Revise Electron-pair sharing reactions for IB Chemistry (R3.4)

A covalent bond is a shared pair of electrons, so making and breaking bonds is about where electron pairs go. A curly (double-headed) arrow shows the movement of a pair of electrons. It has two non-negotiable rules: it must START at a source of electrons — a lone pair on an atom, or the pair of electrons in a bond — and its head must POINT to the atom or bond that receives them. An arrow tail on a positive charge, or an arrow drawn from the electron-poor atom towards the electron-rich one, is chemically wrong and loses marks. In the reactions below, one arrow makes a new bond and another breaks an old one.

Every electron-pair reaction has a donor and an acceptor. A nucleophile is an electron-rich species that DONATES a pair of electrons to form a new bond; it is a negative ion or a neutral molecule carrying a lone pair. An electrophile is an electron-deficient species that ACCEPTS a pair of electrons to form a new bond; it is a positive ion or a molecule with a δ+ atom. These are the same species that acid–base theory calls Lewis bases and Lewis acids — the word 'nucleophile' or 'electrophile' is used when we are talking about a reaction mechanism rather than a proton transfer.

A halogenoalkane has a polar C–X bond (X = Cl, Br, I): because the halogen is more electronegative, the carbon carries a partial positive charge, Cᵟ⁺–Xᵟ⁻. That δ+ carbon is an electrophilic centre, and a nucleophile is attracted to it. When the nucleophile forms a bond to the carbon, the halogen leaves as a halide ion (the leaving group). The nucleophile has SUBSTITUTED for the halogen — hence nucleophilic substitution. With hydroxide ion as the nucleophile the product is an alcohol; the overall equation for bromoethane is:

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

Read the equation as a story: the lone pair on OH⁻ attacks the δ+ carbon bonded to Br, a new C–O bond forms, and the C–Br bond breaks with both electrons leaving on the bromine to give Br⁻. The general pattern, for any nucleophile Nu⁻ and halogenoalkane R–X, is R–X + Nu⁻ → R–Nu + X⁻. Different nucleophiles give different products from the same halogenoalkane: OH⁻ gives an alcohol, CN⁻ gives a nitrile (adding a carbon to the chain), and NH₃ gives an amine.

SL stops at the overall equation. At HL you also need the mechanism, and there are two, distinguished by how many species take part in the rate-determining step. In SN2 (substitution, nucleophilic, bimolecular) the nucleophile attacks the δ+ carbon in the SAME step as the C–X bond breaks — one concerted step, with the leaving group departing on the opposite side. Because two species (the nucleophile and the halogenoalkane) are involved in that single step, the rate depends on both concentrations, and the reaction is fastest for PRIMARY halogenoalkanes, where the small carbon centre is easy to approach. In SN1 (substitution, nucleophilic, unimolecular) the C–X bond breaks FIRST to give a carbocation intermediate (the slow, rate-determining step), and only then does the nucleophile attack. Only the halogenoalkane is in the rate-determining step, so the rate depends on its concentration alone; this route dominates for TERTIARY halogenoalkanes, whose bulky, branched structure both hinders direct attack and stabilises the tertiary carbocation.

Alkenes have a carbon–carbon double bond, C=C, which is a region of high electron density sitting exposed above and below the carbons. That electron-rich π bond is attractive to electrophiles — species short of electrons. When an electrophile approaches, the double bond opens and adds atoms across the two carbons: one new single bond forms to each carbon, the C=C becomes C–C, and NOTHING leaves. This is electrophilic addition, and because every atom of both reactants ends up in one product, it has 100% atom economy in principle. With bromine, ethene gives 1,2-dibromoethane; with hydrogen bromide it gives bromoethane:

$CH_2{=}CH_2 + Br_2 \rightarrow CH_2Br{-}CH_2Br$

The reaction of an alkene with bromine (bromine water) is also the classic TEST for a C=C double bond: orange bromine water is decolourised as the addition takes place. With a symmetrical alkene like ethene there is only one possible product. But with an unsymmetrical alkene and an unsymmetrical reagent like HBr, two products are possible depending on which carbon gets the H and which gets the Br — and that is where Markovnikov's rule comes in.

When an unsymmetrical molecule such as HBr, HCl or H₂O adds to an unsymmetrical alkene, two products are possible, but one forms in greater amount. Markovnikov's rule predicts it: the hydrogen atom adds to the carbon of the double bond that already carries MORE hydrogen atoms, which puts the halogen (or OH) on the more substituted carbon. So propene, CH₃CH=CH₂, plus HBr gives mainly 2-bromopropane, CH₃CHBrCH₃, not 1-bromopropane. (The HL reason is that adding H⁺ this way forms the more stable — more substituted — carbocation intermediate; at SL you apply the rule to name the major product.)

At HL you show the mechanism with curly arrows. Take HBr adding to an alkene. The H–Br bond is polar (Hᵟ⁺–Brᵟ⁻), so the electron-rich C=C π bond attacks the δ+ hydrogen: a curly arrow runs from the double bond to the H, and a second arrow runs from the H–Br bond to the bromine, which leaves as Br⁻. This makes a carbocation on one carbon (the more stable one, following Markovnikov). In the second step a curly arrow runs from the lone pair on Br⁻ to the positive carbon, forming the C–Br bond and completing the addition. For Br₂, the non-polar Br–Br bond becomes polarised as it approaches the electron-rich double bond (an induced dipole), and the same two-step pattern follows via a bromonium-ion/carbocation intermediate.

This is the showcase. In Paper 2 the marks are analytic: each is tied to a specific statement — a method/reasoning mark (M) or an answer/identification mark (A) — and error-carried-forward (ECF) means a wrong earlier step does not automatically cost the marks that depend on it, PROVIDED your reasoning is written down. Study how each of the four marks below is earned by a specific line, and how equivalents are accepted.

Every named organic reaction you meet from here is a variation on these two moves plus radical substitution. Halogenoalkanes go on to make alcohols, nitriles and amines by nucleophilic substitution; alcohols and alkenes interconvert by addition and elimination; and the curly-arrow habit — start electron-rich, point electron-poor — carries straight into the HL mechanisms and into any new reagent you are asked to reason about. Master 'find the δ+ centre and follow the electron pair' and unfamiliar organic questions become recognisable.