JEE MainChemistryAldehydes and Ketones
An aliphatic ketone X undergoes reaction with NaCN followed by heating with ethanol and acid ( H ₃ O ^ ) to yield an intermediate Y. Compound Y reacts with excess methylmagnesium bromide ( MeMgBr ) followed by acidic workup to give 3-ethyl-2-methylpentane-2,3-diol as the major product. Identify the IUPAC names of starting ketone X and intermediate Y.
Options
- AX = Propanone ; Y = Ethyl 2-hydroxy-2-methylpropanoate
- BX = Pentan-2-one ; Y = Ethyl 2-hydroxy-2-methylpentanoate
- CX = Pentan-3-one ; Y = Ethyl 2-ethyl-2-hydroxybutanoate
- DX = Pentan-3-one ; Y = 2-Ethyl-2-hydroxybutanoic acid
Correct answer
C. X = Pentan-3-one ; Y = Ethyl 2-ethyl-2-hydroxybutanoate
Step-by-step solution
The final product is 3-ethyl-2-methylpentane-2,3-diol, which has the structure CH ₃ CH ₂- C ( OH )( CH ₂ CH ₃)- C ( OH )( CH ₃)₂ . This diol is formed by the reaction of an ester with excess methylmagnesium bromide ( MeMgBr ). The - C ( OH )( CH ₃)₂ fragment is generated when the ester group reacts with two equivalents of MeMgBr . The remaining fragment, CH ₃ CH ₂- C ( OH )( CH ₂ CH ₃)- , must have originated from the initial ketone X. This indicates that ketone X is pentan-3-one ( CH ₃ CH ₂ COCH ₂ CH ₃ ). Reaction