Chemistry · Lesson
Carbon bonding, functional groups and structural families
Use bonding patterns and functional groups to organize carbon structures. Treat resonance as electron bookkeeping for one entity, and use degree of unsaturation as a formula constraint rather than a complete structure.
Back to the Chemistry mapBy the end, you can
- Count σ and π contributions in selected elementary bond models.
- Distinguish carbonyl families by the atoms bonded to their carbonyl carbon.
- Separate resonance, proton loss and constitutional changes.
- Calculate degree of unsaturation within its ordinary-valence scope.
A multiple bond still joins one pair of atoms
In an elementary localized two-center description, each bonded atom pair supplies one σ contribution. A double bond additionally supplies one π contribution, and a triple bond two. Include C–H bonds when counting the whole molecule: counting only the heavy-atom skeleton misses much of its σ bonding.
The familiar sp³, sp² and sp hybridization descriptions help explain selected tetrahedral, trigonal and linear local environments. They are orbital models rather than a claim that every real carbon environment has exactly one ideal angle. Benzene has a delocalized π system; a Kekulé drawing with three double bonds is a useful electron-counting representation, not a sequence of three isolated measured π bonds.
| Structure | σ contributions | π contributions | Formula DBE |
|---|---|---|---|
| Ethane, C₂H₆ | 7 | 0 | 0 |
| Ethene, C₂H₄ | 5 | 1 | 1 |
| Ethyne, C₂H₂ | 3 | 2 | 2 |
| Benzene, C₆H₆; Kekulé bookkeeping | 12 | 3 | 4 |
Exact integer bookkeeping for the named neutral structures. σ counts include C–H bonds. Benzene’s Kekulé π count is a representation of a delocalized six-π-electron system; these entries are not measured localized bond populations or ideal-angle data.
Inspect the neighbours of a characteristic group
A functional group is a recognizable bonding motif useful for organizing chemical behaviour. An alkene contains a C=C motif, an alkyne a C≡C motif, an alcohol an OH bonded to a saturated carbon in these simple examples, and an ether an oxygen between carbon groups. A molecule can contain several groups whose behaviour influences one another.
For a carbonyl, first identify C=O and then inspect what else joins that carbon. An aldehyde carbonyl has an attached H; a ketone carbonyl has two carbon attachments. A carboxylic acid has the OH motif at its carbonyl carbon, an ester has OR, and an amide has nitrogen there. Propanal CH₃CH₂CHO and propanone CH₃COCH₃ therefore share C₃H₆O but belong to different constitutional families. The motif is a starting point; charge, conjugation, steric environment and conditions also matter.
Contributors change an electron description without moving nuclei
Resonance contributors preserve the labeled atom framework, attached hydrogens and total electron count. They relocate bond-order and nonbonding-electron bookkeeping. For acetate, CH₃CO₂⁻, the two standard contributors exchange which C–O bond is drawn double and which oxygen bears formal negative charge. The entity has net −1 charge in either representation.
The real electronic structure is not a mixture that spends a measured fraction of time in each contributor. Contributors need not contribute equally; symmetry relates the two equivalent acetate oxygens in the ideal isolated-ion description. Removing the acid proton from CH₃CO₂H changes composition and charge, so it is an acid–base process rather than resonance. Formal charge on a contributor also differs from a measured or calculated partial-charge distribution.
A formula can constrain rings and multiple bonds
For connected neutral ordinary-valence closed-shell formulas with C, H, N, O and halogens, degree of unsaturation is DBE=(2C+2+N−H−X)/2, where the symbols here denote atom counts and X is total halogens. Oxygen does not change this count. A ring or double-bond unit adds one DBE; a triple bond adds two.
DBE reports a hydrogen-deficiency constraint, not the position or identity of every bond. For C₄H₈, DBE=1 allows an alkene or a ring in the stated ordinary examples. Charges, radicals and unusual valences require further analysis rather than blindly applying this neutral formula. A fractional or negative result prompts a check of formula and assumptions; it does not license inventing a plausible drawing.
Worked examples
Include the hydrogens in an ethyne count
Count σ and π contributions for H–C≡C–H in the stated localized model.
- The two C–H atom pairs give two σ contributions.
- The C–C atom pair gives one further σ contribution.
- Its triple-bond representation adds two π contributions.
Three σ and two π contributions; the triple bond does not count as three σ bonds.
Check a nitrile formula against its motif
For neutral ethanenitrile CH₃C≡N, calculate DBE from C₂H₃N.
- Use C=2, H=3, N=1 and X=0.
- DBE=(4+2+1−3)/2=2.
- The depicted C≡N triple bond supplies those two units, with no ring.
DBE=2, consistent with this nitrile graph. The formula constraint alone would not locate the triple bond.
Retain the electrons while changing a contributor
Compare the two standard acetate contributors and neutral ethanoic acid.
- Each acetate contributor has C₂H₃O₂ and net charge −1.
- Its total valence electrons are 2×4+3×1+2×6+1=24.
- Neutral C₂H₄O₂ also has 24 valence electrons, but has one more nucleus and a different net charge.
- Resonance requires the same nuclei and total electrons, not just equal electron totals.
The acetate drawings are contributors of one ion. The acid is a different composition and charge state.
Try it yourself
In CH₃C(=O)NH₂, which selected carbonyl family is present?
Show the answer and reasoning
An amide.
The carbonyl carbon directly joins nitrogen.
Are CH₃CH₂CHO and CH₃COCH₃ two resonance contributors?
Show the answer and reasoning
No; they are constitutional isomers.
Their atom connectivity differs, despite the same formula C₃H₆O.
Find DBE for C₆H₆ and account for it in a Kekulé ring.
Show the answer and reasoning
Four: one ring plus three double-bond units.
The arithmetic is (12+2−6)/2=4; the π electrons remain delocalized in the real aromatic entity.
Can DBE=1 by itself distinguish an alkene from a saturated ring?
Show the answer and reasoning
No.
Both supply one degree of unsaturation under the declared neutral ordinary-valence assumptions.
What to carry forward
- Include C–H bonds in whole-molecule σ counts.
- Read the immediate connectivity of each functional group.
- Resonance preserves nuclei and total electrons.
- Use DBE to constrain a proposed structure, not to replace structural evidence.
References
Teaching resources and reference material supporting this lesson’s explanations and examples.