Organic Chemistry – Some Basic Principles and Techniques · Lesson 8 of 15
Inductive and Resonance Effects
“Explain how attached groups and delocalised electrons shape charge distribution and stability.”
• Explain inductive polarisation and its weakening with distance. • Distinguish electron-withdrawing and electron-donating effects. • Write valid resonance contributors without moving nuclei. • Compare contributor stability and interpret positive and negative resonance effects.
The whole neighbourhood affects a bond
A reactive centre does not exist in isolation. Atoms attached nearby can pull electron density towards themselves or supply it towards another part of a molecule. Some effects are already present in the ground-state molecule; others appear when a reagent approaches. Inductive and resonance effects are permanent features of the electronic distribution and help explain why apparently similar functional groups behave differently.
Electronegativity measures how strongly a bonded atom attracts shared electrons. In a polar bond, the attraction is unequal but the bond has not necessarily split into ions. Partial charges are written delta plus and delta minus. They describe an uneven distribution of a shared cloud, rather than whole electrons being removed from an atom.
Inductive effects act through sigma bonds
In chloroethane, , chlorine attracts the C–Cl bonding electrons, making chlorine partially negative and the attached carbon partially positive. That carbon then draws some density from its neighbouring C–C bond. This induced polarisation spreads through sigma bonds but weakens rapidly with distance; it is usually very small after several intervening bonds.
Permanent displacement of electron density along sigma bonds caused by an electronegative atom, charged centre, or substituent. Its influence decreases rapidly with distance.
Relative to hydrogen, a group that draws density through sigma bonds has a negative inductive effect, written −I; an alkyl group commonly supplies density and is described as +I. Halogens, nitro, cyano, carboxyl, ester, and aryloxy groups illustrate withdrawing groups in the chapter. Methyl and ethyl illustrate donating groups. The sign labels the direction of the effect, not the electrical charge of the entire molecule.
Problem
Compare the C–H and C–Br bonds, then identify the less affected C–C bond in .
- 1.Bromine attracts bonding electrons more strongly than hydrogen, so the C–Br bond is more polar than the corresponding C–H bond. Similarly, C–O is more polar than C–N or C–S in the comparisons here.
- 2.Number from the bromine-bearing carbon. The C1–C2 bond is nearer bromine than C2–C3, so C2–C3 is less affected by induction.
- 3.When comparing only C–C bonds, compare those two bonds. A distant C–H bond is a different comparison and should not be substituted for a C–C bond in the answer.
Induction also helps compare stability and acidity. An electron-withdrawing group can stabilise a nearby negative charge by reducing its concentration. For related carboxylic acids, withdrawing groups tend to stabilise the conjugate base and increase acidity; alkyl donation tends to oppose this. More withdrawing substituents or a shorter intervening distance generally strengthen the effect in comparable molecules.
Why a single Lewis structure can be inadequate
Benzene is often drawn with alternating single and double bonds, yet all six carbon–carbon bond lengths are equal, about 139 pm. This is intermediate between representative single and double values of 154 and 134 pm. The experimental structure is therefore not adequately described by one drawing containing three permanently short and three permanently long bonds.
Resonance uses several Lewis contributors with the same nuclear positions but different electron placements. The actual species is one resonance hybrid with delocalised electrons. It does not jump back and forth between contributors. Nitromethane similarly has equivalent N–O bonds even though a single Lewis contributor draws one N=O and one N–O bond.
The energy difference between the actual resonance hybrid and its lowest-energy individual contributing structure. The hybrid is lower in energy than any single contributor.
To draw valid contributors, keep every nucleus fixed, keep the total charge and electron count unchanged, and do not change the number of unpaired electrons. Move only lone pairs or pi electrons through a conjugated arrangement. A contributor with more covalent bonds, complete octets, less charge separation, and sensible charge placement generally contributes more. Equivalent contributors have equal importance.
Problem
Write the contributors for .
- 1.Draw one oxygen double-bonded to the central carbon and the other single-bonded with a negative charge. Complete the oxygen lone pairs.
- 2.Draw a full curved arrow from a lone pair on the negative oxygen into its C–O bond, and another from the original C=O pi bond onto the other oxygen. Move both pairs together so carbon never exceeds an octet.
- 3.The result swaps the two oxygen roles without moving either atom. Both contributors are equivalent, and the real ion has equivalent C–O bonds with negative charge distributed over the two oxygens.
Problem
Compare resonance contributors of .
- 1.The neutral contributor has C=C and C=O bonds with full octets and no formal charge separation. It is the major contributor.
- 2.A contributor with oxygen negative and the distant carbon positive is less important because it separates charge and leaves a carbon electron deficient, but it places negative charge on an electronegative atom.
- 3.A contributor placing oxygen positive and carbon negative is still less favourable. The comparison uses octets, bonds, and charge placement together, not only a count of drawings.
For methyl ethanoate, , charge-separated contributors are less important than the ordinary neutral ester structure. A contributor that additionally leaves the carbonyl carbon with an incomplete octet is particularly unfavourable. A valid drawing is not automatically a major contributor.
Resonance can donate or withdraw electron density
A conjugated system has adjacent orbitals capable of continuous overlap, often indicated by alternating single and double bonds or a lone pair next to a pi bond. Resonance or mesomeric effects describe the polarity produced by interaction between pi bonds or between a lone pair and a pi bond. Unlike induction, this requires a suitable orbital pathway.
In a +R effect, a substituent donates a lone pair into the conjugated system. Aniline illustrates nitrogen donation into the benzene ring, increasing electron density at particular ring positions. In a −R effect, the substituent draws electron density from the system. Nitrobenzene illustrates withdrawal into the nitro group. Track the curved arrows to identify the electron-rich and electron-poor positions.
| Resonance behaviour | Representative groups | Meaning |
|---|---|---|
| +R | ; halogens | Donation into an adjacent conjugated system |
| +R examples also listed | An attached atom can donate a lone pair | |
| −R | and carbonyl groups | Withdrawal from the conjugated system |
Resonance contributors are not different isomers or rapidly interconverting molecules. Only electrons move between the drawings. A group can show more than one effect: a halogen can withdraw inductively while donating by resonance, so the signs of different effects need not match.
Quiz
Which effect decreases rapidly with distance through sigma bonds?
Which operation is forbidden when drawing resonance contributors?
What best describes the real acetate ion?
A lone pair donated from amino nitrogen into a benzene ring illustrates which effect?
Which contributor is usually favoured, all else comparable?
Practice Problems
- Explain the relative stability of and using induction.
- Use induction to explain the acidity order .
- Draw both acetate contributors with lone pairs and full curved arrows.
- Draw resonance contributors for nitromethane and explain why the N–O bonds are equivalent.
- Draw electron shifts for aniline and nitrobenzene and identify donation or withdrawal from the ring.
- Compare contributors of and using octets and charge separation.
Key Takeaways
• Induction acts through sigma bonds and weakens with distance. • Resonance requires a connected orbital pathway and moves electrons without moving nuclei. • The actual structure is one stabilised hybrid, not a mixture of contributors. • Inductive and resonance effects may act in different directions in the same molecule.