Q. Acetone is treated with excess of ethanol in the presence of hydrochloric acid. The product obtained is:
A) (CH₃)₂C(OC₂H₅)₂
B) CH₃CH₂CH₂–CO–CH₃
C) CH₃CH₂CH₂–CO–CH₂CH₂CH₃
D) (CH₃)₂C(OH)(OC₂H₅)
Explanation
[Solution from source] In dry HCl, acetone adds one ethanol to give a hemiketal, which with excess ethanol loses water and adds a second ethanol to give the ketal 2,2-diethoxypropane, (CH₃)₂C(OC₂H₅)₂. The hemiketal (CH₃)₂C(OH)(OC₂H₅) is only the intermediate, and no C–C bond forming product like a higher ketone is formed. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.