Structure of imidazole and imidazoline

I recently sat for an exam in which I was asked to compare the basicity of the above compounds. I thought imidazole would be more basic because it is aromatic but according to the answer key 2-imidazoline is more basic. The reason provided is that the lone pair on imidazole is a part of aromatic system and hence it's not basic.

Shouldn't the other lone pair which is on nitrogen double-bonded to carbon be basic? After all it's not involved in resonance due to being out of plane of the pi-system. What is the real reason behind higher basicity of 2-imidazoline (if it is really more basic)?

  • $\begingroup$ Are you sure you read the question correctly? Are we just talking about basicity or the basicity of the nitrogen with the hydrogen on it? $\endgroup$
    – Zhe
    Jun 22 '17 at 18:32
  • $\begingroup$ The question asks to compare the basicity of the compunds. It doesn't mention any particular atom/element. I have just checked it again :) $\endgroup$ Jun 22 '17 at 18:34
  • 2
    $\begingroup$ The sp2 carbon should be more electron withdrawing, so I am inclined to agree with your answer without more context. $\endgroup$
    – Zhe
    Jun 22 '17 at 19:07
  • 1
  • $\begingroup$ @Zhe thanks for putting this important concept on sp2 carbon $\endgroup$ Jun 23 '17 at 19:43

I was unable to find experimental $\mathrm{p}K_\mathrm{a}$ values of imidazoline. However, the tools one can use to calculate and thereby predict $\mathrm{p}K_\mathrm{a}$ values are pretty good nowadays as can be seen comparing the predicted and experimental values for imidazole.

$$\textbf{Table 1: }\text{predicted and experimental }\mathrm{p}K_\mathrm{a}\text{ values of imidazole and imidazoline}$$ $$\begin{array}{lcccc}\hline \text{compound} & & \mathrm{p}K_\mathrm{a,1} & \mathrm{p}K_\mathrm{a,2} & \text{ref}\\ \hline \text{imidazole} & \text{(e)} & \phantom{0}7.13 & 12.70 & [1]\\ & \text{(p)} & \phantom{0}6.97 & 13.40 & [2]\\ \text{imidazoline} & \text{(p)} & 10.17 & \text{N/A} & [2]\\ \hline\end{array}$$

Note: $\mathrm{p}K_\mathrm{a,1}$ refers to the $\mathrm{p}K_\mathrm{a}$ values of the protonated heterocycle. The higher this value, the higher the basicity of the compound in question.

Obviously, the answer set’s statement that imidazoline is more basic is correct; the question remains why that is the case.

Both compounds obviously have the same protonated form, namely it is the imino nitrogen that bears a proton and thus a formal positive charge. Both of these protonated forms are $C_\mathrm{2v}$ symmetric. At first sight, the electronics of the two are pretty much equal. Yet obviously, imidazoline’s nitrogen must experience a greater electron density for some reason.

The key difference is the size of the π system which spans five atoms in imidazole but only three in imidazoline. At first sight, it seems like imidazole should be more basic since the positive charge can be delocalised onto five atoms compared to only three in imidazoline. However, this explanation fails to acknowledge that all π electrons are delocalised across more atoms in imidazole’s case.

With the π system being larger and spanning more atoms, we expect less electron density to be concentrated on each of the two nitrogens. On the other hand, the much smaller π system of imidazoline means that a greater electron density can be found on both nitrogen atoms at any time.

Further effects, such as the greater electronegativity of $\mathrm{sp^2}$ carbon atoms or the $+I$ effect of $\ce{C-H}$ bonds should be neglegible.


[1]: The Evans $\mathrm{p}K_\mathrm{a}$ table

[2]: Calculated by ChemAxon’s Chemicalize service.

  • $\begingroup$ I agree with your assessment, but I think the focus is really on the lone pair on the "pyrrole-type" nitrogen (i.e. the singly bonded N). This lone pair isn't the one getting protonated, but it is crucial to stabilisation of the conjugate acid via resonance. In imidazole it is more delocalised (for reasons you stated - more specifically because that nitrogen releases electrons into the π system - if it withdrew electrons then there would be greater electron density on it) and hence contributes to stabilisation less. $\endgroup$
    – orthocresol
    Jan 27 '18 at 18:28

Following this discussion which touched on concepts regarding factors influencing the stability of bases (and leaving groups-though off topic for this question though :)), and incorporating @Zhe's comment above: In order to determine strength of these bases, there are things worth noting in this case:

When evaluating the basicity of a nitrogen-containing organic functional group, the central question we need to take note of is: how reactive (and thus how basic) is the lone pair on the nitrogen? In other words, how much does that lone pair want to break away from the nitrogen nucleus and form a new bond with a hydrogen?

However the corresponding structure is conserved in both compounds and hence cannot be further used to segregate these compounds with regards to basicity (Note: I have put this statement so that it clears doubts)

In this case the possible factors include:

  1. Aromaticity and resonance

    Rings do not necessarily need to be 6-membered in order to have six π electrons. In this case imidizole, has aromatic 5-membered rings with six π electrons (Hückel rule).

    enter image description here

    In both of these molecules one nitrogen atom is $\mathrm{sp^2}$-hybridized. In imidazole, one lone pair occupies a $\mathrm{2p}_z$ orbital and is part of the aromatic sextet, while the second occupies one of the $\mathrm{sp^2}$ orbitals and is not part of the sextet. However, imidazoline has no sextet (no extra 2 electrons to complete aromaticity)

    This means that for imidazole, these electrons are very stable right where they are (in the aromatic system), and are much less available for bonding to a proton (and if they do pick up a proton, the aromatic system is destroyed). For these reasons, imidazole is not strongly basic

    Resonance effects involving aromatic structures can have a dramatic influence on acidity and basicity. The delocalization of electron density has a stabilizing effect, and the greater area over which the delocalization is possible, the greater the stabilisation

    The situation is slightly different in imidazoline, as it has one lacking π bond and the system is not aromatic and thus no greater room for stabilisation due to delocalisation and hence more basic.

    As a side note: The less the stability, the more the basic.

  2. Carbon bond hybridisation

    Another factor contributing to increased basicity of imidazoline lies in the carbon bond hydridisation. This phenomenon can be explained using orbital theory and the inductive effect: the $\mathrm{sp^2}$ orbitals of carbon bond in position 2 is one-part s and two parts p, meaning that it has about 67% s character. On the other hand, this carbon (in same position on imidazoline) has $\mathrm{sp^3}$ orbitals, conversely, are only 25% s character (one-part s, three parts p).

    Because the s atomic orbital holds electrons in a spherical shape, closer to the nucleus than a p orbital, $\mathrm{sp^2}$ hybridisation implies greater electronegative than $\mathrm{sp^3}$ hybridisation. Recall the inductive effect: more electronegative atoms absorb electron density more easily, and thus are less basic (and more acidic).

Remarks, the explanation you noted as the reason for the answer is valid as explained above, and your understanding also is correct, however is it was not enough to explain difference in basicity as other options needed to be taken into consideration (as discussed above) Thus we have sone picture on why imidazoline was chosen as more basic than imidazole.

  • 2
    $\begingroup$ The aromaticity and resonance explanation misses a key point. In both compounds, there are two different nitrogens. In both cases, the nitrogen with the sp2 lone pair is more basic than the one with the 2p lone pair. The explanation is correct that protonating at the nitrogen with 2p lone pair breaks the aromaticity of imidazole. However, a similar effect occurs in imidazoline: the 2p lone pair can be delocalized to the other nitrogen, and this is interrupted by protonation at this lone pair. $\endgroup$
    – jerepierre
    Jun 23 '17 at 20:42
  • 1
    $\begingroup$ Thinking about it on the product side, in both cases, protonating at the sp2 lone pair allows for distributing the positive charge over both nitrogens. Since this occurs for both imidazole and imidazoline, I don't think the fact that one is aromatic and the other isn't has any substantial effect on the basicity difference. $\endgroup$
    – jerepierre
    Jun 23 '17 at 20:49

The answer is right and the explaination given is also perfectly right and you're interpretation of which nitrogen is basic is also right. :) To understand the answer, you need to first understand how nitrogen works here. After the double bonded nitrogen in either of these compounds is protonated it acquires a positive charge which is stabilized by the lone pair of the non double bond nitrogen which is in conjugation with it. Now in imadozole this lone pair (non double bond N) is also part of the aromatic sextet. Hence it's availablity to stabilize the other nitrogen in the conjugate acid is greatly diminished. This is not the case in imadoziline where the line pair on N is localised (not part of aromaticity). So what the answer key refers to as lone pair in aromaticity is the lone pair on the non basic nitrogen (non double bond N). pKa of imadozole= 7.1 pKa of imadziline=11


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.