Methyl propanoate on hydrolysis with dil NaOH | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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:

Similar Questions

The reagent which can perform the conversion $CH_{3}COOH \longrightarrow CH_{3}CH_{2}OH$ is

Arrange the following in order of increasing reactivity (least $\to$ most) towards nucleophilic acyl substitution:
$1. CH_3COCH_2CH_3$,$2. CH_3COCl$,$3. CH_3CONHCH_3$

When ethyl acetoacetate reacts with one mole of methyl magnesium iodide,the product of the reaction will be:

What is the decreasing order of acidic strength for the following acids?
$[I]$ Methanoic acid
$[II]$ Ethanoic acid
$[III]$ Propanoic acid
$[IV]$ Butanoic acid

The reaction between an acid and an alcohol yields:

Vedclass Test Series

Exam Paper Generator

Online Exam Module