COMEDK202510 May 2025Morning ShiftChemistryHaloalkanes and HaloarenesActual
Both reactions (i) and (ii) give the same compound X as the major product. Identify X (i). 3-Methylbut-1-ene + HCl X (ii). Neopentyl alcohol + HCl ( . anh. . ZnCl ₂ ) X
Options
- A( CH ₃ )₂- CCl _ - - CH ₂- CH ₃
- B( CH ₃ )₂- CH - CH ₂- CH ₂ Cl
- C( CH ₃ )₂- CH - CHCl - CH ₃
- DCH ₃- CH ₂- CH ( CH ₃ )- CH ₂ Cl
Correct answer
A. ( CH ₃ )₂- CCl _ - - CH ₂- CH ₃
Step-by-step solution
In reaction (i), the electrophilic addition of HCl to 3-methylbut-1-ene proceeds via the formation of a carbocation. The initial protonation of the double bond yields the secondary carbocation (CH₃)₂CH-CH⁺-CH₃ . This secondary carbocation undergoes a 1,2-hydride shift to form the more stable tertiary carbocation (CH₃)₂C⁺-CH₂-CH₃ . Subsequent attack by the chloride ion gives the major product X as 2-chloro-2-methylbutane, which is (CH₃)₂CCl-CH₂-CH₃ . In reaction (ii), the reaction of neopentyl alcohol (CH₃)₃C-CH₂OH