The increasing order of the pKb of the following compound is
(a) (A)

Q: Why is the pyridine nitrogen considered the basic site instead of the secondary amine nitrogen?

The secondary amine nitrogen ($-NH-$) is flanked by two aromatic rings. Its lone pair is heavily involved in resonance with the $\pi$-systems of both rings, making it highly delocalized and unavailable to bond with a proton. In contrast, the lone pair on the pyridine nitrogen resides in an $sp^2$ hybridized orbital that is perpendicular to the ring's $\pi$-system. Because it cannot overlap with the $\pi$-orbitals, it remains localized and is readily available to act as a base.

Q: Doesn't Fluorine have lone pairs? Shouldn't it show a $+R$ effect and increase basicity?

You are absolutely right that Fluorine has lone pairs and does exhibit a $+R$ (resonance) effect. However, Fluorine is also the most electronegative element on the periodic table, which gives it a very strong $-I$ (inductive) effect. For halogens attached to an aromatic ring, the electron-withdrawing $-I$ effect dominates over the electron-donating $+R$ effect. Therefore, the net result is electron withdrawal, which decreases the basicity.

Q: Why does the $-OCH_3$ group make the molecule a stronger base than the $-CH_3$ group?

The $-OCH_3$ group donates electrons primarily through the $+R$ (resonance) effect, which involves the actual delocalization of its lone pair electrons into the $\pi$-system of the ring. This is a very strong and far-reaching effect. The $-CH_3$ group, on the other hand, donates electrons through the $+I$ (inductive) effect and hyperconjugation. These effects, while significant, are generally weaker than a full resonance effect. Hence, $-OCH_3$ enriches the basic site with more electron density than $-CH_3$.

Q: How does the electron density from the left ring even reach the pyridine nitrogen on the right?

The entire molecule is a conjugated system. The lone pair on the bridging $-NH-$ nitrogen acts as a conduit. When an electron-donating group on the left ring pushes electron density into that ring, it increases the electron density on the bridging nitrogen. This, in turn, reduces the extent to which the bridging nitrogen's lone pair is delocalized into the left ring, allowing it to push more electron density toward the pyridine ring on the right. It's a continuous electronic relay.

Q: What would happen if the $-NO_2$ group was at the meta position instead of the para position?

If the $-NO_2$ group were at the meta position, it would only be able to exert its $-I$ (inductive) effect on the basic site, because resonance effects ($-R$) do not operate effectively from the meta position. While it would still be an electron-withdrawing group and decrease basicity, the decrease would be less severe compared to when it is at the para position (where both $-I$ and $-R$ effects are active).

Question

The increasing order of the $pK_b$ of the following compound is

Options:

(A) (A) < (C) < (D) < (B)

(B) (C) < (A) < (D) < (B)

(C) (B) < (D) < (A) < (C)

(D) (B) < (D) < (C) < (A)

Sigma Percentile · Accepted Answer

### The Battle of Substituents: Decoding $pK_b$ in Substituted Pyridines When you look at a complex organic molecule and are asked to determine its basicity, it can feel like trying to solve a puzzle with too many moving parts. But fear not! The secret lies in breaking the molecule down into its core components and understanding the electronic tug-of-war happening within it. Let's dive into this fascinating problem and decode the $pK_b$ order of these substituted pyridines. ### The Arena: Identifying the Basic Site Before we can even talk about substituents, we must answer a fundamental question: **Where does the proton actually attach?** In our four molecules, we have two distinct nitrogen atoms. One is part of the bridging $-NH-$ group, and the other is embedded within the pyridine ring. Imagine the lone pair on the bridging $-NH-$ nitrogen. It is sandwiched between two aromatic rings. Because of resonance, this lone pair is highly delocalized; it spends its time wandering around the $\pi$-clouds of both rings. A lone pair that is constantly on the move is not very good at grabbing a passing proton. On the other hand, look at the pyridine nitrogen. Its lone pair resides in an $sp^2$ hybridized orbital that sits perpendicular to the aromatic $\pi$-system. Because of this orthogonal geometry, the lone pair cannot participate in resonance. It is localized, stationary, and highly available to accept a proton. Therefore, the **pyridine nitrogen is our primary basic site**. ### The Rule of the Game: Basicity vs. $pK_b$ The question asks for the increasing order of $pK_b$. It is crucial to remember the mathematical relationship here: $$pK_b = -\log K_b$$ A stronger base has a higher $K_b$ value, which translates to a **lower $pK_b$ value**. Therefore, arranging the molecules in increasing order of $pK_b$ is exactly the same as arranging them in **decreasing order of basic strength**. Our mission is to find out which molecule is the strongest base, and which is the weakest. ### The Contenders: Analyzing the Substituents The basic strength of the pyridine nitrogen depends entirely on the electron density around it. The substituents on the far-left phenyl ring act as remote control switches, either pumping electrons toward the basic site or draining them away. **1. The Heavyweight Champion: Molecule (B) with $-OCH_3$** The methoxy group ($-OCH_3$) is a powerhouse of electron donation. The oxygen atom has lone pairs that it readily donates into the aromatic ring via a strong **$+R$ (resonance) effect**. This wave of electron density travels through the conjugated system, significantly enriching the electron density at the pyridine nitrogen. This makes molecule (B) the **strongest base**, and consequently, it has the **lowest $pK_b$**. **2. The Solid Runner-Up: Molecule (D) with $-CH_3$** The methyl group ($-CH_3$) is also an electron donor, but it uses different tactics: the **$+I$ (inductive) effect** and **$+H$ (hyperconjugation)**. While effective, these effects are generally weaker than the full-blown resonance effect of the methoxy group. Thus, molecule (D) is a strong base, but it takes the second place behind (B). **3. The Deceptive Halogen: Molecule (A) with $-F$** Fluorine is a tricky atom. As a halogen, it does possess a $+R$ effect because of its lone pairs. However, fluorine is highly electronegative, meaning its **$-I$ (inductive) effect strongly dominates** over its resonance effect. The net result is that fluorine acts as an electron-withdrawing group, pulling electron density away from the basic site. This makes molecule (A) a weaker base than the unsubstituted version. **4. The Ultimate Drain: Molecule (C) with $-NO_2$** The nitro group ($-NO_2$) is notorious for its electron-withdrawing prowess. It exerts both a strong **$-R$ effect** and a strong **$-I$ effect**. It acts like a vacuum, severely depleting the electron density across the entire molecule, leaving the pyridine nitrogen electron-poor. This makes molecule (C) the **weakest base**, giving it the **highest $pK_b$**. ### The Final Verdict By evaluating the electronic effects, we established the decreasing order of basicity: **(B) > (D) > (A) > (C)** Since the question demands the increasing order of $pK_b$, we simply reverse our basicity trend: **(B) < (D) < (A) < (C)** This logical deduction leads us straight to the correct answer, option **(c)**. By mastering the interplay of inductive and resonance effects, even the most intimidating molecules reveal their secrets!

Animated Solution for Chemistry - Organic Compounds: The increasing order of the $p K_{b}$ of the following compound is

Select Answer:

Visualized Solution

$p K_{b} and Basicity$

$Identifying the Basic Site$

$Effect of - O C H_{3} Group$

$Effect of - C H_{3} Group$

$Effect of - F Atom$

$Effect of - N O_{2} Group$

$Final Order$

The Sigma Insight: Amines

The Battle of Substituents

The Arena

The Rule of the Game

The Contenders

The Final Verdict

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Animated Solution for Chemistry - Organic Compounds: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The increasing order of the pKb​ of the following compound is