Substitution vs elimination – SN1, SN2, E1 and E2 competition in haloalkanes

What it shows

A haloalkane can react with a nucleophile or base in four ways: SN2 (one step, backside attack, inversion), SN1 (via a planar carbocation), E2 (one step, base removes an anti β-hydrogen) and E1 (via a carbocation that loses H⁺). Which one wins depends on the structure of the haloalkane, the strength and size of the reagent, the solvent and the temperature. Elimination forms an alkene: usually the more substituted Zaitsev alkene, but a bulky base gives mainly the less substituted Hofmann alkene. The percentages come from an approximate rate model and show trends, not measured yields.

How to use

Choose the Haloalkane, the Reagent and the Solvent, then drag the Temperature slider and watch the bar chart and the temperature graph change. Press SN2, SN1, E2 or E1 (or tap a bar) to animate that mechanism, with Pause and Reset. Read the product table to see which alkene is the Zaitsev or Hofmann product, and change only one condition at a time.

Parameters you can change

• Haloalkane Primary: 1-bromobutane, Secondary: 2-bromobutane, Tertiary: 2-bromo-2-methylbutane
• Reagent CN⁻: strong nucleophile, weak base, OH⁻: strong nucleophile, strong base, (CH₃)₃CO⁻: strong, bulky base, H₂O: weak nucleophile, weak base
• Solvent Water (polar protic), Ethanol (polar protic), DMSO (polar aprotic)
• Temperature 20–100 °C

Questions to explore

1. Why does 2-bromo-2-methylbutane not react by SN2, even with a strong nucleophile?
2. How does replacing OH⁻ with (CH₃)₃CO⁻ change the main alkene from 2-bromobutane, and why?
3. Does raising the temperature increase or decrease the share of elimination, and how does entropy explain this?