Question

An organic compound (A) with the molecular formula ${C_8}{H_{16}}{O_2}$ was hydrolyzed with dilute sulfuric 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.

Answer 1

The compound that gives a carboxylic acid and an alcohol when hydrolyzed is an ester. Now we know that compound (A) is an ester. Dehydration of alcohol gave $but - 1 - ene$ so the alcohol contains four carbons. Similarly find the number of carbons in other compounds and accordingly the reactions involved.

Complete step-by-step answer: It is given to us that a compound (A) on hydrolysis gave a carboxylic acid and an alcohol. This means that the compound (A) is an ester. It is also given that the molecular formula of compound (A) i.e. ester is ${C_8}{H_{16}}{O_2}$ which means that the ester contains eight carbon atoms. The compound (C) which is an alcohol on dehydration gave $but - 1 - ene$ and this means that the alcohol has four carbon atoms and hence the carboxylic acid has remaining four carbon atoms.In the molecule $but - 1 - ene$ the double bond or functional group lies on the terminal carbon so before hydrolysis, the hydroxyl group must be one the first carbon. Hence we can give the structure for compound (C) as $C{H_3}C{H_2}C{H_2}C{H_2}OH$ It is also given that oxidation of (C) gave compound (B). So let us oxidize the compound (C).
$C{H_3}C{H_2}C{H_2}CHOH\xrightarrow[{Oxidation}]{{Cr{O_3}/C{H_3}COOH}}C{H_3}C{H_2}C{H_2}COOH$
So the compound (B) is Butanoic acid. From the compounds (B) and (C), we can write the structure for compound (A) as $C{H_3}C{H_2}C{H_2}COOC{H_2}C{H_2}C{H_2}C{H_3}$
Now we can write all the reactions as follows.
a) $C{H_3}C{H_2}C{H_2}COOC{H_2}C{H_2}C{H_2}C{H_3}\xrightarrow{{hydrolysis}}C{H_3}C{H_2}C{H_2}C{H_2}OH + C{H_3}C{H_2}C{H_2}COOH$
b) $C{H_3}C{H_2}C{H_2}C{H_2}OH\xrightarrow{{Dehydration}}C{H_3}C{H_2}CH = C{H_2}$
c) $C{H_3}C{H_2}C{H_2}CHOH\xrightarrow[{Oxidation}]{{Cr{O_3}/C{H_3}COOH}}C{H_3}C{H_2}C{H_2}COOH$

Note: In summary, Butyl butanoate on Hydrolysis in a weak acid produced Butanoic acid and Butanol. Butanol on dehydration lost one water molecule to give $but - 1 - ene$. Butanol on oxidation with chromic acid gave Butanoic acid.