The correct order of decreasing pk_a, values of the following compoun…

The correct answer is Option (1) → (B), (C), (D), (A)

Core Concept:

Electron withdrawing groups decrease pKa (increase acidity), while electron donating groups increase pKa (decrease acidity).

Explanation:

Methoxy group (–OCH₃) donates electrons through resonance, reducing acidity and increasing pKa.

Nitro group (–NO₂) withdraws electrons through –I and –R effects, increasing acidity and lowering pKa.

o-Nitrophenol is slightly less acidic than p-nitrophenol due to intramolecular hydrogen bonding reducing ionization.

Step 1: Identify the Compounds

The compounds shown in the question are substituted phenols:

$\textbf{(A):}$ p-nitrophenol (4-nitrophenol)

$\textbf{(B):}$ p-methoxyphenol (4-methoxyphenol)

$\textbf{(C):}$ m-nitrophenol (3-nitrophenol)

$\textbf{(D):}$ o-nitrophenol (2-nitrophenol)

Step 2: Understand the Relationship Between Acidity and pKa

$\textbf{Acidity}$ is the ability of a compound to donate a proton ($H^{+}$).

$\textbf{pKa}$ is the negative logarithm of the acid dissociation constant ($K_{a}$).

The relationship is inverse: a $\textbf{lower pKa}$ value indicates a $\textbf{stronger acid}$, while a $\textbf{higher pKa}$ value indicates a $\textbf{weaker acid}$.

Step 3: Analyze Electronic Effects of Substituents

• Electron-Withdrawing Groups (EWG): Groups like nitro ($-NO_2$) withdraw electron density from the ring. This stabilizes the negative charge on the phenoxide ion (the conjugate base), making the original phenol more acidic (lower $pK_a$).
• Electron-Donating Groups (EDG): Groups like methoxy ($-OCH_3$) donate electron density. This destabilizes the phenoxide ion, making the phenol less acidic (higher $pK_a$).

Step 4: Compare Specific Positions

1. Nitrophenols ($-NO_2$ isomers):

• p-Nitrophenol (A): The nitro group at the para position exerts both a strong inductive effect ($-I$) and a strong resonance effect ($-R$ or $-M$). This makes it the strongest acid among the given nitrophenols, with a $pK_a$ of approximately 7.1.
• o-Nitrophenol (D): Similar to the para isomer, it has both $-I$ and $-R$ effects. However, it is slightly less acidic than p-nitrophenol because of intramolecular hydrogen bonding between the $-OH$ and $-NO_2$ groups, which makes it slightly harder to lose the proton. Its $pK_a$ is approximately 7.2.
• m-Nitrophenol (C): At the meta position, the nitro group only exerts an inductive effect ($-I$) and cannot stabilize the negative charge through resonance. Thus, it is significantly less acidic than the ortho and para isomers, with a $pK_a$ of approximately 8.3.

2. p-Methoxyphenol (B): The methoxy group is an EDG due to its strong resonance effect ($+R$ or $+M$), which destabilizes the phenoxide ion more than any other group in this set. This makes it the weakest acid (highest $pK_a$), with a $pK_a$ of approximately 10.2.

Step 5: Determine the Order of Decreasing $pK_a$

To find the order of decreasing $pK_a$, we arrange them from the weakest acid (highest $pK_a$) to the strongest acid (lowest $pK_a$):

1. p-Methoxyphenol (B): $pK_a \sim 10.2$
2. m-Nitrophenol (C): $pK_a \sim 8.3$
3. o-Nitrophenol (D): $pK_a \sim 7.2$
4. p-Nitrophenol (A): $pK_a \sim 7.1$

Option (C) Phenol

Has no substituent, so it has the highest pKa among the given compounds.

Option (D) o-Nitrophenol

Nitro group increases acidity but intramolecular hydrogen bonding stabilizes neutral form, giving slightly higher pKa than p-isomer.

Option (A) p-Nitrophenol

Strong electron withdrawing nitro group stabilizes phenoxide ion, lowering pKa further.

Option (B) p-Methoxyphenol

Methoxy group donates electrons, destabilizing phenoxide ion, giving highest pKa.