Substitution & elimination reactions
The idea
When a nucleophile or base meets an alkyl halide, two competitions decide the outcome: substitution versus elimination, and stepwise (unimolecular) versus concerted (bimolecular). SN2 and E2 happen in one step with the rate depending on both partners; SN1 and E1 go through a carbocation intermediate whose formation is rate-limiting. You already classify reactions and read structures; the new skill is predicting which of the four pathways dominates from the substrate, the reagent, and the solvent.
Four factors steer the choice. Substrate: methyl and primary centers favor SN2, tertiary centers block backside attack and favor SN1/E1, secondary centers sit on the fence. Reagent: a strong, unhindered nucleophile that is a weak base pushes SN2, a strong bulky base pushes E2, a weak nucleophile in a warm protic solvent invites SN1/E1. Stereochemistry is the fingerprint of mechanism — SN2 inverts the stereocenter through backside attack, while SN1 racemizes because the flat carbocation is attacked from both faces. E2 demands an anti-periplanar hydrogen and tends to give the more substituted (Zaitsev) alkene.
The common trap is treating nucleophilicity and basicity as the same property. They correlate loosely but diverge often: iodide is a superb nucleophile yet a feeble base, while a bulky alkoxide is a strong base but a clumsy nucleophile. That gap is exactly the lever that lets you steer a secondary substrate toward substitution or toward elimination by choosing the reagent.
Worked example
(S)-2-bromobutane is treated with sodium hydroxide in a polar aprotic solvent. Predict the dominant mechanism, name the organic product, and state its stereochemistry relative to the starting material.
- Classify the substrate: C2 of 2-bromobutane is a secondary carbon bearing the leaving group, so it can in principle run any of the four pathways — the reagent and conditions must decide.
- Weigh the reagent: hydroxide is both a strong nucleophile and a strong base, but it is small and unhindered, and the polar aprotic solvent leaves it unencumbered by a solvent shell. On a secondary carbon those conditions favor the concerted SN2 attack over E2 — which prefers a bulky base or heat — and rule out the carbocation routes, which need a weak nucleophile in an ionizing solvent.
- Predict the connectivity: hydroxide replaces bromide at C2, converting CH₃CHBrCH₂CH₃ into CH₃CH(OH)CH₂CH₃, which is butan-2-ol.
- Track the stereochemistry: SN2 proceeds by backside attack, inverting the configuration at the stereocenter. Bromide (priority 1) is replaced by OH (still priority 1), and the ethyl, methyl, and H rankings are unchanged, so the spatial inversion flips the descriptor.
- State the result and sanity-check: starting from (S)-2-bromobutane, inversion gives (R)-butan-2-ol; a single clean inversion rather than a racemic mixture is itself evidence that the reaction went by SN2 and not through a free carbocation.
Answer. The reaction is SN2, giving (R)-butan-2-ol by backside attack with inversion of configuration.
Check your understanding
- Why does SN2 produce inversion while SN1 produces a racemic mixture, and what does each outcome reveal about the intermediate?
- How could you change only the reagent to steer (S)-2-bromobutane toward elimination instead, and what alkene would dominate?
- Why does a tertiary substrate essentially refuse to undergo SN2 even with an excellent nucleophile?
- How would you explain to a friend the difference between a strong base and a strong nucleophile, with a reagent that is one but not the other?
Build the foundations first
Substitution & elimination reactions builds on these concepts. If any feel shaky, start there.