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Substitution Against Elimination: Predicting the Major Product

Four mechanisms compete for the same substrate and the same reagent. Deciding which wins is a repeatable procedure over substrate class, base or nucleophile strength, bulk and temperature.

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What a learner can do afterwards

  • Works through substrate class first and rules out the mechanisms it forbids
  • Separates nucleophile strength from base strength and uses each in the right place
  • Explains why a bulky base pushes a reaction towards elimination and towards the less substituted alkene
  • Predicts the effect of raising the temperature and gives the kinetic reason.

1 · Read

Four mechanisms compete for the same substrate and the same reagent. Start with the substrate class, because it rules routes out before reagent details matter. Ordinary methyl and primary halides forbid SN1 and E1, though allylic and benzylic ones can ionize through resonance-stabilised cations. Methyl has no beta carbon at all, so E2 is closed for it too and only SN2 remains; primary keeps SN2 and E2 open. Tertiary forbid SN2. Secondary keep all four routes open and need the full analysis.

Separate nucleophile strength from base strength and use each in the right place. Nucleophile strength is about attacking carbon, where loose polarizable electrons win. Base strength is about grabbing protons. Iodide is a great nucleophile but a weak base. A small strong nucleophile that is a weak base attacks carbon and substitutes.

Try it together

Give 2-bromo-2-methylbutane a hindered alkoxide in warm alcohol and elimination dominates. The bulky base cannot easily reach carbon past the bulk, so it pulls an exposed proton instead, taking the least hindered one for the less substituted Hofmann alkene. Swap in iodide and the tertiary substrate still cannot take SN2: substitution would need the SN1 route to outrun elimination here, which small size alone does not guarantee in warm alcohol.

Good to know

Let solvent and temperature nudge the balance. Protic solvents favour ionization routes. For heat, use the taught two-path model with elimination climbing the higher barrier at the same attempt rate, both barrier heights fixed over this range: warming helps the harder climb relatively more, so the elimination share rises. Bimolecular routes speed up with more reagent, while unimolecular routes ignore it. When fixed power-law routes run unchanged, a still ratio means shared dependence.

substrate classreagent typesolventtemperature

Filter by substrate first, then read the reagent, then let solvent and heat settle the rest.

The rewritten reading was written for this dataset.

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Substitution Against Elimination: Predicting the Major Product · Science, ages 19 to 21 · LightMySky