An organic compound $(A)$ (molecular formula $C_{8}H_{16}O_{2}$) was hydrolysed with dilute sulphuric acid to give a carboxylic acid $(B)$ and an alcohol $(C)$. Oxidation of $(C)$ with chromic acid produced $(B)$. $(C)$ on dehydration gives but$-1-$ene. Write equations for the reactions involved.

(N/A) An organic compound $(A)$ with molecular formula $C_{8}H_{16}O_{2}$ gives a carboxylic acid $(B)$ and an alcohol $(C)$ on hydrolysis with dilute sulphuric acid. Thus,compound $(A)$ must be an ester.
Further,alcohol $(C)$ gives acid $(B)$ on oxidation with chromic acid. Thus,$(B)$ and $(C)$ must contain an equal number of carbon atoms.
Since compound $(A)$ contains a total of $8$ carbon atoms,each of $(B)$ and $(C)$ contains $4$ carbon atoms.
Again,on dehydration,alcohol $(C)$ gives but$-1-$ene. Therefore,$(C)$ is a straight-chain alcohol,which is butan$-1-$ol.
On oxidation,butan$-1-$ol gives butanoic acid. Hence,acid $(B)$ is butanoic acid.
Thus,the ester with molecular formula $C_{8}H_{16}O_{2}$ is butyl butanoate $(CH_{3}CH_{2}CH_{2}COOCH_{2}CH_{2}CH_{2}CH_{3})$.
The reactions involved are:
$1. \text{Hydrolysis of ester: } CH_{3}CH_{2}CH_{2}COOCH_{2}CH_{2}CH_{2}CH_{3} + H_{2}O$ $\xrightarrow{dil. H_{2}SO_{4}} CH_{3}CH_{2}CH_{2}COOH (B) + CH_{3}CH_{2}CH_{2}CH_{2}OH (C)$
$2. \text{Oxidation of alcohol: } CH_{3}CH_{2}CH_{2}CH_{2}OH \xrightarrow{[O]} CH_{3}CH_{2}CH_{2}COOH (B)$
$3. \text{Dehydration of alcohol: } CH_{3}CH_{2}CH_{2}CH_{2}OH \xrightarrow{H^{+}/\Delta} CH_{3}CH_{2}CH=CH_{2} + H_{2}O$