Types of organic reaction and mechanism

UV light splits Cl₂ homolytically into chlorine free radicals, which then replace an H atom in methane with Cl: a free-radical substitution. Methane has no C=C, so nothing can add to it; electrophilic addition is the reaction of alkenes such as ethene with Br₂. No nucleophile (lone-pair donor) is involved and methane has no polar C-X bond to attack, so it is not nucleophilic substitution. Two products form from two reactants with an H swapped for a Cl, which is substitution, not addition.

In homolytic fission the shared pair is split evenly, one electron to each atom, so both fragments have an unpaired electron: they are free radicals (for example Cl₂ → 2Cl•). Heterolytic fission gives both electrons to one atom, producing a cation and an anion. Saying homolytic fission gives ions is the classic error: one electron each leaves each fragment neutral. Donating a lone pair to form a bond describes dative bond formation or a nucleophile attacking, which makes a bond rather than breaking one.

A nucleophile is an electron-pair donor. It has a lone pair (and often a negative charge) and attacks an electron-deficient δ+ carbon: OH⁻, CN⁻, NH₃ and H₂O all qualify. Accepting an electron pair is the definition of an electrophile, the classic mix-up. A species with an unpaired electron is a free radical. A positive ion attracted to electron-rich regions (such as NO₂⁺) is an electrophile, not a nucleophile.

The nitronium ion, NO₂⁺, is positively charged and electron-deficient, so it accepts an electron pair from an electron-rich region such as a benzene ring: it is an electrophile, like H⁺ and a polarised Br₂ molecule. OH⁻ has lone pairs and a negative charge, so it donates electrons: a nucleophile. NH₃ has a lone pair on N and acts as a nucleophile (and a base). CN⁻ also donates its lone pair to a δ+ carbon, so it is a nucleophile.

The C=C pi bond is electron-rich and induces a dipole in Br₂; the δ+ bromine accepts a pair of electrons from the pi bond, so Br₂ acts as an electrophile. Two molecules combine into one, so it is an addition: electrophilic addition. A nucleophile would be repelled by the electron-rich double bond, so nucleophilic addition is wrong (that occurs at C=O). No UV light or radicals are involved, and nothing is replaced, so it is not free-radical substitution. Electrophilic substitution is typical of benzene, where an H is replaced; here no H is lost.

A full curly arrow shows a pair of electrons moving. Arrow 1 starts at the lone pair on oxygen of OH⁻ and ends at the δ+ carbon, where a new C-O bond forms. Arrow 2 starts at the C-Br bond and ends on the more electronegative Br, so the bond breaks heterolytically and Br⁻ leaves. OH⁻ replaces Br, so the reaction is nucleophilic substitution, giving ethanol. Curly arrows must start from a pair (a lone pair or a bond), not from an atom, and a single-headed (fish-hook) arrow is needed for one electron, so the free-radical reading is wrong. OH⁻ donates an electron pair and attacks carbon, not bromine, so it is a nucleophile. Both bonding electrons go to Br, giving the ion Br⁻, not a radical.

CH₃CH₂⁺ has a positive charge on carbon, so it is a carbocation. The pi electrons form a bond to H and the H-Br bonding pair goes entirely to bromine, giving Br⁻: heterolytic fission, which produces ions. A free radical has an unpaired electron and no charge, so it cannot be CH₃CH₂⁺. Homolytic fission gives each atom one electron and makes radicals, not ions, so it cannot produce a carbocation. A positive species accepts electron pairs, so the carbocation is an electrophile; Br⁻ is the nucleophile that then attacks it.

In (1) hydrogen is added, shown by [H], so ethanal is reduced to ethanol. In (2) a small molecule, water, is removed from a single molecule to make a C=C: elimination (dehydration). In (3) water breaks the ester link into an acid and an alcohol: hydrolysis. Calling (1) oxidation reverses it; oxidation would remove H or add O. A condensation joins two molecules and releases a small molecule such as water, so it is not (2), where only one reactant is present. Reaction (3) is the reverse of a condensation (esterification), because water is used up, not released.