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(C) The compound $A$ is an ester with the molecular formula $C_{6}H_{12}O_{2}$.Reduction of an ester with $LiAlH_{4}$ yields two alcohols.For $A$ to y

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$CH_{3}CH_{2}CH_{2}COOCH_{2}CH_{3}$

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(C) The compound $A$ is an ester with the molecular formula $C_{6}H_{12}O_{2}$.Reduction of an ester with $LiAlH_{4}$ yields two alcohols.For $A$ to yield a $C_{4}$ carboxylic acid upon oxidation of the resulting alcohol $B$,$B$ must be a primary alcohol with four carbon atoms,i.e.,butan$-1-$ol $(CH_{3}CH_{2}CH_{2}CH_{2}OH)$.This implies that the ester $A$ must be propyl propanoate $(CH_{3}CH_{2}COOCH_{2}CH_{2}CH_{3})$ or ethyl butanoate $(CH_{3}CH_{2}CH_{2}COOCH_{2}CH_{3})$.Looking at the options,$CH_{3}CH_{2}CH_{2}COOCH_{2}CH_{3}$ (ethyl butanoate) upon reduction gives butan$-1-$ol and ethanol. Oxidation of butan$-1-$ol yields butanoic acid ($C_{4}$ carboxylic acid).Therefore,$A$ is $CH_{3}CH_{2}CH_{2}COOCH_{2}CH_{3}$.

Column-$I$ (Reaction)	Column-$II$ (Name)
$A$. $RCH_2COOH \rightarrow[H_2O]{X_2/P} RCH(X)COOH$	$i$. Decarboxylation
$B$. $RCOONa \rightarrow[NaOH, CaO, \Delta]{\text{soda lime}} RH$	$ii$. Grignard reaction
$C$. $RCOOH + R'OH \leftrightarrow{H^{+}} RCOOR' + H_2O$	$iii$. Hell-Volhard-Zelinsky reaction
$D$. $RMgX + R'CHO \rightarrow{H_3O^{+}} RCH(OH)R'$	$iv$. Esterification

$a$. $\text{Toluene} \xrightarrow[(ii) H^+/H_2O]{(i) CrO_2Cl_2/CS_2}$	$p$. $\text{Kolbe's electrolysis}$
$b$. $CH_3COOK \xrightarrow{\text{electrolysis}}$	$q$. $\text{Etard reaction}$
$c$. $CH_3Br \xrightarrow{Na, \text{dry ether}}$	$r$. $\text{Wurtz reaction}$
$d$. $\text{Benzene} \xrightarrow{CH_3Cl, AlCl_3}$	$s$. $\text{Friedel-Crafts alkylation}$

In the following sequence of reactions,a compound $A$ (molecular formula $C_{6}H_{12}O_{2}$) with a straight chain structure gives a $C_{4}$ carboxylic acid. $A$ is :
$A$ $\xrightarrow{LiAlH_{4} / H_{3}O^{+}} B$ $\xrightarrow{\text{Oxidation}} C_{4} \text{ carboxylic acid}$

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In the following sequence of reactions,a compound $A$ (molecular formula $C_{6}H_{12}O_{2}$) with a straight chain structure gives a $C_{4}$ carboxylic acid. $A$ is :$A$ $\xrightarrow{LiAlH_{4} / H_{3}O^{+}} B$ $\xrightarrow{\text{Oxidation}} C_{4} \text{ carboxylic acid}$