Compare the basic strength of the following compounds:

four nitrogen heterocycles

  1. d > b > c > a
  2. c > b > a > d
  3. c > a > b > d
  4. a > d > c > b

I know that in compound (c), the π-bond between nitrogen and cyclopropenyl will polarize (due to aromatic stability) such that it increases electron density of the nitrogenous ring. Whereas, in compound (d), the π-bond will polarize reducing the electron density of nitrogenous ring. Hence, comparing compounds with similar nitrogenous rings, I get that the basic strength of (c)>(b)and that of (d)<(a). But I thought the ring containing more number of nitrogen atoms will be more basic as there will be more lone pair donors, so I thought the answer to be option (4), but the correct answer happens to be option (2). I would like to know that how is basic strength of the nitrogenous rings being compared?


It is likely that the electron pair (from the HOMO) being donated to the acid is from a π molecular orbital. Since the π bond is rather weak, the π electrons should be rather high in energy and is thus able to interact well with the LUMO of the acid. When nitrogen atoms are incorporated into the π system, the energy of the HOMO is lowered due to the high electronegativity of N. Thus, incorporation of N atoms into the system would decrease the basicity of the molecular ion. As such, A and D would be inferior bases, compared to B and C.

Choosing between A and D, and between B and C, we would have to base our criteria for the basicity on the extent of delocalisation. In D, the π system is larger, thus greater delocalisation was achieved, with a lowering of the energy of the HOMO. Thus, D is likely to be less basic than A. Similarly, the extent of delocalisation is also larger in C, thus it would have a lower-energy HOMO, compared to B. Thus, B is likely more basic than C.

Therefore, option (2) is likely the correct order of basicity.


Read the answers to this question to see why more no. of lone pairs does not always mean more basicity.

Here, the common feature of all four compounds here is the middle nitrogen atom, so let's focus on that.

For it to be a good base, we need to split the double bond heterolytically to give the nitrogen a negative charge.

Done with step one? Now notice that a and b will have an aromatic ring each, c will have two aromatic rings and d will have one aromatic and one anti aromatic. Based on this, c is the most basic and d the least. Now the confusion is between compounds a and b.

To decide the strength of a and b, notice that in b, there is just one nitrogen pulling electron density away (inductive effect), while in a there are two of them. So, b is more basic.

One thing to keep in mind when deciding b/w compounds a and b –one might say that since there are two nitrogens which can donate lone pairs in a, there will be more resonance stabilisation of the positive ion formed after acid attacks the middle nitrogen which we are focusing on. But remember that the lone pairs of these two nitrogens are not participating in resonance. So, decide based only on inductive effects.

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    $\begingroup$ Why will the middle nitrogen atom act as the basic one? Its lone pairs will be involved in resonance with the nitrogenous ring which will also be responsible for its aromaticity. I think the remaining nitrogen atoms will be donating their lone pairs like in imidazole . That is why I considered the basicity to be dependent on no. of nitrogen atoms present. $\endgroup$ – keenlearner May 18 '18 at 9:40
  • $\begingroup$ @keenlearner $NH_2$- has two lone pairs. $\endgroup$ – Anurag B. May 18 '18 at 9:44
  • $\begingroup$ @keenlearner Also, the middle nitrogen is common in all four options and also bonded to the other ring, which gives me reason enough to focus on it. If you can provide an alternate explanation which does not focus on the middle nitrogen and is satisfactory enough, I'll be happy to accept it. :) $\endgroup$ – Anurag B. May 18 '18 at 9:48
  • $\begingroup$ @keenlearner When solving a question, you will need to look for patterns like this. You will have to ask yourself -- "Why would this question have been created? What is the concept behind it?" Which in this case would be aromaticity. If you see something which can create aromaticity, it is a huge hint for you that the question setter has provided. $\endgroup$ – Anurag B. May 18 '18 at 10:05

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