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E2 eliminations typically form the most substituted (most stable) alkene product—Zaitsev's rule. This is because substitution stabilizes the C=C double bond. However, bulky bases (like t-BuOK) or quaternary ammonium hydroxides undergoing Hofmann elimination form the less substituted, terminal alkene product instead. This anti-Zaitsev selectivity occurs when steric hindrance dominates thermodynamic stability.
From your study of E2 elimination, you know the mechanism: a strong base abstracts a beta-hydrogen while the leaving group departs, forming a new C=C double bond in a single concerted step. But when a substrate has beta-hydrogens on more than one carbon, which hydrogen gets abstracted? This is a question of regioselectivity — the elimination can form different constitutional isomers of the alkene depending on which beta-hydrogen is removed.
Zaitsev's rule states that the major product is the more substituted alkene — the one with more alkyl groups attached to the double bond carbons. Why? Alkyl groups stabilize double bonds through hyperconjugation (overlap of adjacent C–H sigma bonds with the pi system). A trisubstituted alkene is more stable than a disubstituted one, which is more stable than a monosubstituted one. Since E2 transition states have partial double-bond character, the transition state leading to the more substituted product is lower in energy, making it the kinetically and thermodynamically favored pathway. For example, when 2-bromobutane undergoes E2 elimination with sodium ethoxide, the major product is 2-butene (the more substituted, internal alkene), not 1-butene (the less substituted, terminal alkene).
The exception arises when steric effects override thermodynamic preferences. Hofmann elimination produces the less substituted alkene and occurs in two classic situations. First, when you use a bulky base like potassium tert-butoxide (t-BuOK), the large tert-butyl group physically cannot reach the more hindered internal beta-hydrogens. It instead abstracts the more accessible hydrogen on the less substituted carbon, yielding the terminal alkene. Second, the original Hofmann elimination involves heating a quaternary ammonium hydroxide — here the bulky NR₃ leaving group makes the transition state leading to the more substituted alkene sterically crowded, again favoring the less substituted product.
The practical takeaway is a simple decision rule: use a normal-sized base (like NaOEt or NaOH) for Zaitsev products (more substituted alkenes), and switch to a bulky base (t-BuOK) when you want the anti-Zaitsev or Hofmann product (less substituted, terminal alkene). This is one of the clearest examples in organic chemistry of how base choice controls product distribution — the substrate and mechanism are the same, but the size of the base determines which hydrogen is accessible and therefore which alkene forms.