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(A) The reaction of a nitrile $(R-CN)$ with an alcohol $(R'-OH)$ in the presence of water and a strong acid (like conc. $H_2SO_4$) is a Pinner reactio

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Ethyl propionate + Ammonia

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(A) The reaction of a nitrile $(R-CN)$ with an alcohol $(R'-OH)$ in the presence of water and a strong acid (like conc. $H_2SO_4$) is a Pinner reaction or acid-catalyzed hydrolysis followed by esterification.Here,$CH_3CH_2CN$ (propanenitrile) undergoes hydrolysis to form propanoic acid $(CH_3CH_2COOH)$,which then reacts with ethanol $(C_2H_5OH)$ to form ethyl propionate $(CH_3CH_2COOC_2H_5)$ and ammonia $(NH_3)$ as a byproduct.The overall reaction is: $CH_3CH_2CN + C_2H_5OH + H_2O \rightarrow CH_3CH_2COOC_2H_5 + NH_3$.

Column-$I$ (Formula)	Column-$II$ (General Name) and Column-$III$ ($IUPAC$ Name)
$(A)$ $HCOOH$	$(i)$ Succinic acid,$(x)$ Methanoic acid
$(B)$ $CH_3CH_2CH_2COOH$	$(ii)$ Oxalic acid,$(y)$ Ethanedioic acid
$(C)$ $(COOH)_2$	$(iii)$ Formic acid,$(m)$ Butanedioic acid
$(D)$ $(CH_2)_2(COOH)_2$	$(iv)$ Butyric acid,$(n)$ Butanoic acid

What will be the product obtained from the following reaction?
$CH_3CH_2C \equiv N + C_2H_5OH + H_2O \xrightarrow{\text{conc. } H_2SO_4}$

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What will be the product obtained from the following reaction?$CH_3CH_2C \equiv N + C_2H_5OH + H_2O \xrightarrow{\text{conc. } H_2SO_4}$