Identify Structure of Product Y from Reflux of Glyoxal with KOH

To determine the structure of product "$y$", we analyze the steps as follows:

1. Determination of Molecular Formula of Compound "$x$": It is given that the molar ratio of $\text{C}$, $\text{H}$, and $\text{O}$ in compound "$x$" is equal ($1:1:1$).

   • Empirical formula of $x = \text{CHO}$
   • Empirical formula mass $= 12 + 1 + 16 = 29 \text{ g mol}^{-1}$

   The given molar mass of compound "$x$" is $58 \text{ g mol}^{-1}$. $n = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{58}{29} = 2$

   Therefore, the molecular formula of compound "$x$" is $(\text{CHO})_2 = \text{C}_2\text{H}_2\text{O}_2$. The only compound with the molecular formula $\text{C}_2\text{H}_2\text{O}_2$ containing carbonyl functionality is glyoxal ($\text{OHC}-\text{CHO}$).

2. Reaction with $50\%$ $\text{KOH}$ followed by Acidification: Glyoxal ($\text{OHC}-\text{CHO}$) lacks $\alpha$-hydrogen atoms. When treated with a concentrated base ($50\%$ $\text{KOH}$) under reflux, it undergoes an intramolecular Cannizzaro reaction:

   • Hydroxide ion ($\text{OH}^-$) attacks one of the aldehyde carbonyl carbons.
   • Hydride ion ($\text{H}^-$) shifts internally to the adjacent aldehyde carbon.
   • One aldehyde group is oxidized to a carboxylate group ($\text{-COO}^-\text{K}^+$) while the other aldehyde group is reduced to a primary alcohol group ($\text{-CH}_2\text{OH}$), forming potassium glycolate ($\text{HO}-\text{CH}_2-\text{COOK}$).

   Upon subsequent acidification, potassium glycolate is converted into glycolic acid (2-hydroxyethanoic acid): $\text{OHC}-\text{CHO} \xrightarrow[n]{\text{(i) } 50\% \text{ KOH, reflux}} \text{HO}-\text{CH}_2-\text{COOK} \xrightarrow{\text{(ii) } \text{H}^+} \text{HO}-\text{CH}_2-\text{COOH}$

3. Structure of Compound "$y$": The structure of glycolic acid is: $\begin{array}{c} \text{O}=\text{C}-\text{OH} \\ | \\ \text{CH}_2-\text{OH} \end{array}$

Thus, the correct option representing the structure of compound "$y$" is C.