Methyl propanoate on hydrolysis with dil NaOH | vedclass

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Question

(C) $Methyl$ propanoate is $CH_3CH_2COOCH_3$.On alkaline hydrolysis with dil $NaOH$,it undergoes saponification to form sodium propanoate and methanol

Vedclass · Accepted Answer

$CH_3CH_2COOH$

Vedclass · Answer

(C) $Methyl$ propanoate is $CH_3CH_2COOCH_3$.On alkaline hydrolysis with dil $NaOH$,it undergoes saponification to form sodium propanoate and methanol:$CH_3CH_2COOCH_3 + NaOH \rightarrow CH_3CH_2COONa + CH_3OH$Upon acidification with conc. $HCl$,the sodium salt of the carboxylic acid (sodium propanoate) is converted into propanoic acid:$CH_3CH_2COONa + HCl \rightarrow CH_3CH_2COOH + NaCl$Thus,the final product is propanoic acid,which corresponds to option $C$.

$Methyl$ propanoate on hydrolysis with dil $NaOH$ forms a salt that on further acidification with conc. $HCl$ forms:

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