What makes a species nucleophilic?

A nucleophile is a reactant that donates both electrons in a new bonding pair to form a bond with an electron-deficient reaction partner. In classroom terms: it has enough electron density to offer a lone pair.

A nucleophile doesn’t need a negative charge. What it does need is an available lone pair or an electron-rich bond that can be donated. Neutral examples include H₂O and NH₃. Negatively charged examples include OH⁻, CN⁻, Cl⁻, Br⁻ and I⁻. A quaternary ammonium ion such as (CH₃)₄N⁺ is not a nucleophile at nitrogen because the nitrogen has no lone pair left to donate.

In mechanisms, chemists often write Nu: or Nu⁻ for the nucleophile. The colon points to the electron pair. A negative charge, when present, often helps donation, but charge is not what defines a nucleophile.

Recognizing nucleophiles in reactions

Find the species that provides the electron pair for the new bond. In the reaction

CH₃CH₂Cl + OH⁻ → CH₃CH₂OH + Cl⁻

OH⁻ is the nucleophile because an oxygen lone pair forms the new C–O bond. In

CH₃Br + NH₃ → CH₃NH₃⁺ + Br⁻

NH₃ is the nucleophile because the nitrogen lone pair forms the new C–N bond.

Substitution: one group in, one group out

A nucleophilic substitution reaction is a reaction where a nucleophile donates an electron pair to make a new bond, while another bond breaks and a group leaves. A leaving group is an atom or group that leaves a substrate with the electron pair from the broken bond.

A useful general pattern is:

Nu:⁻ + R–X → R–Nu + X⁻

Here, R is an organic group containing the electron-deficient atom being attacked. Nu:⁻ is the nucleophile, and X is the leaving group. For now, focus on the movement of electron pairs rather than the exact timing of the mechanism.

For example, chloroethane reacts with hydroxide ions in aqueous solution:

CH₃CH₂Cl + OH⁻ → CH₃CH₂OH + Cl⁻

Chlorine is more electronegative than carbon, so the C–Cl bond is polar and the carbon bonded to chlorine is δ⁺. A lone pair on OH⁻ forms the new C–O bond. At the same time, the electron pair in the C–Cl bond moves onto chlorine, producing Cl⁻ as the leaving group.

Writing descriptions of electron-pair movement

A curly arrow shows the movement of an electron pair. In nucleophilic substitution, one curly arrow starts at the nucleophile lone pair or negative charge and points to the electron-deficient carbon. A second curly arrow starts at the C–X bond and points to X, showing heterolytic breaking of that bond.

Do not write “the hydroxide attacks the chlorine”. It attacks the carbon. The chlorine leaves because the C–Cl bonding pair moves onto chlorine.

Unequal bond breaking

Heterolytic fission is the breaking of a covalent bond where both bonding electrons go with one of the two fragments formed. Ions are produced: the fragment that takes the electron pair becomes an anion, while the other fragment becomes a cation.

For a polar molecule A–B where B is more electronegative:

A–B → A⁺ + B⁻

The curly arrow begins at the A–B bond and points to B, since the electron pair in the bond moves onto B.

In bromomethane, the C–Br bond is polar, Cδ⁺–Brδ⁻. When heterolytic fission occurs, it forms a methyl carbocation and a bromide ion:

CH₃Br → CH₃⁺ + Br⁻

A carbocation is an organic cation with the positive charge on a carbon atom. Carbocations are usually short-lived intermediates because a positively charged carbon is strongly electron-deficient.

Heterolytic versus homolytic fission

Here’s the main contrast with radical chemistry. Homolytic fission is bond breaking where each atom gets one electron from the bond, producing radicals; chemists show it with single-barbed fish-hook arrows. Heterolytic fission sends an electron pair to one atom, producing ions; it is shown with a double-barbed curly arrow.

In nucleophilic substitution, the leaving group normally takes both electrons from the bond to carbon. So, under these conditions, halogenoalkanes give halide ions rather than halogen radicals.

Electron-pair acceptors in organic reactions

An electrophile is a reactant that bonds to an electron-rich reaction partner by accepting both electrons in the new bonding pair. It is electron-deficient: it may have a full positive charge, or it may contain an atom with a partial positive charge.

Positively charged electrophiles include H⁺, CH₃⁺ and NO₂⁺. Neutral electrophiles include molecules with strongly polar bonds or an incomplete octet, such as BF₃, and molecules containing a polar carbonyl group, such as aldehydes and ketones.

In BF₃, boron is electron-deficient, so it can accept a lone pair. In a carbonyl compound, the C=O bond is polar; the carbonyl carbon is δ⁺ and is the electrophilic site.

Recognizing electrophiles in reactions

Look for the atom that receives the electron pair. In

NH₃ + BF₃ → F₃B←NH₃

BF₃ is the electrophile because boron accepts the nitrogen lone pair. In addition of HBr to an alkene, Hδ⁺ is the first electrophilic atom attacked by the electron-rich C=C bond.

Why alkenes react readily

An alkene is an unsaturated hydrocarbon with at least one carbon–carbon double bond. Because the C=C double bond has high electron density, it attracts electrophiles. The reaction type is electrophilic addition, a reaction in which an electrophile starts the reaction and atoms are added across a multiple bond to give a more saturated product.

That’s why alkenes work well as “starting molecules” in industry. The double bond acts as a reactive handle and can be changed into many different functional groups by addition.

Addition of halogens

With a halogen X₂, the halogen atoms add across the double bond:

alkene + X₂ → dihalogenoalkane

For ethene and bromine:

CH₂=CH₂ + Br₂ → CH₂BrCH₂Br

Bromine water is orange-brown, but alkenes decolourize it even in the dark because the C=C bond reacts by addition. Alkanes don’t decolourize bromine water in the dark: their C–C and C–H σ bonds are comparatively strong and non-polar, so alkane bromination needs radical conditions such as UV light.

Addition of hydrogen halides

Hydrogen halides add across the double bond to produce halogenoalkanes:

alkene + HX → halogenoalkane

For but-2-ene:

CH₃CH=CHCH₃ + HBr → CH₃CHBrCH₂CH₃

A symmetrical alkene gives one structural product with HX. With an unsymmetrical alkene, two products may form; the major product is explained later using carbocation stability.

Addition of water

Water adds across a double bond in acidified conditions, forming an alcohol. This reaction is also called hydration, a reaction in which water is added to a molecule.

For ethene:

CH₂=CH₂ + H₂O → CH₃CH₂OH

For a symmetrical alkene such as hex-3-ene, only one alcohol structure is obtained. For an unsymmetrical alkene, addition can occur in two positions, so the major product is predicted using the carbocation argument introduced later.

A broader acid–base idea

A Lewis acid accepts an electron pair. A Lewis base donates an electron pair. This gives a broader definition than the Brønsted–Lowry model, since proton transfer isn't needed.

A Brønsted–Lowry acid donates a proton, H⁺. A Brønsted–Lowry base accepts a proton. When NH₃ accepts H⁺, the lone pair on nitrogen forms a bond to the proton:

NH₃ + H⁺ → NH₄⁺

In this reaction, NH₃ acts as both a Brønsted–Lowry base and a Lewis base. H⁺ acts as both a Brønsted–Lowry acid and a Lewis acid.

Lewis theory also covers reactions where no proton is involved:

BF₃ + NH₃ → F₃B←NH₃

NH₃ donates an electron pair, so it is the Lewis base. BF₃ accepts that pair at boron, so it is the Lewis acid. Since no H⁺ is transferred, this is not a Brønsted–Lowry acid–base reaction.

Applying the theory

In organic chemistry, nucleophiles behave as Lewis bases because they donate electron pairs. Electrophiles behave as Lewis acids because they accept electron pairs. In inorganic chemistry, many metal cations, such as Cu²⁺, act as Lewis acids by accepting electron pairs from ligands.

Here’s the useful link: every Brønsted–Lowry base must have an electron pair available to bind H⁺, so during protonation it behaves as a Lewis base. Not every Lewis acid or base fits the Brønsted–Lowry definition, because many Lewis reactions do not involve protons.

The bond made when a Lewis base meets a Lewis acid

A coordination bond is a covalent bond where the same atom donates both bonding electrons. Once the bond has formed, it’s just a covalent bond; “coordinate” tells you how the bond was formed, rather than marking it out as a separate type of bond with special properties.

In Lewis formulas, draw the donor with a lone pair, and show the acceptor as electron-deficient or carrying a positive charge. You can draw the coordinate bond as a normal covalent line, or as an arrow pointing from the donor to the acceptor.

For example:

:NH₃ + BF₃ → H₃N→BF₃

Nitrogen in ammonia donates its lone pair to boron in boron trifluoride. So ammonia acts as a nucleophile and Lewis base. Boron trifluoride acts as an electrophile and Lewis acid.

AlCl₃ is a useful inorganic example. Aluminium is electron-deficient, so AlCl₃ can accept a lone pair. Chlorine atoms on another AlCl₃ molecule have lone pairs, allowing two AlCl₃ molecules to form Al₂Cl₆, with bridging chloride atoms donating electron pairs to aluminium centres.

Interpreting Lewis formulas

When you interpret a Lewis acid–base diagram, ask two questions:

• Which atom supplies the electron pair? That species is the Lewis base and nucleophile.
• Which atom receives the electron pair? That species is the Lewis acid and electrophile.

Curly arrows and coordinate-bond arrows are connected, but they don’t mean exactly the same thing. A curly arrow shows electron-pair movement during the reaction; a coordinate-bond arrow in the product shows the donor-to-acceptor origin of the bond.

Complex ions are Lewis acid–base products

A ligand is an ion or molecule that donates an electron pair to a central metal ion, forming a coordination bond. A complex ion is a charged species made up of a central metal ion bonded to surrounding ligands by coordination bonds.

Transition element cations work well as Lewis acids because they carry a positive charge and can accept electron pairs. Common ligands include neutral molecules such as H₂O and NH₃, as well as anions such as Cl⁻, OH⁻ and CN⁻. The ligand acts as the Lewis base/nucleophile; the metal cation acts as the Lewis acid/electrophile.

For example, Cu²⁺ in water forms [Cu(H₂O)₆]²⁺. Six water ligands sit octahedrally around the copper ion.

Deducing charges on complex ions

To find the charge on a complex ion, add the metal ion charge to the total charge from the ligands:

complex ion charge = metal ion charge + total ligand charge

Neutral ligands such as H₂O and NH₃ add zero. Anionic ligands add their charge each time they appear.

In [Co(H₂O)₆]²⁺, all six water ligands are neutral, so the complex ion keeps the same charge as Co²⁺: 2+.

In [CoCl₄]²⁻, cobalt is Co²⁺, while four chloride ligands contribute 4− overall:

2+ + 4(1−) = 2−

So the complex ion is [CoCl₄]²⁻.

You can work backwards too. In [Fe(OH)(H₂O)₅]²⁺, the complex has charge 2+, OH⁻ contributes 1−, and water contributes zero. Iron must therefore be 3+, since 3+ + 1− = 2+.

Examples showing how metal and ligand charges add to give the complex ion charge.

Why halogenoalkanes undergo substitution

A halogenoalkane is an organic compound in which at least one hydrogen atom of an alkane has been replaced by a halogen atom. Since the halogen is more electronegative than carbon, the C–X bond is polar; the carbon bonded to X is δ⁺, so a nucleophile can attack it.

Primary, secondary and tertiary halogenoalkanes don’t all react in the same way. The carbon skeleton affects both steric hindrance and carbocation stability.

The S_N2 mechanism: one concerted step

An S_N2 reaction is a nucleophilic substitution mechanism in which bond formation to the nucleophile and bond breaking to the leaving group happen in one concerted step. “2” means two reacting species are involved in the rate-determining step.

For bromoethane with hydroxide:

CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻

The rate law is:

rate = k₂[RX][Nu⁻], where rate is the change in concentration per time (mol m⁻³ s⁻¹), k₂ is the second-order rate constant (m³ mol⁻¹ s⁻¹), [RX] is the concentration of the halogenoalkane (mol m⁻³), and [Nu⁻] is the concentration of the nucleophile (mol m⁻³).

The nucleophile attacks from the side opposite the leaving group. In the transition state, the C–Nu bond is partly formed and the C–X bond is partly broken. A transition state is the highest-energy arrangement along a reaction pathway; it is not a stable intermediate.

The S_N2 mechanism is stereospecific, meaning that a particular stereochemical arrangement of reactant gives a particular stereochemical arrangement of product. Backside attack at a tetrahedral carbon causes inversion of configuration, like an umbrella turning inside out.

The S_N1 mechanism: two steps through a carbocation

An S_N1 reaction is a nucleophilic substitution mechanism in which the leaving group first departs to form a carbocation intermediate, followed by attack by the nucleophile. “1” means only the halogenoalkane is involved in the rate-determining step.

For 2-chloro-2-methylpropane with hydroxide:

(CH₃)₃CCl + OH⁻ → (CH₃)₃COH + Cl⁻

Step 1, slow: the C–Cl bond breaks heterolytically:

(CH₃)₃CCl → (CH₃)₃C⁺ + Cl⁻

Step 2, fast: the nucleophile attacks the carbocation:

(CH₃)₃C⁺ + OH⁻ → (CH₃)₃COH

The rate law is:

rate = k₁[RX], where k₁ is the first-order rate constant (s⁻¹).

A primary halogenoalkane normally reacts by S_N2 because forming a primary carbocation is unfavourable, and the attacking nucleophile has fairly open access to the carbon. A tertiary halogenoalkane normally reacts by S_N1 because alkyl groups stabilize the tertiary carbocation, while backside attack is crowded. Secondary halogenoalkanes can follow either model, depending on conditions.

Carbocation stability and inductive effects

A positive inductive effect is electron donation through σ bonds from an alkyl group toward an electron-deficient centre. Alkyl groups help spread out the positive charge on a carbocation. Carbocation stability therefore follows:

tertiary > secondary > primary > methyl

This explains why tertiary halogenoalkanes favour S_N1.

Energy profiles and models

An S_N2 energy profile has one maximum because there is one transition state and no intermediate. An S_N1 energy profile has two maxima with a carbocation intermediate between them; the first maximum is usually higher because carbocation formation is the slow step.

S_N1 and S_N2 are models. They are useful because they connect structure, rate law, stereochemistry and mechanism. They aren’t magic labels: real secondary halogenoalkanes, mixed solvents and competing reactions may not behave as a perfect textbook case.

The C–X bond controls the trend

The identity of the leaving group affects the rate of nucleophilic substitution. For comparable halogenoalkanes, use the usual rate order:

RI > RBr > RCl

Iodoalkanes react fastest; chloroalkanes are the slowest of these three.

This mainly comes down to bond strength. The C–I bond is weaker than the C–Br bond, and the C–Br bond is weaker than the C–Cl bond, so C–I undergoes heterolytic fission more easily. When the bond to carbon is weaker, the halide ion leaves more readily.

Comparison of leaving group effects on substitution rate for RCl, RBr and RI.

I⁻ is a better leaving group than Cl⁻ because iodide is larger and more polarizable. Its negative charge is spread over a larger ion, making it more stable. Chloride is smaller and holds the charge more densely, so it is less stable as the departing ion.

The syllabus comparison is RCl, RBr and RI. Solvent effects and detailed mechanism effects on the rate do exist, but they’re not the focus here. For this topic, use the leaving group trend above when predicting relative rates.

General pattern

In electrophilic addition, the electron-rich C=C bond donates an electron pair to an electrophile. One carbon makes a new bond to the electrophile; the other carbon sits in the carbocation intermediate. A nucleophile attacks that carbocation next.

When you draw the mechanism, start the first curly arrow at the C=C bond, not at a carbon atom label. Point it towards the electron-deficient atom in the electrophile.

Halogens and symmetrical alkenes

Bromine, Br₂, is non-polar. As it gets close to the high electron density of a C=C bond, though, it becomes temporarily polarized. The nearer bromine becomes δ⁺, so that bromine atom acts as the electrophile.

For ethene:

CH₂=CH₂ + Br₂ → CH₂BrCH₂Br

In the mechanism, the C=C bond attacks Brδ⁺. The Br–Br bond breaks heterolytically to Br⁻, then Br⁻ attacks the carbocation to form 1,2-dibromoethane.

Hydrogen halides and symmetrical alkenes

Hydrogen halides are already polar: Hδ⁺–Xδ⁻. The alkene attacks Hδ⁺, the H–X bond breaks heterolytically to X⁻, and X⁻ attacks the carbocation.

For but-2-ene:

CH₃CH=CHCH₃ + HBr → CH₃CHBrCH₂CH₃

But-2-ene is symmetrical, so adding H and Br in either direction gives the same product: 2-bromobutane.

Water and symmetrical alkenes

Alkenes hydrate in acidified solution. The proton, H⁺, acts as the electrophile. First the C=C bond is protonated, giving a carbocation. H₂O then attacks as a nucleophile to form an oxonium ion. Finally, loss of a proton gives the alcohol and regenerates H⁺.

For hex-3-ene:

CH₃CH₂CH=CHCH₂CH₃ + H₂O → CH₃CH₂CH(OH)CH₂CH₂CH₃

The acid works as a catalyst: H⁺ is used in the first step and regenerated in the final step.

Predicting the major product

An unsymmetrical alkene has the two carbon atoms of the C=C bond attached to different groups. When it reacts with HX or water, either carbon may form a carbocation, so two products may be possible.

The major product comes from the more stable carbocation. The order of carbocation stability is:

tertiary > secondary > primary > methyl

Alkyl groups donate electron density through the positive inductive effect, which reduces the intensity of the positive charge.

Markovnikov’s rule as a shortcut

Markovnikov’s rule predicts that, when H–X or H–OH adds across an unsymmetrical alkene, H usually bonds to the double-bond carbon already bearing more hydrogen atoms. That pathway forms the more stable carbocation.

With propene and HBr, protonation can produce either a primary or a secondary carbocation. The secondary carbocation is more stable, so Br⁻ attacks that carbocation and the major product is 2-bromopropane:

CH₃CH=CH₂ + HBr → CH₃CHBrCH₃

1-bromopropane is the minor product because it would form through the less stable primary carbocation.

The same reasoning applies to hydration. When propene reacts with water in acid, the major product is propan-2-ol, since protonation first gives the more stable secondary carbocation before water attacks:

CH₃CH=CH₂ + H₂O → CH₃CHOHCH₃

Don’t memorize “the OH goes to the middle” as a rule. Ask instead: which first step gives the more stable carbocation?