Alkenes & Alkynes · Section 41 of 64

Addition reactions

Practice this — interactive lesson

An addition reaction converts a pi bond into two new sigma bonds. It is thermodynamically favourable almost every time — you trade roughly 65 kcal/mol of pi bond for two sigma bonds worth substantially more — so the interesting questions are not whether addition happens but where each group ends up (regiochemistry) and from which face it arrives (stereochemistry).

The alkene is the nucleophile now

This is a genuine role reversal from Module 6. There, carbon was the electrophile and something else attacked it. Here the pi bond's exposed electron density, sitting above and below the sigma framework and loosely held, makes the alkene itself the nucleophile — it attacks an electrophile rather than waiting to be attacked.

The general pattern is electrophilic addition, and it has a consistent shape: the pi bond attacks an electrophile, generating a positively charged intermediate, which a nucleophile then captures. What changes from reaction to reaction is the nature of that intermediate, and that is what controls the stereochemistry.

Hydrohalogenation (HX addition)

the alkene attacksCH₂CH₃HHHCCHBrthe π bond is the nucleophile —it reaches up and grabs the protona carbocation formsCH₃CH₂CH₃HC+the proton went to the end that leavesthe BETTER cation — secondary, not primarybromide closes the dealCCH₃CH₂CH₃HBrthis carbon has four different groups, andthe flat cation is hit from either face: racemic
The reversal that makes Module 7 easy if you have done Module 6. Everywhere so far the alkene has been something to make; here it is the reagent, and it is a nucleophile — those exposed pi electrons reach up and take a proton. What is left is an ordinary carbocation, with all the behaviour you already know: it is flat, so the bromide can arrive on either face, and it will rearrange if a more stable cation is within reach.Everything about this is Module 6 run backwards. There the cation lost a leaving group; here it gains one. Same intermediate, same stability ladder, same flat geometry — so the same racemisation, and the same risk of a rearrangement if a better cation is one hydride shift away.

With HBr: the pi bond attacks the hydrogen of H–Br — tail on the pi bond, head on H — while the H–Br bond breaks heterolytically, both electrons going to bromine. This gives a carbocation on one alkene carbon and a free bromide ion. Bromide then attacks the cation, forming the C–Br bond.

Which carbon becomes the cation is not arbitrary: it is whichever gives the more stable carbocation, which is the more substituted one. That is Markovnikov selectivity, and the next section takes it up properly.

Because the carbocation is flat, bromide can attack either face, so a new stereocenter formed this way is racemic. Reactivity follows HI > HBr > HCl, tracking the acid strengths.

the bromine caps this faceso the bromide has one way in:underneath, on the other sideANTI addition, every timethe bromonium ion, turned solid
The step the flat drawing can only assert. The bromine does not sit beside the two carbons, it bridges them, and in doing so it covers one whole face of what used to be the alkene. There is no room for anything to arrive on that side, so the incoming bromide has exactly one approach left — underneath — and the two bromines end up on opposite faces. That is what anti addition means, and it is why bromination gives one diastereomer rather than the mixture an open, flat carbocation would have given.
The carbocation intermediate here is the same species as in SN1 and E1 — which means it can rearrange by hydride or alkyl shift if a more stable cation is reachable. Adding HCl to 3-methyl-1-butene gives mainly 2-chloro-2-methylbutane, because the initial secondary cation shifts a hydride to become tertiary before chloride arrives. Every time you draw a carbocation, check for the shift.

Halogenation: no carbocation, no rearrangement

no open carbocation hereRHRHCCBrBrthe alkene attacks the near brominea BROMONIUM ion — a three-membered ringCCBr+RRthe bromine bridges BOTH carbons, so onewhole face of the alkene is now blockedBrattack only from the far sideCCBrBrone bromine up, one down:ANTI addition, every time
Bromination looks like it should go through a carbocation and does not — and the stereochemistry is how we know. The bromine bridges both carbons into a three-membered bromonium ion, sealing off the face it sits on, so the incoming bromide has exactly one way in: the other side. The result is clean anti addition. A flat open cation would have given a mixture, and no rearrangement is possible either, because there is no open cation to rearrange.This is the experiment that proves the bridged ion is real. An open carbocation would be flat and would let the second bromide in on either face, giving a mixture. Bromination gives ANTI addition cleanly instead — which only makes sense if something was already sitting on one face, blocking it.

With Br₂ or Cl₂ the mechanism differs in a way that changes everything downstream. The pi bond attacks one bromine of Br–Br, but instead of releasing a free carbocation the departing bromide bridges back onto both carbons, forming a three-membered bromonium ion. That cyclic, positively charged intermediate shields one entire face of the molecule.

A separate bromide then attacks a carbon of the bromonium ion from the opposite face, backside, exactly as in SN2 — and the ring opens. Because the second attack is forced to the far face, the two bromines end up on opposite faces: anti addition, every time.

This is a genuinely testable prediction rather than a preference. Adding Br₂ to cis-2-butene gives the meso dibromide; adding it to trans-2-butene gives the racemic (R,R)/(S,S) pair. Same reagent, same mechanism, different alkene geometry, different products — which is exactly what "stereospecific" means.

Running the reaction in water instead of an inert solvent gives a useful variation: water, present in vast excess, opens the bromonium ion instead of bromide, producing a halohydrin with OH and Br anti to each other. Water attacks the more substituted carbon, because that carbon carries more of the positive charge in the unsymmetrical bridged ion.

Worked example — proving the bromonium ion exists

If bromination went through a free carbocation, the two bromines could end up syn or anti in roughly equal measure, since a flat cation can be attacked from either face.

What is observed is exclusively anti addition, with no syn product at all. A free cation cannot explain that; a bridged ion that blocks one face completely can.

The halohydrin result confirms it from another angle: a free cation would let water attack either carbon at random, but water goes specifically to the more substituted one — which means the bridged ion is unsymmetrical, with more positive charge on the carbon better able to bear it.

Acid-catalyzed hydration

Water adds across the double bond with catalytic acid. The pi bond attacks a proton from H₃O⁺, giving a carbocation with Markovnikov selectivity; water attacks the cation as the nucleophile; and a final deprotonation by another water molecule regenerates the catalyst and gives the neutral alcohol.

Three steps, one catalyst, consumed and regenerated. This is the most direct route from an alkene to an alcohol and the bridge into Module 8 — and because it goes through a carbocation, it carries the same rearrangement risk as hydrohalogenation. When you need the Markovnikov alcohol without that risk, oxymercuration–demercuration does the same job through a bridged mercurinium ion that cannot rearrange.

It is worth noticing that hydration is the exact reverse of the E1 dehydration from Module 6, sharing every intermediate. Which direction the reaction runs is controlled by conditions: dilute acid and excess water give the alcohol, while concentrated acid and heat with the alkene distilled off give the alkene. Le Châtelier's principle, applied deliberately.

Match the mechanism to the stereochemistry. A reaction that goes through a free carbocation cannot give a single diastereomer, because the flat cation is attacked from both faces. A reaction that gives clean anti or clean syn addition must be going through something else — a bridged ion, or a concerted step. When a question tells you the stereochemical outcome, it has told you what the intermediate was.

The additions worth tabulating

ReagentAddsRegioStereo
HXH, XMarkovnikovnone (racemic)
X₂X, Xanti
X₂ / H₂OOH, XOH to more subst.anti
H₃O⁺H, OHMarkovnikovnone
BH₃ then H₂O₂/HO⁻H, OHanti-Markovnikovsyn
H₂ / PdH, Hsyn

What carries forward

Electrophilic addition is the reactivity pattern of every pi system in the course. Aromatic rings undergo the same first step in Module 13 and then, uniquely, eliminate rather than add — a difference that is the whole point of aromaticity. Bridged-ion opening reappears with epoxides in Module 8. And the alkene-to-alcohol conversions here are the standard entry into the functional group chemistry of the next four chapters.