See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: The reaction of a ketone with Al(Hg) (amalgamated aluminum) in benzene under heat, followed by hydrolysis with H2O and heat, is a classic Pinacol coupling (reductive coupling). Al(Hg) is a single-electron reductant that generates ketyl radicals from the carbonyl compound. Two ketyl radicals from two molecules of cyclopentanone couple to form a 1,2-diol (pinacol product) after workup with water. Step 1 – Reagent identification: Al(Hg) in benzene/heat is a bimetallic reductant (similar to Mg or Zn for pinacol couplings). It donates one electron to the carbonyl of cyclopentanone, forming a radical anion (ketyl radical) on the carbonyl carbon. Step 2 – Radical coupling: Two cyclopentanone ketyl radicals combine at their respective carbonyl carbons (C1 of each cyclopentanone ring), forming a C–C bond between the two former carbonyl carbons and generating an aluminum alkoxide intermediate. Step 3 – Hydrolysis: Treatment with H2O (and heat) cleaves the Al–O bonds, releasing the diol. Each former C=O becomes C–OH. The product is 1,1'-bicyclopentane-1,1'-diol: two cyclopentane rings joined at C1–C1', each with an –OH at the junction. Step 4 – Note on heat in step 2: With some pinacol diols, acid-catalyzed or thermal pinacol rearrangement can occur, but in this case the question indicates the product is the diol itself (option b), which is the direct pinacol coupling product before any rearrangement. The heat here assists hydrolysis of the metal alkoxide. Why other options fail: - (a) The diketo-dimer would result from oxidative coupling at alpha carbons, not reductive coupling at carbonyl carbons. - (c) The bis-cyclopentadienyl-type dimer would require elimination/dehydration steps not indicated. - (d) A mixed aldol-type or enol coupling product; not consistent with Al(Hg) reductive pinacol mechanism. Therefore, the correct answer is B.