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

Here you will find the additional higher level part of Reactivity 3.4, “Electron-pair sharing reactions”, for IB DP Chemistry, taught from first principles with worked examples. Numbering matches the IB’s February 2023 Diploma Programme Chemistry guide, first assessed in 2025, and the page covers understandings Reactivity 3.4.6 to 3.4.13. The guide places all eight under additional higher level, so everything below is HL only. It is written for examination sessions from 2025 onwards.

Short revision notes and marked practice questions accompany this page. The SL ideas this builds on (nucleophiles, electrophiles, heterolytic fission) are taught on the Reactivity 3.4.1 to 3.4.5 page and summarised in the Reactivity 3 mechanisms overview. Other units are on the DP Chemistry hub and the printable DP Chemistry checklist; a free diagnostic quiz shows where your gaps are.

Scope and level

Understanding	Skills expected	Level
Reactivity 3.4.6	Identify Lewis acids and Lewis bases in inorganic and organic reactions	HL only
Reactivity 3.4.7	Draw and interpret Lewis formulas that show a coordination bond forming; link nucleophiles and electrophiles to Lewis theory	HL only
Reactivity 3.4.8	Deduce the charge on a complex ion from its formula and ligands	HL only
Reactivity 3.4.9	Describe and explain SN1 and SN2 mechanisms for tertiary and primary halogenoalkanes, including the stereospecific SN2 route	HL only
Reactivity 3.4.10	Predict and explain relative substitution rates for RCl, RBr and RI	HL only
Reactivity 3.4.11	Describe and explain electrophilic addition to symmetrical alkenes with halogens, water and hydrogen halides	HL only
Reactivity 3.4.12	Use carbocation stability to predict the major product with unsymmetrical alkenes	HL only
Reactivity 3.4.13	Describe and explain the mechanism of benzene reacting with a charged electrophile, E⁺	HL only

Every curly arrow starts at an electron pair (a lone pair or a bond) and ends where that pair goes.

Reactivity 3.4.6 Lewis acids and Lewis bases

• A Lewis acid accepts an electron pair.
• A Lewis base donates an electron pair.

A Lewis base needs a lone pair (or a π bond). A Lewis acid needs room for a new pair: an empty orbital (BCl₃, metal cations) or a multiple bond whose π pair can move away (CO₂).

Link to Brønsted–Lowry theory. A Brønsted–Lowry base accepts H⁺ by donating a lone pair to it, so every Brønsted–Lowry base is also a Lewis base, and H⁺ is a Lewis acid. The Lewis definition is wider: BCl₃ and Fe³⁺ are Lewis acids but have no H⁺ to donate.

Worked example 1: assigning roles

Reaction	Lewis acid	Lewis base	Why
AlCl₃ + Cl⁻ → AlCl₄⁻	AlCl₃	Cl⁻	Al has only six outer electrons; Cl⁻ supplies a lone pair
CO₂ + OH⁻ → HCO₃⁻	CO₂	OH⁻	O of OH⁻ donates to the carbon; one C=O π pair moves onto oxygen
Fe³⁺ + 6H₂O → [Fe(H₂O)₆]³⁺	Fe³⁺	H₂O	each water donates an oxygen lone pair to the cation
(CH₃)₃C⁺ + H₂O → (CH₃)₃COH₂⁺	the carbocation	H₂O	an organic example: the positive carbon accepts an oxygen lone pair

Reactivity 3.4.7 Coordination bonds in Lewis acid–base reactions

In any Lewis acid–base reaction both electrons of the new bond come from the base, so that bond is a coordination bond. So every nucleophile is a Lewis base, and every electrophile is a Lewis acid.

Worked example 2: dimethyl ether and boron trichloride

Draw Lewis formulas to show the reaction between (CH₃)₂O and BCl₃.

Oxygen in (CH₃)₂O has two lone pairs; boron in BCl₃ has only six outer electrons. A curly arrow goes from one oxygen lone pair to B.

(CH₃)₂O (two lone pairs on O) + BCl₃ (empty orbital on B) → (CH₃)₂O→BCl₃ (new O–B coordination bond; B now has eight outer electrons)

The arrow (O → B) marks the coordination bond. In the full Lewis formula, show three lone pairs on each Cl and the one lone pair left on O. Formal charges in the adduct are +1 on O and −1 on B, and the shape around boron changes from trigonal planar to tetrahedral.

Once formed, a coordination bond behaves exactly like an ordinary covalent bond, so in NH₄⁺ the four N–H bonds cannot be told apart.

Reactivity 3.4.8 Ligands and complex ions

A ligand donates a lone pair to a transition element cation, forming a coordination bond, so ligands are Lewis bases and the cation is a Lewis acid. The product is a complex ion.

Charge = charge of metal ion + sum of ligand charges. H₂O and NH₃ add 0; Cl⁻, F⁻, CN⁻ and OH⁻ each add −1.

Worked example 3: charges on complex ions

Complex	Working	Charge
Fe³⁺ with six F⁻	+3 + 6(−1)	[FeF₆]³⁻
Cr³⁺ with two Cl⁻ and four H₂O	+3 + 2(−1) + 0	[CrCl₂(H₂O)₄]⁺
Ni²⁺ with four CN⁻	+2 + 4(−1)	[Ni(CN)₄]²⁻

The same sum works backwards. For [Mn(CN)₆]³⁻: x + 6(−1) = −3, so x = +3 and the metal ion is Mn³⁺.

Reactivity 3.4.9 SN1 and SN2 in halogenoalkanes

The carbon bonded to the halogen is δ+ and is attacked by nucleophiles such as OH⁻, H₂O, NH₃ and CN⁻; the halide leaves as X⁻.

SN2: primary halogenoalkanes (one step)

Substitution, nucleophilic, bimolecular. A concerted mechanism: the new bond forms as the old one breaks.

• Arrow 1: nucleophile lone pair to the carbon, attacking from the side opposite X. Arrow 2: C–X bond to X.
• Transition state: partial bonds to both Nu and X, other three groups in one plane, in square brackets with ‡ (and a negative charge if Nu was an anion).
• Rate = k[halogenoalkane][nucleophile]. The energy profile has one hump and no intermediate.

Primary carbons are uncrowded, and a primary carbocation would be too unstable to form.

Stereospecific. Back attack flips the other three groups through a plane, like an umbrella blowing inside out, so the configuration is inverted. One enantiomer of a chiral halogenoalkane gives one enantiomer of product.

SN1: tertiary halogenoalkanes (two steps)

“Substitution, nucleophilic, unimolecular”.

1. Slow: C–X breaks heterolytically (arrow from the bond to X), giving X⁻ and a tertiary carbocation.
2. Fast: the nucleophile donates a lone pair to the positive carbon.

• Rate = k[halogenoalkane]; the nucleophile is not in the slow step.
• Energy profile: two humps, carbocation in the dip, first hump higher.
• Three alkyl groups block back attack and stabilise the carbocation by releasing electron density (positive inductive effect).
• The planar carbocation can be attacked on either face, so SN1 is not stereospecific.

Secondary halogenoalkanes can react by both mechanisms.

How useful are these models? SN1 and SN2 are mechanistic models, supported by evidence such as rate equations and the stereochemistry of products. They predict well for primary and tertiary substrates, but a secondary substrate often shows features of both, so a mechanism is a well-supported model rather than something directly observed.

Worked example 4: choosing and describing the mechanism

(a) 1-bromo-5-methylhexane + OH⁻: primary, so SN2. OH⁻ attacks C1 opposite Br, through one transition state, giving 5-methylhexan-1-ol.

(b) 3-chloro-3-methylpentane + water: tertiary, so SN1. C–Cl breaks first (carbocation on C3); water bonds to C3; the ion loses H⁺, giving 3-methylpentan-3-ol.

(c) One enantiomer of 2-chloropentane + OH⁻ by SN2 gives one enantiomer of pentan-2-ol, with OH on the side opposite to where Cl was.

Worked example 5: using the rate equations

An SN2 reaction has an initial rate of 6.0 × 10⁻⁶ mol dm⁻³ s⁻¹. The halogenoalkane concentration is doubled and the hydroxide concentration is halved.

• SN2: rate = k[RX][OH⁻], so the new rate = 6.0 × 10⁻⁶ × 2 × 0.5 = 6.0 × 10⁻⁶ mol dm⁻³ s⁻¹ (unchanged).
• If the same changes were made to an SN1 reaction with the same starting rate, rate = k[RX] gives 1.2 × 10⁻⁵ mol dm⁻³ s⁻¹.

Reactivity 3.4.10 The leaving group

Compare RCl, RBr and RI with the same R group. The rate order is:

RI > RBr > RCl

• C–X breaks in both mechanisms. Down group 17 the bond gets longer and weaker, so C–I breaks most easily.
• I⁻ is the best leaving group: the largest halide ion, with its charge spread most widely, so it is the most stable once released.
• C–Cl is the most polar bond yet reacts slowest, so bond strength, not polarity, decides the rate.

The guide states that the effects of solvent and of the mechanism on rate are not assessed here.

Reactivity 3.4.11 Electrophilic addition to symmetrical alkenes

The π bond of C=C is electron-rich and attacks electrophiles. Take symmetrical cyclooctene, C₈H₁₄, a ring of eight carbons with one C=C.

With a hydrogen halide (HBr). Arrows go from the π bond to H, and from the H–Br bond to Br, giving a carbocation and Br⁻; a Br⁻ lone pair then bonds to the positive carbon. Product: bromocyclooctane.

With Br₂. The π electrons repel the electrons in an approaching Br₂, so the nearer Br becomes δ+. The π bond attacks it, Br–Br breaks heterolytically, and Br⁻ then bonds to the carbocation. Product: 1,2-dibromocyclooctane.

With water (acid catalyst). The π bond attacks H⁺; water donates a lone pair to the carbocation; the protonated alcohol loses H⁺, regenerating the catalyst. Product: cyclooctanol.

Reactivity 3.4.12 Unsymmetrical alkenes and carbocation stability

With an unsymmetrical alkene, H⁺ can add to either end, giving two different carbocations. Stability: tertiary > secondary > primary, because alkyl groups release electron density towards the positive carbon (positive inductive effect). The more stable carbocation forms faster and gives the major product: H joins the C=C carbon that already holds more hydrogens.

Worked example 6: predicting the major product

3-methylpent-2-ene, CH₃CH=C(CH₃)CH₂CH₃, reacts with HCl.

• H on C3 would give a secondary carbocation on C2.
• H on C2 gives a tertiary carbocation on C3, which is more stable.
• Cl⁻ attacks C3. Major product: 3-chloro-3-methylpentane; minor product: 2-chloro-3-methylpentane.

With water and an acid catalyst the same tertiary carbocation forms, so the major product is 3-methylpentan-3-ol.

Reactivity 3.4.13 Electrophilic substitution in benzene

Addition would destroy benzene’s delocalised π system, which is energetically costly, so benzene reacts by substitution. The guide states that formation of E⁺ is not assessed.

1. Arrow from inside the delocalised ring to E⁺.
2. Carbocation intermediate: a broken “horseshoe” with + over five carbons; the sixth carbon carries H and E and is outside the horseshoe.
3. The C–H bonding pair moves back into the π system (curly arrow), H⁺ departs, and the delocalised ring is regained.

Overall: C₆H₆ + E⁺ → C₆H₅E + H⁺.

Linking example: nitration. Sulfuric acid protonates nitric acid, so here HNO₃ acts as a base: HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻. NO₂⁺ is the E⁺, and the product is nitrobenzene.

Common errors

• Calling BCl₃ a Brønsted–Lowry acid. It has no H⁺ to give; it is a Lewis acid only.
• Drawing the coordination-bond arrow from the acceptor to the donor. It points from the atom that supplies the pair.
• Giving a neutral ligand such as NH₃ a charge, or forgetting to multiply an anionic ligand’s charge by the number of those ligands.
• Writing the nucleophile into the SN1 rate equation.
• Explaining RI > RCl by bond polarity.
• Showing the hydrogen halide adding to give the less stable carbocation.
• Closing the horseshoe over all six carbons of the benzene intermediate.

Official syllabus

This guide follows 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).