Mix hydroxide ion with bromomethane, CH₃Br. One of the oxygen's lone pairs reaches the carbon and becomes a new C–O bond. At the same moment the C–Br bond breaks, and bromide leaves with both of its electrons. The product is methanol.
Hydroxide brought the electron pair for the new bond, so hydroxide is the nucleophile. The carbon that received the pair is the electrophile, covered on the next page, Electrophiles. Bromide left with a pair, so it is the leaving group (see Leaving groups). This one-step swap is the SN2 reaction, which Substitution & Elimination covers in full.
Nearly every reaction in this course pairs a species that brings electrons with a species that receives them, and this page is about the first of the two. Once you can look at a flask of reagents and say which one is the nucleophile, you know where the tail of your first curved arrow goes.
What it is
A nucleophile is any species with an electron pair it can donate to form a new bond. The name means "nucleus-loving": the pair is drawn to an electron-poor atom, one with a full or partial positive charge. In the language of Lewis acids/bases, a nucleophile is a Lewis base. Chemists usually say nucleophile when the pair goes to an atom other than a proton, most often carbon, and base when it goes to a proton. The line is not strict: an alkene giving its π pair to the H of HBr is also called a nucleophile.
The pair can come from three places, and each one gives a family of nucleophiles.
- Anions. Hydroxide (HO⁻), alkoxides (RO⁻), cyanide (⁻CN), thiolates (RS⁻), halides, and hydride donors, which are reagents that deliver H⁻, a hydrogen with a pair of electrons and a negative charge. The most reactive anions in the course are carbanions, carbons that carry a lone pair and a full negative charge. Carbon is less electronegative than oxygen, nitrogen or a halogen, so it holds a negative charge poorly and gives its pair up readily.
- Neutral molecules with a lone pair. Water, alcohols, ammonia, amines and phosphines.
- π bonds. Alkenes, alkynes and aromatic rings. They are weak nucleophiles, yet they drive most of the reactivity of Alkenes & Alkynes and Aromatic Chemistry.
Beginners tend to miss the π family, because a π bond has no charge and no lone pair. Its electrons sit above and below the plane of the molecule, where an electron-poor atom can reach them. The π-bond figures in Bonding show this.
What makes one stronger
Nucleophilicity, or nucleophile strength, is how fast a species uses its pair to form a new bond. Six factors control it. Four do most of the work: charge, electronegativity, polarizability, and crowding around the donating atom, called sterics. Two matter in closer comparisons: delocalization and hybridization. The solvent can also change the outcome, so it gets its own section below.
Charge. Let methoxide, CH₃O⁻, and methanol, CH₃OH, each attack bromomethane. Methoxide reacts far faster. The rule is that a negatively charged species is a stronger nucleophile than its own neutral partner: HO⁻ beats H₂O, RO⁻ beats ROH, and H₂N⁻ beats NH₃. Between different atoms or structures, the other factors below can outweigh charge.
Here is why. When methanol loses its O–H proton, the pair that held the hydrogen stays on oxygen as a third lone pair. The oxygen keeps the same nuclear charge but now carries a full negative charge, so it holds its pairs less tightly, and they sit higher in energy. Less energy is then needed to start the new bond, so the reaction runs faster. The Energy diagrams page, later in this chapter, draws that energy as a hill the reaction climbs.
So chemists often remove the proton from an alcohol before using it. Methanol is a middling nucleophile, while sodium methoxide, NaOMe, is a strong one; that is why NaOMe appears in so many reaction conditions.
Electronegativity. Compare the amide ion, H₂N⁻, with hydroxide, HO⁻. The amide ion is not the amide functional group; it is ammonia with one proton removed. Each carries one negative charge, and nitrogen and oxygen sit side by side in the same row. Oxygen is the more electronegative atom (3.44 against 3.04), so it holds its lone pairs more tightly and shares them less readily. The amide ion is the stronger nucleophile.
Across the whole row the order is H₃C⁻ > H₂N⁻ > HO⁻ > F⁻: within a row, nucleophilicity falls as electronegativity rises. (The values are the ones from Electronegativity.) The solvent does not change this order. The rule works because atoms in one row are similar in size. Down a column the size changes a lot, and polarizability takes over.
Polarizability. Iodide, I⁻, and fluoride, F⁻, sit at opposite ends of one column. In water or methanol, iodide is by far the better nucleophile, and the column runs I⁻ > Br⁻ > Cl⁻ > F⁻. Part of the reason is that iodide is polarizable.
An atom is polarizable when its electron cloud is easy to push out of shape. Iodine is a large atom that holds its outer electrons loosely and far from the nucleus, so its cloud deforms easily. As iodide nears a δ+ carbon, its cloud stretches toward the carbon and starts to bond before the atoms are close. Fluorine is small and holds its electrons tightly and close in, so its cloud barely changes. Big and loose means polarizable; small and tight means not.
The same holds for sulfur against oxygen. Sulfur is the larger, more polarizable atom, so a thiolate, RS⁻, is a better nucleophile than an alkoxide, RO⁻. Polarizability comes back in leaving-group ability, and it already appeared in the acidity trend down a group (Factors affecting acidity).
The rest of iodide's advantage comes from the solvent. In a solvent that cannot donate hydrogen bonds, the halide order reverses, so polarizability alone cannot explain the order. The solvent section below shows why.
Sterics. A nucleophile has to physically reach the atom it attacks, so bulky groups near the donating atom slow it down, sometimes drastically. tert-Butoxide, (CH₃)₃CO⁻, attacks through the same O⁻ as methoxide, but its three methyl groups keep it away from a crowded carbon. Crowding is also what separates nucleophilicity from basicity, so the last section of this page draws this case.
Delocalization. Compare acetate, CH₃COO⁻, with methoxide, CH₃O⁻. Acetate spreads its charge over two oxygens by resonance, so neither oxygen carries all of it. Methoxide keeps the whole charge on its one oxygen. Acetate is a mediocre nucleophile, and methoxide is a strong one.
The rule is that an anion whose charge is spread by resonance is a weaker nucleophile than one whose charge sits on a single atom. With less electron density at any one site, it has less to donate, and for the same reason it is also a weaker base. The idea comes from Resonance: whatever stabilizes an anion also makes it less eager to attack.
Hybridization of the donating pair. Trimethylamine, pyridine and acetonitrile all donate from a nitrogen lone pair. In trimethylamine, (CH₃)₃N, the nitrogen is sp³ and the pair has 25% s character. In pyridine, a benzene-like ring with one nitrogen in it, the ring nitrogen is sp² and its pair has 33% s character. In acetonitrile, CH₃C≡N, the nitrogen is sp and its pair has 50% s character.
More s character holds a pair closer to the nucleus, so the pair is less available. Trimethylamine is the best nucleophile of the three, and acetonitrile is the weakest. Factors affecting acidity used the same s-character argument for anions. Hybridization is a smaller effect than charge or sterics, but it explains several comparisons that otherwise look arbitrary.
Solvent matters too, and it can reverse the order
A protic solvent has an O–H or N–H bond, so it can donate hydrogen bonds. A hydrogen bond is the attraction between a hydrogen on O or N and a lone pair or negative charge nearby. Water, methanol, ethanol and acetic acid are protic. An aprotic solvent has no O–H or N–H bond. Acetone, DMSO (dimethyl sulfoxide, (CH₃)₂S=O), DMF (dimethylformamide), acetonitrile and THF (tetrahydrofuran) are aprotic. Aprotic solvents can still accept hydrogen bonds; they just cannot donate them. Both kinds can be highly polar, and the choice between them is one of the most consequential in Substitution & Elimination.
Dissolve a fluoride salt in methanol. Each fluoride ion ends up inside a tight shell of methanol molecules, all pointing their O–H hydrogens at the ion. Fluoride is small and its charge is packed into a small volume, so these hydrogen bonds are short and strong. Fluoride has to shed part of that shell before it can bond. That costs energy up front, so fluoride reacts slowly, and its nucleophilicity collapses. Iodide is surrounded by methanol too, but its charge is spread over a large ion, so each hydrogen bond is longer and weaker, and the shell holds it only loosely. So in methanol or water the halide order is I⁻ > Br⁻ > Cl⁻ > F⁻, the exact reverse of the basicity order.
In a polar aprotic solvent such as DMSO or DMF, there is no O–H to cage anything. These solvents surround cations well, through their oxygen atoms, and anions poorly, so the anion is left bare and much more reactive. The halide order flips back to the basicity order, F⁻ > Cl⁻ > Br⁻ > I⁻, and every anion is a stronger nucleophile than it was in methanol, often by several orders of magnitude.
Sodium cyanide reacts with 1-bromobutane: cyanide replaces bromide, the same kind of swap as hydroxide and bromomethane at the top of this page. In DMSO, a polar aprotic solvent, the reaction is fast. In methanol it is much slower.
Nothing about the substrate changed. In methanol, cyanide, a small anion, sits in a shell of hydrogen-bonded solvent that it must partly shed before it can attack. In DMSO there is no such shell. In SN2 chemistry the solvent works like a reagent, not a background detail.
The size of the gain depends on the anion. Iodide gains the least from the switch, because methanol holds it loosely to begin with; small anions such as chloride and cyanide gain far more. The classic iodide reaction, sodium iodide with 1-bromobutane, is run in acetone for a second reason: acetone dissolves NaI but not NaBr, so NaBr precipitates and pulls the reaction forward.
Rank CH₃S⁻, CH₃O⁻, CH₃OH and CH₃COO⁻. Apply the factors one at a time, and stop as soon as one of them separates a pair.
Charge first. CH₃OH is the only neutral species, and its own anion, CH₃O⁻, is in the set, so it ranks below methoxide. Here it is also below the other two anions: its oxygen has no charge, and its lone pairs are held tightly. Methanol goes last.
Delocalization next, among the three anions. CH₃COO⁻ spreads its charge over two oxygens, while CH₃O⁻ and CH₃S⁻ each keep it on one atom. Acetate is the weakest of the three anions and goes third.
Then the column, between the two localized anions. CH₃S⁻ and CH₃O⁻ differ only in the donating atom, and sulfur sits below oxygen. Sulfur is larger and more polarizable, and methanol cages it far less tightly. The thiolate wins.
Sterics does not come into it, because each species has a single methyl group and none of the donating atoms is crowded. Naming the factor you did not need is part of the answer.
Final order: CH₃S⁻ > CH₃O⁻ > CH₃COO⁻ > CH₃OH. This is not the basicity order, since methoxide is a much stronger base than the thiolate. The last section of this page is about that difference.
When a nucleophile has two ends: ambident nucleophiles
Cyanide, ⁻C≡N, has a lone pair on carbon and another on nitrogen, so it could attack through either atom. A nucleophile with two atoms that can attack is called ambident. With bromomethane, cyanide gives acetonitrile, CH₃C≡N.
Cyanide almost always attacks through carbon, although you might expect the charge, and so the attack, to sit on nitrogen, the more electronegative atom. Because carbon is less electronegative, its lone pair is the more available and the more polarizable of the two, and the C–C bond it forms is stronger than the C–N bond that attack through nitrogen would give. So cyanide adds a carbon, not a nitrogen, which makes it a simple way to add one carbon to a chain.
Enolates raise the same question more sharply. An enolate is the anion left when a base removes a hydrogen from the carbon next to a C=O; the figure draws the enolate of acetone. One resonance form puts the negative charge on oxygen, and the other puts it on carbon. The form with the charge on oxygen contributes more, yet the enolate usually reacts at carbon. Which end reacts depends on the electrophile and the conditions, and Enolate Chemistry takes that up properly.
For now, the useful habit is to notice when a nucleophile has more than one atom that could attack, and to ask which one does, rather than assuming the charge marks the spot.
Nucleophilicity is not basicity
Hydroxide and tert-butoxide show the difference. tert-Butoxide is the stronger base: its conjugate acid, tert-butanol, is a weaker acid than water. Yet toward a crowded carbon it is a far weaker nucleophile than hydroxide.
The two ideas often rise and fall together, which is why they are easy to confuse, but they measure different things. Basicity is thermodynamic. It describes where a proton-transfer equilibrium settles, and pKa measures it (Acids & Bases). Nucleophilicity is kinetic. It describes how fast a species attacks an electron-poor atom, usually carbon.
Crowding pulls the two apart. A proton is tiny and sits on the outside of a molecule, so tert-butoxide can reach it past its own methyl groups. A carbon that carries other groups is buried among them, and the bulky tert-butyl group cannot get close enough to it. Chemists use this on purpose. When Substitution & Elimination wants elimination (removing an H and a leaving group from neighboring carbons to make a C=C) rather than substitution, it picks a bulky base for exactly this reason.
Polarizability pulls them apart in the other direction. Iodide is an excellent nucleophile and a very weak base. A thiolate, RS⁻, is a stronger nucleophile than an alkoxide, RO⁻, and at the same time a much weaker base; Lewis acids/bases flagged this pair under hard and soft. Any reagent that is a good nucleophile but a weak base tends to favor substitution over elimination.
What carries forward
Telling a good nucleophile that is a weak base from a strong base that is bulky decides SN2 against E2 in Substitution & Elimination, and it is one of the most heavily tested judgments in the course. The nucleophile decides the product in nucleophilic addition (Carbonyl Chemistry) and in acyl substitution (Carboxylic Acids & Derivatives). And most alkene reactions, along with electrophilic aromatic substitution, start from the fact that a π bond is a nucleophile.