Question

The major product obtained from the following reaction is :

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

Hg2+/H+ hydration of an unsymmetrical diaryl alkyne: water adds to the carbon better able to carry positive charge (the one on the p-methoxyphenyl side), giving a ketone there after tautomerisation.

Answer 1

The Correct Option is C

Reactant: 4-O₂N-C₆H₄-C\(\equiv\)C-C₆H₄-4-OCH₃, treated with Hg²⁺/H⁺, H₂O (hydration of the alkyne).
Step 1: Hg²⁺ forms a bridged (mercurinium-like) intermediate; water attacks the alkyne carbon which carries more positive charge.
Step 2: The positive charge is stabilised by the electron-donating p-OCH₃ ring (+M effect), while the p-NO₂ ring destabilises it (-M, -I). So the nucleophile (water) attacks the carbon attached to the methoxyphenyl ring.
Step 3: The enol formed tautomerises to the ketone: the C=O is on the carbon attached to the anisyl ring and CH₂ is attached to the nitrophenyl ring.
Product: 4-NO₂-C₆H₄-CH₂-CO-C₆H₄-4-OCH₃. Options (2) and (4) contain a phenolic OH (the OCH₃ group is not cleaved under these conditions) and are wrong; option (1) has the carbonyl on the nitrophenyl side, the less favoured regiochemistry.
Answer: (3) 4-NO₂-C₆H₄-CH₂-CO-C₆H₄-4-OCH₃