Animated Solution for Chemistry - Organic Compounds Containing Nitrogen: In the following compounds, the decreasing order of basic strength will be

When we talk about the basic strength of amines, we are essentially discussing their ability to accept a proton ($H^{+}$) from water. The reaction looks like this:

The stronger the base, the more the equilibrium shifts to the right. This shift is entirely dependent on the stability of the conjugate acid formed—the ammonium cation ($R-N{H}_3^{+}$). If the cation is highly stable, the amine will readily accept a proton, making it a strong base. In an aqueous solution, the stability of this cation is governed by a fascinating tug-of-war between three distinct factors: the Inductive Effect, the Solvation Effect, and Steric Hindrance.

The Inductive Effect (+I)

Alkyl groups, such as the ethyl group ($-C_2{H}_5$), are electron-donating. Through the sigma bonds, they push electron density towards the positively charged nitrogen atom. This $+I$ effect helps to disperse the positive charge, thereby stabilizing the cation.

If we look purely at the $+I$ effect, more alkyl groups should mean more stability. Therefore, diethylamine (a secondary amine with two ethyl groups) should form a more stable cation than ethylamine (a primary amine with one ethyl group), which in turn should be more stable than ammonia (which has no alkyl groups).

Based on the $+I$ effect alone, the expected order of basicity is: $2^{\circ }>1^{\circ }>N{H}_3$.

The Solvation Effect

However, we are not in a vacuum; we are in an aqueous solution. Water molecules are highly polar and love to interact with charges. They surround the ammonium cations and stabilize them through hydrogen bonding. This is known as the solvation effect.

The extent of hydrogen bonding depends on the number of hydrogen atoms attached directly to the positively charged nitrogen. The ammonium ion ($N{H}_4^{+}$) has four hydrogens, allowing for extensive hydrogen bonding. The ethylammonium ion ($C_2{H}_{5}N{H}_3^{+}$) has three, and the diethylammonium ion ($(C_2{H}_5{)}_{2}N{H}_2^{+}$) has only two.

Based on the solvation effect alone, the expected order of basicity is completely reversed: $N{H}_3>1^{\circ }>2^{\circ }$.

Steric Hindrance (The Tie-Breaker)

We have a direct conflict! The $+I$ effect favors the secondary amine, while the solvation effect favors ammonia. To resolve this, we must introduce the third factor: steric hindrance.

Ethyl groups are bulky. As you add more ethyl groups around the nitrogen atom, they physically block the incoming water molecules. This steric crowding severely restricts the ability of water to form hydrogen bonds with the nitrogen's protons, effectively diminishing the solvation effect.

For ethyl-substituted amines, the $+I$ effect and the steric hindrance dominate over the solvation effect. The secondary amine, diethylamine, strikes the optimal balance. It benefits from the strong $+I$ effect of two ethyl groups, and while its solvation is hindered, the electronic stabilization is significant enough to make it the strongest base.

Ethylamine comes next, balancing one $+I$ group with decent solvation. Ammonia, lacking any $+I$ stabilization, ends up being the weakest base in this series.

Therefore, the final decreasing order of basic strength for ethylamines in aqueous solution is:

A fascinating twist: If we were dealing with methyl groups instead of ethyl groups, the smaller size of the methyl group reduces steric hindrance, allowing the solvation effect to play a much larger role. For methylamines, the order changes to $2^{\circ }>1^{\circ }>3^{\circ }>N{H}_3$. Always remember that in chemistry, context and subtle structural changes can completely flip the rules!