See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: Hydride donor ability in the context of these compounds refers to the Cannizzaro-type or Meerwein–Ponndorf–Verley hydride transfer. The compounds are geminal diolate anions (hydrated aldehyde dianions) of the form Ar-CH(O⁻)(O⁻). These can donate a hydride from the C–H bond. The reference compound is Ph-CH(O⁻)(O⁻) (the geminal diolate of benzaldehyde). A better hydride donor means the C–H bond is weaker/more electron-rich at the carbon, making hydride transfer more favorable. Step 1 – Identify the reference: The reference is PhCH(O⁻)(O⁻), the geminal diolate of benzaldehyde. Step 2 – Analyze each compound: (a) 4-NO2-C6H4-CH(O⁻)(O⁻): The nitro group is strongly electron-withdrawing (EWG). It withdraws electron density from the ring and from the CH carbon, making the C–H bond stronger (less hydridic). This compound is a WORSE hydride donor than the reference. (b) 4-CH3O-C6H4-CH(O⁻)(O⁻): The methoxy group is electron-donating (EDG) by resonance. It increases electron density at the CH carbon, weakening the C–H bond and making hydride transfer more favorable. This compound is a BETTER hydride donor than the reference. (c) 4-CH3-C6H4-CH(O⁻)(O⁻): The methyl group is a weak electron-donating group (EDG) by induction/hyperconjugation. It slightly increases electron density at the CH carbon relative to Ph alone, making it a BETTER hydride donor than the reference. (d) 4-CH3-C6H4-C(CH3)(O⁻)(O⁻): This compound has no C–H bond at the geminal diolate carbon (it is replaced by CH3). Without an H on that carbon, it cannot act as a hydride donor from that position in the same manner. This compound is NOT a hydride donor from that carbon. Step 3 – Count: Compounds (b) and (c) are better hydride donors than the reference. x = 2. Step 4 – Why (a) fails: EWG nitro group destabilizes the developing negative charge on carbon after hydride loss, making it harder to donate hydride. Why (d) fails: No transferable hydride on the diolate carbon. Therefore, the correct answer is B,C.