Consider the following molecules : (p) CH_3- | vedclass

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(D) The rate of hydrolysis of carboxylic acid derivatives depends on the leaving group ability of the group attached to the carbonyl carbon. Better le

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$p > q > r > s$

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(D) The rate of hydrolysis of carboxylic acid derivatives depends on the leaving group ability of the group attached to the carbonyl carbon. Better leaving groups make the carbonyl carbon more electrophilic and facilitate the nucleophilic attack by water. The leaving group ability order is $Cl^- > CH_3COO^- > CH_3CH_2O^- > NH_2^-$. Therefore,the rate of hydrolysis follows the order $:$ $(p) \ (acid \ chloride) > (q) \ (anhydride) > (r) \ (ester) > (s) \ (amide)$. Thus,the correct order is $p > q > r > s$.

Consider the following molecules $:$
$(p) \ CH_3-CH_2-COCl$
$(q) \ CH_3-CH_2-CO-O-COCH_3$
$(r) \ CH_3-CH_2-CO-OCH_2-CH_3$
$(s) \ CH_3-CH_2-CONH_2$
The correct order of rate of hydrolysis is $:$

Similar Questions

The reaction of $RCOOH$ with $R'OH$ takes place in the presence of $H_2SO_4$ to form an ester. The intermediate in the reaction is:

Arrange the following in order of their reactivity toward decarboxylation reaction:
$I. CH \equiv C - COOH$
$II. CH_2 = CH - COOH$
$III. CH_3 - CH_2 - COOH$
$IV. CH_3 - CH(CH_3) - COOH$

What is $A$ in the following reaction?

$RCOOH + NaHCO_3 \to RCOONa + H_2O + CO_2$. In this reaction,the $CO_2$ evolved comes from:

The descending order of acidity among the following compounds is:

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Consider the following molecules $:$ $(p) \ CH_3-CH_2-COCl$ $(q) \ CH_3-CH_2-CO-O-COCH_3$ $(r) \ CH_3-CH_2-CO-OCH_2-CH_3$ $(s) \ CH_3-CH_2-CONH_2$ The correct order of rate of hydrolysis is $:$