Basicity of nitrogen heterocyclic compounds

Question

Question :

The order of basicity for the following compounds is :

The logic I've applied is that since 4 has a conjugated system, the lone pair of N will be delocalised and hence would be the least basic. Conjugate systems/Resonance is not seen in the other three compounds

I'm not entirely sure about how to differentiate between the other three. For morpholine 2, If we see the stability of the conjugate acid formed, the O would exert (very less) -I effect, which should destabilize the positive charge on N (on formation of conjugate acid). Now between 1 and 3, 1 has a five other C atoms attached to the N, while 3 has four. Now CH3 is a +I group, hence more CH3 in 1 should stabilize the positive charge on N more than 3.

I don't think Inductive Effect should matter much, considering Inductive Effect becomes very ineffective with increase in chain length. But this is the only reasoning that strikes me. So ultimately, order of basicity becomes 4 < 2 < 3 < 1. Is my reasoning and the final answer correct?

Answer 1

Nitrogen inversion is not relevant to this discussion. A faster rate of inversion does not correspond to a greater delocalisation in the electronic structure, as lone pair inversion is not equivalent to resonance. The lone pair does not exist as a delocalised smear; it exists in two different localised forms.

The PDF on nitrogen inversion in the comments does not even bring up the topic of basicity, and quite rightly so.

Inductive effects explain why morpholine 2 is less basic, and resonance effects explain why pyrrolidone 4 is even less basic. However, since the pK_aH's of piperidine 1 and pyrrolidine 3 are so similar (11.22 and 11.27 respectively), I think it is pointless to try and qualitatively rationalise a difference in their basicities. It is like trying to explain why the temperature in Spain one day is 11.22 °C and the temperature the next day is 11.27 °C. The relative basicity could well be different if you switch the solvent, which would invalidate any reasoning based on - for example - s-orbital character of the nitrogen lone pair, which I think is the only plausible qualitative argument here.

Simply looking at the data, though, the order you want is 4 < 2 < 1 < 3.

Answer 2

To compare between 1 and 3, let's take into account the interaction between the non-bonding electron pair on nitrogen and the anti-bonding MO of the C-H bond (σ*C-H).

In order for the interaction to be effective, the lone pair on nitrogen and the C-H bond must be planar. In the six-membered ring, there is always a C-H bond next to the nitrogen lays in the parallel position against nitrogen's lone pair, which is the axial one. In the five-membered ring, the C-H bond does not totally lie on the same plane with the lone pair of nitrogen, so the interaction is less effective.

This interaction decreases the energy of the non-bonding MO, which consequently decreases the basicity of the nitrogen. As this interaction is more stressed in the six-membered ring, its basicity is less then that of the five-membered ring.

Basicity: 4<2<1<3

Answer 3

I would like to answer it,

KEY POINT: More electron density on N, more availability of the lone pair, more good base.

$(1)$ The nitrogen is directly connected inside the ring, surrounded by lots of carbon atoms. N being more electronegative than C, snatches the electrons towards itself as the nearest carbon atoms shows +I effect, it drops down as we move away from N. So, electron density on N increases, which means the lone pair is more available for donation. Hence, it's a good base!

$(2)$ The ring is the same but this time, a highly electronegative O came now. As it is more electronegative than N, it will snatches the electrons towards itself from the downside ring. Thus, decreasing the electron density on N. Hence, it's now no more willing to donate its lone pair. It is more tightly bound to the N as the repulsions from the electrons around it are decreased. Hence, not so good base.

$(3)$ Everything still the same, but the number of carbon atoms reduces to $4$ around it. Hence, +I effect reduces compared to $(1)$. So, electron denisty reduces comapred to $(1)$. Basic character reduces compared to $(1)$.

$(4)$ Less basic than $(3)$ as N got a $\ce{sp^2}$ hybridized carbon around it which is more electronegative than the previous. Hence, that +I effect reduces. So, a weak base than $(3)$ . But, the O here isn't connected directly to the ring. So, it will not effect any other C atom to perform their +I effect. Hence, the density on N reduces a bit, whereas in $(2)$, it reduces a lot as the O here will effect every C to not to give their best at +I effect, it will also snatches electron density around N a bit. Hence, it's the least Basic. So, correct order:

$\ce{(1)>(3)>(4)>(2)}$

Actually I do not at all like chemistry, but i don't know why i posted this as my ever first unexpected answer on this site. This is probably because this topic is my favourite by mistake from my class $12$ chapter: Amines.