Animated Solution for Chemistry - Organic Chemistry: Considering the basic strength of amines in aqueous solution, which one has the smallest $p{K}_b$ value?

Imagine you are a proton ($H^{+}$) swimming in an aqueous solution, desperately looking for a pair of electrons to bond with. The molecules that offer you these electrons are called bases. In the world of chemistry, we measure the strength of these bases using a value called $p{K}_b$.

Before we dive into the molecules, let's clear up a common point of confusion: the relationship between basic strength and $p{K}_b$. The base dissociation constant, $K_b$, directly measures how well a base grabs a proton from water. A larger $K_b$ means a stronger base. However, $p{K}_b$ is defined as the negative logarithm of $K_b$ ($p{K}_b=-\log K_b$). Because of that negative sign, the relationship flips! A larger $K_b$ results in a smaller $p{K}_b$.

Therefore, when the question asks for the molecule with the smallest $p{K}_b$ value, it is simply asking: "Which of these is the strongest base?"

Let's line up our competitors. We have four distinct amines: 1. $(C{H}_3{)}_{2}NH$: Dimethylamine, a $2^{\circ }$ (secondary) aliphatic amine. 2. $C{H}_{3}N{H}_2$: Methylamine, a $1^{\circ }$ (primary) aliphatic amine. 3. $(C{H}_3{)}_{3}N$: Trimethylamine, a $3^{\circ }$ (tertiary) aliphatic amine. 4. $C_6{H}_{5}N{H}_2$: Aniline, an aromatic amine.

Our job is to evaluate how willingly the nitrogen atom in each of these molecules will donate its lone pair of electrons.

Let's start with aniline ($C_6{H}_{5}N{H}_2$). In aniline, the nitrogen atom is directly attached to a benzene ring. The benzene ring is a massive, electron-hungry pi-system. The lone pair of electrons on the nitrogen atom doesn't just sit there; it gets drawn into the ring and participates in resonance.

Because this lone pair is busy delocalizing around the benzene ring, it is largely unavailable to bond with an incoming proton. This makes aromatic amines significantly weaker bases than their aliphatic counterparts. Consequently, aniline is the weakest base in our lineup and will have the largest $p{K}_b$ value. We can safely eliminate it.

A Three-Way Tug of War

Now we are left with the three aliphatic methyl amines. If this reaction were happening in the gas phase (where there are no solvent molecules to interfere), the basicity would depend solely on the Inductive Effect (+I). Alkyl groups like methyl ($-C{H}_3$) are electron-donating. They push electron density onto the nitrogen, making the lone pair more available. In the gas phase, the $3^{\circ }$ amine, with three methyl groups, would be the undisputed champion.

However, the question explicitly states we are in an aqueous solution. Here, water changes the rules of the game entirely. The basic strength now depends on a delicate balance of three competing factors:

1. The Inductive Effect (+I): As mentioned, more alkyl groups mean more electron density on nitrogen, favoring higher basicity ($3^{\circ }>2^{\circ }>1^{\circ }$). 2. Solvation Effect (Hydration): When an amine accepts a proton, it forms a positively charged conjugate acid (an ammonium ion). For this ion to be stable, it needs to form hydrogen bonds with surrounding water molecules. The more hydrogen atoms attached to the positively charged nitrogen, the more hydrogen bonds it can form, and the more stable it becomes. This factor favors less substituted amines ($1^{\circ }>2^{\circ }>3^{\circ }$). 3. Steric Hindrance: Alkyl groups are bulky. If there are too many of them around the nitrogen, they physically block water molecules from approaching and solvating the conjugate acid. This severely penalizes highly substituted amines.

When we combine these three factors for methyl-substituted amines, a fascinating compromise emerges.

The $3^{\circ }$ amine ($(C{H}_3{)}_{3}N$) has a great +I effect, but its conjugate acid has only one hydrogen atom for hydrogen bonding, and the three methyl groups create massive steric hindrance. Its solvation is incredibly poor, dropping its basicity significantly.

The $1^{\circ }$ amine ($C{H}_{3}N{H}_2$) has excellent solvation but a weak +I effect.

The $2^{\circ }$ amine ($(C{H}_3{)}_{2}NH$) hits the "Goldilocks zone." It has a strong +I effect from two methyl groups, and its conjugate acid still has two hydrogen atoms available for effective hydrogen bonding with minimal steric hindrance.

Because it strikes the perfect balance, the experimental order of basicity for methyl amines in aqueous solution is: $2^{\circ }>1^{\circ }>3^{\circ }>N{H}_3$

Based on our analysis, dimethylamine ($(C{H}_3{)}_{2}NH$) is the strongest base among the given options in an aqueous solution. Since it is the strongest base, it mathematically possesses the smallest $p{K}_b$ value.

Therefore, the correct option is (a).