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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-$I$ (Reactant)	Column-$II$ (Product)
$A$. $RCH_2OH \xrightarrow[(ii) H_3O^{+}]{(i) NaOH, KMnO_4} ?$	$i$. $C_6H_5COOH + C_2H_5OH$
$B$. $CH_3(CH_2)_8CH_2OH \xrightarrow{Jones reagent} ?$	$ii$. $RCOOH$
$C$. $RCONH_2 \xrightarrow{H_3O^{+}, \Delta} ?$	$iii$. $CH_3(CH_2)_8COOH$
$D$. $C_6H_5COOEt \xrightarrow{H_3O^{+}} ?$	$iv$. $RCOOH + NH_3$

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?$CH_3CH_2C \equiv N + C_2H_5OH + H_2O \xrightarrow{\text{conc. } H_2SO_4}$