Difficulty: Intermediate | Prerequisites: Lewis structures, electronegativity, orbital basics
Resonance structures show how electrons (especially lone pairs) can be delocalised across a molecule, which directly affects stability, acidity and basicity. Acids are ranked by how well their conjugate base is stabilised after losing a proton; bases are ranked by how available the lone pair is to accept one. The three big stabilising factors to keep straight are resonance, electronegativity and inductive effects.
Resonance structures
Alternative Lewis structures for the same molecule that differ only in the placement of electrons (not atoms). Think of them as snapshots of where electrons could be; the real molecule is a blend of all of them.
Resonance stabilisation
The extra stability a molecule gains when its electrons are delocalised across multiple atoms rather than pinned to one. In simple terms, spreading electrons out lowers energy and makes the species more stable.
Conjugate acid / conjugate base
When a base accepts a proton it becomes the conjugate acid; when an acid donates a proton the leftover species is the conjugate base. Every acid-base reaction has two conjugate pairs.
pKa
A numerical measure of acid strength. Lower pKa = stronger acid, meaning it gives up its proton more readily. The scale is logarithmic, so a difference of one unit is a tenfold difference in acid strength.
Electronegativity
The tendency of an atom to pull electron density towards itself in a bond. More electronegative atoms stabilise negative charge better, which matters when you compare conjugate bases.
Inductive effect
The pull (or push) of electron density through sigma bonds caused by electronegative or electropositive substituents. In simple terms, electron-withdrawing groups tug electrons towards themselves along the carbon chain, stabilising nearby negative charges.
Electron delocalization
The spreading of electrons (typically a lone pair or pi electrons) across multiple atoms via overlapping p orbitals. This is the physical reality that resonance structures represent on paper.
Aniline has a nitrogen lone pair sitting in a p orbital that overlaps with the aromatic ring's pi system. This lets the lone pair delocalise into the ring, which you show by drawing four resonance structures:
Structure 1: The lone pair on nitrogen pushes into the C-N bond, creating a double bond between N and the ring carbon. The ring now has a negative charge at the ortho position.
Structure 2: The negative charge moves to the other ortho position (the one across the ring from the first).
Structure 3: The negative charge sits at the para position.
Structure 4: A fourth structure places the charge at the remaining ortho carbon.
In each case, nitrogen carries a formal positive charge (it has donated its lone pair into the ring), and one ring carbon carries a formal negative charge.
This delocalisation is why aniline is a weaker base than cyclohexylamine: the nitrogen lone pair is partially tied up in the ring rather than fully available to accept a proton.
Basicity measures how readily a species donates its lone pair to a proton. The more available the lone pair, the stronger the base.
Factors that decrease basicity (make the lone pair less available):
Resonance with a carbonyl or aromatic ring. If the lone pair is delocalised into a pi system, it is less available. Amides (RC(O)NH2) are far less basic than amines because the lone pair is delocalised into the C=O.
Increased electronegativity of the atom bearing the lone pair. Oxygen is more electronegative than nitrogen, so it holds its electrons more tightly and is less willing to donate them.
Positive charge. A species already bearing a negative charge (such as CH3NH-) is more basic than its neutral counterpart because it has more electron density to share.
Worked example from the exam:
Rank in order of increasing basicity: CH3NH- (A), PhNH2 (B), CH3O- (C), PhNH2 with no charge (D), CH3C(O)NH2 (E).
Answer: C < E < B < D < A
CH3O- (C): oxygen is more electronegative than nitrogen, making the alkoxide a weaker base than nitrogen bases (but note it carries a negative charge, so context matters; here it is compared against nitrogen species where the lone pair availability on N wins overall).
CH3C(O)NH2 (E): the amide. The nitrogen lone pair is delocalised into the carbonyl, dramatically reducing basicity.
PhNH2 (B): aniline. The nitrogen lone pair delocalises into the aromatic ring, reducing basicity compared to a simple alkylamine.
PhNH2 (D): same as B but with different substituent context in the exam figure. Where the ring is electron-donating rather than withdrawing, basicity rises.
CH3NH- (A): the methylamide anion carries a full negative charge, making it extremely basic.
Acidity measures how easily a compound loses a proton. The key question is always: how stable is the conjugate base that forms?
Factors that increase acidity (stabilise the conjugate base):
Resonance stabilisation of the conjugate base. Carboxylate anions (RCO2-) spread the negative charge over two oxygens, making carboxylic acids much more acidic than alcohols.
Electronegativity. The more electronegative the atom bearing the negative charge, the more stable the conjugate base. HCl is a strong acid partly because chlorine happily bears the negative charge.
Inductive effects. Electron-withdrawing groups near the acidic proton stabilise the conjugate base. CF3CO2H is more acidic than CH3CO2H because the three fluorines pull electron density away from the carboxylate.
Element effects (size and electronegativity down a group). Larger atoms stabilise negative charge better. HCl is more acidic than HF despite fluorine being more electronegative, because chloride is much larger and distributes the charge over a bigger volume.
Worked example from the exam:
Rank in order of increasing acidity: CH3CO2H (A), H2O (B), PhOH (C), CF3CO2H (D), HCl (E), CH3OH (F).
Answer: F < B < C < A < D < E
CH3OH (F): an alkyl alcohol. No special stabilisation of the conjugate base (methoxide). Least acidic here.
H2O (B): slightly more acidic than methanol because the methyl group is weakly electron-donating, destabilising the conjugate base in methanol relative to hydroxide.
PhOH (C): phenol. The phenoxide conjugate base is resonance-stabilised by the aromatic ring, making phenol notably more acidic than simple alcohols.
CH3CO2H (A): acetic acid. The carboxylate conjugate base has two equivalent resonance structures spreading the charge over two oxygens.
CF3CO2H (D): trifluoroacetic acid. The three fluorines withdraw electron density through induction, further stabilising the carboxylate.
HCl (E): a strong acid. Chloride is large and electronegative, making it a very stable conjugate base.
Students often assume that a negative charge automatically makes a species a strong base. It does not. What matters is how stable that charge is. Chloride has a negative charge but is a terrible base because the charge is very stable on the large, electronegative chlorine atom.
Students frequently confuse acidity with basicity. A strong acid has a weak conjugate base, and vice versa. If you are asked to rank basicity, think about the lone pair's availability; if acidity, think about conjugate base stability.
Drawing resonance structures does not mean the molecule flips between them. The molecule exists as one structure (the resonance hybrid) at all times. Resonance structures are a bookkeeping tool, not a description of molecular motion.
Students often forget that resonance delocalisation of a lone pair reduces basicity. If you see an amine attached to a carbonyl or an aromatic ring, that nitrogen is a weaker base than a simple alkylamine.
⚠️ Ranking acidity and basicity questions appear on nearly every organic chemistry exam. They test whether you understand the underlying principles, not just memorised pKa values.
⚠️ You will almost certainly be asked to draw resonance structures for a molecule with a heteroatom (N, O) bonded to a pi system. Aniline is the classic example. Be sure you can draw all four structures showing the nitrogen lone pair moving into the ring at ortho and para positions.
⚠️ The inductive effect of halogens (especially fluorine) on carboxylic acid acidity is a favourite exam comparison. Know that CF3CO2H >> CH3CO2H in acidity.
⚠️ Amide vs. amine basicity is a commonly tested contrast. The amide lone pair is delocalised into the carbonyl, making amides far less basic than amines.
True or false: Aniline is a stronger base than cyclohexylamine. (False: the nitrogen lone pair delocalises into the aromatic ring, reducing basicity.)
Fill in the blank: A lower pKa value means a ______ acid. (Stronger.)
True or false: The conjugate base of a strong acid is a strong base. (False: the conjugate base of a strong acid is a weak base.)
Fill in the blank: CF3CO2H is more acidic than CH3CO2H because of the ______ effect of the fluorine atoms. (Inductive / electron-withdrawing.)
True or false: Resonance structures show a molecule switching between different arrangements of its atoms. (False: only the electrons move; atoms stay put.)
Q: Draw four resonance structures for aniline (PhNH2) that show the nitrogen lone pair delocalised into the aromatic ring. What charge does nitrogen carry in each?
A: In each of the four structures, nitrogen carries a formal +1 charge (it has "donated" its lone pair into a pi bond with the ring carbon). The negative charge appears at the ortho position (two structures, one for each ortho carbon) and the para position (one structure). A fourth structure places the charge at the remaining ortho carbon. In every case, the nitrogen lone pair forms a double bond to the adjacent ring carbon and one ring carbon bears a formal -1 charge.
Q: Rank the following in order of increasing basicity: CH3NH- , PhNH2, CH3O- , CH3C(O)NH2. Explain your reasoning.
A: CH3O- < CH3C(O)NH2 < PhNH2 < CH3NH-. Methoxide is less basic than the nitrogen bases because oxygen is more electronegative and holds its electrons more tightly. The amide (CH3C(O)NH2) has its nitrogen lone pair delocalised into the carbonyl. Aniline (PhNH2) has its lone pair partially delocalised into the ring but less so than the amide. CH3NH- carries a full negative charge, making it the strongest base.
Q: Why is phenol (PhOH) significantly more acidic than cyclohexanol (C6H11OH)?
A: The conjugate base of phenol (phenoxide, PhO-) is resonance-stabilised: the negative charge delocalises into the aromatic ring across multiple resonance structures. The conjugate base of cyclohexanol has no such resonance stabilisation, so losing the proton costs more energy.
Q: Place these in order of increasing acidity: CH3OH, H2O, HCl, CH3CO2H. Explain the key factor for each step in the ranking.
A: CH3OH < H2O < CH3CO2H < HCl. Methanol vs water: the methyl group is weakly electron-donating, slightly destabilising the conjugate base. Water vs acetic acid: the carboxylate conjugate base has two-oxygen resonance stabilisation. Acetic acid vs HCl: chloride is a very stable anion due to chlorine's size and electronegativity.
Resonance and acid-base principles feed directly into reaction mechanisms. Whether a nucleophile attacks a carbonyl or an electrophile adds to a double bond, the driving force often comes down to electron availability and conjugate-base stability. Understanding why aniline is less basic than cyclohexylamine also explains why aniline is less nucleophilic, which matters for electrophilic aromatic substitution and nucleophilic substitution reactions.
Stereochemistry connects here too: the planar sp2 nitrogen in an amide (due to resonance) vs. the pyramidal sp3 nitrogen in an amine affects the geometry and reactivity of the molecule.
Resonance structures, electron delocalization, conjugate acid, conjugate base, pKa, pKb, Bronsted-Lowry acid, Bronsted-Lowry base, Lewis acid, Lewis base, inductive effect, electronegativity, resonance stabilization, aniline basicity, phenol acidity, carboxylic acid acidity, amide basicity, trifluoroacetic acid, acid strength ranking, base strength ranking, organic chemistry acid-base, CHEM 2700