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NEET Chemistry QUESTION #1111
Question 1
The major product of the reaction: PhCOCH\(_2\)CN \(\xrightarrow{\text{(i) excess CH}_3\text{MgBr, (ii) H}_3\text{O}^+}\)
  • Ph–C(CH\(_3\))(OH)–CH(OH)–CH\(_3\)CH\(_3\)✔️
  • Ph–CO–CH\(_2\)–CO–CH\(_3\)
  • Ph–C(CH\(_3\))(OH)–CH\(_2\)–CN
  • Ph–C(CH\(_3\))(OH)–CH\(_2\)–C(CH\(_3\))\(_2\)–OH... no. Product from ketone and nitrile with excess Grignard
Correct Answer Explanation
The compound Ph–CO–CH\(_2\)–CN has both a ketone and nitrile group. Excess CH\(_3\)MgBr reacts with: (i) ketone → tertiary alcohol Ph–C(CH\(_3\))(OH)–CH\(_2\)–CN. (ii) nitrile → imine → after H\(_3\)O\(^+\) → ketone... with excess Grignard, the ketone from nitrile hydrolysis reacts again → Ph–C(CH\(_3\))(OH)–CH\(_2\)–C(CH\(_3\))\(_2\)(OH). But since the CN converts via Grignard to imine then hydrolysis gives ketone which reacts with another Grignard. Final: Ph–C(OH)(CH\(_3\))–CH\(_2\)–C(OH)(CH\(_3\))\(_2\) = option (1) structure with two OH groups.