Question:
A carbonyl compound X (C\(_8\)H\(_8\)O) undergoes disproportionation with conc. KOH on heating. Product of X with Zn-Hg/HCl is Y and product of X with NaBH\(_4\) is Z. What are Y and Z respectively?
A carbonyl compound X (C\(_8\)H\(_8\)O) undergoes disproportionation with conc. KOH on heating. Product of X with Zn-Hg/HCl is Y and product of X with NaBH\(_4\) is Z. What are Y and Z respectively?
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NaBH\(_4\) reduces carbonyls to alcohols; Zn-Hg/HCl (Clemmensen reduction) converts ketones to hydrocarbons.
Updated On: Jun 6, 2025
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The Correct Option is B
Solution and Explanation
Step 1: The compound X is an aromatic methyl ketone, most likely acetophenone (C\(_6\)H\(_5\)COCH\(_3\)) with molecular formula C\(_8\)H\(_8\)O.
Step 2: When heated with concentrated KOH, X undergoes Cannizzaro-type reaction (if it were formaldehyde) or more likely aldol condensation-disproportionation to form Y, a hydrocarbon with two alkyl groups (isopropyl benzene).
Step 3: Reduction of X with NaBH\(_4\) gives a secondary alcohol, i.e., \chemfig{Ph-CH(OH)CH3}.
Step 4: So, Y is isopropyl benzene and Z is 1-phenylethanol.
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