There are several examples where the negative inductive effect of a substituent gets increased when a hydrogen atom on that substituent is replaced by an alkyl group.

Some particular examples:

$$\ce{-NR3+} > \ce{-NH3+} \qquad \ce{-OR} > \ce{-OH}$$

I don't understand why it should be that way, as an alkyl group shows more positive inductive effect than a hydrogen atom which should, in turn, reduce the negative inductive effect of the substituent.

The answer here gives the reason that a carbon atom is more electronegative than a hydrogen atom and hence it should increase the −I effect (the answer also talks about bond length but I think it's irrelevant to mention that). But wouldn't the inductive effect shown by the whole group be a better parameter than the electronegativity of its primary atom to judge its electron withdrawing/donating ability?

Then there are also examples where replacing hydrogen atom by an alkyl group reduces the negative inductive effect of the substituent, as with $\ce{-CHO}$ and $\ce{-CRO}$ groups. I would usually explain it using the +I effects of $\ce{-R}$ and $\ce{-H}$ groups, but since that logic doesn't hold for the previous case, there should be some other reason for it.

So my question is: How does replacing an H atom in a substituent by an alkyl group affect the inductive effect of the substituent?

P.S. There is a related question, but it is too specific and does not have an answer to my question.


1 Answer 1


There are several examples where the negative inductive effect of a substituent gets increased when a hydrogen atom on that substituent is replaced by an alkyl group.

Some particular examples: \begin{align} \ce{-NR3^+} &> \ce{-NH3^+}& \ce{-OR} &> \ce{-OH} \end{align}

About comparing $\ce{-NR3^+}$ and $\ce{-NH3^+}$, at first, the electron-donating inductive effect of the methyl groups (+I) makes us feel that the electron-withdrawing inductive effect trend should be the opposite. Neither, could I find any reliable sources with an explanation. So, I came up with a few of my own.

Part I: The Bent's Rule Interpretation (More Reliable)

To analyse the inductive effect, let's take the groups attached to a hydrocarbon chain and then see.

enter image description here

The $\ce{C-H}$ bond orbitals will have more s-character than in the $\ce{N-C}$ bond orbitals because there is more electron-density in the former due to the greater electronegativity of carbon over hydrogen and nitrogen over carbon, which requires the presence of more, lower-energy s-character in them to lower the system's overall energy. Thus, an implication of this would be the increase of p-character in the three $\ce{N-C_{Me}}$ bond-orbitals and the subsequent decrease of p-character in the $\ce{N-C_{chain}}$. This results in an increase of electronegativity of $\ce{N}$, thus the (-I) trend is:

$$\ce{-NR3^+} > \ce{NH3^+}$$

Part II: I don't know what to call this (Need help with this)

Let's consider the examples of phenol and anisole. It is considered that phenol donates more electron density to the benzene ring than anisole because it exists in equilibrium with the phenoxide ion, which is a far-more electron-donating group, despite anisole's $\ce{-OCH3}$ group being a better electron-donating group.

If we apply this to our case, we can assume that $\ce{-NH3^+}$ may be in equilibrium with $\ce{-NH2}$ which makes it a better (+I) group and poorer (-I) group, while $\ce{-NMe3^+}$ can't deprotonate itself and establish an equilibrium like that, making it a better (-I) group.

(somebody please help here)

Part III: Should we really consider $\ce{-CH3}$ as a (+I) group here? (Just thinking, very reliable*)

It is considered that the inductive effect of $\ce{-CH3}$ is mainly attributed to its hyperconjugative effect. However, hyperconjugation requires an empty p orbital carrying a positive charge into which it donates its $\sigma_{\ce{C-H}}$ bond-electron density. However, nitrogen's orbitals are sp³ hybridized here, with no empty orbital carrying the charge as such. So, I think (not really sure*), we should only consider carbon's greater electronegativity over hydrogen, thus making $\ce{-NMe3^+}$ more inductively electron-withdrawing than $\ce{-NH3^+}$. You can also apply this to $\ce{-OR}$ and $\ce{-OH}$.

* Just saw Jan's answer in the middle of my writing process. That makes this a valid point. Great answer, a must check-out.

  • $\begingroup$ Although the last point can provide solution to all the questions that I have asked, it's not a very popular or widely known fact. Even orthocresol, whose answer is just below Jan's answer to the same question, didn't mention anything like that. $\endgroup$ Commented Feb 28, 2020 at 15:44
  • $\begingroup$ I admit it's not well-heard. Just on that account, I mentioned my uncertainty on it. My motive was to provide different ways in which I could think about the situation. Although if you see the first part, you can also use that for answering both your questions. It applies to the OH/OR comparison as well. Edit: If you find the interpretations helpful, an upvote would help :) $\endgroup$ Commented Feb 28, 2020 at 16:22
  • $\begingroup$ I didn't understand the first point. Why should N-C bond have lower electron density? In fact, I read this on SE Chemistry only, that the electron density in NR3+ is more on N and C atoms, while the positive charge is spread on H atoms. $\endgroup$ Commented Feb 29, 2020 at 1:17
  • $\begingroup$ May I have a link to the SE Chem answer that states otherwise? $\endgroup$ Commented Feb 29, 2020 at 3:54
  • 1
    $\begingroup$ Also, I really don't buy the Bent's rule explanation: there are a few holes in the logic here. Firstly, the shading doesn't make sense, because s- or p-character is a characteristic of orbitals, not bonds; and there are two orbitals to a bond. So each bond should have two colours. The second thing is that the orbitals that C uses to bond to N has no influence on the orbitals that N uses to bond to C (Bent's rule only deals with orbitals on the same atom). So there isn't any "chain of reasoning" that lets you figure out the orbitals that atom Z uses, based on the orbitals that atom X uses. $\endgroup$ Commented Mar 7, 2020 at 0:07

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.