An organic compound x where the molar ratio o | vedclass

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Question

(C) Let the empirical formula be $(COH)_{n}$.The molar mass is $12n + 16n + n = 29n = 58 \Rightarrow n = 2$.Thus, the molecular formula is $C_{2}H_{2}

Vedclass · Accepted Answer

$HO - CH_{2} - COOH$

Vedclass · Answer

(C) Let the empirical formula be $(COH)_{n}$.The molar mass is $12n + 16n + n = 29n = 58 \Rightarrow n = 2$.Thus, the molecular formula is $C_{2}H_{2}O_{2}$, which corresponds to glyoxal $(CHO-CHO)$.Glyoxal contains two aldehyde groups and lacks $\alpha$-hydrogen atoms, so it undergoes the Cannizzaro reaction in the presence of $50\% \ KOH$.In this reaction, one aldehyde group is oxidized to a carboxylate group and the other is reduced to a primary alcohol group.$CHO-CHO + KOH \rightarrow HO-CH_{2}-COOK$.Upon acidification, the salt converts to glycolic acid: $HO-CH_{2}-COOH$.Therefore, the correct structure of $y$ is $HO-CH_{2}-COOH$.

An organic compound $x$ where the molar ratio of $C$, $O$, and $H$ are equal, on treatment with $50\% \ KOH$ under reflux followed by acidification produced $y$. The most likely structure of $y$ is: [Molar mass of $x$ is $58 \ g \ mol^{-1}$]

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