The strongest acid among the following is:
(A) o-methoxyphenol
(B) m-methoxyphenol
(C) p-methoxyphenol
(D) phenol
Answer
273.9k+ views
Hint: The acidity of the derivatives of phenol depends upon the stability of the corresponding phenoxide ion. Presence of EWG on phenyl rings stabilizes the phenoxide ion and presence of EDG destabilizes the phenoxide ion.
Complete step by step solution:
Here, we are being asked to compare the acidic strength of the given phenol derivatives.
-We know that phenol loses the proton from –OH group and forms phenoxide ions. Upon the stability of that phenoxide ion, the stability of the corresponding acid is measured. Let’s see the formation of phenoxide ion and how it is stabilized.

Here, we can see that the negative charge on oxygen is stabilized by resonance. We can see four canonical structures of the phenoxide ion.
-Now, depending upon the substituent groups the acidity of the phenol derivative may differ.
-If there is an electron-withdrawing group (EWG) attached to the phenyl ring, then the acidity of the corresponding phenol increases and if any electron releasing group (EDG) is present at the phenyl ring, then the acidity of the corresponding phenol decreases.
-So, we can say that if we substitute methoxy group which is an electron-donating group, then it will decrease the acidity of the corresponding phenol because the negative charge is delocalized and the methoxy group being electron donating in nature will destabilize the phenoxide ion and as a result, the acidity of the corresponding phenol will be decreased.
-Thus, we can conclude that acidity of ortho, meta and para derivatives of phenol will be less than phenol.
So, the correct answer to this question is (D).
Note: The effect of the presence of substituent group at ortho and para positions have a more strong effect than at meta position. Do not forget that an alkoxy group is an electron-donating group.
Complete step by step solution:
Here, we are being asked to compare the acidic strength of the given phenol derivatives.
-We know that phenol loses the proton from –OH group and forms phenoxide ions. Upon the stability of that phenoxide ion, the stability of the corresponding acid is measured. Let’s see the formation of phenoxide ion and how it is stabilized.

Here, we can see that the negative charge on oxygen is stabilized by resonance. We can see four canonical structures of the phenoxide ion.
-Now, depending upon the substituent groups the acidity of the phenol derivative may differ.
-If there is an electron-withdrawing group (EWG) attached to the phenyl ring, then the acidity of the corresponding phenol increases and if any electron releasing group (EDG) is present at the phenyl ring, then the acidity of the corresponding phenol decreases.
-So, we can say that if we substitute methoxy group which is an electron-donating group, then it will decrease the acidity of the corresponding phenol because the negative charge is delocalized and the methoxy group being electron donating in nature will destabilize the phenoxide ion and as a result, the acidity of the corresponding phenol will be decreased.
-Thus, we can conclude that acidity of ortho, meta and para derivatives of phenol will be less than phenol.
So, the correct answer to this question is (D).
Note: The effect of the presence of substituent group at ortho and para positions have a more strong effect than at meta position. Do not forget that an alkoxy group is an electron-donating group.
Recently Updated Pages
Classification of Drugs in Chemistry: Types, Examples & Exam Guide

Difference Between Alcohol and Phenol: Structure, Tests & Uses

Wheatstone Bridge – Principle, Formula, Diagram & Applications

Mass vs Weight: Key Differences Explained for Students

Area vs Volume: Key Differences Explained for Students

Mutually Exclusive vs Independent Events: Key Differences Explained

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Derivation of Equation of Trajectory Explained for Students

Understanding the Different Types of Solutions in Chemistry

Other Pages
JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

CBSE Class 12 Chemistry Question Paper 2026 PDF Download (All Sets) with Answer Key

NCERT Solutions For Class 12 Chemistry Chapter 2 Electrochemistry - 2025-26

NCERT Solutions For Class 12 Chemistry Chapter 1 Solutions - 2025-26

NCERT Solutions For Class 12 Chemistry Chapter 3 Chemical Kinetics - 2025-26

