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

This page teaches the opening understandings of Reactivity 3.4 in IB DP Chemistry, the part on nucleophiles, electrophiles and substitution. Its reference is the IB’s DP Chemistry guide, first assessed in 2025 (the February 2023 edition), and it is written for exam sessions from 2025. We cover Reactivity 3.4.1 to 3.4.5. All five are SL and HL content, so every DP chemist needs them. The HL-only understandings (3.4.6 onwards) have their own page.

Use the IB DP chemistry hub for the rest of the course, tick topics off on our checklist, and find gaps quickly with a free diagnostic. Afterwards, use the revision notes and try the practice questions. For a view of how all four parts of Reactivity 3 fit together, see the Reactivity 3 overview.

What this unit covers

Reactivity 3.4 is built around one guiding question: what goes on when a reactant shares an electron pair with another species?

Understanding	Skills you need	Who sits it
Reactivity 3.4.1	Recognise nucleophiles in reactions, both neutral and negatively charged	SL and HL
Reactivity 3.4.2	Deduce equations for nucleophilic substitution and describe and explain how the electron pairs move	SL and HL
Reactivity 3.4.3	Explain, with equations and curly arrows, how heterolytic fission forms ions	SL and HL
Reactivity 3.4.4	Recognise electrophiles in reactions, both neutral and positively charged	SL and HL
Reactivity 3.4.5	Explain why alkenes undergo electrophilic attack; deduce equations for their reactions with water, halogens and hydrogen halides	SL and HL

Two depth limits matter. The guide says further mechanism detail for nucleophilic substitution is not required at SL, and the mechanisms of alkene addition are not assessed at SL. HL students meet those mechanisms later, in Reactivity 3.4.9 and 3.4.11 (HL only).

The key idea: who gives the electron pair?

Every reaction in this unit makes a new covalent bond in which both electrons come from one reactant. That reactant is the nucleophile; the partner that receives the pair is the electrophile. The two always come together.

A curly arrow with a full, double-barbed head shows one electron pair on the move. Its tail sits on the pair’s present home: a lone pair, a negative charge or a bond. Its head shows the pair’s destination: an atom that receives it, or the gap in which a new bond will form.

Reactivity 3.4.1: Nucleophiles

A nucleophile gives both electrons of a new covalent bond to its partner, the electrophile. It needs a spare electron pair, usually a lone pair, and it may be neutral or a negative ion. The guide asks you to recognise both kinds.

Negatively charged nucleophiles	Neutral nucleophiles
hydroxide, OH⁻	water, H₂O (lone pairs on O)
cyanide, CN⁻ (lone pair on C)	ammonia, NH₃ (lone pair on N)
halide ions, Cl⁻, Br⁻, I⁻	alcohols, e.g. CH₃OH (lone pairs on O)
alkoxide ions, e.g. CH₃O⁻	amines, e.g. CH₃NH₂ (lone pair on N)

Worked example 1: spotting the nucleophile

Identify the nucleophile in each reaction.

(a) CH₃CH₂Br + NH₃ → CH₃CH₂NH₂ + HBr (b) CH₃I + CN⁻ → CH₃CN + I⁻ (c) (CH₃)₂CHCl + H₂O → (CH₃)₂CHOH + HCl

Method: find the species whose atom ends up bonded to carbon using its own electron pair. (a) The nitrogen of NH₃ ends up bonded to carbon. It used its lone pair, so ammonia is the nucleophile (neutral). (b) The carbon of CN⁻ forms the new C–C bond with its lone pair, so the cyanide ion is the nucleophile (negative). (c) The oxygen of H₂O bonds to carbon, so water is the nucleophile (neutral).

In each case the halogenoalkane is the electrophile.

Reactivity 3.4.2: Nucleophilic substitution

In a nucleophilic substitution, the nucleophile uses its electron pair to make a new bond to carbon, and a different bond to that carbon breaks at the same time. Whatever departs, carrying the electrons of the broken bond, is the leaving group.

Halogenoalkanes are the standard substrates. A halogen pulls bonding electrons towards itself more strongly than carbon does, so in C–X the carbon is δ+ and the halogen is δ−. The nucleophile is drawn to that δ+ carbon.

Describing the movement of electron pairs

For the reaction of 1-bromo-4-methylpentane with hydroxide ions:

(CH₃)₂CHCH₂CH₂CH₂Br + OH⁻ → (CH₃)₂CHCH₂CH₂CH₂OH + Br⁻

1. A lone pair on the oxygen of OH⁻ is attracted to the δ+ carbon bonded to bromine.
2. That lone pair is donated to the carbon, forming a new C–O bond (curly arrow from the O lone pair to the carbon).
3. At the same time the C–Br bonding pair shifts wholly onto bromine (a second curly arrow, drawn from the C–Br bond onto the bromine atom).
4. Bromine leaves as a bromide ion, Br⁻, the leaving group. The product is 4-methylpentan-1-ol.

Worked example 2: deducing products

Write the equation for 1-chloro-2-methylpropane, (CH₃)₂CHCH₂Cl, reacting with (a) hydroxide ions, (b) cyanide ions, (c) ammonia.

Method: swap the halogen for the nucleophile’s bonding atom, then balance charges.

(a) (CH₃)₂CHCH₂Cl + OH⁻ → (CH₃)₂CHCH₂OH + Cl⁻ (2-methylpropan-1-ol) (b) (CH₃)₂CHCH₂Cl + CN⁻ → (CH₃)₂CHCH₂CN + Cl⁻ (3-methylbutanenitrile: the chain gains a carbon, and the nitrile carbon is C1 when naming) (c) (CH₃)₂CHCH₂Cl + NH₃ → (CH₃)₂CHCH₂NH₂ + HCl (2-methylpropan-1-amine)

In (c) the neutral nucleophile must also lose an H⁺, which leaves with the chloride as HCl. With excess ammonia this becomes NH₄Cl, so the equation uses 2NH₃.

Reactivity 3.4.3: Heterolytic fission

Heterolytic fission is the breaking of a covalent bond in which both bonding electrons stay with one of the two fragments. The fragment that keeps the pair becomes a negative ion. The other becomes a positive ion. It is usually the more electronegative atom that keeps the pair.

Show it with one full curly arrow from the middle of the bond to the atom that keeps the electrons:

(CH₃)₃C–I → (CH₃)₃C⁺ + I⁻ (curly arrow: from the C–I bond to I)

The carbon fragment has lost its share of the bonding pair, so it now has only six outer electrons and a positive charge. An ion with a positively charged carbon like this is a carbocation. Iodine leaves with eight outer electrons and a negative charge.

Contrast with homolytic fission (linking question, Reactivity 3.3)

	Homolytic fission	Heterolytic fission
Electrons go	one to each fragment	both to one fragment
Products	two radicals	a cation and an anion
Arrow	single-barbed (fish-hook), one per electron	double-barbed, one per pair
Typical setting	Cl₂ in UV light	polar bonds such as C–Cl, C–I, H–Br

The bond breaking in nucleophilic substitution is heterolytic: the leaving group always departs as an ion carrying the former bonding pair.

Worked example 3: heterolysis of H–Br

H–Br → H⁺ + Br⁻. The arrow runs from the centre of the H–Br bond to Br. Bromine is more electronegative, so it keeps the pair and becomes Br⁻; hydrogen is left with no electrons, as H⁺.

Reactivity 3.4.4: Electrophiles

An electrophile receives both electrons of the new bond from its partner, the nucleophile. It has, or develops, a region of low electron density. The guide asks you to recognise both positive and neutral electrophiles.

Positively charged electrophiles	Neutral electrophiles
H⁺ (often as H₃O⁺)	hydrogen halides, H–X (the δ+ hydrogen)
carbocations, e.g. (CH₃)₃C⁺	halogens, Br₂ and Cl₂ (a dipole is induced near a C=C bond)
the nitronium ion, NO₂⁺	halogenoalkanes (the δ+ carbon)

Worked example 4: classifying both partners

In CH₂=CH₂ + HBr → CH₃CH₂Br, identify the nucleophile and the electrophile.

Br is more electronegative than H, so HBr is polar: Hδ+–Brδ−. The δ+ hydrogen accepts an electron pair, so HBr is the electrophile (neutral). The pair it accepts comes from the C=C double bond, so ethene acts as the nucleophile. It donates both electrons of the new C–H bond.

Reactivity 3.4.5: Alkenes and electrophilic addition

The C=C double bond is a sigma bond plus a pi bond. The pi electrons lie above and below the plane of the molecule, so electrons are concentrated around the double bond (high electron density), and electron-poor species are pulled towards it. The pi bond breaks, two new single bonds form and nothing is lost: this is electrophilic addition.

Learn the three reactions with symmetrical alkenes (where both ends of the C=C are the same, so there is only one possible product):

Reagent	Product type	Example
Halogen, X₂ (Cl₂, Br₂)	dihalogenoalkane	cyclopentene + Br₂ → 1,2-dibromocyclopentane
Hydrogen halide, HX (HCl, HBr, HI)	halogenoalkane	dec-5-ene + HCl → 5-chlorodecane
Water (as steam, with an acid catalyst such as phosphoric acid)	alcohol	hex-3-ene + H₂O → hexan-3-ol

Worked example 5: deducing addition equations

Write equations for the reaction of hex-3-ene, CH₃CH₂CH=CHCH₂CH₃, with (a) bromine, (b) hydrogen iodide, (c) water.

Method: break the double bond and put one fragment of the reagent on each carbon.

(a) CH₃CH₂CH=CHCH₂CH₃ + Br₂ → CH₃CH₂CHBrCHBrCH₂CH₃ (3,4-dibromohexane) (b) CH₃CH₂CH=CHCH₂CH₃ + HI → CH₃CH₂CH₂CHICH₂CH₃ (3-iodohexane) (c) CH₃CH₂CH=CHCH₂CH₃ + H₂O → CH₃CH₂CH₂CH(OH)CH₂CH₃ (hexan-3-ol)

Check: product formula = alkene + reagent (C₆H₁₂ + H₂O gives C₆H₁₄O).

If the alkene is not symmetrical (propene, for example), HX and H₂O can add two ways round and give two isomeric products. At SL you only need to see that two are possible. Predicting the major one is Reactivity 3.4.12 (HL only).

Why bromine water is decolourised by alkenes in the dark but not by alkanes (linking question)

Orange-brown bromine water turns colourless with an alkene, even in the dark: the electron-rich C=C bond attacks bromine and uses it up by addition. An alkane offers only C–C and C–H bonds, which are strong and barely polar. It reacts with bromine only by radical substitution, which needs UV light to break Br–Br homolytically (Reactivity 3.3), so in the dark the colour stays.

Why alkenes are “starting molecules” in industry (linking question, Structure 2.4)

The reactive double bond turns one feedstock into many products: alcohols (with water), halogenoalkanes (with HX, then converted further by nucleophilic substitution) and addition polymers.

Worked example 6: a yield from an addition reaction

A student adds excess bromine to 3.50 g of cyclopentene, C₅H₈, and obtains 9.40 g of 1,2-dibromocyclopentane, C₅H₈Br₂. Calculate the percentage yield. (Use Ar: C 12.01, H 1.01, Br 79.90.)

1. M(C₅H₈) = 5(12.01) + 8(1.01) = 68.13 g mol⁻¹
2. n(C₅H₈) = 3.50 / 68.13 = 0.0514 mol
3. The ratio is 1 : 1, so n(C₅H₈Br₂) = 0.0514 mol.
4. M(C₅H₈Br₂) = 68.13 + 2(79.90) = 227.93 g mol⁻¹
5. Theoretical mass = 0.0514 × 227.93 = 11.7 g
6. Percentage yield = (9.40 / 11.7) × 100 = 80.3%

Common errors

• Starting a curly arrow at a positive charge or at an atom’s symbol instead of at a lone pair or a bond. Arrows show where electrons go, so they start at electrons.
• Calling the halogenoalkane the nucleophile because it “reacts”. The halogenoalkane supplies the δ+ carbon; it is the electrophile.
• Forgetting that neutral species (H₂O, NH₃, Br₂, HBr) can be nucleophiles or electrophiles. The guide asks for both charged and neutral examples.
• Leaving the HX product off substitution by a neutral nucleophile, which leaves the equation unbalanced in hydrogen.
• Naming a nitrile without counting the CN carbon: (CH₃)₂CHCH₂CN is a butanenitrile, not a propanenitrile.
• Calling alkene + halogen a substitution. Nothing leaves; it is addition.

Official syllabus

Source: the Chemistry guide for the IB Diploma Programme (International Baccalaureate, published February 2023; first assessment 2025), Reactivity 3.4, understandings 3.4.1 to 3.4.5, which are standard level and higher level content.