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An acyclic alcohol, 2,2,4-trimethylpentan-3-ol, is heated with concentrated H ₂ SO ₄ . What is the IUPAC name of the major alkene product formed?

Options

  1. A2,4,4-trimethylpent-2-ene
  2. B2,4,4-trimethylpent-1-ene
  3. C2,3,4-trimethylpent-2-ene
  4. D2,3,4-trimethylpent-1-ene

Correct answer

C. 2,3,4-trimethylpent-2-ene

Step-by-step solution

Protonation of 2,2,4-trimethylpentan-3-ol and subsequent loss of a water molecule generates a secondary carbocation at C-3: CH ₃- C ( CH ₃)₂- CH ^+- CH ( CH ₃)₂ . This carbocation can undergo either a 1,2-hydride shift from C-4 or a 1,2-methyl shift from C-2. A 1,2-hydride shift would yield a tertiary carbocation that eliminates to form 2,4,4-trimethylpent-2-ene, a trisubstituted alkene. However, a 1,2-methyl shift from the highly sterically hindered tert-butyl group yields a different tertiary carbocation: CH ₃- C

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