Basic Principles of Organic Chemistry

Ethane has a $p{K}_a$ near $50$ — hopelessly unacidic. Cyclopentadiene has a $p{K}_a$ of $16$, about the same as water. Why is one hydrocarbon $\text{C}-\text{H}$ bond thirty orders of magnitude more acidic than another?

Because of what is left behind.

Removing a proton from cyclopentadiene gives a planar, fully conjugated five-membered ring carrying six $\pi$ electrons — a Hückel number. The cyclopentadienyl anion is aromatic, and that stabilisation is worth an enormous amount. An ethyl anion gets nothing at all.

Every comparison in this chapter reduces to the same question: what stabilises the species produced? Acidity, basicity, carbocation stability, tautomer position and reaction rate are all answered by looking at the product rather than the reactant.

1. The four electronic effects

When inductive and resonance effects oppose, resonance almost always wins. Chlorine is inductively withdrawing yet still directs substitution to the ortho and para positions, because its lone pair donates by resonance to exactly those carbons.

Hyperconjugation is delocalisation of a $\sigma$ bonding pair into an adjacent empty or antibonding orbital, sometimes called no-bond resonance. Its effects are cumulative, so more alkyl groups mean more stabilisation — which is why tertiary carbocations and more substituted alkenes are the more stable.

Illustration 1

Explain why chlorobenzene is less reactive than benzene towards electrophiles yet still directs substitution to the ortho and para positions.

Chlorine is strongly electronegative, so its inductive withdrawal pulls density from the whole ring and deactivates it.

But chlorine also carries lone pairs, which donate by resonance specifically to the ortho and para carbons.

The inductive effect governs the overall rate, which falls; the resonance effect governs the position, which stays ortho and para.

The two effects answer different questions. A substituent can be deactivating and ortho-para directing at once, and the halogens are the standard case.

2. Aromaticity

A ring is aromatic only if it satisfies all four conditions: cyclic, planar, fully conjugated, and carrying $\left(4n+2\right)$ $\pi$ electrons.

Cyclooctatetraene is the instructive case. With eight $\pi$ electrons it would be antiaromatic if planar, so it escapes by puckering into a tub in which the double bonds no longer conjugate. Antiaromaticity is avoided at the cost of losing conjugation entirely.

Charged rings matter because gaining or losing a proton or an electron can move a ring into or out of a Hückel count — which is exactly what happens in the hook.

Illustration 2

Classify cyclopropene, the cyclopropenyl cation and the cyclopropenyl anion.

Cyclopropene has an $s{p}^3$ carbon, so the ring is not fully conjugated: non-aromatic.

The cation loses a hydride from that carbon, leaving an empty $p$ orbital that completes the conjugation. Two $\pi$ electrons is $4n+2$ with $n=0$: aromatic, and remarkably stable for so strained a ring.

The anion has four $\pi$ electrons, a $4n$ count: antiaromatic, and correspondingly very unstable.

One ring spans all three categories. Which applies depends entirely on the $\pi$ electron count, never on the ring size alone.

Illustration 3

Rank the three resonance contributors of the nitrite ion and of an amide by importance, stating the rule used in each case.

For nitrite, the two structures with one double and one single bond are equivalent and contribute equally. A structure with charge separation and an incomplete octet would contribute negligibly.

For an amide, the neutral structure with a $\text{C}=\text{O}$ double bond contributes most, since it has no charge separation.

The charge-separated structure with $\text{C}=\text{N}$ and a negative oxygen contributes substantially, because the negative charge sits on the more electronegative atom.

The rules in order: complete octets first, then least charge separation, then negative charge on the more electronegative atom. Applying them in that order settles almost every ranking question.

3. Acidity: look at the conjugate base

An acid is strong when its conjugate base is stable, and four things stabilise an anion: the electronegativity and size of the atom bearing the charge, resonance delocalisation, and inductive withdrawal nearby.

$\text{RCOOH}\left(p{K}_a\approx 5\right)>\text{ArOH}\left(10\right)>\text{ROH}\left(16\right)>\text{RC}\equiv \text{CH}\left(25\right)>\text{RH}\left(50\right)$

Carboxylic acids beat phenols because carboxylate spreads its charge over two equivalent oxygens, while phenoxide pushes it onto carbon atoms, which bear it far less comfortably.

The ortho effect is the exception worth memorising: every ortho-substituted benzoic acid is stronger than benzoic acid, whether the substituent donates or withdraws. The cause is steric — the substituent twists the carboxyl group out of the ring plane, breaking its conjugation and destabilising the acid more than the anion.

Illustration 4

Arrange in order of increasing acidity: phenol, $p$-nitrophenol, $p$-cresol, $2,4,6$-trinitrophenol.

$p$-Cresol carries an electron-donating methyl group, which destabilises the phenoxide: weakest.

Phenol is the reference at $p{K}_a$ $10$.

$p$-Nitrophenol has one strongly withdrawing group conjugated with the oxygen, stabilising the anion considerably.

Picric acid has three, and at $p{K}_a$ $0.4$ is stronger than most carboxylic acids.

Order: $p$-cresol $<$ phenol $<$ $p$-nitrophenol $<$ picric acid.

A nitro group at the para position acts by resonance as well as induction, which is why it is far more effective there than at the meta position, where only induction operates.

4. Basicity: the gas phase and water disagree

In the gas phase, amine basicity follows induction alone and rises steadily with substitution:

${\text{R}}_3\text{N}>{\text{R}}_2\text{NH}>{\text{RNH}}_2>{\text{NH}}_3$

In water the order is scrambled, with the secondary amine usually strongest:

${\text{R}}_2\text{NH}>{\text{R}}_3\text{N}\approx {\text{RNH}}_2>{\text{NH}}_3$

The reason is solvation. A protonated amine is stabilised by hydrogen bonds to water, and the more alkyl groups it carries, the fewer $\text{N}-\text{H}$ bonds remain to form them — and the more the water is crowded away sterically. Two opposing trends meet, and the secondary amine sits at the optimum.

Aromatic amines are far weaker: aniline's lone pair is delocalised into the ring and less available to a proton, so its $p{K}_b$ is about $9$ against roughly $4$ for an aliphatic amine.

Illustration 5

Explain why aniline is a much weaker base than cyclohexylamine, and why $p$-nitroaniline is weaker still.

In aniline the nitrogen lone pair is conjugated with the ring and delocalised over the ortho and para carbons, so it is much less available for protonation.

Cyclohexylamine has no such conjugation and its lone pair is fully available.

In $p$-nitroaniline the nitro group withdraws that delocalised density further by resonance, leaving the lone pair still less available.

Protonation destroys the conjugation altogether, so the cost of protonating aniline includes losing the resonance stabilisation the neutral molecule enjoyed.

5. Intermediates and rearrangement

$\text{carbocation stability}:\text{benzylic}\approx \text{allylic}>3^{\circ }>2^{\circ }>1^{\circ }>\text{methyl}>\text{vinyl}$

Benzylic and allylic cations are stabilised by resonance, the alkyl series by hyperconjugation and induction. A vinyl cation is exceptionally unstable, its positive charge sitting on an $sp$ carbon whose greater $s$ character holds electrons more tightly.

Carbanions run the opposite way, since alkyl groups destabilise a negative charge:

$\text{methyl}>1^{\circ }>2^{\circ }>3^{\circ }$

Free radicals follow the carbocation order, being electron deficient, but with smaller differences.

Rearrangement occurs whenever a $1,2$-shift of hydride or alkyl converts a cation into a more stable one, and it is the standard trap in any reaction proceeding through a carbocation.

Illustration 6

Predict the major product when $3,3$-dimethylbutan-2-ol is treated with concentrated sulphuric acid, and explain.

Protonation and loss of water gives a secondary carbocation at carbon two.

A methyl group migrates from carbon three, converting it into a tertiary cation at carbon three.

Loss of a proton from an adjacent carbon gives $2,3$-dimethylbut-2-ene, the more substituted and more stable alkene.

The carbon skeleton has changed. Any reaction going through a carbocation must be checked for rearrangement, which is precisely why such routes are avoided when the skeleton must be preserved.

6. Stereochemistry: counting and assigning

Conformations interconvert by rotation about single bonds and cannot be separated. In butane the anti arrangement is lowest, gauche about $3.8$ kJ mol${}^{-1}$ higher, and the eclipsed forms are maxima.

Configurations cannot interconvert without breaking bonds. A compound with $n$ stereocentres has at most $2^n$ stereoisomers, but fewer if a meso form exists, since a meso compound is superimposable on its own mirror image.

Geometrical isomerism needs restricted rotation and two different groups on each doubly bonded carbon. The $E$ and $Z$ labels use Cahn-Ingold-Prelog priorities and remain unambiguous where cis and trans do not.

Illustration 7

How many stereoisomers exist for tartaric acid, which has two stereocentres?

The upper bound is $2^2=4$.

But the two stereocentres carry identical substituents, so one arrangement has an internal mirror plane and is superimposable on its own mirror image.

That meso form is a single achiral compound, not a pair.

Total: three stereoisomers — one pair of enantiomers and one meso form.

Whenever two stereocentres bear the same four groups, expect a meso form and subtract. This is why $2^n$ is an upper bound rather than a count.

Illustration 8

Assign $E$ or $Z$ to the alkene $\text{CHCl}=\text{CHBr}$ in its two forms, and explain why cis and trans labels would be ambiguous for $\text{CFCl}=\text{CFBr}$.

On each carbon compare the two attached atoms by atomic number. Chlorine outranks hydrogen on one carbon, bromine outranks hydrogen on the other.

If chlorine and bromine lie on the same side the alkene is $Z$; on opposite sides, $E$.

For $\text{CFCl}=\text{CFBr}$ neither carbon carries a hydrogen, so there is no pair of identical groups to call cis or trans. The words simply do not apply.

Cahn-Ingold-Prelog priorities work in every case, which is why the $E$ and $Z$ system replaced the older labels entirely.

7. Tautomerism

Keto and enol tautomers are structural isomers in equilibrium, differing by the position of a proton and a double bond. Their proportions vary enormously:

Acetylacetone is heavily enolised because its enol is stabilised both by conjugation with the second carbonyl and by an intramolecular hydrogen bond closing a six-membered ring. Phenol exists entirely as the enol because the keto form would destroy the ring's aromaticity.

Tautomerism requires an $\alpha$-hydrogen. A ketone with none, such as benzophenone, cannot enolise at all.

Illustration 9

Explain why the enol content of acetylacetone falls sharply in water but is high in hexane.

The enol's stability comes largely from an internal hydrogen bond between its hydroxyl and the second carbonyl.

In hexane nothing competes for that bond, so it forms and the enol is strongly favoured.

In water the solvent hydrogen bonds to both forms, removing the enol's special advantage and stabilising the more polar keto form instead.

Tautomer ratios are solvent-dependent in general, which is why a quoted enol percentage must always specify the medium.

Illustration 10

Arrange in order of increasing acidity: ethanol, phenol, acetic acid, cyclopentadiene.

Ethanol at $p{K}_a$ $16$ has an alkoxide stabilised only by oxygen's electronegativity.

Cyclopentadiene is comparable at $16$, despite being a hydrocarbon, because its anion is aromatic.

Phenol at $10$ delocalises its charge over the ring.

Acetic acid at $4.8$ shares its charge between two equivalent oxygens.

Order: ethanol $\approx$ cyclopentadiene $<$ phenol $<$ acetic acid.

Aromatic stabilisation is worth about as much as an oxygen atom here, which is a striking measure of how large the effect is.

Illustration 11

Explain why $o$-toluic acid is stronger than benzoic acid, although a methyl group is electron donating.

An electron-donating group should destabilise the carboxylate and weaken the acid, and at the para position that is exactly what happens.

At the ortho position the methyl group crowds the carboxyl group and twists it out of the ring plane.

That breaks the conjugation between carboxyl and ring, destabilising the acid more than its anion, so the acid becomes stronger.

This is the ortho effect, and it operates for every ortho substituent regardless of electronic character — which is what makes it worth remembering as a rule rather than deriving each time.

Illustration 12

Rank these carbocations by stability: ${\text{CH}}_3^{+}$, ${\text{CH}}_2=\text{CH}-{\text{CH}}_2^{+}$, ${\left({\text{CH}}_3\right)}_3{\text{C}}^{+}$, ${\text{C}}_6{\text{H}}_5{\text{CH}}_2^{+}$.

Benzylic is most stable, its charge delocalised over the whole aromatic ring.

Allylic is next, delocalised over three carbons.

Tertiary follows, stabilised by hyperconjugation from nine $\text{C}-\text{H}$ bonds and by induction.

Methyl is least stable, with nothing at all.

Order: ${\text{CH}}_3^{+}<{\left({\text{CH}}_3\right)}_3{\text{C}}^{+}<{\text{CH}}_2=\text{CH}-{\text{CH}}_2^{+}<{\text{C}}_6{\text{H}}_5{\text{CH}}_2^{+}$

Resonance beats hyperconjugation, which is why even a primary benzylic cation outranks a tertiary alkyl one.

Illustration 13

Explain why trichloroacetic acid ($p{K}_a$ $0.7$) is far stronger than acetic acid ($4.8$), while $4$-chlorobutanoic acid is barely stronger than butanoic acid.

Three chlorines on the $\alpha$ carbon withdraw electron density inductively and strongly stabilise the carboxylate, raising the acidity by four orders of magnitude.

The inductive effect falls off sharply with distance, roughly by a factor of three per bond.

In $4$-chlorobutanoic acid the chlorine is three bonds from the carboxyl group, so almost nothing reaches it.

Induction dies out over three or four bonds, which is why the position of a substituent matters as much as its identity.

Summary

• Every comparison here asks what stabilises the product: the conjugate base, the cation, the enol, the transition state.
• Cyclopentadiene has $p{K}_a$ $16$ because its anion is aromatic; ethane has $50$ because its anion gains nothing.
• Resonance usually beats induction where they conflict, so chlorine deactivates the ring yet still directs ortho and para.
• Hyperconjugation is cumulative: more alkyl groups mean more stabilisation of cations and of alkenes.
• Aromatic needs all four: cyclic, planar, fully conjugated, $\left(4n+2\right)$ $\pi$ electrons.
• Cyclobutadiene is antiaromatic; cyclooctatetraene escapes by puckering into a non-planar tub.
• Resonance ranking: complete octets first, then least charge separation, then negative charge on the more electronegative atom.
• $E$ and $Z$ from Cahn-Ingold-Prelog priorities work where cis and trans have no meaning at all.
• Acidity: $\text{RCOOH}>\text{ArOH}>\text{ROH}>\text{RC}\equiv \text{CH}>\text{RH}$.
• The ortho effect strengthens every ortho-substituted benzoic acid, whatever the substituent, by breaking conjugation sterically.
• Induction dies out over three to four bonds, so substituent position matters as much as identity.
• Amine basicity rises with substitution in the gas phase; in water the secondary amine usually wins, through solvation.
• Carbocations: benzylic $\approx$ allylic $>3^{\circ }>2^{\circ }>1^{\circ }>$ methyl $>$ vinyl. Carbanions run the opposite way.
• Any carbocation reaction must be checked for a $1,2$-shift to a more stable cation.
• $2^n$ stereoisomers is an upper bound; a meso form reduces it, as for tartaric acid's three.
• Enol content runs from $10^{-4}\%$ for acetone to $80\%$ for acetylacetone to $100\%$ for phenol, and depends on solvent.