Animated Solution for Chemistry - Organic Compounds Containing Nitrogen: A. phenyl methanamine B. N,N-dimethylaniline C. N-methyl aniline D. Benzenamine Choose the correct order of basic nature of the above amines.

What Makes an Amine Basic?

To master the basicity of amines, we must first understand what basicity truly means in organic chemistry. According to the Lewis theory, a base is an electron pair donor. In amines, the nitrogen atom possesses a lone pair of electrons. The fundamental rule is simple: the more available this lone pair is for donation to a proton ($H^{+}$), the stronger the base.

When evaluating basicity, we are essentially acting as detectives, looking for any structural features that either increase or decrease the electron density on the nitrogen atom. Two primary suspects usually emerge: the resonance effect ($R$) and the inductive effect ($I$).

Phenyl Methanamine (A)

Let's begin our investigation with molecule A, phenyl methanamine (commonly known as benzylamine). At first glance, it has a benzene ring, so you might be tempted to treat it like an aromatic amine. But look closer!

The nitrogen atom is not directly attached to the benzene ring. Instead, it is bonded to a $-C{H}_2-$ group, which is an $s{p}^3$ hybridized carbon. This carbon acts as an insulator. Because of this physical separation, the lone pair on the nitrogen cannot overlap with the $\pi$-electron cloud of the benzene ring.

Since it cannot participate in resonance, the lone pair remains completely localized on the nitrogen atom. It is highly available for donation, making phenyl methanamine the strongest base among the four options.

The Power of Resonance

Now, let's shift our focus to molecules B, C, and D. In all three of these compounds, the nitrogen atom is directly attached to the benzene ring. This direct connection changes the game entirely.

The lone pair of electrons on the nitrogen atom is in conjugation with the $\pi$-system of the benzene ring. Through the $+R$ effect (positive resonance effect), the lone pair delocalizes into the ring, increasing the electron density at the ortho and para positions.

However, this comes at a cost for basicity. Because the electrons are busy delocalizing around the ring, they are much less available to accept an incoming proton. Therefore, all three of these aromatic amines are inherently less basic than our aliphatic amine, molecule A.

Ranking the Aromatic Amines

Since B, C, and D all suffer from resonance delocalization, how do we rank them? We must look at the substituents attached to the nitrogen atom. Alkyl groups, such as methyl ($-C{H}_3$), are electron-donating groups. They push electron density through the $\sigma$-bonds via the $+I$ effect (positive inductive effect).

In molecule B, N,N-dimethylaniline, there are two methyl groups attached to the nitrogen. Their combined $+I$ effect pushes a significant amount of electron density back onto the nitrogen atom. This partially compensates for the electrons lost to the ring via resonance, making the lone pair somewhat more available.

In molecule C, N-methylaniline, there is only one methyl group. It benefits from a single $+I$ effect, making it less basic than molecule B, but still more basic than an unsubstituted aromatic amine.

Finally, we look at molecule D, benzenamine (aniline). There are no methyl groups attached to the nitrogen—only hydrogen atoms. There is absolutely no $+I$ effect to compensate for the strong resonance delocalization into the ring. As a result, its lone pair is the least available for donation.

Synthesizing all our observations, we arrive at the final decreasing order of basicity:

1. A (Phenyl methanamine): Localized lone pair (Strongest base). 2. B (N,N-dimethylaniline): Delocalized lone pair, but compensated by two $+I$ groups. 3. C (N-methylaniline): Delocalized lone pair, compensated by one $+I$ group. 4. D (Benzenamine): Delocalized lone pair, no $+I$ compensation (Weakest base).

Therefore, the correct order is $A>B>C>D$, which corresponds perfectly to option (d).