The product (C) of the above reaction is:Male | vedclass

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(C) $1$. Hydrolysis of maleic anhydride with $H_3O^+$ yields maleic acid $(A)$,which is $cis$-but$-2-$enedioic acid.$2$. Reaction of maleic acid with

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$HC \equiv CH$

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(C) $1$. Hydrolysis of maleic anhydride with $H_3O^+$ yields maleic acid $(A)$,which is $cis$-but$-2-$enedioic acid.$2$. Reaction of maleic acid with $NaOH$ gives the sodium salt of maleic acid $(B)$,which is sodium maleate.$3$. Kolbe electrolysis of sodium maleate involves the decarboxylation of the carboxylate ions at the anode,leading to the formation of a triple bond between the carbons that were originally attached to the carboxylate groups.$4$. The reaction is: $cis-NaOOC-CH=CH-COONa \xrightarrow{	ext{electrolysis}} HC \equiv CH + 2CO_2 + H_2 + 2NaOH$.$5$. Thus,the final product $(C)$ is ethyne $(HC \equiv CH)$.

The product $(C)$ of the above reaction is:
Maleic anhydride $\xrightarrow{H_3O^+}$ $(A)$ $\xrightarrow{NaOH}$ $(B)$ $\xrightarrow{\text{Kolbe electrolysis}}$ $(C)$

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The product $(C)$ of the above reaction is:Maleic anhydride $\xrightarrow{H_3O^+}$ $(A)$ $\xrightarrow{NaOH}$ $(B)$ $\xrightarrow{\text{Kolbe electrolysis}}$ $(C)$