7HLLewis acids and bases

SL students can skip to section 14.

A Lewis acid is an electron-pair acceptor; a Lewis base is an electron-pair donor. When a Lewis base reacts with a Lewis acid, a coordination bond forms: a covalent bond in which both shared electrons came from the same atom. Figure 7 draws two examples.

Ammonia has a lone pair on nitrogen. Boron trifluoride has only six electrons in boron's outer shell, so boron can accept a pair. The lone pair on N forms an N→B coordination bond: NH₃ is the Lewis base, BF₃ the Lewis acid. The arrow points from donor to acceptor. With H⁺, ammonia forms NH₄⁺ the same way. Once formed, a coordination bond is an ordinary covalent bond: the four N–H bonds in NH₄⁺ are identical.

The last row is the organic link the guide wants: nucleophiles are Lewis bases and electrophiles are Lewis acids.

Linking question: how do Brønsted–Lowry and Lewis theory relate? A Brønsted–Lowry base accepts H⁺ by giving it a lone pair, so it is also a Lewis base, and H⁺ is a Lewis acid. But BF₃, AlCl₃ and metal ions have no proton to donate, so Lewis theory is the wider one, with Brønsted–Lowry as a special case.

Nature of science: why has the definition of acid evolved? Each definition was widened when evidence it could not explain appeared: Lavoisier's acids containing oxygen (until hydrochloric acid was shown to contain none), Arrhenius's H⁺ in water, Brønsted–Lowry proton transfer in any medium, and Lewis's electron-pair acceptance, which needs no protons at all. Each is still used where it is most convenient.

8HLComplex ions

Ligands are molecules or ions with a lone pair that they donate to a transition element cation, forming coordination bonds. The result is a complex ion. Transition element ions can accept these pairs because they have empty orbitals of suitable energy, so the metal ion is the Lewis acid and each ligand a Lewis base. Figure 8 shows six water ligands around a copper(II) ion, the pale blue ion in copper(II) sulfate solution.

Deducing the charge. The charge on a complex ion is the charge on the metal ion plus the charges of all the ligands. Neutral ligands (H₂O, NH₃) count zero; anionic ligands (Cl⁻, CN⁻, OH⁻) count −1 each.

The same sum runs backwards to find the metal's oxidation state. In [CuCl₄]²⁻, x + 4(−1) = −2, so x = +2 and the ion contains copper(II).

9HLSN2 and SN1: two mechanisms for nucleophilic substitution

Which of two mechanisms a halogenoalkane uses depends on how many alkyl groups surround the carbon bearing the halogen.

SN2: primary halogenoalkanes. The name means substitution, nucleophilic, bimolecular: two species are involved in the one step. Figure 9 draws it for 1-bromobutane and hydroxide.

The hydroxide attacks the δ+ carbon from the side opposite the bromine. As the O–C bond forms, the C–Br bond breaks, in a single concerted step through a transition state: carbon partly bonded to both O and Br, the other three groups flattened into a plane, the whole drawn in square brackets with ‡ and −. It is not an intermediate; it exists only at the top of the energy barrier.

Because both species take part in that one step, the rate depends on both concentrations:

SN2 is stereospecific. Attack from the back pushes the other three groups through to the far side, like an umbrella turned inside out: inversion of configuration. If the carbon is chiral (S3.2), one enantiomer of the halogenoalkane gives only one enantiomer of the alcohol, with the opposite arrangement.

SN1: tertiary halogenoalkanes. Substitution, nucleophilic, unimolecular: only one species in the slow step. Figure 10 draws it for 2-bromo-2-methylpropane.

Step 1, slow: the C–Br bond breaks heterolytically on its own, giving a tertiary carbocation and Br⁻. Step 2, fast: the hydroxide's lone pair bonds to C⁺. The slow rate-determining step involves only the halogenoalkane, so the rate does not depend on [OH⁻]:

The carbocation is trigonal planar, so either face is attacked with equal probability; if the carbon is chiral the product is a racemic mixture. SN1 is not stereospecific.

Why the split between primary and tertiary? Two reasons, and a full answer gives both.

• Carbocation stability. Alkyl groups release electron density towards a positive carbon (a positive inductive effect). A tertiary carbocation has three alkyl groups to spread its charge; a primary one has only one. The tertiary carbocation is stable enough to form in step 1, the primary one is not.
• Steric hindrance. In a tertiary halogenoalkane three bulky alkyl groups crowd the back of the carbon, so the nucleophile cannot reach it for an SN2 attack. A primary carbon has two small hydrogen atoms there, so backside attack is easy.

Secondary halogenoalkanes, such as 2-bromopropane, sit in between and react by both mechanisms.

Linking questions: which mechanism has an intermediate, and how do the energy profiles differ? Figure 11 answers both. SN2 has one step, so one hump, whose peak is the transition state. SN1 has two steps, so two humps; the dip between them is the carbocation, a genuine intermediate that exists for a short time. The first hump is higher, because step 1 is the slow, rate-determining step.

Nature of science: how useful are mechanistic models such as SN1 and SN2? Nobody watches the electrons move. A mechanism is a model, accepted because its predictions match what can be measured: rate equations and product stereochemistry. SN1 and SN2 are clean limits; many real reactions, especially of secondary compounds, fall between them. Isotope tracers test such models: an atom labelled with, say, oxygen-18 can be tracked by mass spectrometry to show which bonds broke and formed.

10HLThe leaving group decides the rate

For 1-chloro-, 1-bromo- and 1-iodobutane, same skeleton and mechanism, the rate is RI > RBr > RCl.

The reason is the strength of the carbon–halogen bond, which must break in every mechanism (typical data-booklet values; use your own booklet's figures):

The C–I bond is the weakest, so it breaks most easily and the activation energy is lowest. Polarity does not explain it: C–Cl is the most polar bond, yet chloroalkanes react slowest.

Linking question: why is the iodide ion a better leaving group than the chloride ion? The iodide ion is much larger (S3.1), so its charge is spread over a bigger volume and it is more stable on its own.

A classic way to see it is to warm each halogenoalkane with aqueous silver nitrate in ethanol. Water acts as the nucleophile, and each halide ion released precipitates with Ag⁺: AgCl white, AgBr cream, AgI yellow. The yellow precipitate appears first, the white one last.

11HLElectrophilic addition mechanisms

All three additions follow one pattern, shown in Figure 12 for ethene.

With hydrogen bromide (a). An arrow from the C=C bond to the δ+ H shows the π pair forming a C–H bond; a second arrow from the H–Br bond to Br shows it breaking heterolytically. That gives the carbocation CH₃CH₂⁺ and Br⁻, and a lone pair on Br⁻ then bonds to C⁺: bromoethane.

With bromine (b). Br₂, polarized by the π cloud, is attacked at its δ+ end; the Br–Br bond breaks heterolytically, giving BrCH₂CH₂⁺ and Br⁻, which then bond: 1,2-dibromoethane. The intermediate is drawn here as a carbocation, the simpler model; further reading may show a three-membered ring containing bromine instead.

With water and an acid catalyst (c). The acid catalyst supplies H⁺, which the π pair bonds to, making the carbocation. A lone pair on water bonds to C⁺, giving an oxonium ion, CH₃CH₂OH₂⁺, which loses H⁺ to give ethanol and return the catalyst.

12HLUnsymmetrical alkenes: which product wins

When HBr adds to propene, the H⁺ can attach to either carbon of the double bond, and each choice gives a different carbocation. Figure 13 follows both routes.

If H adds to the end CH₂ carbon, the charge sits on the middle carbon: a secondary carbocation, CH₃C⁺HCH₃, with two alkyl groups pushing electron density towards it. If H adds to the middle carbon, the result is a primary carbocation, CH₃CH₂C⁺H₂, with only one. The secondary carbocation is more stable and forms far more often, so 2-bromopropane is the major product and 1-bromopropane the minor one.

Carbocation stability: tertiary > secondary > primary. The major product comes from the most stable carbocation, so H adds to the carbon of the double bond that already carries more H atoms.

The second sentence is Markovnikov's rule: a shortcut, while the carbocation argument is the explanation that "explain" questions want. Water behaves the same way: propene gives mainly propan-2-ol, and 2-methylpropene with HBr gives mainly 2-bromo-2-methylpropane, via a tertiary carbocation.

13HLElectrophilic substitution in benzene

Benzene, C₆H₆, is highly unsaturated on paper, yet it does not decolourize bromine water. Its six π electrons are delocalized round the ring (S2.2), all six C–C bonds are the same length, and the ring is much more stable than three separate double bonds would be. Estimates from enthalpies of hydrogenation put this extra stability at roughly 150 kJ mol⁻¹. Addition would destroy the delocalized ring and throw that stability away; substitution keeps it.

Linking question: what features of benzene favour electrophilic substitution? The delocalized ring is electron-rich above and below the plane, so it attracts electrophiles; after the first step it loses H⁺ rather than adding a nucleophile, because that restores the delocalized system. Figure 14 draws the mechanism for a charged electrophile, E⁺.

1. A curly arrow from the delocalized ring to E⁺: two of the π electrons form a new C–E bond.
2. The intermediate is a positive ion. The carbon bearing H and E is tetrahedral and out of the delocalized system; the remaining four π electrons spread over the other five carbons, drawn as a horseshoe with a + inside, open towards that carbon.
3. A curly arrow from the C–H bond into the ring: that pair rejoins the π system, H⁺ leaves, and the full delocalized ring is restored.

The standard example is nitration, with the nitronium ion, NO₂⁺, from concentrated nitric and sulfuric acids (how it forms is not assessed), giving nitrobenzene:

Linking question: how can the acid–base behaviour of HNO₃ in the nitrating mixture be described? Sulfuric acid, the stronger acid, donates a proton to nitric acid, which is therefore acting as a Brønsted–Lowry base; the protonated nitric acid loses water, leaving NO₂⁺. Overall, HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻.