatic substitution.