Question

The correct decreasing order of acidic strength is

• A. \( \text{C}_6\text{H}_5\text{OH} > p\text{CH}_3\text{C}_6\text{H}_4\text{OH} > m\text{CH}_3\text{C}_6\text{H}_4\text{OH} > \text{C}_2\text{H}_5\text{OH} \)
• B. \( \text{C}_2\text{H}_5\text{OH} > m\text{CH}_3\text{C}_6\text{H}_4\text{OH} > p\text{CH}_3\text{C}_6\text{H}_4\text{OH} > \text{C}_6\text{H}_5\text{OH} \)
• C. \( m\text{CH}_3\text{C}_6\text{H}_4\text{OH} > \text{C}_6\text{H}_5\text{OH} > p\text{CH}_3\text{C}_6\text{H}_4\text{OH} > \text{C}_2\text{H}_5\text{OH} \)
• D. \( \text{C}_6\text{H}_5\text{OH} > m\text{CH}_3\text{C}_6\text{H}_4\text{OH} > p\text{CH}_3\text{C}_6\text{H}_4\text{OH} > \text{C}_2\text{H}_5\text{OH} \)
• E. \( \text{C}_6\text{H}_5\text{OH} > p\text{CH}_3\text{C}_6\text{H}_4\text{OH} > \text{C}_2\text{H}_5\text{OH} > m\text{CH}_3\text{C}_6\text{H}_4\text{OH} \)

Hint:

In aromatic compounds, electron-donating groups (like \( \text{CH}_3 \)) decrease acidity, while electron-withdrawing groups increase acidity.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The acidic strength of an alcohol or phenol is determined by the stability of its conjugate base (alkoxide or phenoxide ion) after donating a proton (H\(^+\)). Factors that stabilize the conjugate base increase the acidity.
- Phenols vs. Alcohols: Phenols are much more acidic than aliphatic alcohols because the negative charge on the phenoxide ion is delocalized over the benzene ring through resonance, making it very stable. The negative charge on an alkoxide ion (like C\(_2\)H\(_5\)O\(^-\)) is localized on the oxygen atom and is actually destabilized by the electron-donating (+I) effect of the alkyl group.
- Substituent Effects on Phenols: - Electron-donating groups (EDGs) like -CH\(_3\) decrease acidity by destabilizing the phenoxide ion. - Electron-withdrawing groups (EWGs) increase acidity by stabilizing the phenoxide ion. The effect of a substituent depends on its position (ortho, meta, para). The methyl group (-CH\(_3\)) is an EDG. It has a +I (inductive) effect and a +H (hyperconjugation) effect, which is similar to a +R effect.
Step 2: Detailed Explanation:
Let's compare the given compounds:
1. C\(_2\)H\(_5\)OH (Ethanol): An aliphatic alcohol. Its conjugate base, ethoxide (C\(_2\)H\(_5\)O\(^-\)), is destabilized by the +I effect of the ethyl group. It will be the least acidic.
2. C\(_6\)H\(_5\)OH (Phenol): The benchmark phenol. Its phenoxide ion is resonance-stabilized.
3. p-CH\(_3\)-C\(_6\)H\(_4\)OH (p-cresol): The -CH\(_3\) group is at the para position. It exerts a +I effect and a +H (hyperconjugation) effect. Both effects are electron-donating, which destabilize the phenoxide ion by increasing the negative charge density on the ring and on the oxygen. This makes p-cresol less acidic than phenol.
4. m-CH\(_3\)-C\(_6\)H\(_4\)OH (m-cresol): The -CH\(_3\) group is at the meta position. At the meta position, the resonance/hyperconjugation effect does not operate. Only the +I effect is significant. The +I effect is electron-donating and destabilizes the phenoxide ion, making m-cresol less acidic than phenol.
Comparing the Cresols:
- In p-cresol, both the +I and +H effects are working to destabilize the phenoxide ion.
- In m-cresol, only the +I effect is working.
Since the combined donating effect (+I and +H) in p-cresol is stronger than the donating effect (+I only) in m-cresol, p-cresol is less acidic than m-cresol.
Final Order:
- Phenol is the most acidic among the phenols because it has no destabilizing EDG.
- m-cresol is next, as it is only destabilized by the weaker +I effect.
- p-cresol is less acidic than m-cresol, as it is destabilized by both +I and +H effects.
- Ethanol is the least acidic of all, as it's an alcohol.
The decreasing order of acidic strength is:
C\(_6\)H\(_5\)OH>m-CH\(_3\)-C\(_6\)H\(_4\)OH>p-CH\(_3\)-C\(_6\)H\(_4\)OH>C\(_2\)H\(_5\)OH
Step 3: Final Answer:
The correct decreasing order of acidic strength is Phenol>m-cresol>p-cresol>Ethanol. This corresponds to option (D).