Q. In the following reactions, (a) CH₃–CH(CH₃)–CH(OH)–CH₃ —(H⁺/heat)→ A (major product) + B (minor product) (b) A —(HBr, dark, in absence of peroxide)→ C (major product) + D (minor product) The major products (A) and (C) are respectively
A) CH₂=C(CH₃)–CH₂–CH₃ and CH₃–CBr(CH₃)–CH₂–CH₃
B) CH₂=C(CH₃)–CH₂–CH₃ and BrCH₂–CH(CH₃)–CH₂–CH₃
C) CH₃–C(CH₃)=CH–CH₃ and CH₃–CBr(CH₃)–CH₂–CH₃
D) CH₃–C(CH₃)=CH–CH₃ and CH₃–CH(CH₃)–CHBr–CH₃
Explanation
[Solution from source] Protonation and loss of water from 3-methylbutan-2-ol gives a secondary carbocation, which undergoes a 1,2-hydride shift to the tertiary cation; loss of H⁺ gives the more substituted alkene 2-methylbut-2-ene (Saytzeff). HBr adds by Markovnikov's rule without peroxide, giving 2-bromo-2-methylbutane. The 2-methylbut-1-ene or anti-Markovnikov products are the minor/trap options. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.