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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.

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Aumkaar Pranav
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1 Answers1

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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.

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Vijay Bharadwaj
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  • 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. – Aumkaar Pranav Feb 28 '20 at 15:44
  • 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 :) – Vijay Bharadwaj Feb 28 '20 at 16:22
  • 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. – Aumkaar Pranav Feb 29 '20 at 01:17
  • May I have a link to the SE Chem answer that states otherwise? – Vijay Bharadwaj Feb 29 '20 at 03:54
  • Sorry, I just made a mistake in my comment. The Bent's Rule suggests in the case of CH₃F that the fluorine will lessen the electron density in the CF bond orbital, thus increasing p character in it, and in turn increasing s-character in the CH bonds where there is more electron density than in CF, where F has pulled it substantially. That leads to somewhat acidic hydrogen due to increased electronegativity in CH bond orbitals. But, you know about Bent's Rule so no worries. :) The proposition that the electron density is more on N and C atoms doesn't seem to be valid because the carbon directly – Vijay Bharadwaj Feb 29 '20 at 04:07
  • attached to the positive carrying nitrogen should have less electron density. I do feel that hydrogens at beta positions would have a tendency to protonate (like you said, positive charge on H atoms) to undergo Hoffmann Elimination however I don't feel it would be with the hydrogens on the methyl, which are at alpha. – Vijay Bharadwaj Feb 29 '20 at 04:10
  • https://chemistry.stackexchange.com/a/72216/43333 – Aumkaar Pranav Feb 29 '20 at 04:34
  • @VijayBhardwaj if we attribute the inductive effect of an alkyl group to its hyperconjugative effect, then how would one explain the higher acidic strength of butan-1-ol than that of tert butyl alcohol? – Aumkaar Pranav Mar 06 '20 at 15:38
  • I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details. – Martin - マーチン Mar 06 '20 at 17:44
  • @AumkaarPranav the most straightforward explanation here is that tBuO- is poorly solvated because of steric crowding around the oxygen. Less stable conjugate base = weaker acid. – orthocresol Mar 07 '20 at 00:01
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    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. – orthocresol Mar 07 '20 at 00:07