# Is the inductive effect always measured relative to hydrogen?

Wikipedia has defined the Inductive effect thus:

"In Chemistry and Physics, the inductive effect is an experimentally observable effect of the transmission of charge through a chain of atoms in a molecule, resulting in a permanent dipole in a bond."

Recently, I learned from a teacher that the inductive effect of any atom/group is always measured with respect to hydrogen.

Under this definition, we would ideally attach the atom/group with an atom of hydrogen and observe the direction of the dipole moment caused by the difference in electronegativity between hydrogen and the atom/group in question. If the electrons-initially positioned at the middle of the bond-are pushed toward hydrogen, the atom/group is termed as a "$+I$", electron-donating group. On the other hand, if the electrons are pulled toward the atom/group, it is termed as a "$-I$", electron-withdrawing group. Thereafter, this atom/group will always exhibit this type of inductive effect. Even if a situation does arise, in which it appears as though an atom/group that exhibits the $-I$ effect with hydrogen is exhibiting the $+I$ effect with another atom/group, this must be the cause of other dominating effects such as the mesomeric effect and in those cases also, it is said that the atom exhibits the $-I$ and not the $+I$ effect.

To illustrate the clear distinction between the definition put forward by the teacher and the one on Wikipedia, let us consider a molecule of $\ce{Cl2}$.

Under the definition on Wikipedia, it is clear that $\ce{Cl2}$ is a non-polar molecule (i.e. No permanent dipole moment has been experimentally observed in this molecule). Hence, no inductive effect is in play and neither of the two $\ce{Cl}$ atoms show either the $+I$ or $-I$ effect.

On the other hand, if we work with the definition put forward by the teacher, we observe that an atom of $\ce{Cl}$ exhibits the $-I$ effect when attached to an atom of hydrogen, being more electronegative than it. Therefore, in the $\ce{Cl2}$ molecule, no permanent dipole exists as a result of the $-I$ effect of both chlorine atoms cancelling away from opposite directions.

It appears that only one of these two definitions is definitely correct and I am confused as to which one that is.

• Why should one Cl atom exert inductive effect on the other? They are both equally electronegative, so they will pull the shared electron pair with equal 'force' and so there would be no inductive effect. – Shoubhik Raj Maiti Nov 26 '16 at 6:29
• I think this is in principle a problem in terminology. You will always have an (a net) inductive effect when you have different electronegativities. The terms +I/-I should only be used in organic chemistry, and there you would indeed compare them to the standard CH bond. – Martin - マーチン Nov 30 '16 at 13:13
• @Martin-マーチン Is it possible for u to perhaps elaborate your point and write an answer? – user33789 Dec 1 '16 at 0:29

Your definition is falling apart because you are over-applying it. In the definition, you can see that it says:

...transmission of charge through a chain of atoms in a molecule...

It doesn't make sense to apply this in the case of a two-atom molecule, because the inductive effect is used to measure how one bond effects the electron density of other bonds in the molecule. You can assign any bond a dipole moment, but inductive effects are used specifically to explain phenomena observed in organic chemistry, where chains of molecules are (exceedingly) common.

To answer your other question: yes, inductive effects are always measured relative to hydrogen. Linear free-energy relationships are a quantifiable way of measuring a reaction's sensitivity with respect to a certain parameter. The Hammett equation is used to study how sensitive a molecule is in a given reaction to a change in its subsituent's inductive and resonance effects, for which the reference subsituent in all cases is a hydrogen atom. This is perhaps because the $\ce{C-H}$ is the most common bond in all of organic chemistry.

Linear free-energy relationships are particularly useful because similar as to what you said, substituent groups will always have the same relative inductive effects no matter the molecule. A quaternary ammonium group will always have a larger $-I$ effect (and larger $\sigma$ value) than a nitro group than a trifluoromethyl group. A molecule's response to these changes can give great mechanistic insight into organic reactions and the pathways through which they proceed.

A table of the different types of Hammett relations$^{[1]}$:

Their reference reactions:

$\hspace{3.25cm}$

$[1]$ Anslyn, E. V.; Dougherty, D. A. Modern Physical Organic Chemistry; University Science: Sausalito, CA, 2006.

• I'm afraid I don't quite understand the last bit completely: "Linear free energy relationships are a quantifiable way of measuring a reaction's sensitivity with respect to a certain parameter." Is it possible to simplify this so that a highschool student can also understand it? – user33789 Dec 2 '16 at 0:14
• The Hammett equation, for instance, measures how sensitive the ionization of various substances are to changing the substituent para to the carboxyl group. There are other relationships that measure things such as how leaving groups or nucleophiles are affected by changing substituents, how the choice of solvent effects the reaction... the list goes on. If you are interested you can read more in the link I added. – ringo Dec 2 '16 at 2:53

I myself am having a little bit of trouble wrapping my head around the Wikipedia definition which they cite from an organic chemistry textbook. I am specifically troubled by the phrase ‘resulting in a permanent dipole moment.’ According to all definitions of the inductive effect I know, a methyl group is electron-donating ($+I$). Yet, if you consider para-xylene, the molecule suddenly has a centre of inversion (point group $C_\mathrm{2h}$). Therefore, there is no permanent dipole in para-xylene. (Depending on the methyl rotamers, it might also have $C_\mathrm{2v}$, in which case the dipole would be perpendicular to the benzene ring plane — and since there is almost no vertical structure, ignoring the three aliphatic $\ce{C-H}$ bonds, it is predicted to be very small.)

There is no reason to assume that methyl groups suddenly lose their $+I$ effect just because two of them are bonded to two different sides of a benzene ring. The NMR supports the assignment of a double $+I$-substitution with the aromatic protons of para-xylene displaying chemical shifts of $7.05~\mathrm{ppm}$ (compared to $7.36~\mathrm{ppm}$ in benzene; same solvent $\ce{CDCl3}$). Obviously there is additional shielding present in spite of no observable permanent dipole, which can only be attributed to the methyl groups’ $+I$ effect. The same case can be made for other symmetrically para-substituted benzenes such as para-nitrobenzene ($8.41~\mathrm{ppm}$ in $\ce{C6D12 + C3D6O}$), para-dichlorobenzene ($7.26~\mathrm{ppm}$ in $\ce{CDCl3}$) and others. In these cases, especially dichlorobenzene, the point group is actually predicted to strictly be $D_\mathrm{2h}$ — strictly containing inversion.

Thus, when having to choose between your teacher’s and Wikipedia’s definions, I want to go with your teacher’s because it better explains what we discuss in basically every seminar.

In Pauling electro negativity scale H is standard or reference atom.

The electron pairs shared between carbon atoms in a carbon chain enjoy a certain excess of electron charge from C-H sigma bonded MOs . Now the other atom or groups influence on this excess share can be either strengthening or weakening the electron cloud density on carbon chain. All +I GROUPS INCREASE SUCH DENSITY to a considerable extent on immediately bonded fellow to stranger , which dissipates rapidly as you go farther in carbon chain. Similarly - I groups have opposite effect.

Hope i conveyed the idea !