Question

Arrange the following in decreasing order of their acidic strength: 
A. \( {CH}_3 {COOH} \) 
B. \( {CICH}_2 {COOH} \) 
C. \( {Cl}_2 {COOH} \) 
D. \( {Cl}_3 {COOH} \) 
E. \( {F}_3 {COOH} \) 
Choose the correct answer from the options given below:

• A. \(A>B>C>D>E\)
• B. \(E>D>C>B>A\)
• C. \(A>E>D>C>B\)
• D. \(B>C>D>A>E\)

Hint:

The presence of electron-withdrawing groups (like \( {Cl} \) or \( {F} \)) increases the acidity of carboxylic acids, while electron-donating groups (like \( {CH}_3 \)) decrease their acidity.

Answer 1

The Correct Option is B

Step 1: Understanding the Effect of Substituents on Acidity.
The acidic strength of carboxylic acids is influenced by the electron-withdrawing or electron-donating nature of substituents attached to the benzene ring (in case of substituted carboxylic acids). Electron-withdrawing groups (like halogens) increase the acidity of carboxylic acids by stabilizing the negative charge on the conjugate base (the carboxylate anion).
Electron-withdrawing groups (like \( {Cl}, {F} \)) make the acid stronger.
Electron-donating groups (like \( {CH}_3 \)) reduce the acidity.
Step 2: Analyze the compounds.
A. \( {CH}_3 {COOH} \): The methyl group is an electron-donating group, which decreases the electron density on the carboxyl group, reducing its acidity. Hence, this is the weakest acid.
B. \( {CICH}_2 {COOH} \): The \( {Cl} \) group is electron-withdrawing but is attached to the second carbon. It increases the acidity slightly compared to A.
C. \( {Cl}_2 {COOH} \): Two chlorine atoms attached to the carboxyl group significantly withdraw electron density, making the acid stronger than A and B.
D. \( {Cl}_3 {COOH} \): With three chlorine atoms, this compound is even more acidic than C due to a greater electron-withdrawing effect.
E. \( {F}_3 {COOH} \): Fluorine is a very strong electron-withdrawing group, and having three fluorine atoms attached to the carboxyl group makes this compound the most acidic among the options.
Step 3: Conclusion.
The order of acidic strength is:
\[ E > D > C > B > A \]
Thus, the correct answer is (2) E > D > C > B > A.