See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: Acid-catalyzed hydration of an internal alkyne (Markovnikov addition of water) using HgSO4/dil. H2SO4 proceeds via Markovnikov's rule to give the more stable enol intermediate, which tautomerizes to the ketone. For an unsymmetrical internal alkyne Ar1-C≡C-Ar2, water adds such that the OH attaches to the carbon bearing the more electron-rich (more stabilizing) group, giving the enol that tautomerizes to the ketone with C=O on that carbon. Step 1 – Identify the alkyne: The starting material is Ph-C≡C-(4-MeO-C6H4), an unsymmetrical diaryl alkyne. The two aryl groups are phenyl (Ph) and para-methoxyphenyl (4-MeO-C6H4). Step 2 – Regioselectivity of water addition: The para-methoxyphenyl group is more electron-donating than phenyl due to the -OCH3 group. In Markovnikov hydration, water (as OH-) adds to the carbon that can better stabilize positive charge (or equivalently, the more electron-rich carbon is protonated/activated). The -OCH3 group donates electrons to its attached alkyne carbon, making it the one that receives the nucleophilic OH. Thus OH ends up on the carbon bearing the 4-methoxyphenyl group, giving the enol: Ph-CH=C(OH)-(4-MeO-C6H4). Step 3 – Tautomerization: The enol Ph-CH=C(OH)-(4-MeO-C6H4) tautomerizes to the keto form: Ph-CH2-C(=O)-(4-MeO-C6H4), which is Ph-CH2-CO-C6H4-OCH3 (para). Step 4 – Match to options: This corresponds to option (c): Ph-CH2-C(=O)-C6H4-OCH3(para). Why other options fail: - Option (a): Ph-CO-CH2-C6H4-OCH3 would arise if OH added to the phenyl-bearing carbon (anti-Markovnikov with respect to electron density), giving the less stable enol; this is the minor product. - Option (b): A secondary alcohol would require incomplete reaction or reduction, not simple hydration/tautomerization. - Option (d): An alcohol product (Ph-CH2-CH(OH)-C6H4-OCH3) would require hydrogenation, not acid hydration. Therefore, the correct answer is C.