IB DP Chemistry Reactivity 3.4: Electron-pair sharing: Lewis acids, complexes and mechanisms – Revision Notes

For full explanations and longer worked examples, read the electron-pair sharing study guide first. These notes compress IB DP Chemistry understandings Reactivity 3.4.6 to 3.4.13 from the IB’s February 2023 Diploma Programme Chemistry guide, first assessed in 2025, into a form you can review in one sitting. The guide lists all of these understandings as additional higher level, so the whole page is HL only, for examinations from 2025 onwards.

Check your recall with the practice questions with worked answers. The DP Chemistry hub lists every unit, the DP Chemistry checklist tracks your progress, and the free diagnostics help you pick what to revise next.

Definitions

Term	Meaning
Lewis acid	electron-pair acceptor (Reactivity 3.4.6)
Lewis base	electron-pair donor (Reactivity 3.4.6)
Coordination bond	covalent bond where one atom supplies both shared electrons (Reactivity 3.4.7)
Ligand	species that donates a lone pair to a transition element cation (Reactivity 3.4.8)
Complex ion	central metal cation plus coordinately bonded ligands (Reactivity 3.4.8)
Leaving group	the species that departs with the bonding pair, here X⁻ (Reactivity 3.4.10)
Carbocation	ion with a positive carbon holding only three bonds (Reactivity 3.4.9, 3.4.12)
Stereospecific	gives one particular stereoisomer from one particular reactant (Reactivity 3.4.9)

Equivalences to learn: nucleophile = Lewis base; electrophile = Lewis acid.

Reactivity 3.4.6 and 3.4.7 Lewis theory and coordination bonds

Method in steps: who is the Lewis acid?

1. Find the new bond in the product.
2. Ask which reactant supplied both of its electrons. That one is the Lewis base.
3. The other reactant, with an empty orbital or a π pair that can move away, is the Lewis acid.

Small reminder. SnCl₄ + 2Cl⁻ → SnCl₆²⁻. Each chloride donates a lone pair to tin, so Cl⁻ is the Lewis base and SnCl₄ the Lewis acid. Two new coordination bonds form.

Must-know distinction: Lewis vs Brønsted–Lowry

	Brønsted–Lowry	Lewis
Acid	proton donor	electron-pair acceptor
Base	proton acceptor	electron-pair donor
Scope	needs H⁺ transfer	any pair donation
Example only covered by Lewis	–	Fe³⁺, BCl₃, CO₂ as acids

Every Brønsted–Lowry base is a Lewis base, because accepting H⁺ means donating a lone pair to it.

Lewis formulas. Show every lone pair on the donor atom. Draw the coordination bond as an arrow from donor to acceptor, or as a normal line once formed. A coordination bond is no different from other covalent bonds after it has formed.

Reactivity 3.4.8 Charge on a complex ion

Method in steps

1. Write the charge on the metal ion.
2. Add the charge of each ligand multiplied by the number of that ligand.
3. Neutral: H₂O, NH₃ (0). Anionic: Cl⁻, Br⁻, F⁻, CN⁻, OH⁻ (−1 each).

Worked reminder. Find the oxidation state of iron in [Fe(H₂O)₅(OH)]²⁺: x + 5(0) + (−1) = +2, so x = +3.

Reactivity 3.4.9 SN1 vs SN2

Feature	SN2	SN1
Typical substrate	primary halogenoalkane	tertiary halogenoalkane
Steps	one (concerted)	two
Species in rate-determining step	RX and nucleophile	RX only
Rate equation	rate = k[RX][Nu]	rate = k[RX]
Energy profile	one hump, transition state	two humps, carbocation intermediate
Direction of attack	opposite the leaving group	either face of planar carbocation
Stereochemistry	inversion of configuration (stereospecific)	not stereospecific

Secondary halogenoalkanes react by both mechanisms.

SN2 arrows: lone pair on Nu → δ+ carbon; C–X bond → X. Transition state in square brackets with ‡, partial bonds to Nu and X, overall charge shown.

SN1 arrows: C–X bond → X (slow); lone pair on Nu → C⁺ (fast).

Worked reminder. Halve [RX] and triple [Nu]. SN2 rate changes by 0.5 × 3 = × 1.5; SN1 rate changes by × 0.5, because only [RX] appears in its rate equation. A change in rate when [Nu] alone changes is the experimental sign of SN2.

Why the split? Three alkyl groups crowd the back of a tertiary carbon and stabilise a carbocation by releasing electron density. A primary carbon has neither problem nor benefit, so it reacts in one step.

Reactivity 3.4.10 Leaving group

Rate: RI > RBr > RCl (same R group).

• C–I is the longest and weakest of the three bonds, so it breaks fastest.
• I⁻ is the best leaving group: largest ion, charge most spread out, most stable.
• Do not argue from polarity. C–Cl is the most polar but the slowest.

Reactivity 3.4.11 Electrophilic addition (symmetrical alkenes)

Method in steps

1. Arrow from the C=C π bond to the electrophilic atom (H of HX, Brδ+ of a polarised Br₂, or H⁺ from the acid catalyst).
2. Arrow from the breaking bond (H–X or Br–Br) to the atom that leaves as an anion.
3. Carbocation forms.
4. Arrow from a lone pair on the nucleophile (X⁻, Br⁻ or H₂O) to C⁺.
5. For water only: the protonated alcohol loses H⁺, regenerating the catalyst.

Reagent	Electrophile	Product type
HX	Hδ+ of HX	halogenoalkane
X₂	induced Xδ+	dihalogenoalkane
H₂O with acid catalyst	H⁺	alcohol

Reactivity 3.4.12 Unsymmetrical alkenes

• Carbocation stability: tertiary > secondary > primary (positive inductive effect of alkyl groups).
• Major product comes from the more stable carbocation.
• Short cut: H adds to the C=C carbon that already has more H atoms.

Small reminder. 2-methylhex-1-ene, CH₂=C(CH₃)CH₂CH₂CH₂CH₃, with HBr: H adds to C1, giving a tertiary C⁺ on C2. Major product 2-bromo-2-methylhexane; minor product 1-bromo-2-methylhexane.

Reactivity 3.4.13 Electrophilic substitution in benzene

1. Arrow from the delocalised ring to E⁺.
2. Intermediate: horseshoe over five carbons with + inside; the sp³ carbon bearing H and E sits outside it.
3. The C–H bonding pair returns to the π system; H⁺ departs and delocalisation returns.

Why substitution, not addition? Keeping the delocalised ring is energetically favoured; addition would lose it. The guide says formation of E⁺ is not assessed. In nitration, HNO₃ behaves as a base towards H₂SO₄, giving NO₂⁺.

Quick self-test

1. In Zn²⁺ + 4OH⁻ → [Zn(OH)₄]²⁻, which species is the Lewis acid?
2. Why is BCl₃ a Lewis acid but not a Brønsted–Lowry acid?
3. Give the charge on a complex of Ni²⁺ with four H₂O and two Cl⁻ ligands.
4. Give the charge on a complex of Co³⁺ with five NH₃ and one H₂O.
5. Give the charge on a complex of Cu²⁺ with four Cl⁻ ligands.
6. Write the rate equation for the hydrolysis of a tertiary halogenoalkane, RX, with water.
7. How many humps does the energy profile of an SN2 reaction have, and why?
8. Rank CH₃CH₂CH₂Cl, CH₃CH₂CH₂Br and CH₃CH₂CH₂I by rate of reaction with OH⁻.
9. A single enantiomer of a secondary halogenoalkane reacts only by SN2. Describe the product.
10. Name the major product of but-1-ene with steam and an acid catalyst.
11. In the benzene mechanism, why does the intermediate lose H⁺ rather than gain a nucleophile?
12. Which species acts as a Brønsted–Lowry base when NO₂⁺ is generated from nitric and sulfuric acids?

Answers

1. Zn²⁺: it accepts a lone pair from each OH⁻.
2. It accepts a lone pair into an empty orbital on boron but has no H⁺ to donate.
3. +2 + 2(−1) + 0 = 0, a neutral complex, [Ni(H₂O)₄Cl₂].
4. +3 + 0 = 3+, [Co(NH₃)₅(H₂O)]³⁺.
5. +2 + 4(−1) = 2−, [CuCl₄]²⁻.
6. Rate = k[RX]; water is not in the slow step.
7. One: the single step passes through one transition state and there is no intermediate.
8. CH₃CH₂CH₂I > CH₃CH₂CH₂Br > CH₃CH₂CH₂Cl, following bond strength.
9. A single enantiomer of product with inverted configuration: the nucleophile sits on the side opposite where the halogen was.
10. Butan-2-ol, from the secondary carbocation.
11. Losing H⁺ restores the delocalised ring, which is more stable than any addition product.
12. HNO₃ accepts H⁺ from H₂SO₄.

Where marks are usually lost

• Naming the acceptor as the Lewis base, by mixing up “donor” and “acceptor”.
• Drawing a coordination arrow from the metal to the ligand.
• Writing the complex charge as the metal charge alone, ignoring anionic ligands.
• Showing SN2 attack on the same side as the halogen, so the inversion is lost.
• Missing the partial bonds, brackets or charge on the SN2 transition state.
• Putting the nucleophile in the SN1 rate equation.
• Using bond polarity to explain why iodoalkanes react fastest.
• Forgetting to show that Br₂ is polarised by the π electrons before it reacts.
• Choosing the product from the less stable carbocation with an unsymmetrical alkene.
• Drawing the benzene intermediate with a full circle or with + outside the horseshoe.

Official syllabus

Aligned to the Diploma Programme Chemistry guide (International Baccalaureate Organization), first assessment 2025, published February 2023, sub-topic Reactivity 3.4, understandings 3.4.6 to 3.4.13 (additional higher level).