Compound P($\ce{C7H6O}$) on refluxing with aqueous ethanolic $\ce{KCN}$ solution for 1 hour, forms Q($\ce{C14H12O2}$). Q on oxidation gives R($\ce{C14H10O2}$). R on refluxing with aqueous $\ce{NaOH}$ solution followed with acidification forms an $\alpha-$hydroxyacid S ($\ce{C14H12O3}$). What are the structures of the compounds?

My effort:

P has DBE (double bond equivalent) equal to 8, so 2 benzene rings (not possible with 7 carbons) or some cyclic or many double bonds might be necessary. Now some enolate reaction might be happening with $\ce{KCN}$, like Aldol or Perkin or Knoevnagel or Stobbe or Claisen or Benzion etc. Oxidation doesn't affect the formula so it might be becoming an isomer (that too a functional one). After that a Cannizaro might be the thing.


You weren't far from resolution. The substrate is benzaldehyde, first reaction benzoin condensation, so Q is benzoin. It is later oxidated to benzil - R ($\ce{C14H10O2}$) (not ($\ce{H12}$) - in oxidation formula must change). Finally benzil undergoes benzilic acid rearrangement, so S is benzilic acid.

As for how one could get to the solution... Compounds having 6-8 carbons and highly unsaturated are usually benzene derivatives. In this case benzaldehyde is only common thing with this formula (it was unlikely to be for example tropone). First reaction must be dimerisation, next oxidation, and hydration.


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