Basic Principles of Organic Chemistry (GOC)

Chlorine on a benzene ring. One substituent, two questions.

Does it make the ring react faster or slower? Chlorine is electronegative, so it pulls electron density out of the ring by induction, leaving the ring poorer and less attractive to an incoming electrophile. Slower. Chlorobenzene nitrates about 30 times more slowly than benzene.

Where does the incoming group go? Chlorine has lone pairs it can push into the ring by resonance, and that donation lands specifically at the ortho and para positions. Ortho and para. Chlorobenzene nitrates to give about 30 per cent ortho and 70 per cent para, with almost no meta.

So chlorine is a deactivating, ortho-para directing group: it slows the reaction down and still steers the product to the positions it enriched.

Almost every student meets that as a contradiction to be memorised. It is not a contradiction at all.

$-I$ and $+R$ are two separate accounts of what chlorine does, and they are allowed to point opposite ways because they are answering different questions. Induction wins on rate because it operates on the whole ring; resonance wins on position because it is the only effect that has a position.

That is the discipline this chapter teaches. Name the question before reaching for an effect.

Everything else follows from two more questions that run through the whole of organic chemistry.

Nearly every organic reaction is something electron-rich attacking something electron-poor. This chapter carries more weight than its mark count suggests, because every organic chapter after it depends on it, and a candidate who memorised named reactions without GOC will find that almost nothing transfers.

1. Tetravalency and Hybridisation

The quick rule is to count sigma bonds plus lone pairs: four gives $s{p}^3$, three gives $s{p}^2$, two gives $sp$.

What s character does

An s orbital holds its electrons closer to the nucleus than a p orbital does. More s character therefore means a shorter, stronger bond, a wider angle, and a more electronegative carbon.

Bond lengths fall accordingly: 154 pm for C-C, 134 pm for C=C, 120 pm for a triple bond.

Illustration 1

Ethyne has $\mathrm{p}{K}_a$ 25 and ethane has 50. Both are ordinary C-H bonds. Account for a difference of $10^{25}$.

The bonds are not ordinary in the same way. Ethyne's C-H uses an $sp$ orbital with 50 per cent s character; ethane's uses $s{p}^3$ with 25 per cent.

The consequence is not really about the acid at all. It is about the anion left behind.

Removing a proton leaves the electron pair on carbon. In the acetylide ion that pair sits in an $sp$ orbital, held close to the nucleus where it is comfortable. In the ethyl carbanion it sits in $s{p}^3$, far out and poorly held.

Trap. Acidity is a property of the conjugate base, not of the acid. Every acidity question in this chapter is answered by asking what the anion looks like once the proton has gone, and the compound with the more comfortable anion is the stronger acid.

That single habit answers the alkyne question, the trichloroacetic acid question and the phenol question, all of which appear later in this chapter with entirely different-looking explanations.

2. Classification and Nomenclature

Organic compounds are classified by functional group, the atom or group that determines the chemistry. A homologous series shares one functional group, differs by $\mathrm{C}{\mathrm{H}}_2$, and shows a regular gradation of physical properties.

IUPAC naming

A name is built from word root for the longest chain, suffix for the principal functional group, and prefixes for everything else.

Number the chain to give the principal functional group the lowest possible locant. With no functional group, give the lowest locant to the first point of difference.

Seniority order:

$\text{acid}>\text{anhydride}>\text{ester}>\text{acid halide}>\text{amide}>\text{nitrile}>\text{aldehyde}>\text{ketone}>\text{alcohol}>\text{amine}>\text{alkene}>\text{alkyne}>\text{halide, nitro}$

Anything below the chosen suffix is named as a prefix instead. A molecule containing both an alcohol and an aldehyde is named as an al with a hydroxy prefix, never the reverse.

Illustration 2

Name the compound $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2-\mathrm{C}(\mathrm{C}{\mathrm{H}}_3)(\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3)-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$, drawn in the figure above.

The horizontal row holds six carbons and looks like a hexane. It is not the parent chain.

Trace instead from the end of the lower propyl group, through the central carbon, and out along the upper propyl: that path holds seven carbons. The parent is heptane, and the longest chain is under no obligation to be the one drawn in a straight line.

Numbering from either propyl end puts the branch point at C4, so the locants tie and both substituents sit on the same carbon. Cite them alphabetically:

$\text{4-ethyl-4-methylheptane}$

The ethyl group only became a substituent once the seven-carbon chain was chosen. Pick the six-carbon row instead and you would report a propyl group that does not exist in the correct name at all — which is why chain selection comes before everything else.

Illustration 3

Name $\mathrm{HOOC}-\mathrm{C}{\mathrm{H}}_2-\mathrm{CH}(\mathrm{OH})-\mathrm{C}{\mathrm{H}}_3$, and say why only one of the two functional groups appears as a suffix.

Both a carboxylic acid and an alcohol are present, and only the senior group takes the suffix. The priority order runs

$\mathrm{COOH}>\mathrm{S}{\mathrm{O}}_3\mathrm{H}>\mathrm{ester}>\mathrm{acid}\mathrm{halide}>\mathrm{amide}>\mathrm{nitrile}>\mathrm{CHO}>\mathrm{C}=\mathrm{O}>\mathrm{OH}>\mathrm{N}{\mathrm{H}}_2$

so the acid wins and the alcohol is demoted to a hydroxy prefix.

The chain is four carbons, and the carboxyl carbon is always C1 — its position is not negotiable, so no comparison of locants is needed:

$\text{3-hydroxybutanoic acid}$

Note the two separate jobs the priority table does. It decides which group is named as the suffix, and it fixes the direction of numbering, since the principal group must get the lowest possible locant. Alphabetical order settles only the writing sequence of prefixes, never the numbering, and confusing those two rules is the commonest naming error at this level.

3. Isomerism

Structural isomerism

Stereoisomerism

Geometrical isomerism requires restricted rotation, from a pi bond or a ring, and two different groups on each of the two carbons involved.

Cis and trans describe the arrangement when each carbon carries one identical pair. Where all four groups differ, the E and Z system applies, using Cahn-Ingold-Prelog priorities.

Optical isomerism requires chirality, the absence of any plane of symmetry. The commonest source is a carbon bearing four different groups.

Illustration 4

Tartaric acid, $\mathrm{HOOC}-\mathrm{CHOH}-\mathrm{CHOH}-\mathrm{COOH}$, has two stereocentres, so $2^n$ predicts four stereoisomers. It has three. Which one went missing, and why is one of the survivors optically inactive?

Label the two centres $R$ or $S$. The four candidates are $RR$, $SS$, $RS$ and $SR$.

$RR$ and $SS$ are a genuine pair of enantiomers, non-superimposable mirror images, each optically active.

$RS$ and $SR$ turn out to be the same molecule. The two halves of the chain are identical, so a mirror plane runs through the centre of the molecule, and reflecting it simply exchanges the two ends. Rotate it 180 degrees and it lands on itself.

That is the meso form. It has two stereocentres and is still optically inactive, because the rotation caused by one centre is cancelled exactly by the other. This is internal compensation, and it differs from a racemic mixture, where the cancellation happens between two separate molecules and can in principle be undone by separating them.

So $2^n$ is an upper bound, not a count. Look for an internal mirror plane whenever the molecule has identical halves, and expect $2^n-1$ isomers when you find one.

Illustration 5

Assign the configuration of $\mathrm{C}{\mathrm{H}}_3-\mathrm{CH}=\mathrm{CBrCl}$, and explain why the cis and trans labels cannot be used here at all.

Rank the two groups on each doubly bonded carbon by atomic number.

On the substituted carbon, bromine (35) outranks chlorine (17), so Br is senior. On the other carbon, the methyl group outranks hydrogen, so CH₃ is senior. If the two senior groups lie on the same side of the double bond the alkene is Z, and if they lie on opposite sides it is E.

Now the reason cis and trans fail. Those labels ask whether two identical groups sit on the same side, and here neither carbon carries a duplicate of anything on the other. There is no pair to be cis or trans about, so the question the label asks has no answer.

The CIP rules replace "same group" with "senior group", which is always decidable. That is why E and Z are not merely a modern relabelling of cis and trans: they cover cases the older system simply cannot describe.

4. Electronic Effects

Four effects redistribute electron density, and telling them apart is the core skill of the chapter.

Inductive effect

Electron density pulled or pushed along a chain of sigma bonds by an electronegativity difference. Permanent, and it weakens rapidly with distance, becoming negligible beyond three carbons.

$-I:\mathrm{N}{\mathrm{O}}_2>\mathrm{CN}>\mathrm{COOH}>\mathrm{F}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}>\mathrm{OH}>\mathrm{OR}>{\mathrm{C}}_6{\mathrm{H}}_5$

Alkyl groups show $+I$, pushing density towards whatever they are attached to.

Electromeric effect

A temporary complete transfer of a pi electron pair to one atom, occurring only in the presence of an attacking reagent and reverting when the reagent leaves.

It operates only where a pi bond exists, and it is always the stronger effect when it and the inductive effect oppose one another.

Resonance or mesomeric effect

Delocalisation of pi or lone pair electrons over more than two atoms, giving a hybrid that no single drawing represents.

Rules for judging contributing structures: more covalent bonds is better, complete octets are better, charge separation is destabilising, and a negative charge sits best on the more electronegative atom.

Two conditions must hold. The structures must differ only in where the electrons are, never in where the nuclei are, which is what separates resonance from tautomerism. And the system must be planar and conjugated, so the p orbitals are parallel and overlap continuously.

Resonance is permanent like induction, but it works through pi systems and it does not die away with distance, which is why a para substituent still matters.

Hyperconjugation

Delocalisation of the electrons of a sigma C-H bond into an adjacent empty p orbital or pi system, sometimes called no-bond resonance.

Its strength counts alpha hydrogens, those on the carbon next to the electron-deficient centre: nine for the tert-butyl cation, six for isopropyl, three for ethyl, none for methyl, which is exactly the observed order of carbocation stability.

Illustration 6

Nitrobenzene nitrates about $10^7$ times more slowly than benzene, and the product is meta. Phenol nitrates so fast it needs dilute acid, and the product is ortho and para. Explain both facts with one framework, and say why chlorine sits in neither camp.

Ask the two questions separately, exactly as in the hook.

Nitro group. It is $-I$ and $-R$: electronegative nitrogen pulls through the sigma bond, and the pi system drains ring density into the group. Both point the same way, so the ring is heavily depleted, hence the $10^7$ slowdown. Resonance withdrawal empties the ortho and para positions specifically, so the only position with any density left is meta.

Hydroxyl group. It is $-I$ but strongly $+R$, and here resonance wins on both counts, because oxygen's lone pair donates powerfully into the ring. The ring is enriched, so the reaction is fast, and enriched specifically at ortho and para.

Chlorine. $-I$ and $+R$, like hydroxyl, but the resonance donation is far weaker, because chlorine's 3p lone pair overlaps a carbon 2p orbital poorly, being the wrong size. So induction wins on rate and resonance still wins on position.

Only chlorine has its two effects disagreeing, and only chlorine gives a split answer. Groups whose effects agree give consistent answers; groups whose effects disagree give a rate from one and a position from the other. That is the whole rule, and it is worth more than the table.

Illustration 7

Rank but-1-ene, cis-but-2-ene, trans-but-2-ene and 2-methylpropene by stability using hyperconjugation, then check the ranking against measured heats of hydrogenation.

Count the α-hydrogens, meaning hydrogens on sp³ carbons bonded directly to a doubly bonded carbon:

• but-1-ene, $\mathrm{C}{\mathrm{H}}_2=\mathrm{CH}-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$: one $\mathrm{C}{\mathrm{H}}_2$ attached, so 2.
• but-2-ene, $\mathrm{C}{\mathrm{H}}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{C}{\mathrm{H}}_3$: two methyls, so 6.
• 2-methylpropene, $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{C}=\mathrm{C}{\mathrm{H}}_2$: two methyls on the substituted carbon, so 6.

The count puts but-1-ene last, and the measured heats of hydrogenation in kJ mol⁻¹ agree: 126.8 for but-1-ene against 119.7, 115.5 and 118.8 for cis-but-2-ene, trans-but-2-ene and 2-methylpropene. Lower heat released means a more stable starting alkene.

But notice where the count runs out. It gives three of the four the same score of six and cannot separate them, whereas the measurements order them clearly, with trans-but-2-ene most stable of all.

Hyperconjugation explains the gap between mono- and disubstituted alkenes and nothing finer. The remaining spread is steric: the cis isomer forces its two methyl groups against each other and pays for it, which is why trans beats cis by 4.2 kJ mol⁻¹. Counting is a first pass, not the whole answer.

5. Reaction Intermediates

Homolytic fission gives each fragment one electron, producing free radicals, and is favoured by non-polar solvents, heat and light. Heterolytic fission gives both electrons to one fragment, producing ions, and is favoured by polar solvents.

Stability orders

$\text{Carbocations: benzyl}>\text{allyl}>3^{\circ }>2^{\circ }>1^{\circ }>\text{methyl}$

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

Free radicals follow the carbocation order for the same reasons, being electron-deficient at carbon.

Illustration 8

The two orders above are exact reverses. Rather than memorising both, derive the second from the first.

There is one principle underneath, and it is not about alkyl groups at all.

A charge is stabilised by anything that spreads it out.

An alkyl group is electron-releasing, by $+I$ and by hyperconjugation. Now ask what that does in each case.

The alkyl group did exactly the same thing in both cases. The sign of the charge it was donating into changed, so the consequence reversed.

The same reasoning gives the corollary that questions actually test: electron-withdrawing groups stabilise carbanions. This is why the hydrogens next to a carbonyl are acidic enough to be removed by ordinary base, and why a nitro group makes a carbanion so accessible that nitroalkanes have $\mathrm{p}{K}_a$ around 10.

Getting this reversal right is worth several marks a year, and deriving it takes ten seconds while memorising two lists takes longer and fails under pressure.

6. Nucleophiles, Electrophiles and Reaction Types

A nucleophile is electron-rich and attacks electron-poor centres. It is a Lewis base: $\mathrm{O}{\mathrm{H}}^{-}$, $\mathrm{C}{\mathrm{N}}^{-}$, $\mathrm{N}{\mathrm{H}}_3$, ${\mathrm{H}}_2\mathrm{O}$, alkenes.

An electrophile is electron-poor and attacks electron-rich centres. It is a Lewis acid: ${\mathrm{H}}^{+}$, $\mathrm{N}{\mathrm{O}}_2^{+}$, $\mathrm{B}{\mathrm{F}}_3$, $\mathrm{AlC}{\mathrm{l}}_3$, carbocations.

Every nucleophile is a Lewis base and every electrophile a Lewis acid, but the emphasis differs: nucleophilicity is about how fast a species attacks, and basicity about how strongly it holds a proton at equilibrium. The two orders do not always agree.

Illustration 9

In water, ${\mathrm{I}}^{-}$ is a far better nucleophile than ${\mathrm{F}}^{-}$, yet ${\mathrm{F}}^{-}$ is by far the stronger base. Both are halide anions. How can one order reverse the other?

Because the two words measure different things and the solvent affects them differently.

Basicity is thermodynamic: how firmly does the anion hold a proton at equilibrium? Fluoride, small and highly charge-dense, holds one very firmly, so HF is the weakest of the hydrogen halides and ${\mathrm{F}}^{-}$ the strongest base.

Nucleophilicity is kinetic: how quickly does the anion reach a carbon and attack? Here fluoride's charge density becomes a liability. Water hydrogen bonds to it so tightly that it is buried inside a shell of solvent, and stripping that shell off costs energy before any attack can begin. Iodide is large and diffuse, weakly solvated, and its outer electrons are polarisable enough to reach out towards the carbon early.

$\text{basicity: }{\mathrm{F}}^{-}>\mathrm{C}{\mathrm{l}}^{-}>\mathrm{B}{\mathrm{r}}^{-}>{\mathrm{I}}^{-}\text{nucleophilicity in water: }{\mathrm{I}}^{-}>\mathrm{B}{\mathrm{r}}^{-}>\mathrm{C}{\mathrm{l}}^{-}>{\mathrm{F}}^{-}$

The clinching evidence is that in a polar aprotic solvent, which cannot hydrogen bond to anions, the nucleophilicity order flips back to match basicity. The intrinsic ranking never changed; the solvent was voting.

Four reaction types

Illustration 10

Classify each of these as substitution, addition, elimination or rearrangement.

• (a) $\mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+\mathrm{B}{\mathrm{r}}_2\to \mathrm{C}{\mathrm{H}}_2\mathrm{Br}-\mathrm{C}{\mathrm{H}}_2\mathrm{Br}$
• (b) $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{Br}+\mathrm{O}{\mathrm{H}}^{-}\to \mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}+\mathrm{B}{\mathrm{r}}^{-}$
• (c) $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{Br}+\mathrm{KOH}(\mathrm{alcoholic})\to \mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+\mathrm{KBr}+{\mathrm{H}}_2\mathrm{O}$
• (d) $\mathrm{C}{\mathrm{H}}_3-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2^{+}\to \mathrm{C}{\mathrm{H}}_3-\mathrm{C}{\mathrm{H}}^{+}-\mathrm{C}{\mathrm{H}}_3$

(a) Addition — the π bond opens and both bromine atoms join, with nothing leaving.

(b) Substitution — hydroxide replaces bromide, one group for one group.

(c) Elimination — HBr is removed across two adjacent carbons and a π bond forms.

(d) Rearrangement — a hydride migrates within the same species, converting a primary carbocation into the more stable secondary one. Nothing enters or leaves.

Now compare (b) with (c). Same substrate, same reagent, and the products are not even the same class of compound. The only difference is the medium: aqueous hydroxide acts mainly as a nucleophile and attacks carbon, while alcoholic hydroxide acts mainly as a base and removes a β-hydrogen.

That is the practical content of the nucleophile-versus-base distinction from the previous section. A species does not have one fixed role; conditions decide which of its two capacities dominates, and exam questions signal the intended path through the solvent named in the equation.

7. Putting the Tools Together

The value of GOC is that these ideas combine to answer questions that look like they need memorisation. Every one of the following is one electronic effect applied to a conjugate species.

Illustration 11

Rank ethanol, water, phenol and acetic acid by acidity, and account for the whole spread with one argument.

Read the column on the right and the whole order falls out of one sentence: the more thinly the negative charge is spread, and the more electronegative the atoms it is spread over, the stronger the acid.

Ethanol is worse than water because an ethyl group pushes electrons towards an oxygen that already has a surplus. Phenol beats both because the ring accepts some of the charge, though it parks it on carbon, which is not where a negative charge is happiest. Acetic acid wins because its two oxygens are equivalent by resonance and each carries only half a charge, on the element best suited to hold it.

Add a $-I$ group and the same logic extends. Trichloroacetic acid has $\mathrm{p}{K}_a$ 0.7 against acetic acid's 4.76, a factor of $10^4$, because three chlorines drain density away from the carboxylate and spread the charge further still. The chlorines are three bonds from the acidic hydrogen and not attached to it at all, which is induction reaching exactly as far as it does before dying out.

Summary

Name the question before reaching for an effect. Chlorine on benzene is deactivating and ortho-para directing at once, because $-I$ answers the rate question and $+R$ answers the position question, and two separate accounts are allowed to disagree.

More s character means a shorter, stronger bond and a more electronegative carbon, so ethyne has $\mathrm{p}{K}_a$ 25 and ethane 50. Acidity is always a property of the conjugate base, not of the acid.

IUPAC naming runs word root, suffix from the senior functional group and prefixes for the rest, with the seniority order deciding which group gets the suffix.

$2^n$ is an upper bound on stereoisomers, not a count. Tartaric acid has two stereocentres and three isomers, because $RS$ and $SR$ are the same meso compound, inactive by internal compensation, which unlike a racemate cannot be resolved.

Four electronic effects: induction through sigma bonds, permanent and dying past three carbons; the electromeric effect through a pi bond, temporary and reagent-triggered; resonance through a conjugated planar pi system, permanent and not dying with distance; and hyperconjugation from alpha C-H bonds, counted by how many there are.

Groups whose $I$ and $R$ effects agree give consistent answers, as the nitro group does. Groups whose effects disagree give a rate from one and a position from the other, as chlorine does.

Carbocation and carbanion stability orders are reverses of each other because an alkyl group donates electrons in both cases and only the sign of the charge changed. The corollary, that electron-withdrawing groups stabilise carbanions, is what questions actually test.

Nucleophilicity is kinetic and basicity thermodynamic, so iodide beats fluoride as a nucleophile in water while fluoride is the far stronger base. In an aprotic solvent the order flips back, which proves the solvent was doing the work.

Acidity across ethanol, water, phenol and acetic acid is one argument: the more thinly the charge is spread and the more electronegative the atoms holding it, the stronger the acid.