Question
Predict the major product when 2-methylbutan-2-ol is treated with concentrated at . Explain the mechanism in two lines. Also predict whether Saytzeff or Hofmann’s rule applies. NEET 2023 pattern.
Solution — Step by Step
2-methylbutan-2-ol is . Concentrated at high temperature performs acid-catalysed dehydration — an E1 elimination producing an alkene.
Protonation of gives a leaving group (water). Loss of water generates a tertiary carbocation: . Tertiary carbocations are stable, so no rearrangement occurs.
Saytzeff (Zaitsev) says: the more substituted alkene is the major product. We have two β-positions: a (with 3 H’s) and a (with 2 H’s, leading to a more substituted double bond).
Eliminating from gives 2-methyl-2-butene (trisubstituted). Eliminating from gives 2-methyl-1-butene (disubstituted).
Major product: 2-methyl-2-butene (trisubstituted alkene, Saytzeff product). Hofmann’s rule does not apply here — that’s for elimination with bulky bases (E2 with ), not acid-catalysed E1.
Why This Works
E1 dehydration always favours the most stable alkene because the rate-determining step is carbocation formation, but the product distribution is controlled by the relative stability of the alkenes — and more substituted alkenes are more stable (hyperconjugation, ground-state effect).
This is exactly Saytzeff’s rule: most substituted alkene wins. Hofmann’s rule (least substituted) only applies in special cases like quaternary ammonium thermolysis or E2 with bulky bases that can’t reach the more hindered β-hydrogen.
Alternative Method
Use the carbocation stability check. For tertiary alcohols + acid, you always get E1 with no rearrangement. Just identify the most stable alkene from the two β-elimination options. No need to write the full mechanism for a 1-mark MCQ.
Students often pick Hofmann’s product (less substituted) for acid-catalysed dehydration. Wrong context: Saytzeff applies to E1 and standard E2 with small bases. Hofmann applies to E2 with bulky bases or thermolysis of quaternary ammonium hydroxides.
Memorise: alcohol + conc. acid + heat = E1 = Saytzeff. Alkyl halide + bulky base (KOtBu) = E2 with Hofmann tendency. Know the conditions that flip the rule.