1. A strong unhindered nucleophile drives SN2 substitution, swapping the halide for the nucleophile in one step.
Bromoethane + azide → ethyl azide: backside attack, clean inversion.
2. Other strong nucleophiles react the same way, so cyanide and hydroxide give a nitrile and an alcohol on the same 1° carbon.
Bromoethane + cyanide → propanenitrile: a new C–C bond by SN2.
Bromoethane + hydroxide → ethanol: substitution wins on an unhindered 1° halide.
3. A strong bulky base abstracts a β-hydrogen instead, forcing E2 elimination toward the less-substituted alkene.
2-Bromopropane + KOtBu → propene: bulky base = Hofmann elimination.
4. A strong but small base also eliminates, but its E2 follows Zaitsev toward the more-substituted, more-stable alkene.
tert-Butyl bromide + ethoxide → isobutylene: E2 on a 3° halide.
5. A weak nucleophile in a hot protic solvent lets a 3° halide ionize, giving a mix of SN1 substitution and E1 elimination.
tert-Butyl bromide + hot water → tert-butanol (SN1) alongside isobutylene (E1).
Strong unhindered Nu, aprotic → SN2 · bulky base → E2 (Hofmann) · small base → E2 (Zaitsev) · weak Nu, protic, heat, 3° → SN1/E1.
Quiz yourself
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Propanenitrile (CCC#N) by SN2 — cyanide is a strong, small nucleophile and the carbon is unhindered 1°.
KOtBu is bulky, so it grabs the most accessible β-H (Hofmann, less-substituted alkene); a small base can reach the internal β-H and follows Zaitsev.
tert-Butanol (SN1) plus some isobutylene (E1) — the 3° cation forms first, then water traps it or a β-H is lost.
Reagent bulk/basicity: a strong bulky base (KOtBu) favors E2, while a strong unhindered nucleophile (NaN3, NaCN, NaOH) favors SN2.
Draw this on the whiteboard
Open the OChem Board whiteboard — benzene rings, wedge/dash bonds, and a clickable periodic table built in. No account needed.