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Are the inductive effect and hyperconjugation both different ways of looking at the same phenomenon - i.e. methyl groups donate negative charge for example. Inductively we can argue that carbon is of greater electronegativity than hydrogen, and so carbon withdraws negative charge from hydrogen, and thus negative charge is "donated" to whatever that the methyl is attached to.

From a hyperconjugative effect we can argue that there is some interaction between the C-H orbitals and some antibonding orbital ... how is this supposed to be stabilizing if it involves an antibonding orbital?

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Are the inductive effect and hyperconjugation both different ways of looking at the same phenomenon

I think so, both inductive and resonance (that's what hyperconjugation is after all) effects move electrons around. I tend to separate the two effects as follows:

  • inductive effects are associated with electron movement through sigma bonds due to electronegativity differences
  • resonance effects are associated with electron movement through p orbitals due to electronegativity differences

For example, we often say that a methyl group stabilizes a double bond (carbon-carbon or carbonyl) or a carbocation. The methyl group is roughly $\ce{sp^3}$ hybridized while the double bond carbon or carbocation carbon is roughly $\ce{sp^2}$ hybridized. An $\ce{sp^2}$ orbital is generally lower in energy than an $\ce{sp^3}$ orbital because it contains more s-character. Therefore electrons will generally prefer to flow from $\ce{sp^3}$ to $\ce{sp^2}$ orbitals. The direction of this electron flow also indicates that $\ce{sp^2}$ orbitals are more electronegative than $\ce{sp^3}$ orbitals. How can the $\ce{sp^3}$ methyl group shift electrons to the lower energy, more electronegative $\ce{sp^2}$ carbon? It can do it inductively (through sigma bonds) or through resonance (through p orbitals).

I've added a picture showing resonance structures involving hyperconjugation, just to be clear as to what hyperconjugation involves. enter image description here

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  • $\begingroup$ why would increased electron density in a double bond stabilize it? $\endgroup$ – Dissenter Nov 1 '14 at 19:40
  • $\begingroup$ see this link $\endgroup$ – ron Nov 1 '14 at 19:59
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Inductive effect and hyperconjugation are two distinct phenomena.

Inductive effects are purely through sigma bonds. For example, in an organofluorine compound, the fluorine withdraws electrons from carbon by not sharing the electrons of the sigma bond equally. We explain this using the rationale that fluorine is more electronegative than carbon.

Similarly, in a carbocation, the electron-deficient carbon is more electronegative than any substituent carbons. Thus, the carbocation will withdraw electrons from substituents through the sigma bonds.

Hyperconjugation is electron donation of a filled bonding orbital to a nearby low-lying, unfilled orbital. Considering t-butyl cation as an example, a filled C-H bonding orbital can donate electrons to the empty p-orbital on the electron-deficient carbon. From a molecular orbital view, this essentially serves to lower the energy of the C-H sigma orbital and raise the energy of the p-orbital. With only two electrons in this system, the effect is stabilizing.

enter image description here

Notice that this means there are two effects that explain why greater substitution increases carbocation stability: inductive effect and hyperconjugation.

Hyperconjugation can also be invoked for the substitution of alkenes. In this case, the filled C-H sigma orbital overlaps with the empty C-C pi-antibonding orbital.

enter image description here

A good rule of thumb is that a filled orbital overlapping with an unfilled orbital is a stabilizing interaction.

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  • $\begingroup$ How does overlapping with an anti bonding orbital lend something to stability? $\endgroup$ – Dissenter Nov 3 '14 at 21:14
  • $\begingroup$ @Dissenter When two orbitals (of any type) are combined, two new orbitals are generated, one of lower energy and one of higher energy. If there are only two electrons available to fill the two new orbitals, they will go into an orbital that is lower in energy than either of the two originals. So if we overlap an empty anti-bonding orbital with a filled orbital,a higher energy anti-bonding orbital will result. But that orbital is empty anyway. The important part is that the electrons can be placed in a lower energy orbital. $\endgroup$ – jerepierre Nov 3 '14 at 21:21
  • $\begingroup$ @Dissenter Perhaps you are thinking of a situation where the end result places electrons in an anti-bonding orbital. That would be destabilizing. $\endgroup$ – jerepierre Nov 3 '14 at 21:22
  • $\begingroup$ @Dissenter Keep in mind that many reactions require that a filled orbital (i.e. nucleophile) overlap with an unfilled orbital (i.e. electrophile) which is often an anti-bonding orbital. $\endgroup$ – jerepierre Nov 3 '14 at 21:23
  • $\begingroup$ Right ... The c-c anti bonding orbital is only antibonding with respect to the alkene. $\endgroup$ – Dissenter Nov 3 '14 at 21:28

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