Q. In the following reaction: (CH₃)₃C–CH=CH₂ —(H₂O/H⁺)→ A (major product) + B (minor product) The major product is
A) (CH₃)₃C–CH₂–CH₂OH
B) (CH₃)₂C(OH)–CH(CH₃)₂
C) HOCH₂–C(CH₃)(OH)–CH₂CH₃
D) (CH₃)₃C–CH(OH)–CH₃
Explanation
[Solution from source] Protonation of the alkene gives the secondary carbocation (CH₃)₃C–C⁺H–CH₃, which undergoes a 1,2-methyl shift to the more stable tertiary carbocation (CH₃)₂C⁺–CH(CH₃)₂. Water attacks this to give 2,3-dimethylbutan-2-ol, (CH₃)₂C(OH)–CH(CH₃)₂. The unrearranged alcohol (CH₃)₃C–CH(OH)–CH₃ is the minor product. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.