Question

Arrange the following compounds in increasing order of acidic strength: \[ \text{Phenol, Water, Propan-1-ol, Propan-2-ol} \]

• A. \(\text{Water} < \text{Propan-2-ol} < \text{Propan-1-ol} < \text{Phenol}\)
• B. \(\text{Propan-2-ol} < \text{Propan-1-ol} < \text{Water} < \text{Phenol}\)
• C. \(\text{Propan-1-ol} < \text{Propan-2-ol} < \text{Water} < \text{Phenol}\)
• D. \(\text{Phenol} < \text{Water} < \text{Propan-1-ol} < \text{Propan-2-ol}\)

Hint:

Resonance stabilization increases acidity, while electron donating \(+I\) effect decreases acidity.

Answer 1

The Correct Option is B

Concept: Acidic strength depends upon the ease with which a compound can donate a proton \((H^+)\). Greater stability of the conjugate base formed after loss of proton means stronger acidity.

• Alkyl groups show \(+I\) effect and decrease acidity.
• Branching increases electron donation and further decreases acidity.
• Phenoxide ion formed from phenol is resonance stabilized, making phenol highly acidic.

Step 1: Compare propan-1-ol and propan-2-ol. In propan-2-ol, the carbon attached to \(-OH\) group has two alkyl groups, producing stronger \(+I\) effect. This destabilizes the alkoxide ion formed after removal of proton. Hence: \[ \text{Propan-2-ol} < \text{Propan-1-ol} \]

Step 2: Compare alcohols with water. Water does not contain electron donating alkyl groups, so the hydroxide ion formed is comparatively more stable. Therefore: \[ \text{Propan-2-ol} < \text{Propan-1-ol} < \text{Water} \]

Step 3: Compare phenol with water. Phenol forms phenoxide ion after losing proton. \[ \text{C}_6\text{H}_5\text{OH} \rightarrow \text{C}_6\text{H}_5\text{O}^- + H^+ \] Phenoxide ion is resonance stabilized because the negative charge is delocalized over the benzene ring. Hence phenol is most acidic. \[ \boxed{ \text{Propan-2-ol} < \text{Propan-1-ol} < \text{Water} < \text{Phenol} } \]