Organic Chemistry – Some Basic Principles and Techniques · Lesson 9 of 15
Electromeric Effect, Hyperconjugation, and Reaction Patterns
“Connect reagent-induced electron shifts and sigma-electron delocalisation to stability and broad reaction outcomes.”
• Distinguish temporary electromeric shifts from permanent electron effects. • Explain positive and negative electromeric effects using attacking species. • Relate hyperconjugation to adjacent C–H bonds and carbocation stability. • Recognise substitution, addition, elimination, and rearrangement patterns.
An approaching reagent can redirect a pi pair
Inductive polarisation and resonance describe features already associated with a molecular structure. The electromeric effect describes a complete shift of a pi pair in a multiple bond under the influence of an attacking reagent. It is a temporary response: without the approaching reagent, the induced displacement described by this model is not maintained. The concept focuses attention on where a pi pair goes during the interaction.
A reagent-induced complete displacement of a shared pi-electron pair towards one atom of a multiple bond. It is represented with a full curved arrow and is temporary.
In a positive electromeric effect, +E, the pi pair moves towards the atom that bonds to the attacking reagent. When a proton approaches an alkene, the pi pair can form a new carbon–hydrogen bond. One carbon gains the proton and the other is left electron deficient. The arrow begins at the pi bond because that pair supplies the new bond; it does not begin at the proton.
In a negative electromeric effect, −E, the pi pair moves towards the atom to which the attacking reagent does not attach. When cyanide attacks a carbonyl carbon, the C=O pi pair moves to oxygen while the new C–C bond forms at carbon. Carbon does not retain both its original full carbonyl bond and an extra bond: that would give it too many valence electrons.
Problem
Follow the electron changes when attacks the carbonyl carbon of a ketone.
- 1.Cyanide donates an available electron pair to the electrophilic carbonyl carbon, forming a new bond.
- 2.At the same time, the C=O pi pair shifts to oxygen. Oxygen receives negative charge in the immediate addition intermediate.
- 3.The attacking reagent bonds to carbon while the pi pair moves to oxygen, so this is the −E pattern. Subsequent proton transfer can convert the oxygen anion to OH.
| Effect | Present without attacking reagent? | Main electronic pathway |
|---|---|---|
| Inductive | Yes | Polarisation through sigma bonds |
| Resonance | Yes, for a suitable conjugated structure | Delocalisation through adjacent orbitals |
| Electromeric | No; induced during attack | Complete displacement of a pi pair |
| Hyperconjugation | Yes, when orbitals are suitably arranged | Adjacent sigma-bond donation into an orbital system |
If inductive and electromeric displacements are described as opposing during an attack, the complete pi-pair displacement dominates that immediate bond-change description. This is not a rule that induction becomes irrelevant to every property. Always state whether you are comparing a ground-state distribution or following a reacting electron pair.
Sigma electrons can also delocalise
Hyperconjugation is a stabilising interaction in which an adjacent sigma bond, commonly C–H, overlaps with an empty or partly occupied p orbital or a pi system. It requires appropriate neighbouring bonds and alignment. In an ethyl carbocation, the positively charged carbon has an empty p orbital. A neighbouring methyl C–H bond can overlap with it and supply electron density, spreading the positive charge.
The effect is permanent in a suitable molecular arrangement. Alternative drawings sometimes express it as “no-bond resonance”, but they are a model for electron delocalisation rather than evidence that hydrogen ions have physically detached. Hyperconjugation is also possible in alkenes such as propene and in alkylarenes, where an adjacent C–H sigma bond interacts with the unsaturated system.
Problem
Why is tert-butyl carbocation more stable than ethyl carbocation, and methyl carbocation least stable in this comparison?
- 1.The tert-butyl cation has three adjacent methyl groups and nine neighbouring C–H bonds capable of contributing hyperconjugative interactions as appropriate conformations are considered.
- 2.Ethyl cation has one adjacent methyl group and three such C–H bonds. Alkyl induction also helps stabilise the positive centre.
- 3.Methyl cation has no adjacent carbon supplying these C–H bonds. The simple alkyl order is tertiary > secondary > primary > methyl; resonance or other structural features require a separate comparison.
Look at the net change to recognise a reaction pattern
A mechanism describes the sequence of steps, whereas a reaction category describes the broad change. In substitution, one group replaces another. In addition, atoms or groups add across a multiple bond. In elimination, groups leave and a multiple bond is created. In rearrangement, the atom connectivity changes within the framework, sometimes during a reaction whose full equation also includes substitution or another change.
Problem
Classify bromide replacement by HS⁻, HCl addition to an alkene, and removal of HBr from a bromoalkane.
- 1.When HS⁻ replaces bromide on a saturated carbon, one bond is exchanged for another: substitution.
- 2.When H and Cl attach to the two carbons of an alkene, the pi bond is consumed and new sigma bonds form: addition.
- 3.When hydrogen and bromide are removed from neighbouring carbons and C=C forms, the pattern is elimination. If the carbon skeleton migrates during a conversion, identify the rearrangement as well.
Hyperconjugation does not mean that all adjacent C–H bonds overlap maximally at one instant. Orbital alignment matters. Also, a reaction type does not by itself specify the detailed mechanism, reagent conditions, or every intermediate.
Quiz
Which effect is temporary and induced by an attacking reagent?
During cyanide attack on a carbonyl carbon, where does the C=O pi pair move?
What interaction is central to hyperconjugation in an ethyl carbocation?
Converting an alkene into a product bearing H and Cl on its two former double-bond carbons is which pattern?
Which change is characteristic of elimination?
Practice Problems
- Draw curved arrows for proton attack on an alkene and identify the +E displacement.
- Draw curved arrows for cyanide attack on a carbonyl group and identify the −E displacement.
- Sketch an adjacent C–H bond and empty p orbital in ethyl carbocation and explain the stabilising overlap.
- Explain why propene can show hyperconjugation while an isolated saturated molecule without an adjacent accepting orbital is a different situation.
- Classify one substitution, one addition, one elimination, and one rearrangement example by the bonds changed.
- Compare the meanings of a resonance contributor, a reactive intermediate, and a reaction product.
Key Takeaways
• Electromeric displacement follows a pi pair during reagent attack. • Hyperconjugation spreads electron density through suitable adjacent orbital overlap. • Simple alkyl carbocation stability reflects hyperconjugation and induction. • Reaction patterns describe net bond changes; mechanisms explain their sequence.