IB DP Chemistry Reactivity 3.4: Electron-pair sharing: nucleophiles, electrophiles and substitution – Revision Notes

These notes boil down Reactivity 3.4.1 to 3.4.5 from the IB DP Chemistry guide (issued in February 2023; first assessment 2025), so they suit any session from 2025 on. All five understandings are SL and HL. The longer explanations live in the study guide, and original exam-style questions with marked answers are in the practice set.

Course links: IB DP chemistry hub, printable checklist, free diagnostics, and the Reactivity 3 overview notes. Radicals and homolytic fission are in Reactivity 3.3; the HL-only part of Reactivity 3.4 is on its own page.

Scope at a glance

Guide reference	Core skill	Level
Reactivity 3.4.1	Recognise nucleophiles (neutral and negative)	SL and HL
Reactivity 3.4.2	Equations for nucleophilic substitution, with electron-pair movement described and explained	SL and HL
Reactivity 3.4.3	Heterolytic fission: equations and curly arrows	SL and HL
Reactivity 3.4.4	Recognise electrophiles (neutral and positive)	SL and HL
Reactivity 3.4.5	Electrophilic addition: alkenes with water, halogens, hydrogen halides	SL and HL

Depth limits from the guide: detailed substitution mechanisms are not needed at SL, and alkene addition mechanisms are not assessed at SL. HL students learn both later in Reactivity 3.4.9 and 3.4.11 (HL only).

Key definitions

• Nucleophile: a species that supplies both electrons of a new covalent bond to its partner, the electrophile.
• Electrophile: a species that takes in both electrons of a new covalent bond from its partner, the nucleophile.
• Nucleophilic substitution: the nucleophile’s pair makes a new bond to carbon as another bond to that carbon breaks, so a leaving group is released.
• Leaving group: the atom or group that departs with the electron pair of the broken bond (in halogenoalkanes, a halide ion).
• Heterolytic fission: bond breaking in which one fragment keeps the whole bonding pair, giving an anion and a cation.
• Electrophilic addition: an electrophile attacks the electron-rich C=C bond and the reagent adds across it, leaving a saturated product.
• Curly arrow (double-barbed): shows one electron pair moving, from a lone pair or bond to an atom or new bond position.

Recognising the two partners

	Nucleophile	Electrophile
Electron density	high (electron-rich)	low (electron-poor), or made low by a nearby electron-rich site
What it has	a lone pair, a negative charge, or pi electrons	a positive charge, a δ+ atom, or a bond that can be polarised
Charged examples	OH⁻, CN⁻, Cl⁻, Br⁻, I⁻, CH₃O⁻	H⁺, NO₂⁺, carbocations such as (CH₃)₂CH⁺
Neutral examples	H₂O, NH₃, CH₃OH, CH₃NH₂, the C=C bond of an alkene	HCl, HBr, HI (δ+ H), Cl₂, Br₂, the δ+ carbon of a halogenoalkane
Curly arrow	starts here	ends here

Quick test: which species ends up sharing a pair it did not have to give? That is the electrophile. Which species gave its own pair? That is the nucleophile.

Equation patterns (learn these shapes)

Reaction	General equation	Product
Halogenoalkane + hydroxide	R–X + OH⁻ → R–OH + X⁻	alcohol
Halogenoalkane + water	R–X + H₂O → R–OH + HX	alcohol
Halogenoalkane + cyanide	R–X + CN⁻ → R–CN + X⁻	nitrile (chain one carbon longer)
Halogenoalkane + ammonia	R–X + NH₃ → R–NH₂ + HX (or + 2NH₃ → R–NH₂ + NH₄X)	amine
Alkene + halogen	C=C + X₂ → X–C–C–X	dihalogenoalkane
Alkene + hydrogen halide	C=C + HX → H–C–C–X	halogenoalkane
Alkene + water (steam, acid catalyst)	C=C + H₂O → H–C–C–OH	alcohol

Here R is an alkyl group and X is Cl, Br or I.

Method in steps: describing a substitution

1. Mark the polar C–X bond: Cδ+ and Xδ−.
2. The nucleophile’s lone pair is attracted to Cδ+ and is donated to it: new bond forms.
3. The C–X bonding pair moves onto X: the C–X bond breaks.
4. X leaves as X⁻ (the leaving group).
5. If the nucleophile was neutral (H₂O, NH₃), an H⁺ is lost too, so HX appears as a product.

Small reminder: 1-bromobutane + CN⁻ → CH₃CH₂CH₂CH₂CN + Br⁻. The product is pentanenitrile, because the CN carbon counts in the chain.

Method in steps: heterolytic fission

1. Decide which atom is more electronegative: it keeps the pair.
2. Draw one curly arrow from the centre of the bond to that atom.
3. That fragment becomes an anion; the other becomes a cation.
4. A carbon that loses its share of the pair is a carbocation (six outer electrons, positive charge).

Small reminder: (CH₃)₂CHBr → (CH₃)₂CH⁺ + Br⁻.

Method in steps: alkene addition equations

1. Draw the alkene so the C=C bond is visible.
2. Turn C=C into C–C.
3. Attach one piece of the reagent to each of those two carbons (Br and Br; H and Br; H and OH).
4. Check the product formula equals alkene + reagent.
5. Name it, numbering from the end that gives the lowest locant.

Small reminder: but-2-ene + Br₂ → CH₃CHBrCHBrCH₃, 2,3-dibromobutane.

Must-know distinctions

Pair	The difference
Homolytic vs heterolytic fission	One electron to each fragment, giving radicals and fish-hook arrows; or both electrons to one fragment, giving ions and double-barbed arrows
Substitution vs addition	Substitution swaps one group for another and releases a leaving group; addition joins the whole reagent on and releases nothing
Nucleophile vs electrophile	Donates the pair vs accepts the pair
Halogenoalkane vs alkene	Halogenoalkane has an electron-poor δ+ carbon (attacked by nucleophiles); alkene has an electron-rich C=C (attacked by electrophiles)
Bromine water with alkene vs alkane	Alkene: decolourised in the dark by addition. Alkane: no change in the dark; radical substitution needs UV light (Reactivity 3.3)
Symmetrical vs unsymmetrical alkene	Symmetrical gives one product with HX or H₂O; unsymmetrical can give two (choosing the major one is Reactivity 3.4.12, HL only)

Linking questions from the guide

• Reactivity 3.3: radical bond-breaking (homolytic) differs from the heterolytic bond-breaking in nucleophilic substitution, where the leaving group keeps the pair.
• Reactivity 3.3: bromine water and alkenes react in the dark; alkanes need UV light for radical substitution.
• Structure 2.4: alkenes are industrial “starting molecules” because the C=C bond converts readily into alcohols, halogenoalkanes and addition polymers.

Quick self-test

1. Define an electrophile.
2. From OH⁻, NO₂⁺, CH₃NH₂ and H⁺, pick the nucleophiles.
3. Write an equation for 1-iodopropane reacting with hydroxide ions and name the organic product.
4. Name the leaving group when 1-bromopropane reacts with cyanide ions.
5. 8.50 g of 1-iodopropane, C₃H₇I, reacts completely with excess hydroxide ions. Calculate the mass of propan-1-ol formed. (Ar: C 12.01, H 1.01, O 16.00, I 126.90)
6. Write an equation for the heterolytic fission of the C–Br bond in 2-bromopropane.
7. Give the product of cyclohexene with hydrogen bromide.
8. Explain why the alkene, not the HBr, is the nucleophile in question 7.
9. Write the equation for ethene with steam and state the type of catalyst.
10. State the arrow used for homolytic fission and the arrow used for heterolytic fission.
11. How many organic products can form when pent-2-ene reacts with HCl? Name them.
12. Why does bromine water stay orange-brown with hexane in the dark?

Answers

1. A species that accepts both electrons of a new covalent bond from its partner, the nucleophile.
2. OH⁻ and CH₃NH₂ (both have lone pairs to donate). NO₂⁺ and H⁺ are electrophiles.
3. CH₃CH₂CH₂I + OH⁻ → CH₃CH₂CH₂OH + I⁻; propan-1-ol.
4. The bromide ion, Br⁻.
5. M(C₃H₇I) = 170.00 g mol⁻¹, so n = 8.50 / 170.00 = 0.0500 mol. Ratio 1 : 1, M(C₃H₇OH) = 60.11 g mol⁻¹, mass = 0.0500 × 60.11 = 3.01 g.
6. CH₃CHBrCH₃ → CH₃CH⁺CH₃ + Br⁻ (arrow from the C–Br bond to Br).
7. Bromocyclohexane, C₆H₁₁Br.
8. The C=C bond donates the electron pair that forms the new C–H bond; HBr’s δ+ hydrogen accepts it.
9. CH₂=CH₂ + H₂O → CH₃CH₂OH; an acid catalyst (for example phosphoric acid).
10. Single-barbed (fish-hook) for one electron in homolytic fission; double-barbed for an electron pair in heterolytic fission.
11. Two: 2-chloropentane and 3-chloropentane, because the alkene is unsymmetrical.
12. Hexane has no C=C bond, only strong, non-polar sigma bonds; without UV light no radicals form, so no reaction occurs.

Where marks are usually lost

• Writing that a nucleophile must carry a negative charge. Water and ammonia are neutral nucleophiles; what counts is giving away an electron pair.
• Defining an electrophile as “a positive ion”. Br₂ and HBr are neutral electrophiles.
• Drawing a curly arrow from Cδ+ towards the nucleophile, which is backwards. Arrows run from electrons to where they go.
• Using a fish-hook arrow for heterolytic fission, or showing radicals as its products.
• Leaving HX out of the equation when water or ammonia is the nucleophile, so hydrogen does not balance.
• Calling the reaction of an alkene with bromine a substitution.
• Miscounting the chain in a nitrile name: the CN carbon is carbon 1.
• Forgetting the acid catalyst when writing the conditions for hydrating an alkene with steam.
• Claiming one definite product from an unsymmetrical alkene at SL instead of recognising that two are possible.
• Writing “alkanes do not react with bromine” without saying the reaction needs UV light.

Official syllabus

Source: the Chemistry guide for the IB Diploma Programme (International Baccalaureate, published February 2023; first assessment 2025). These notes cover Reactivity 3.4, understandings 3.4.1 to 3.4.5, for SL and HL.