 Nitrogenous organic molecules make good Bronsted Lowry base. In this video, we'll be looking at Nitrogenous organic molecules and their aromatic derivatives and understanding their basic strength. Let's start with Ammonia, the most simple Nitrogenous organic molecule. It's a good base basically because it has a lone pair which is available and can be donated to an H plus ion to make a bond with it. An N compound that can accept an H plus easily makes a good Bronsted Lowry base. What if I remove this H and replace it with a benzene ring? Here I have a molecule called Aniline. The idea of the basicity of Aniline will depend on the availability of the lone pair which is available on the nitrogen here. Simply because this lone pair will enter into the delocalization with the benzene ring, the lone pair will become less available. The lesser the lone pair is available, the lesser is the basicity of the molecule because it's less probable for the H plus to make a bond with this nitrogen atom. The case is worsened if I remove one of the hydrogens and replace it with another benzene ring. The lone pair on this nitrogen is not only in delocalization with this benzene ring but it is also in delocalization with this benzene ring. Which means that an H plus will be even less probable to make a bond with this nitrogen atom. It's easy to predict that if I input another benzene ring by removing this hydrogen, the basicity should decrease even more, right? This lone pair goes into resonance with this benzene. After it completes an entire cycle, this lone pair goes into resonance with this benzene ring. And after its cycle is complete, this lone pair goes up to this benzene ring. So in this series of molecules, we can say if we... So in this series of molecules, if we number ammonia as 1, aniline as 2, and the double substituted ammonia as 3 and this as 4, the basic strength should be 1 is greater than 2, is greater than 3 and it is greater than 4. Alright. What if I add a substituent right next to the NH2 group? The lone pair is still there and we should now dig deeper as to how this ethyl group affects the basic strength of an aniline molecule. As we have started in our benzoic acid videos, there is something called steric inhibition. Let's look at the spatial arrangement of this molecule or how this molecule looks in 3D to understand what is the effect of this ethyl group in the aniline molecule. If we consider this a plane and our benzene ring is planar with this ethyl group and this NH2 group, it's easy to see that there should be some sort of electron cloud on NH2 and another electron cloud on ethyl group. Because both of these groups are bulky, they acquire a position where there is minimum repulsion. But now, if I try to protonate this nitrogen atom by making a bond between a hydrogen and the nitrogen, this should make the NH3 acquire a position where it will create electronic repulsions with the ethyl group. These electronic repulsions makes this molecule unstable. Because this molecule is unstable, a compound will not want to accept the hydrogen because the product is unstable. A base, after it acquires an H+, becomes a BH+, this is called a conjugate acid. If our conjugate acid is not sufficiently stable, this reaction does not tend to go in this direction. This only happens when two of the groups are really bulky and acquire a position which creates electronic repulsion. This is called steric inhibition of protonation. Steric inhibition of protonation decreases the basicity. So essentially, the basic strength of this molecule will be even lesser than an aniline molecule. Great. But now, let's introduce the nitrogen atom inside the cyclic molecule. This molecule is called pyrrole. It's an aromatic molecule. I want you folks to pause the video and judge if this molecule is a good base or not. The answer to this lies in how available is this lone pair to be donated to an H+, ion. To maintain the aromaticity of this molecule, this lone pair has to participate in delocalization. This means two things. A, this lone pair is less available for protonation. Second, in the case that it does get protonated, pyrrole molecule loses aromaticity. As aromaticity provides great stabilization, no atom wants to be protonated on the cost of aromaticity. Hence, to protonate a pyrrole molecule is a difficult task, which means the basicity decreases. After all, for a nitrogenous aromatic organic molecule, the lone pair on the nitrogen should be available and should not disturb the aromaticity of the molecule.