SN2 with hydroxide: practice and mechanism
A primary haloalkane with hydroxide reacts via SN2: OH replaces the halogen at the same carbon and the chain length stays the same.
Example
Primary haloalkanes react via SN2: hydroxide attacks the carbon from the back side and the halide leaves. A primary alcohol forms; the chain length stays the same.
Mechanism step by step
- 1
The lone pair of OH⁻ attacks the carbon bearing the halogen, exactly opposite the C–halogen bond (backside attack).
- 2
In the transition state the carbon is five-coordinate: the new C–O bond forms while the C–halogen bond breaks (one concerted step).
- 3
The halide leaves; the three remaining groups flip over like an umbrella in the wind (Walden inversion).
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- That is the elimination product. With primary haloalkanes and aqueous hydroxide substitution wins; elimination would need a strong, bulky base and heat.
Often confused with: SN2 with cyanide (chain extension) · Williamson ether synthesis · SN2 with ammonia
Frequently asked questions
Why do primary haloalkanes prefer SN2?
The backside attack needs space. At primary carbons the back is barely shielded; at tertiary ones three alkyl groups block the way.
What does the rate depend on?
SN2 is bimolecular: the rate depends on the concentrations of both haloalkane and nucleophile (rate = k·[RX]·[Nu⁻]).