2
$\begingroup$

In an exercise from an exam several years ago, they ask to order the following three compounds from highest to lowest basicity:

enter image description here

I would not know how to start the exercise without going to a database, which is what I have done as indicated below.

In the CRC Handbook of Chemistry and Physics, the following pKa values are given for these compounds:

Pyridine: pKa = 5.23

Piperidine: pKa = 11.22

1-ethylpiperidine: pKa = 10.45

So, I have deduced that the order would be: II > III > I.

But, even so, this method would not be valid for me for an exam, since I cannot resort to any reference book.

So, I would be interested in what I should look for in order to be able to answer the question satisfactorily, since I can hardly find any information in basic books on the subject.

$\endgroup$
3
  • 1
    $\begingroup$ No problem; it happens. The next thing to consider is the hybridisation of the nitrogen lone pair, and what that tells you about its basicity. $\endgroup$ Jan 27, 2022 at 18:46
  • $\begingroup$ The hybridization of the nitrogen atom in pyridine is sp2. For the rest of the compounds, I believe it presents an sp3 hybridization. @orthocresol $\endgroup$
    – Carlos
    Jan 27, 2022 at 20:03
  • $\begingroup$ Indeed. Which has higher energy? $\endgroup$ Jan 27, 2022 at 20:09

1 Answer 1

4
$\begingroup$

Some general points of comparison of basic strengths of organic compounds are the following (The order of the points is important, check for condition 1. at first and then check the others in subsequent order)

  1. Delocalized and Localized lone pairs of electrons - In case of the presence of any N,O,S or halogen atom, we first compare whether the lone pair on the atom is a part of Resonance. Incase if the lp resonates, then delocalization of the lp occurs decreasing Basic character. A localized lp offers greater basic strength because by the Arrhenius definition of a base, a stronger base is the one which has greater tendency to donate electrons.

  2. Hybridization - as the s character reduces, the electronegativity decreases and the hybridization gets more Basic. sp hybridization being most electronegative is least basic. (basic nature of hybridizations is $sp < sp^2 < sp^3$ …)

  3. Inductive Effect - A +I group or an Electron Donating Group (EDG) increases the basicity whereas a -I group or Electron Withdrawing Group (EWG) decreases Basicity.

Some other factors which may sometimes play a role in determining Basic strength

--> Cross Conjugation - Incase if the lp is delocalized between multiple rings due to Resonance, the Basic strength reduces to a great extent.

--> Amines are generally more basic than Amides and Alcohols are generally more basic than Carboxylic Acids

--> Ortho Effect - Incase if a bulky group like any alkyl group,-NO2,-SO3H,-COOH etc. is attached at ortho position of the cyclic compound, then also basicity reduces.

--> Steric Hindrance - Incase if around the atom containing the localized lps, a lot of groups are attached, it creates Steric Hindrance due to which the atom is not able to donate its lp freely and the basicity falls.

It is not necessary that these factors will always be followed. They may be present but may not play a role in determining the Basicity. There will be several exceptions and corresponding reasons because of the experimental nature of the subject. So finally, for accurate results, you must refer to the $\ce{pka}$ values

In the question that u have asked, all the 3 compounds have localized lps. So we now check hybridization. Pyridine has $sp^2$ hybridization while the other 2 compounds have $sp^3$ hybridization making Pyridine least basic. In my opinion 1-ethylpiperidine should have been more basic than piperidine due to a +I group or EDG attached to the former. But according to the pKa values that you have furnished, Piperidine is more basic than 1-ethylpiperidine . I'm not completely certain about this but I believe that due to too many bulky groups being attached to the N atom in 1-ethylpiperidine, it experiences Steric Inhibition which brings down its basicity to some extent. This probably provides a reason for the order of basic strength to be $II > III > I$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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