Organic Chemistry · Alkenes: Structure and Reactivity

Electrophilic Addition Reactions of Alkenes

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Why this matters
The college version
Eli explains
Worked example
Key takeaway
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In 30 seconds

The double bond is the most reactive part of an alkene, and its signature reaction is electrophilic addition: a reagent adds across the double bond, the π bond breaks, and two new single bonds form. The name says it all — an (electron lover, e.g., H+ from HBr) is attacked by the electron-rich π bond, then a (nucleus lover, e.g., Br−) completes the addition.

This topic establishes the general two-step mechanism behind a family of alkene reactions: hydrogen halide addition (HCl, HBr, HI), acid-catalyzed , and halogen addition. Once you can describe it, every specific addition is a variation: an electrophile attacks the π bond, then a nucleophile captures the . The regiochemical question — which carbon the electrophile lands on (Markovnikov's rule) — is the next topic; here the goal is mechanism, energetics, and products.

Why this matters

Electrophilic addition turns cheap, abundant alkenes into valuable functionalized molecules. Industrial acid-catalyzed hydration of ethene produces ethanol — millions of tons per year. HX addition gives alkyl halides, the workhorse intermediates of later chapters (Chapter 11). Addition polymerization of ethene and propene — the same π-bond reactivity repeated thousands of times — gives polyethylene and polypropylene, the most-produced plastics on Earth (Chapter 31). The same π-bond attack and carbocation chemistry drives biosynthesis of fats, steroids, and terpenes. On exams, this is the most heavily tested reaction type in the alkene chapters: master the two-step story for HBr and you can derive most additions.

The college version

Core Concepts

The π bond is a nucleophile

The π electrons sit above and below the molecular plane, far from the nuclei — loosely held, electron-rich, ready to donate into an empty orbital or polarized bond of an electron-poor species, an electrophile. In H–Br the bond is polarized (bromine more electronegative), so hydrogen carries a partial positive charge: it is the electrophile. The alkene is the nucleophile that starts every addition.

The general two-step mechanism for HX addition

Take HBr + ethene. Step 1 (slow, rate-determining): the π electrons attack the hydrogen of H–Br, forming a C–H bond at one alkene carbon; simultaneously the H–Br bond breaks heterolytically — both electrons stay with bromine — giving a bromide ion (Br−) and a carbocation (a carbon with only three bonds and a positive charge) at the other alkene carbon. Step 2 (fast): the bromide ion donates an electron pair to the carbocation, forming the C–Br bond. Net result: the π bond is gone (the σ bond remains) and two new σ bonds have appeared. The carbocation is a true — a real, if short-lived, species — and this stepwise character is the central fact of alkene addition chemistry. In arrow notation: the π pair curves to the H of H–Br, the H–Br pair curves onto bromine (heterolysis), and the bromide lone pair curves to the carbocation — always in the order: nucleophile attacks electrophile, leaving group departs with the bond pair, nucleophile captures the cation.

Energetics: why addition happens at all

Addition converts a π bond plus an H–X bond into two C–X σ bonds. The π bond is weak (~63–65 kcal/mol) and H–X bonds modest, while C–H and C–X σ bonds are strong, so the reaction is exothermic (negative ΔH). Entropy is unfavorable (two molecules become one), but for HX and H2O additions enthalpy dominates: the reactions are favorable and typically irreversible. Hydrogenation (Topic 6) is the same story with H–H in place of H–X.

The same mechanism, different electrophiles

HCl, HBr, HI: identical mechanism; acidity increases down the group (HI most reactive).
Acid-catalyzed hydration: the electrophile is H+ (from H2SO4 or H3O+); the nucleophile in step 2 is water, giving an oxonium ion that loses a proton to solvent, delivering an alcohol. Water alone cannot add — the acid is required to generate H+.
Halogens (Br2, Cl2): preview — a cyclic halonium ion forms (no free carbocation) and the second halogen attacks from the opposite face, giving anti addition (Chapter 8).
HX with peroxides: a special case where HBr adds against Markovnikov's rule by a radical mechanism — also a later chapter.

Where regiochemistry comes from

In unsymmetrical alkenes, step 1 can put the positive charge on either alkene carbon. The reaction takes the path giving the more stable carbocation (tertiary > secondary > primary, Topic 9) — step 1 is rate-determining, and its transition state resembles the carbocation (Hammond postulate, Topic 10). That choice fixes the regiochemistry — Markovnikov's rule — the subject of the next topic.

Common Confusions

Do Not Confuse	With	Difference
Addition (alkenes)	substitution (alkanes)	Alkenes add across the π bond; alkanes substitute H by radicals/light.
Markovnikov addition	anti-Markovnikov HBr + peroxides	Normal HBr adds H to the H-rich carbon; with peroxides, Br• adds first and the product is inverted (radical mechanism, later chapter).
"The π bond breaks"	net bond loss	Two strong σ bonds form as the π bond breaks — the molecule gains bonding overall (exothermic).
Stepwise (carbocation)	concerted (one-step)	HX addition passes through a discrete carbocation; rearrangements (Topic 11) rule out a one-step path.
H2O adds alone	acid-catalyzed hydration	Neutral water cannot generate H+; H2SO4 or H3O+ is required.
Product of 2-butene + HCl	regiochemical prediction	The alkene is symmetric, so both carbocations are identical — no Markovnikov decision needed.

Eli explains

The same idea, in plain words

Explain it like I’m 10

The double bond is like friends holding hands loosely — the π electrons are easy to grab. A hungry H+ grabs one end, the other end is left positive and empty-handed (the carbocation), and a shy Br− rushes in to grab the free hand. Two firm handholds where one loose rope was.

Worked example

Example 1 — Mechanism of 2-methylpropene + HBr, in words. Reactant: CH2=C(CH3)2. Step 1: the π electron pair attacks the partially positive hydrogen of H–Br, forming a C–H bond at the CH2 carbon, while the H–Br bond pair moves onto bromine as Br−. The positive charge lands on the central carbon, already bonded to two methyls: a tertiary carbocation, (CH3)3C+. Step 2: the bromide lone pair forms the C–Br bond. Product: 2-bromo-2-methylpropane, (CH3)3C–Br. Step 1 is fast because it makes the most stable kind of carbocation, and regiochemistry follows stability (Markovnikov, Topic 8). Net change: π bond + H–Br → C–H + C–Br, exothermic.

Example 2 — Stoichiometry with dimensional analysis: how much HBr reacts with 5.00 g of 1-butene? The balanced equation is 1:1:

$C 4 H 8 + HBr \to C 4 H 9 Br$

Write the conversion formula first, then substitute:

$n(C 4 H 8) = m M = 5.00 g 56.11 g mol -1 = 0.0891 mol$

(Molar mass check: C4H8 = 4(12.011) + 8(1.008) = 56.11 g/mol.) Because the mole ratio is 1:1,

$n(HBr) = 0.0891 mol, m(HBr) = n \times M = 0.0891 mol \times 80.91 g mol -1 = 7.21 g$

Units cancel (mol × g/mol = g): 7.21 g of HBr is required. Product check: 0.0891 mol C4H9Br (M = 137.02 g/mol) = 12.2 g, matching 5.00 + 7.21 g by conservation of mass.

Example 3 — Hydration of 1-methylcyclohexene. A six-membered ring with a double bond between C1 (bearing the methyl) and C2. Step 1: H+ (from H2SO4) adds to C2, the less substituted carbon; the positive charge forms at C1, stabilized by its methyl (tertiary carbocation). Step 2: water's oxygen lone pair attacks the carbocation, forming a C–O bond and an oxonium ion. Step 3: a water molecule removes a proton from the oxonium oxygen, regenerating the catalyst. Product: 1-methylcyclohexanol — a tertiary alcohol with Markovnikov orientation (OH on the more substituted carbon). Three steps, one net addition of H–OH.

Example 4 — Symmetric alkene, no regiochemistry question. 2-Butene (CH3–CH=CH–CH3) + HCl: the two alkene carbons are equivalent (each bears CH3 and H), so protonation of either gives the same secondary carbocation. Product: 2-chlorobutane, CH3–CH(Cl)–CH2–CH3. Always check for symmetry before invoking regiochemistry — here Markovnikov's rule has nothing to decide.

Key takeaways

Alkenes react by electrophilic addition: π bond breaks, two new σ bonds form.
Two-step mechanism for HX: (1) π electrons + H+ → carbocation (slow, rate-determining); (2) X− + carbocation → C–X bond (fast).
Addition is exothermic: the two new σ bonds more than repay breaking the π bond plus H–X.
Regiochemistry is set by carbocation stability (Markovnikov) — full treatment in Topic 8.
Hydration needs an acid catalyst (H2O alone does not add); halogen addition goes through a cyclic halonium ion (anti addition, Chapter 8).
Rate-determining step = carbocation formation: whatever stabilizes carbocations speeds up addition.

Check yourself

5 review questions from the chapter. Try each one, then open the answer.

Write the two steps of HX addition to an alkene in words, and identify the .
Show answer
Step 1: the π electrons attack H+ of H–X, forming a C–H bond and a carbocation (X− is released by heterolysis); this step is slow and rate-determining. Step 2: X− donates a lone pair to the carbocation, forming the C–X bond.
Why is electrophilic addition exothermic even though a bond (the π bond) is broken?
Show answer
The π bond (~63–65 kcal/mol) and the H–X bond are weak, but two strong σ bonds (C–H, C–X) form in their place; net ΔH is negative despite unfavorable entropy (two molecules become one).
What is the product of the hydration of ethene, and why is an acid catalyst required?
Show answer
Ethanol (CH3CH2OH): ethene + H2O under H2SO4 catalysis. The acid generates H+, the electrophile that attacks the π bond; water alone cannot start the addition.
In 2-methylpropene + HBr, where does the positive charge form in step 1, and what kind of carbocation results?
Show answer
The positive charge forms at the central carbon already bearing two methyl groups — a tertiary carbocation, (CH3)3C+ — because that is the most stable arrangement; H+ adds to the CH2 carbon.
A reaction converts 5.00 g of 1-butene with excess HBr. How many grams of HBr are consumed, and how many grams of 2-bromobutane form?
Show answer
HBr consumed: 7.21 g (0.0891 mol, from 5.00 g / 56.11 g mol−1). 2-Bromobutane (C4H9Br, M = 137.02 g/mol) formed: 0.0891 mol × 137.02 g mol−1 = 12.2 g, matching the 5.00 + 7.21 g mass balance.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

electrophile: An electron-poor species (H+, Br+, etc.) that accepts an electron pair.
nucleophile: An electron-rich species (Br−, H2O) that donates an electron pair.
carbocation: A carbon bearing a positive charge and only three bonds (R3C+).
heterolytic bond breaking: The bond pair goes entirely to one fragment (here, bromine).
rate-determining step: The slowest step, which controls the overall rate.
hydration: Addition of water across a double bond, acid-catalyzed, giving an alcohol.
intermediate: A species that exists transiently between steps (the carbocation).

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