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(C) The given reaction is an intramolecular Cannizzaro reaction. Aldehydes that do not have $\alpha$-hydrogen atoms undergo self-oxidation and reducti

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(C) The given reaction is an intramolecular Cannizzaro reaction. Aldehydes that do not have $\alpha$-hydrogen atoms undergo self-oxidation and reduction (disproportionation) in the presence of concentrated alkali. In the given molecule,$2,2',6,6'$-tetraformylbiphenyl,the aldehyde groups are present on the benzene rings. Upon treatment with excess $NaOH$,two of the $-CHO$ groups are reduced to $-CH_2OH$ (alcohol) groups,and the other two $-CHO$ groups are oxidized to $-COO^-$ (carboxylate) groups. Subsequent acidification with $H^+/H_2O$ converts the carboxylate groups into $-COOH$ (carboxylic acid) groups. Thus,the final product contains two $-CH_2OH$ groups and two $-COOH$ groups.

Predict the product of the following reaction:
$2,2',6,6'$-tetraformylbiphenyl $\xrightarrow[(ii)\,H^{+}/H_2O, 0^{\circ}C]{(i)\,NaOH (excess)}$ ?

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Predict the product of the following reaction:$2,2',6,6'$-tetraformylbiphenyl $\xrightarrow[(ii)\,H^{+}/H_2O, 0^{\circ}C]{(i)\,NaOH (excess)}$ ?