Mix but-1-ene, CH2=CH–CH2CH3, with hydrogen bromide and the two molecules combine into one, 2-bromobutane. The H bonds to one carbon of the C=C, the Br bonds to the other, and the double bond becomes a single bond. A reaction like this, in which two groups join the two carbons of a π bond and nothing leaves, is an addition reaction.
Addition is almost always downhill in energy. The reaction above breaks two bonds, the π bond and the H–Br bond, and forms two σ bonds, a C–H and a C–Br. The π bond is weak, only about 65 kcal/mol (see Alkene structure), so the new C–H and C–Br bonds release more energy than breaking the π bond and the H–Br bond costs. Since addition is favorable, the interesting questions are where each new group ends up and which face of the alkene it arrives from. The first is a question of regiochemistry, meaning which of the possible positions each group takes. The second is a question of stereochemistry.
The alkene is the nucleophile now
In Substitution & Elimination, the carbon bearing the leaving group was the electrophile, and a nucleophile attacked it. Here the roles are reversed, and the alkene itself is the nucleophile. Its π electrons are loosely held, and they sit above and below the plane of the atoms rather than between the two nuclei. An electrophile can reach them from either side.
Every reaction on this page follows one pattern, called electrophilic addition. The π bond attacks an electrophile to give a positively charged intermediate. A nucleophile then bonds to that intermediate. The name comes from the first step, in which an electrophile adds to the alkene. What changes from one reaction to the next is the kind of intermediate, and the intermediate is what decides the stereochemistry.
Hydrohalogenation (HX addition)
HBr and but-1-ene show the pattern in its simplest form. In the first step the π bond attacks the hydrogen of H–Br. At the same moment the H–Br bond breaks heterolytically: both of its electrons go to bromine, which leaves as bromide ion, Br−. C1 now carries the new H. C2 has lost its share of the π electrons, so it is a carbocation. In the second step a lone pair on bromide attacks the cation and forms the C–Br bond.
The H goes to C1, not C2, because of the cation it leaves behind. With the H on C1, the positive charge sits on C2, which carries two carbon groups, so the cation is secondary. With the H on C2, the charge would sit on C1 as a primary cation. The secondary cation is more stable, so it also forms faster (see Carbocations), and nearly all of the product comes from it. In general, the H goes to whichever alkene carbon leaves the more stable cation behind. This preference is called Markovnikov’s rule, and the next topic, Markovnikov / anti-Markovnikov, takes it further.
The cation is flat, so bromide can reach it from either face. Attack from one face gives one enantiomer of 2-bromobutane, and attack from the other face gives its mirror image, in equal amounts. So when addition to a flat cation creates a new stereocenter, the product is a racemic mixture.
Among the hydrogen halides, HI reacts fastest and HCl slowest: HI > HBr > HCl. The order follows their acid strengths, because a stronger acid hands its proton to the π bond more readily.
Take 3-methylbut-1-ene and HCl. Protonation gives a secondary cation on C2. The next carbon, C3, carries one H and two methyl groups. When that hydrogen moves to C2 with its bonding pair, the charge moves to C3, which is tertiary. Chloride captures both kinds of cation, so the product is a mixture of 2-chloro-3-methylbutane, from the unshifted cation, and 2-chloro-2-methylbutane, from the shifted one. Every time you draw a carbocation, check for a shift.
Halogenation: no carbocation, no rearrangement
Bromine, Br2, adds to cis-but-2-ene to give 2,3-dibromobutane. The first step starts like the HBr case. The π bond attacks one bromine atom of Br–Br, the Br–Br bond breaks, and the other bromine leaves as bromide.
This time, though, no open carbocation forms. As the π bond bonds to one bromine, a lone pair on that same bromine bonds to the second carbon. The bromine ends up bonded to both carbons at once, in a three-membered ring with the positive charge on bromine. This ring is a bromonium ion. Chlorine, Cl2, behaves the same way and gives a chloronium ion; the general name for either is a halonium ion.
The bromine sits over one face of the old double bond and blocks it. So in the second step a bromide ion attacks one of the ring carbons from the opposite face. It comes in from the back of the C–Br bond that breaks, just as a nucleophile does in SN2, and the ring opens. The two bromines therefore end up on opposite faces. Adding two groups to opposite faces is called anti addition (see Diastereomers). For simple alkenes such as but-2-ene, bromination is always anti.
Since no open cation ever forms, there is nothing to rearrange. Halogenation keeps the carbon skeleton of the alkene exactly as it was.
The two but-2-enes show why the anti rule matters. Adding Br2 to trans-but-2-ene gives meso-2,3-dibromobutane, a single achiral compound. Adding it to cis-but-2-ene gives (2R,3R)- and (2S,3S)-2,3-dibromobutane in equal amounts, a racemic mixture. The reagent and the mechanism are the same, and only the geometry of the alkene differs, yet that geometry decides which stereoisomer forms. A reaction in which the stereoisomer you start with fixes the stereoisomer you get is stereospecific.
These two results are easy to swap, so derive them rather than memorize them. Take cis-but-2-ene and put the bromonium ion on the top face. Bromide can now attack C2 or C3 from below, and it attacks each one equally often. Attack at C2 gives (2S,3S), and attack at C3 gives (2R,3R). The two products are enantiomers, formed 50:50. A bromonium ion on the bottom face gives the same pair again. Worked example (a) below does the same for trans-but-2-ene, and the Meso compounds section reaches both results another way.
Run the same reaction in water instead of an inert solvent and the product changes. Water is present in huge excess, so a water molecule, rather than bromide, usually opens the bromonium ion. Once the oxygen from water loses a proton, the product has an OH on one carbon and a Br on the next. A compound with a halogen and an OH on neighboring carbons is a halohydrin. The ring still opens from the far face, so the OH and the Br add anti.
With 2-methylpropene, (CH3)2C=CH2, water bonds to C2, the carbon with the two methyl groups, and not to the CH2 end. That surprises anyone who expects SN2-style attack at the less crowded carbon, and it is the detail of this reaction that exams test most.
The reason is that the bromonium ion is not the symmetrical triangle it is usually drawn as. C2 carries two methyl groups, so it can hold positive charge far better than C1, for the same reason that a tertiary cation beats a primary one. So the bromine sits closer to C1 than to C2. The C2–Br bond grows long and weak, and C2 takes on much of the positive charge, while the C1–Br bond stays short. Water attacks the carbon that is already partway to being a cation, because that path has the lower barrier. Here charge matters more than crowding.
Suppose bromination went through an open, flat carbocation. Bromide could then attack from either face, and a simple alkene such as but-2-ene would give a mixture of syn addition, with both bromines on the same face, and anti addition.
Only anti addition is observed. A flat cation cannot explain that result, but a bridged ion that blocks one face can.
Halohydrin formation adds a detail. Water still adds anti, so the ion is still bridged. But water goes to the more substituted carbon, so the bridge must be uneven, with more of the positive charge on that carbon.
The method is the same every time: draw the intermediate, then ask what the intermediate allows.
(a) trans-But-2-ene + Br2. First, bromine bridges C2 and C3 on one face. Second, bromide attacks either carbon from the other face. Third, assign the new stereocenters. One comes out R and the other S: attack at C2 gives (2S,3R), and attack at C3 gives (2R,3S). Numbering the chain from its other end turns one name into the other, so the two attacks give the same molecule. That molecule has a mirror plane between C2 and C3, so it is meso-2,3-dibromobutane, a single achiral compound, and there is no racemic pair here.
(b) 1-Methylcyclohexene + Br2 in water. First, the bromonium ion forms across C1 and C2. Second, ask which carbon carries more of the positive charge. C1, which bears the methyl group, is the more substituted, so water attacks it. Third, the ring opens from the opposite face, so the OH and the Br add anti, which on a ring means trans. The product is 2-bromo-1-methylcyclohexan-1-ol, with the OH on C1 and the Br on C2, trans to each other. It forms as a racemic mixture, because bromine is equally likely to bridge either face of the ring.
The trap in (b). The bromine ends up on the less substituted carbon, which can look like the reverse of Markovnikov’s rule. It is not. The electrophile, Br, went to the less substituted carbon and the nucleophile, water, went to the more substituted one, exactly as the H and the Br do in HBr addition.
Acid-catalyzed hydration
Propene in dilute aqueous acid, such as H2SO4 in water, turns into propan-2-ol: water adds across the double bond. The acid supplies hydronium ion, H3O+, and the reaction runs in three steps.
- Protonation. The π bond takes a proton from H3O+. The H goes to C1 and a secondary cation forms on C2, the same choice as in HBr addition.
- Capture. A water molecule attacks the cation with one of its lone pairs. This gives an oxonium ion, an oxygen with three bonds and a positive charge.
- Deprotonation. A second water molecule removes a proton from that oxygen, which leaves the neutral alcohol and makes H3O+ again.
The acid used in the first step comes back in the last one, so it is a catalyst: it speeds the reaction up without being used up. Hydration is the most direct route from an alkene to an alcohol, and it leads into Alcohols, Ethers & Related Chemistry. It passes through a free carbocation, and that cation can rearrange, just as in HBr addition.
Hydration is the E1 dehydration from Substitution & Elimination run backwards, through the same intermediates. Alkene plus water and the alcohol sit in equilibrium, and the conditions decide which way it runs. Dilute acid and plenty of water push it toward the alcohol. Concentrated acid and heat, with the alkene distilled off as it forms, push it toward the alkene. That is Le Châtelier’s principle used on purpose: adding a reactant or removing a product shifts an equilibrium toward the product side.
Oxymercuration–demercuration: Markovnikov without the rearrangement
Hydrate 3,3-dimethylbut-1-ene with aqueous acid and its carbon skeleton changes. Protonation gives a secondary cation on C2, and a methyl group on C3 then shifts to C2 with its bonding pair, which leaves a tertiary cation on C3. Water bonds there, and the product is 2,3-dimethylbutan-2-ol. Oxymercuration–demercuration adds water to the same alkene without any shift and gives 3,3-dimethylbutan-2-ol.
The first step uses mercuric acetate, Hg(OAc)2, in water; OAc is shorthand for acetate, CH3CO2−. It ionizes to give +Hg(OAc), a large, soft electrophile, meaning its electron cloud is easily distorted. The π bond attacks the mercury. Just as bromine did, the mercury bonds to both carbons instead of leaving an open cation, and the result is a three-membered ring called a mercurinium ion.
Water opens that ring at the more substituted carbon, for the same reason as in halohydrin formation: that carbon carries more of the positive charge. So the OH ends up on the more substituted carbon, and the H added in the next step goes to the other one: the Markovnikov placement. Water attacks from the face opposite the mercury, so this step adds the OH and the mercury anti.
In the second step, sodium borohydride, NaBH4, replaces the C–Hg bond with a C–H bond and removes the mercury. This step goes through radicals, which scramble whatever stereochemistry the first step set, so the overall reaction is not stereospecific.
The additions worth tabulating
Two rows of the table look ahead. Hydroboration–oxidation, which puts the OH on the less substituted carbon, is taught in the next topic, Markovnikov / anti-Markovnikov. Catalytic hydrogenation, in which a metal surface delivers both hydrogens of H2 to one face, comes later in this chapter. They appear here only so that the table is complete; nothing on this page relies on them.
| Reagent | Adds | Regio | Stereo |
|---|---|---|---|
| HX | H, X | Markovnikov | none (racemic) |
| X₂ | X, X | — | anti |
| X₂ / H₂O | OH, X | OH to more subst. | anti |
| H₃O⁺ | H, OH | Markovnikov | none |
| BH₃ then H₂O₂/HO⁻ (next topic) | H, OH | anti-Markovnikov | syn |
| Hg(OAc)₂ then NaBH₄ | H, OH | Markovnikov | none |
| H₂ / Pd (later this chapter) | H, H | — | syn |
What carries forward
Electrophilic addition is how every π system in the course meets an electrophile. A benzene ring takes the same first step in Aromatic Chemistry, but then it loses a proton instead of adding a nucleophile, which keeps the ring aromatic. The same backside opening of a three-membered ring returns with epoxides in Alcohols, Ethers & Related Chemistry. And the routes from alkenes to alcohols on this page are the usual way into the alcohol and carbonyl chemistry of the chapters ahead.