Animated Solution for Chemistry - Organic Chemistry: The increasing basicity order of the following compounds is (A) ${\text{CH}}_3{\text{CH}}_2{\text{NH}}_2$ (B) $\begin{array}{c}{\text{CH}}_3{\text{CH}}_2\text{NH}\\ |\\ {\text{CH}}_2{\text{CH}}_3\end{array}$ (C) $\begin{array}{c}{\text{H}}_3\text{C}-\text{N}-{\text{CH}}_3\\ |\\ {\text{CH}}_3\end{array}$ (D) $\begin{array}{c}\text{Ph}-\text{N}-\text{H}\\ |\\ {\text{CH}}_3\end{array}$

Welcome to one of the most classic and conceptually rich topics in organic chemistry: the basicity of amines! This isn't just about memorizing a sequence; it's about understanding the beautiful tug-of-war between different physical and electronic effects. Let's dive into the problem and decode the logic step by step.

We are given four different amines and asked to arrange them in increasing order of basicity. Let's first identify our contenders: - (A) Ethylamine (${\text{CH}}_3{\text{CH}}_2{\text{NH}}_2$): A primary ($1^{\circ }$) aliphatic amine. - (B) Diethylamine (${\text{CH}}_3{\text{CH}}_2{\text{NHCH}}_2{\text{CH}}_3$): A secondary ($2^{\circ }$) aliphatic amine. - (C) Trimethylamine ($({\text{CH}}_3{)}_3\text{N}$): A tertiary ($3^{\circ }$) aliphatic amine. - (D) N-methylaniline (${\text{PhNHCH}}_3$): A secondary ($2^{\circ }$) aromatic amine.

Basicity is fundamentally about how easily the nitrogen atom can donate its lone pair of electrons to a proton ($H^{+}$). The more available the lone pair, the stronger the base.

Let's look at compound (D), N-methylaniline. Because the nitrogen atom is directly attached to a benzene ring, its lone pair is not sitting idle. Instead, it is delocalized into the $\pi$-electron system of the benzene ring through resonance.

Because the lone pair is "busy" participating in resonance, it is much less available to accept an incoming proton. Therefore, aromatic amines are generally much weaker bases than aliphatic amines. This immediately tells us that (D) is the least basic of the group.

Now we are left with the three aliphatic amines: (A), (B), and (C). Since the problem doesn't specify a solvent, we must assume the reaction is happening in an aqueous solution. In water, the basicity of aliphatic amines is governed by a delicate balance of three competing factors:

1. The $+I$ Effect (Inductive Effect): Alkyl groups are electron-donating. More alkyl groups mean more electron density pushed onto the nitrogen, making it more basic. Based on this alone, the order would be $3^{\circ }>2^{\circ }>1^{\circ }$. 2. Solvation Effect (Hydrogen Bonding): When the amine accepts a proton, it forms a conjugate acid. In water, this conjugate acid is stabilized by hydrogen bonding. A primary amine's conjugate acid has three hydrogens to form bonds, a secondary has two, and a tertiary has only one. Based on solvation alone, the order would be $1^{\circ }>2^{\circ }>3^{\circ }$. 3. Steric Hindrance: Bulky alkyl groups physically block water molecules from approaching and stabilizing the conjugate acid. This severely hurts the basicity of tertiary amines.

When we combine these factors, the secondary amine (B) hits the "sweet spot." It has a strong $+I$ effect from two ethyl groups, pushing plenty of electron density onto the nitrogen. At the same time, its conjugate acid still has two hydrogen atoms, allowing for excellent stabilization via hydrogen bonding with water.

Because it perfectly balances electronic push and aqueous stabilization, the secondary amine (B) emerges as the most basic compound in our list.

Now, the final showdown is between the primary amine (A) and the tertiary amine (C). You might think that trimethylamine (C) would be more basic because it has three electron-donating methyl groups. However, this is a classic trap!

In an aqueous solution, the three bulky methyl groups create massive steric hindrance. When trimethylamine accepts a proton, the resulting conjugate acid is so crowded that water molecules struggle to get close enough to form hydrogen bonds. With only one hydrogen available for bonding and severe crowding, the solvation energy is very low.

As a result, the primary ethylamine (A), which has much better solvation and less steric hindrance, actually beats the tertiary amine (C) in basicity.

Bringing all our logical deductions together: - (D) is the least basic due to resonance. - (B) is the most basic due to the perfect balance of $+I$ effect and solvation. - (A) is more basic than (C) because steric hindrance ruins the solvation of the tertiary amine.

Therefore, the final increasing order of basicity is:

This perfectly matches option (d).

It's crucial to understand why we got this answer, because the rules change if the environment changes. Imagine if this exact same question was asked, but the solvent was a non-polar liquid like benzene, or if the reaction was in the gaseous phase.

In the gas phase, there is no water to provide solvation or hydrogen bonding. The only factor that matters is the $+I$ inductive effect. In that scenario, the tertiary amine would proudly take the crown, and the order would strictly be $3^{\circ }>2^{\circ }>1^{\circ }>{\text{NH}}_3$. Always pay attention to the phase when dealing with amine basicity!