I am not able to find any examples of the inductive effect operating through multiple bonds. Why can't electron displacement take place in multiple bonds?

  • 1
    $\begingroup$ it might be beneficial if you could post an example of two systems that you are comparing, or if you are talking about some "process" or context in particular. $\endgroup$
    – Eric Brown
    Commented May 5, 2013 at 11:53
  • $\begingroup$ @EricBrown I have just started studying Organic chemistry , I am confused why electron displacement doesn't occur in bonds like C=O . $\endgroup$
    – Vivek
    Commented May 5, 2013 at 13:52
  • $\begingroup$ @VivekKhandewal The physics of "single bonds" and "double bonds" are the same, and so both would experience an "inductive effect" for whatever that term means. It will take me some time to come up with a comprehensive answer to a very good question. $\endgroup$
    – Eric Brown
    Commented May 6, 2013 at 13:46
  • $\begingroup$ en.wikipedia.org/wiki/Inductive_effect $\endgroup$
    – Vivek
    Commented May 6, 2013 at 17:34

2 Answers 2


The argument is false. Inductive effects happen through any covalent bond.

It is true that as far as systems with multiple bonds are concerned, conjugation, hyper conjugation, or resonance effects are often more significant than inductive effects. These effect may reach a much larger distance and have richer chemistry, and frankly are more interesting. As a result, text books focus mainly on these effects in double bonded systems. Presence of a myriad of strong effects also makes it harder to identify the role of induction in system with multiple bonds.

However, this by no means eliminates the presence of inductive effect and in fact induction often do have significant contribution to reactivity in such systems.

The best example might be fluorobenzene (Ph-F). Electrophilic substitutions of fluorobenzene is significantly slower than those of benzene, and almost predominantly the para- product is obtained. This is caused by the induction of F atom. This can be attributed solely to inductive effects because the conjugation effect of F atom is electron donating and would cause an increase in reactivity instead, and would cause near statistical distribution between para and ortho substitution products(which is the case in chlorobenzene and bromobenzene).

In fact inductive effect also occur through C=O. You could not find an example only because indicator of electron charge on the oxygen atom of C=O is harder to come by. The examples do exist. For example, pKa of cation acid $\textrm{R2C=O-H}^+{\longleftrightarrow}\textrm{R2C=O + H}^+$ is strongly influenced by substitutions on the R groups.


If I'm not mistaken a good example of the inductive effect through multiple bonds is trifluoro-acetic-acid, or $\ce{F3CCOOH}$. The proton is so electron deficient that it is easily abstracted by anything that can act as a base. The not 'fluorinated' equivalent, acetic acid, is not nearly as strong an acid.

The reason for the difference in acidity, is the difference in electro negativity between hydrogen atoms and fluorine. Due to the very large electro-negativity of fluorine atoms the electrons in the acid protons are 'pulled' towards it, leading to a clear example of the inductive effect over multiple bonds.

A comment on one of the comments below the question: Electron displacement does occur in double bonds. A good example is the carbonyl functional group (C=O) in a molecule. The inductive effect results in a slightly more positive charge on the Carbon atom and slightly higher at the O-atom. This is also due to electro-negativity and results in the reactivity of the C=O functional group.

Hope this helped!

  • $\begingroup$ The example using F3CCOOH is not pertinent, it is not about inductive effect propagation through a double bond. Well it does on the conj base, but then there is not really a double bond. $\endgroup$
    – Alchimista
    Commented Nov 9, 2021 at 12:52

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.