What makes a species nucleophilic?

A nucleophile is a reactant that bonds to an electron-deficient reaction partner by donating both electrons in the new bonding pair. In class terms: it has enough electron density to offer a lone pair.

A nucleophile doesn't need to carry a negative charge. What matters is that it has an available lone pair or an electron-rich bond it can donate. Neutral examples include $H_2O$ and $NH_3$. Negatively charged examples include $OH^-$, $CN^-$, $Cl^-$, $Br^-$ and $I^-$. A quaternary ammonium ion such as $(CH_3)_4N^+$ is not a nucleophile at nitrogen because the nitrogen has no lone pair left to donate.

In mechanisms, chemists often write $\text{Nu:}$ or $Nu^-$ for the nucleophile. The colon shows the electron pair; a negative charge, when present, often helps donation, but charge itself is not the definition.

Recognizing nucleophiles in reactions

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

$CH_3CH_2Cl + OH^- \to CH_3CH_2OH + Cl^-$

$OH^-$ is the nucleophile because an oxygen lone pair becomes the new C–O bond. In

$CH_3Br + NH_3 \to CH_3NH_3^+ + Br^-$

$NH_3$ 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 happens when a nucleophile donates an electron pair to make a new bond, while another bond breaks and a group leaves. A leaving group is the atom or group that leaves a substrate with the electron pair from the broken bond.

A useful general pattern is:

$$Nu:^ - + R-X \to R-Nu + X^-$$

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 electron-pair movement rather than the exact timing of the mechanism.

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

$$CH_3CH_2Cl + OH^- \to CH_3CH_2OH + Cl^-$$

The C–Cl bond is polar because chlorine is more electronegative than carbon, so the carbon bonded to chlorine is $\delta^+$. 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 breaks a covalent bond so that both bonding electrons stay 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\text{–}B \to A^+ + B^-\

The curly arrow starts at the $A$–$B$ bond and ends on $B$, because the electron pair in the bond moves to $B$.

In bromomethane, the $C$–$Br$ bond is polar, $C^{\delta+}\text{–}Br^{\delta-}$. Heterolytic fission forms a methyl carbocation and a bromide ion:

CH_3Br \to CH_3^+ + Br^-\

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

Heterolytic versus homolytic fission

This is the main contrast with radical chemistry. Homolytic fission breaks a bond so that each atom receives one electron from the bond, giving radicals; it is shown with single-barbed fish-hook arrows. In heterolytic fission, an electron pair moves to one atom, giving 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. That’s why halogenoalkanes give halide ions, not halogen radicals, under these conditions.

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 $H^+$, $CH_3^+$ and $NO_2^+$. Neutral electrophiles include molecules with strongly polar bonds or an incomplete octet, such as $BF_3$, and molecules with a polar carbonyl group, such as aldehydes and ketones.

In $BF_3$, the boron atom is short of electrons and can accept a lone pair. In a carbonyl compound, the $C=O$ bond is polar, so the carbonyl carbon is $\delta^+$ and acts as the electrophilic site.

Recognizing electrophiles in reactions

Look for the atom that receives the electron pair. In

$$NH_3 + BF_3 \to F_3B \leftarrow NH_3$$

$BF_3$ is the electrophile because boron accepts the nitrogen lone pair. When HBr adds to an alkene, $H^{\delta+}$ 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. The $C=C$ double bond has high electron density, so it attracts electrophiles. Alkenes therefore react by electrophilic addition, a reaction in which an electrophile starts the reaction and atoms add across a multiple bond, producing a more saturated product.

That double bond makes alkenes useful “starting molecules” in industry. It acts as a reactive handle and can be changed into many different functional groups by addition.

Addition of halogens

For ethene and bromine:

$CH_2=CH_2 + Br_2 \to CH_2BrCH_2Br$

Bromine water is orange-brown, but alkenes decolourize it, even in the dark, because the $C=C$ bond undergoes addition. Alkanes don’t decolourize bromine water in the dark. Their C–C and C–H $\sigma$ 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 and form halogenoalkanes:

$\text{alkene} + HX \to \text{halogenoalkane}$

For but-2-ene:

$CH_3CH=CHCH_3 + HBr \to CH_3CHBrCH_2CH_3$

A symmetrical alkene gives one structural product with $HX$. An unsymmetrical alkene may give two possible products. The major product is explained later using carbocation stability.

Addition of water

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

For ethene:

$CH_2=CH_2 + H_2O \to CH_3CH_2OH$

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

A broader acid–base idea

A Lewis acid is a chemical species that accepts an electron pair. A Lewis base is a chemical species that donates an electron pair. This definition goes beyond the Brønsted–Lowry model, since it does not require proton transfer.

A Brønsted–Lowry acid is a species that donates a proton, $H^+$. A Brønsted–Lowry base is a species that accepts a proton. When $NH_3$ accepts $H^+$, the lone pair on nitrogen forms a bond to the proton:

$$NH_3 + H^+ \to NH_4^+$$

So, in this reaction, $NH_3$ is both a Brønsted–Lowry base and a Lewis base. $H^+$ is both a Brønsted–Lowry acid and a Lewis acid.

Lewis theory can also describe reactions where no proton is involved:

$$BF_3 + NH_3 \to F_3B\leftarrow NH_3$$

Here, $NH_3$ donates an electron pair, making it the Lewis base. $BF_3$ accepts that pair at boron, making it the Lewis acid. It is not a Brønsted–Lowry acid–base reaction, because no $H^+$ is transferred.

Applying the theory

In organic chemistry, nucleophiles act as Lewis bases because they donate electron pairs. Electrophiles act as Lewis acids because they accept electron pairs. In inorganic chemistry, many metal cations such as $Cu^{2+}$ act as Lewis acids by accepting electron pairs from ligands.

One useful link is worth remembering: every Brønsted–Lowry base must have an electron pair available to bind $H^+$, so during protonation it behaves as a Lewis base. The reverse does not always work. 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 made, not that it has a special set of properties.

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

For example:

: $NH_3 + BF_3 \to H_3N\to BF_3$

The 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.

$AlCl_3$ is a useful inorganic example. Aluminium is electron-deficient, so $AlCl_3$ can accept a lone pair. Chlorine atoms on another $AlCl_3$ molecule have lone pairs; as a result, two $AlCl_3$ molecules can form $Al_2Cl_6$, 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 show exactly the same thing. A curly arrow shows electron-pair movement during the reaction; a coordinate-bond arrow in the product shows where the bond came from, donor to acceptor.

Complex ions are Lewis acid–base products

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

Transition element cations act as good Lewis acids: they’re positively charged, so they can accept electron pairs. Common ligands include neutral molecules such as $H_2O$ and $NH_3$, and anions such as $Cl^-$, $OH^-$ and $CN^-$. The ligand is the Lewis base/nucleophile; the metal cation is the Lewis acid/electrophile.

For example, $Cu^{2+}$ in water forms $[Cu(H_2O)_6]^{2+}$, with six water ligands arranged octahedrally around the copper ion.

Deducing charges on complex ions

Neutral ligands such as $H_2O$ and $NH_3$ add zero. Anionic ligands add their charge each time they appear.

In $[Co(H_2O)_6]^{2+}$, all six water ligands are neutral, so the complex ion has the same charge as $Co^{2+}$: $2^+.$

In $[CoCl_4]^{2-}$, cobalt is $Co^{2+}$ and four chloride ligands contribute $4^-$ overall:

$2^+ + 4((-1)^-) = 2^-$

So the complex ion is $[CoCl_4]^{2-}$.

You can work backwards too. In $[Fe(OH)(H_2O)_5]^{2+}$, the complex has charge $2^+$, $OH^-$ contributes $1^-$, and water contributes zero. Iron must therefore be $3^+$ because $3^+ + 1^- = 2^+.$

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. Because the halogen is more electronegative than carbon, the C–X bond is polar; the carbon bonded to X is $\delta^+$, so a nucleophile can attack it.

Primary, secondary and tertiary halogenoalkanes do not react in exactly the same way. The carbon skeleton affects steric hindrance and the stability of any carbocation that might form.

The $S_{N}2$ mechanism: one concerted step

An $S_{N}2$ reaction is a nucleophilic substitution mechanism in which the bond to the nucleophile forms as the bond to the leaving group breaks, all in one concerted step. “2” means two reacting species are involved in the rate-determining step.

For bromoethane with hydroxide:

$$CH_3CH_2Br + OH^- \to CH_3CH_2OH + Br^-$$

The rate law is:

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 $S_{N}2$ 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, rather like an umbrella turning inside out.

The $S_{N}1$ mechanism: two steps through a carbocation

An $S_{N}1$ reaction is a nucleophilic substitution mechanism in which the leaving group departs first to form a carbocation intermediate, then the nucleophile attacks. “1” means only the halogenoalkane is involved in the rate-determining step.

For 2-chloro-2-methylpropane with hydroxide:

$$(CH_3)_3CCl + OH^- \to (CH_3)_3COH + Cl^-$$

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

$$(CH_3)_3CCl \to (CH_3)_3C^+ + Cl^-$$

Step 2, fast: the nucleophile attacks the carbocation:

$$(CH_3)_3C^+ + OH^- \to (CH_3)_3COH$$

The rate law is:

$$rate = k_1[RX]$$

where $k_1$ is the first-order rate constant $(\mathrm{s^{-1}})$.

A primary halogenoalkane usually reacts by $S_{N}2$: forming a primary carbocation is unfavourable, and the nucleophile has relatively open access to the carbon. A tertiary halogenoalkane usually reacts by $S_{N}1$ 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 $\sigma$ bonds from an alkyl group toward an electron-deficient centre. Alkyl groups help spread the positive charge on a carbocation. Carbocation stability therefore follows:

$$\text{tertiary} > \text{secondary} > \text{primary} > \text{methyl}$$

This is why tertiary halogenoalkanes favour $S_{N}1$.

Energy profiles and models

An $S_{N}2$ energy profile has one maximum, since there is one transition state and no intermediate. An $S_{N}1$ energy profile has two maxima with a carbocation intermediate between them; the first maximum is usually higher because carbocation formation is the slow step.

$S_{N}1$ and $S_{N}2$ are models. They’re useful because they link structure, rate law, stereochemistry and mechanism. They are not magic labels: real secondary halogenoalkanes, mixed solvents and competing reactions may not behave like 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, the usual rate order is:

$RI > RBr > RCl$

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 heterolytic fission of C–I happens more easily. When the bond to carbon is weaker, the halide ion is a better leaving group.

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, so it is more stable. Chloride is smaller and holds charge more densely, making it less comfortable as the departing ion.

The syllabus comparison is $RCl$, $RBr$ and $RI$. Solvent effects and detailed mechanism effects on the rate are real, but they are 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 becomes part of a carbocation intermediate. A nucleophile then attacks the carbocation.

When you draw these mechanisms, start the first curly arrow at the $C=C$ bond, not at a carbon atom label. Point the arrow to the electron-deficient atom of the electrophile.

Halogens and symmetrical alkenes

Bromine, $Br_2$, 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 $\delta^+$ and acts as the electrophilic atom.

For ethene:

$CH_2=CH_2 + Br_2 \to CH_2BrCH_2Br$

In the mechanism, the $C=C$ bond attacks $Br^{\delta+}$, the $Br-Br$ bond breaks heterolytically to $Br^-$, and then $Br^-$ attacks the carbocation to form 1,2-dibromoethane.

Hydrogen halides and symmetrical alkenes

Hydrogen halides are already polar: $H^{\delta+}-X^{\delta-}$. The alkene attacks $H^{\delta+}$, the $H-X$ bond breaks heterolytically to $X^-$, and $X^-$ attacks the carbocation.

For but-2-ene:

$CH_3CH=CHCH_3 + HBr \to CH_3CHBrCH_2CH_3$

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

Water and symmetrical alkenes

Hydration of alkenes takes place in acidified solution. The proton, $H^+$, is the electrophile. First the $C=C$ bond is protonated, giving a carbocation. Next, $H_2O$ attacks as a nucleophile to form an oxonium ion. A proton is then lost, giving the alcohol and regenerating $H^+$.

For hex-3-ene:

$CH_3CH_2CH=CHCH_2CH_3 + H_2O \to CH_3CH_2CH(OH)CH_2CH_2CH_3$

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

Predicting the major product

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

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

$\text{tertiary} > \text{secondary} > \text{primary} > \text{methyl}$

Alkyl groups help stabilise the carbocation by donating 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 that already has more hydrogen atoms. That route 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:

$$CH_3CH=CH_2 + HBr \to CH_3CHBrCH_3$$

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

The same reasoning applies to hydration. When propene reacts with $H_2O$ in acid, propan-2-ol is the major product, since protonation first gives the more stable secondary carbocation before water attacks:

$$CH_3CH=CH_2 + H_2O \to CH_3CHOHCH_3$$

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

Why benzene substitutes rather than adds

Benzene, $C_6H_6$, looks highly unsaturated from its formula, but it does not react like an alkene. Its six $\pi$ electrons are delocalized around the ring, which gives an aromatic system with unusual stability. An addition reaction would permanently destroy that aromatic stability. Benzene therefore tends to react by substitution: one hydrogen is replaced, and aromaticity is restored.

An electrophilic substitution reaction is a reaction in which an electrophile replaces an atom or group in a molecule. For benzene, the key electrophile is usually written as $E^+$.

General mechanism with $E^+$

Step 1: the delocalized $\pi$ electrons of benzene attack $E^+$, forming a C–E bond and a carbocation intermediate. This intermediate is often drawn with an incomplete dashed circle and a positive charge on the ring, to show that aromaticity has been disrupted and the charge is delocalized.

Step 2: a base removes $H^+$ from the carbon bearing E. The electron pair from the C–H bond returns to the ring, restoring aromaticity and forming substituted benzene.

Overall:

$C_6H_6 + E^+ \to C_6H_5E + H^+$

The first step is normally the rate-determining step, because it breaks the stable aromatic $\pi$ system. The second step is favourable because it restores aromaticity.

Nitration as the standard example

In nitration, the electrophile is the nitronium ion, $NO_2^+$. Its formation is not assessed here, but the acid–base idea is useful: in a mixture of concentrated nitric acid and sulfuric acid, $HNO_3$ behaves as a Brønsted–Lowry base by accepting a proton from the stronger acid, $H_2SO_4$. This produces $NO_2^+$.

The substitution step is:

$C_6H_6 + NO_2^+ \to C_6H_5NO_2 + H^+$

If water is shown as the base in the final step, the released proton can be written as $H_3O^+$ instead of $H^+$.

Drawing the arrows correctly

For benzene electrophilic substitution, the first curly arrow starts from the ring of delocalized electrons and points to $E^+$. In the intermediate, show the positive charge on the ring and the interrupted delocalization. The curly arrow for loss of $H^+$ starts at the C–H bond and points back into the ring. If a base such as $H_2O$ is shown, add a curly arrow from its lone pair to H to complete the proton-transfer step.