JEE MainChemistryAldehydes and Ketones
An unknown carbonyl compound 'X' undergoes reaction with HCN to form a cyanohydrin. This intermediate, upon heating with 95% H ₂ SO ₄ , yields 2-phenylbut-2-enoic acid as the major product. The compound 'X' is:
Options
- A1-phenylpropan-1-one
- B1-phenylpropan-2-one
- C1-phenylethan-1-one
- D2-phenylpropanal
Correct answer
A. 1-phenylpropan-1-one
Step-by-step solution
The final product is 2-phenylbut-2-enoic acid, which has the structure CH ₃- CH = C ( C ₆ H ₅)- COOH . Working backwards, the double bond is formed by the dehydration of an -hydroxy acid. Adding water across the double bond places the - OH group at the -carbon, giving 2-hydroxy-2-phenylbutanoic acid, CH ₃- CH ₂- C ( OH )( C ₆ H ₅)- COOH . The - COOH group is formed by the acidic hydrolysis of a nitrile ( - CN ) group. Replacing - COOH with - CN gives the cyanohydrin intermediate, CH ₃- CH ₂- C ( OH )( C ₆ H ₅)- CN