Question

Arrange the following in the increasing order of pKa values. 

• A. \( III < IV < II < I \)
• B. \( II < III < IV < I \)
• C. \( IV < II < I < III \)
• D. \( IV < III < II < I \)

Hint:

Electron-withdrawing groups (like NO\(_2\)) increase acidity by stabilizing the conjugate base, leading to a lower pKa value.

Answer 1

The Correct Option is C

Step 1: Understanding pKa Values 
The pKa value of a compound determines its acidity. Lower pKa values correspond to stronger acids. 

Step 2: Analyzing the Given Structures 
- Compound IV (p-Nitrobenzoic Acid) has the lowest pKa due to the strong electron-withdrawing NO\(_2\) group stabilizing the carboxylate ion. 
- Compound II (p-Nitrophenol) is more acidic than phenol due to the NO\(_2\) group stabilizing the phenoxide ion. 
- Compound I (Phenol) has a higher pKa than nitrophenol because it lacks an additional electron-withdrawing group.
- Compound III (Ethanol) has the highest pKa as alcohols are weaker acids compared to phenols and carboxylic acids. 

Step 3: Arranging in Increasing Order of pKa 
Thus, the correct order is: \[ IV<II<I<III \]