6.4 An Example of a Polar Reaction: Addition of HBr to Ethylene

Addition of HBr to Ethylene

Adding HBr across a carbon-carbon double bond is one of the first full reaction mechanisms you'll encounter in organic chemistry. It demonstrates how electron-rich alkenes react with electrophiles and introduces the two-step pattern of electrophilic addition.

This reaction also highlights why alkenes are so much more reactive than alkanes. The pi electrons in a double bond are exposed and available, making alkenes natural targets for electron-poor species like H$^+$.

Addition of HBr to Alkenes

Mechanism of HBr addition to ethylene

The addition of HBr to ethylene is a two-step electrophilic addition reaction. H and Br end up on adjacent carbons where the double bond used to be.

Step 1: Protonation of the alkene

The pi electrons of the double bond act as a nucleophile, attacking the electrophilic H of HBr. This breaks the H–Br bond and forms a new C–H bond, generating a carbocation intermediate. The carbocation carbon is sp2-hybridized, with a planar geometry, an empty p orbital, and a formal positive charge.

Step 2: Nucleophilic attack by Br$^-$

Br$^-$, now free with its lone pairs, donates a pair of electrons into the empty p orbital of the carbocation. This forms the new C–Br bond and gives the final alkyl bromide product.

For ethylene specifically, both carbons of the double bond are equivalent, so there's no regiochemical choice to make. With unsymmetrical alkenes, though, Markovnikov's rule applies:

• H$^+$ adds to the less substituted carbon of the double bond.
• Br$^-$ adds to the more substituted carbon.

This selectivity exists because the carbocation intermediate is more stable when the positive charge sits on a more substituted carbon (tertiary > secondary > primary).

Reactivity of alkenes vs. alkanes

Why do alkenes react with HBr while alkanes don't? It comes down to electron accessibility.

• In an alkene, the pi bond holds electron density above and below the plane of the molecule. These pi electrons are loosely held and exposed, making them easy for an electrophile to reach.
• In an alkane, all electrons sit in sigma bonds directly between nuclei. They're tightly held and much less available for reaction.

The hybridization difference matters too. Alkene carbons are sp2 hybridized with a p orbital available for pi bonding, while alkane carbons are sp3 hybridized with no pi system. Without that accessible electron density, alkanes don't undergo electrophilic addition under normal conditions.

Products of HBr addition to alkenes

For ethylene, the product is straightforward: bromoethane ($CH_3CH_2Br$). But with unsymmetrical alkenes, Markovnikov's rule determines the major product.

• Cyclohexene + HBr → bromocyclohexane. The ring carbons of the double bond are equivalently substituted, so only one product forms.
• 2-Methylpropene + HBr → 2-bromo-2-methylpropane. H adds to the terminal (less substituted) carbon, and Br adds to the internal carbon. This pathway is favored because it goes through a tertiary carbocation, which is more stable than the alternative primary carbocation.

Carbocation stability follows this order: 3° > 2° > 1° > methyl. Each additional alkyl group stabilizes the positive charge through hyperconjugation and inductive effects. Markovnikov's rule is really just a consequence of this stability trend: the reaction proceeds through whichever carbocation intermediate is more stable.

Factors Affecting the Reaction

• Rate-determining step: The slow step is Step 1, formation of the carbocation. Once the carbocation forms, Br$^-$ attacks quickly.
• Solvent effects: Polar protic solvents (like water or alcohols) help stabilize the carbocation intermediate and Br$^-$ through solvation, which can lower the activation energy for the rate-determining step.
• Temperature: Higher temperatures generally increase the reaction rate, as more molecules have sufficient energy to overcome the activation barrier for carbocation formation.