I was trying to compare the basicity of o-phenanthroline and ammonia.

According to me, the factors that are seen in o-phenanthroline are:

  1. o-Phenanthroline has two nitrogen atoms at the 1,10 positions.
  2. It has a free lone pair on one nitrogen that is not delocalized due to resonance.
  3. The nitrogens are $\mathrm{sp^2}$ hybridized which is more electronegative compared to the $\mathrm{sp^3}$ lone pair on ammonia.

However, would there be hydrogen bonding in its conjugate base? If so, then how do we compare its basicity with a base like ammonia?


A quote from here explains and give a good answer to your wondering:

Amines are the most basic of the common organic functional groups, but are still fairly weak bases. Protonation occurs on the non-bonded electron pair exclusively. The basicity of amines is directly dependent on the “electron density” at the nitrogen atom. Both inductive and resonance effects can alter the basicity of a nitrogen atom.

Hybridization on the $\ce{N}$ also affects basicity. An increase in $\mathrm{s}$ character on an atom increases the electronegativity of that atom which favors acidity and therefore disfavors basicity. Hence $\mathrm{sp^3}$-hybridized nitrogen is more basic than either $\mathrm{sp^2}$ or $\mathrm{sp}$ hybridized nitrogen.

The availability of this non-bonding lone pair is a factor of basisity. The 1,10-Phenanthroline is a pyridine derivative. Thus, both lone pairs of two $\ce{N}$ atoms in the ring system are contributed to the system's aromaticity. This make them not 100% available to incoming protons while the lone pair in ammonia is 100% available. Therefore, in aqueous solutions, the basicity of $\ce{NH3} \ (\mathrm{p}K_\mathrm{a} = 9.3)\gt$ basicity of 1,10-Phenanthroline $(\mathrm{p}K_\mathrm{a} \approx 4.9)$. This is similar to $\mathrm{p}K_\mathrm{a}$ of pyridine, which is $5.2$ in water (comparison with piperidine is depicted in following scheme):

Pyridine and piperidine

Keep in mind that there are some other factors are effecting the basicity of amine as well. For example, the ring sizes of cyclic amines (Ref.1):

pKa of cyclic amines

The order is 5-membered $\ge$ 4-membered $\gt$ 6-membered $\gg$ 3-membered.


  1. Scott Searles, Milton Tamres, Frank Block, Lloyd A. Quarterman, "Hydrogen Bonding and Basicity of Cyclic Imines," J. Am. Chem. Soc. 1956, 78(19), 4917–4920 (https://doi.org/10.1021/ja01600a029).

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.