G in the above sequence of reactions is | vedclass

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(C) The reaction sequence involves the alkaline hydrolysis (saponification) of an ester $G$ $(C_6H_{12}O_2)$ to form an alcohol $(H)$ and a carboxylat

Vedclass · Accepted Answer

$CH_3 CH_2 COOCH_2 CH_2 CH_3$

Vedclass · Answer

(C) The reaction sequence involves the alkaline hydrolysis (saponification) of an ester $G$ $(C_6H_{12}O_2)$ to form an alcohol $(H)$ and a carboxylate salt,which upon acidification gives a carboxylic acid $(I)$.$H$ is then oxidized by $CrO_3/H^+$ to form the same carboxylic acid $(I)$.This indicates that the alcohol $H$ is a primary alcohol,which oxidizes to a carboxylic acid with the same number of carbon atoms.Looking at the options,we need an ester that hydrolyzes into a primary alcohol and a carboxylic acid such that the alcohol can be oxidized to the acid.Option $C$ is $CH_3CH_2COOCH_2CH_2CH_3$ (propyl propionate).Hydrolysis: $CH_3CH_2COOCH_2CH_2CH_3 + OH^- \rightarrow CH_3CH_2COO^- + CH_3CH_2CH_2OH$.Acidification: $CH_3CH_2COO^- + H^+ \rightarrow CH_3CH_2COOH$ (propanoic acid).Oxidation of $H$ $(CH_3CH_2CH_2OH)$: $CH_3CH_2CH_2OH \xrightarrow{CrO_3/H^+} CH_3CH_2COOH$.Since the acid formed from hydrolysis and the acid formed from oxidation of the alcohol must be the same for the cycle to work as shown,the ester must be $CH_3CH_2COOCH_2CH_2CH_3$.

$G$ in the above sequence of reactions is

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