Reactivity 3.4.12 – Carbocation Stability in the Electrophilic Addition to Unsymmetrical Alkenes

When hydrogen halides (e.g. HBr, HCl) add to unsymmetrical alkenes (like propene), more than one product is theoretically possible. However, one major product dominates due to the formation of the more stable carbocation intermediate during the reaction mechanism. This explains the observed regioselectivity of the reaction.

What is an Unsymmetrical Alkene?

An alkene where the two carbon atoms of the double bond are not bonded to the same groups.

For example:

• CH₃-CH=CH₂ (propene)
• CH3-CH=CHCH3 (but-2-ene is symmetrical, but but-1-ene is not)

Electrophilic Addition to Unsymmetrical Alkenes:

In reactions like propene + HBr, there are two possible sites for H⁺ to add, forming different carbocations:

• 1° carbocation: if H⁺ adds to the CH₂ end → less stable
• 2° carbocation: if H⁺ adds to the CH group → more stable

Carbocation Stability Order:

The stability of carbocations follows this trend due to inductive and hyperconjugation effects:

3° > 2° > 1° > methyl

• 3° (tertiary): Carbon with 3 alkyl groups stabilizing the positive charge
• 2° (secondary): Carbon with 2 alkyl groups
• 1° (primary): Carbon with only 1 alkyl group

Mechanism Explanation (e.g. Propene + HBr):

1. The π electrons from the double bond attack H⁺ from HBr.
2. This can give either a 1° or 2° carbocation intermediate.
3. The more stable 2° carbocation forms faster and is lower in energy.
4. Br^– attacks the carbocation → major product formed from more stable carbocation.

Markovnikov’s Rule:

When a hydrogen halide or water adds to an unsymmetrical alkene, the hydrogen atom bonds to the carbon that already has more hydrogen atoms (i.e. the less substituted carbon), and the other group bonds to the more substituted carbon. This occurs because it leads to a more stable carbocation intermediate.

Rationale: This pathway leads to the formation of the most stable (usually 2° or 3°) carbocation intermediate.

Example – Propene + HBr:

Two possible products:

• Minor product: 1-bromopropane (via 1° carbocation)
• Major product: 2-bromopropane (via 2° carbocation)

Mechanism favours 2° carbocation → major product is 2-bromopropane

Predicting the Major Product in Electrophilic Addition to Unsymmetrical Alkenes

In electrophilic addition reactions involving unsymmetrical alkenes, more than one product may form, but one is often dominant. The major product is determined by the stability of the intermediate carbocation formed during the reaction. This explains Markovnikov’s Rule.

Steps for Predicting Major Product:

1. Identify the alkene: Locate the double bond and note asymmetry.
2. Add H⁺ to both possible carbon atoms: Draw both possible carbocation intermediates.
3. Determine stability: Use the trend 3° > 2° > 1° to choose the more stable intermediate.
4. Add nucleophile: Add halide ion (e.g. Br^–) or water to the more stable carbocation.

Example: Propene + HBr

• • Two possible products: 1-bromopropane (via 1° carbocation), 2-bromopropane (via 2° carbocation)
  • 2° carbocation is more stable → major product is 2-bromopropane

Equation: \( \text{CH}_3\text{CH} = \text{CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{CHBrCH}_3 \)

Example: But-1-ene + HCl

• • H⁺ can add to either end of double bond → gives 1° or 2° carbocation
  • 2° carbocation (on C-2) is more stable → major product is 2-chlorobutane

Equation: \( \text{CH}_2 = \text{CHCH}_2\text{CH}_3 + \text{HCl} \rightarrow \text{CH}_3\text{CHClCH}_2\text{CH}_3 \)

Example: But-1-ene + H₂O (acid-catalysed)

• • Mechanism involves formation of carbocation from alkene + H⁺
  • 2° carbocation → water attacks → forms butan-2-ol

Equation: \( \text{CH}_2 = \text{CHCH}_2\text{CH}_3 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3\text{CHOHCH}_2\text{CH}_3 \)

Key Notes:

• Only the H⁺ adds first in the mechanism – the halide/water acts second as the nucleophile.
• Product distribution depends entirely on carbocation stability, not just initial placement of atoms.