Question

Explain the mechanism of substitution reactions of Haloalkane.

Hint:

Tertiary haloalkanes follow $S_N1$, primary haloalkanes follow $S_N2$, and secondary haloalkanes may follow either depending on conditions.

Answer 1

Haloalkanes undergo nucleophilic substitution reactions because the carbon atom attached to the halogen is electron-deficient (due to the electronegativity of halogen). The C–X bond is polarized (C$^{\delta+}$ – X$^{\delta-}$), making carbon susceptible to attack by nucleophiles.
There are two main mechanisms:
1. $S_N1$ mechanism (Unimolecular Nucleophilic Substitution):
- Two-step process.
- Step 1: Slow ionization of haloalkane to form carbocation (rate-determining).
\[ R{-}X \;\longrightarrow\; R^+ \;+\; X^- \]
- Step 2: Fast attack of nucleophile on carbocation.
\[ R^+ + Nu^- \;\longrightarrow\; R{-}Nu \]
- Rate depends only on haloalkane concentration: Rate = $k$[R–X].
- Favoured in tertiary haloalkanes (stable carbocation).
2. $S_N2$ mechanism (Bimolecular Nucleophilic Substitution):
- One-step, concerted process.
- Nucleophile attacks carbon from the side opposite to the leaving group (backside attack).
- Transition state has partial bonds to both nucleophile and leaving group.
\[ R{-}X + Nu^- \;\longrightarrow\; [R{-}(Nu)(X)]^\ddagger \;\longrightarrow\; R{-}Nu + X^- \]
- Rate depends on both haloalkane and nucleophile concentrations: Rate = $k$[R–X][Nu$^-$].
- Favoured in primary haloalkanes (less steric hindrance).