JEE MainChemistryAldehydes and Ketones
An unknown carbonyl compound 'X' gives a yellow precipitate when treated with I ₂ and NaOH . When 'X' is treated with one equivalent of phenyl magnesium bromide ( PhMgBr ) followed by acidic hydrolysis, it yields 2-phenylbutan-2-ol. The IUPAC name of compound 'X' is:
Options
- AButan-2-one
- BPentan-2-one
- CButanal
- DAcetophenone
Correct answer
A. Butan-2-one
Step-by-step solution
The product is 2-phenylbutan-2-ol, whose structure is CH ₃- CH ₂- C ( Ph )( OH )- CH ₃ . In the Grignard reaction, the phenyl group ( Ph ^- ) comes from the Grignard reagent, phenyl magnesium bromide ( PhMgBr ). Disconnecting the C - Ph bond from the product reveals that the original carbonyl compound must be butan-2-one ( CH ₃- CH ₂- CO - CH ₃ ). Butan-2-one contains a methyl ketone group ( CH ₃- CO - ), which reacts with I ₂ and NaOH to give a yellow precipitate of iodoform ( CHI ₃ ). This confirms the identity o