3.4 Electron-pair sharing reactions

A nucleophile is an electron-rich species that donates an electron pair to an electron-deficient centre.

Look for an available electron pair: OH⁻ and CN⁻ use a negative charge and lone pair, while NH₃ uses a lone pair without being an anion. A curly arrow must start at that pair and point toward the atom where the new bond forms.

The nucleophile donates a pair to carbon while the leaving group departs with its bonding pair. Deduce the product by replacing the leaving group with the nucleophile.

For CH₃CH₂Br + OH⁻, the C–O bond forms as the C–Br bond breaks, giving CH₃CH₂OH + Br⁻. Account for charge and every atom in the product; the leaving group takes the bonding pair rather than departing as a neutral bromine atom.

In heterolytic fission both bonding electrons remain with one fragment, producing ions. Curly arrows show movement of an electron pair.

Place the curly-arrow tail on the bond being broken and its head on the fragment receiving both electrons. Then assign charges from electron ownership: heterolysis creates ions, unlike homolysis, which gives one electron to each radical.

An electrophile is an electron-deficient species that accepts an electron pair from a nucleophile.

Identify the electron-poor atom, not merely a positive-looking formula. H⁺ and carbocations are electrophiles, and the δ⁺ carbon in a polar C–X bond can also accept a pair. The incoming curly arrow ends at this acceptor.

The electron-rich C=C attacks an electrophile. Deduce addition products with water, halogens or hydrogen halides within the SL mechanism boundary.

Treat the C=C as the reactive site and place the two added groups on its two carbon atoms. Bromine addition removes the double bond and forms a dibromoalkane; hydration forms an alcohol. At SL, deducing these products does not require a mechanism.

A Lewis acid accepts an electron pair; a Lewis base donates an electron pair. Nucleophiles correspond to Lewis bases and electrophiles to Lewis acids.

In BF₃ + NH₃ → F₃B←NH₃, NH₃ donates the pair and is the Lewis base; BF₃ accepts it and is the Lewis acid. Classify the roles from electron-pair movement rather than from whether H⁺ appears.

A ligand acts as a Lewis base and donates an electron pair to a Lewis-acid transition-metal cation, forming a coordinate bond.

Show a coordination bond with an arrow from a ligand lone pair to the metal ion. The arrow records the origin of the shared pair; after formation the bond is not a different electrostatic species from other covalent bonds.

Identify the central transition-metal cation and the surrounding ligands. Each ligand donates an electron pair to the metal centre.

Read [Cu(NH₃)₄]²⁺ as one Cu centre with four NH₃ ligands and coordination number 4. Use ligand charges and the overall bracket charge to deduce the metal oxidation state; do not confuse coordination number with oxidation state.

SN2 is a concerted backside attack: the nucleophile bonds as the carbon–halogen bond breaks. SN1 first forms a carbocation; the nucleophile then attacks the planar intermediate. Primary substrates generally favour SN2 and tertiary substrates generally favour SN1 because steric access and carbocation stability differ. Secondary substrates can support either mechanism, so use the conditions as well as the substrate class.

Do not treat SN1 and SN2 as labels that follow substrate class with no exceptions, or confuse inversion in SN2 with the mixture possible after planar SN1 attack.

Leaving-group identity affects substitution rate through carbon–halogen bond enthalpy and polarity: a weaker, more easily broken bond generally permits faster departure.

For comparable halogenoalkanes, the weaker C–I bond usually makes iodide a better leaving group and substitution faster than for chloride. Explain the trend with bond enthalpy and the actual rate-determining bond change, not polarity alone.

Use curly arrows from the alkene π bond to the electrophile, then from the intermediate to the nucleophile. Halogens may pass through a halonium intermediate; HX or water may pass through a carbocation.

Begin with an arrow from the alkene π bond to the electrophile. HX or acid-catalysed hydration can form a carbocation, whereas halogen addition uses a bridged halonium ion; the next arrow must start from the nucleophile's pair.

For HX or water addition to an unsymmetrical alkene, compare the possible carbocations and select the pathway through the more stable carbocation as the major product.

For an unsymmetrical alkene, draw both protonation routes and compare the carbocations: tertiary is generally more stable than secondary, then primary. Choose the product from the more stable intermediate; do not apply this shortcut to a halonium pathway.

Benzene attacks the charged electrophile E+, forms a charged intermediate, then loses H+ to restore aromaticity. The formation of E+ is outside this assessed mechanism boundary.

The first step temporarily disrupts aromatic delocalization; loss of H⁺ then restores the ring and replaces H by E. An addition product would leave aromaticity broken, which explains why the assessed outcome is substitution.