Correct order of pK_{a} of given compounds is | vedclass

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Question

(B) The $pK_{a}$ value is inversely proportional to the acidic strength of the carboxylic acid.Acidic strength increases with the presence of electron

Vedclass · Accepted Answer

$Q < R < P < S$

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(B) The $pK_{a}$ value is inversely proportional to the acidic strength of the carboxylic acid.Acidic strength increases with the presence of electron-withdrawing groups $(EWG)$ and decreases with electron-donating groups $(EDG)$.Comparing the compounds:$(S) \ C_{2}H_{5}COOH$ has an ethyl group $(EDG)$,which decreases acidity,making it the weakest acid.$(P) \ CH_{3}COOH$ has a methyl group $(EDG)$,which is less donating than the ethyl group,making it stronger than $(S)$.$(Q) \ FCH_{2}COOH$ and $(R) \ ClCH_{2}COOH$ have halogen atoms $(EWG)$. Since $F$ is more electronegative than $Cl$,$(Q)$ is a stronger acid than $(R)$.Thus,the order of acidic strength is: $(S) Since $pK_{a} = -\log(K_{a})$,the order of $pK_{a}$ is the reverse of the acidic strength order: $(Q) < (R) < (P) < (S)$.

Correct order of $pK_{a}$ of given compounds is $:-$
$(P) \ CH_{3}COOH$
$(Q) \ FCH_{2}COOH$
$(R) \ ClCH_{2}COOH$
$(S) \ C_{2}H_{5}COOH$

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Correct order of $pK_{a}$ of given compounds is $:-$$(P) \ CH_{3}COOH$$(Q) \ FCH_{2}COOH$$(R) \ ClCH_{2}COOH$$(S) \ C_{2}H_{5}COOH$