Hydrocarbons · Lesson 7 of 14
Alkene Addition and Product Direction
“Predict addition products by following electron movement and intermediate stability.”
• Connect accessible pi electrons to electrophilic addition. • Predict hydrogenation, halogenation, hydrogen halide addition, and hydration products. • Explain Markovnikov orientation through carbocation stability. • Contrast ionic HBr addition with its peroxide-initiated radical route.
Accessible pi electrons create a reactive site
Alkenes resemble alkanes in many physical properties. The lower members are gases, intermediate members liquids, and higher members solids. They are poorly soluble in water and more soluble in non-polar solvents. Boiling points generally increase with chain length and decrease with branching among comparable isomers. Ethene is colourless and has a faint sweet odour, but odour is not a safe identification test.
Their main chemical difference comes from the double bond. Its exposed pi electrons can interact with an electron-seeking reagent, an electrophile. In addition, the pi part is consumed and new sigma bonds join incoming atoms or groups to the two carbons. The carbon skeleton normally remains connected, unlike oxidative cleavage.
Hydrogenation adds hydrogen with nickel, platinum, or palladium to give an alkane. Bromine and chlorine add to give vicinal dihalides. Bromine solution loses its reddish-orange colour as the reagent is consumed, providing an observation associated with unsaturation. Iodine does not normally add under these simple conditions. Halogen addition proceeds through a halonium-ion pathway; its detailed treatment is not required for the product patterns here.
An unsymmetrical alkene offers two orientations
Hydrogen chloride, bromide, and iodide add to give alkyl halides. Their usual reactivity order is HI greater than HBr greater than HCl. With a symmetrical alkene such as ethene, either orientation gives the same constitutional product. In propene, however, the two ends of the double bond are different, so the possible intermediates must be compared.
In the usual ionic addition of a hydrogen halide to a simple unsymmetrical alkene, hydrogen adds predominantly to the double-bond carbon already carrying more hydrogens. Halide then attaches to the other carbon.
The rule describes a product pattern; the mechanism explains it. The pi electrons form a bond to electrophilic hydrogen while the H–Br bond breaks towards bromine. Protonation at the terminal carbon leaves a secondary carbocation at the middle carbon. The alternative leaves a primary carbocation at the terminal carbon. In this comparison, the secondary carbocation is more stable and its formation pathway is favoured. Bromide then supplies an electron pair to bond at the positively charged carbon.
Problem
Predict the principal constitutional product from hex-1-ene and HBr without peroxide.
- 1.Hydrogen attachment to the terminal carbon puts positive charge at carbon 2. That carbon is bonded to two other carbons, so it is a secondary carbocation.
- 2.The reverse protonation puts positive charge on a primary terminal carbon. This pathway is less favourable in the comparison.
- 3.Bromide attacks carbon 2, producing 2-bromohexane. The position follows the intermediate rather than an arbitrary preference for the left or right end of the drawing.
Peroxide makes bromine radical add first
A peroxide contains a relatively weak oxygen–oxygen bond that can split homolytically to initiate radicals. Benzoyl peroxide, for example, gives benzoyloxy radicals, which can lose carbon dioxide to form phenyl radicals. Initiator-derived radicals abstract hydrogen from HBr to generate bromine radicals. These attack the double bond before hydrogen adds.
When bromine radical attaches to propene’s terminal carbon, the unpaired electron remains at a secondary carbon. The reverse orientation leaves a primary carbon radical. The secondary radical is favoured. It then abstracts hydrogen from another HBr molecule, giving 1-bromopropane while regenerating bromine radical. This is a radical propagation chain, without a carbocation intermediate.
This is called the peroxide, Kharash, or anti-Markovnikov effect. In the reactions considered here it is observed with HBr, not HCl or HI. The radical chain must have energetically favourable propagation steps. Breaking the strong H–Cl bond is unfavourable in the hydrogen-abstraction step, while iodine-radical addition to the alkene is unfavourable and reversible. A weaker H–I bond alone therefore cannot make the entire sequence work.
Problem
Compare the stable intermediate selected with and without peroxide.
- 1.Without peroxide, hydrogen adds first. Its terminal attachment leaves the more stable secondary carbocation at the middle carbon; bromide attaches there afterward.
- 2.With peroxide, bromine radical adds first. Its terminal attachment leaves the more stable secondary radical at the middle carbon; hydrogen attaches there afterward.
- 3.Both comparisons favour the more stable carbon intermediate. The first incoming species differs, so bromine finishes at a different carbon.
Sulphuric acid addition and hydration
Cold concentrated sulphuric acid adds across a double bond to form an alkyl hydrogen sulphate. Hydrogen follows the usual Markovnikov orientation. With water and an acid catalyst, an alkene instead forms an alcohol, predominantly following the same orientation in these simple examples. The catalyst is regenerated; it is not permanently attached as a sulphate group in the final alcohol.
Problem
What forms on acid-catalysed hydration of 2-methylpropene?
- 1.Hydrogen attachment to its CH2 carbon leaves a tertiary carbocation on the carbon bearing the two methyl groups.
- 2.Water bonds at that carbon and subsequent proton loss produces a neutral alcohol.
- 3.The product is 2-methylpropan-2-ol. Terminal hydroxyl attachment would ignore the favoured ionic pathway.
The peroxide effect is not a general instruction to reverse every addition. Apply it here to HBr under radical-initiating conditions. Also interpret bromine decolourisation as evidence of reagent consumption, not as a unique identification of one molecular structure.
Quiz
What attracts an electrophile to an alkene?
What is the major product of propene and HBr without peroxide?
Which intermediate explains HBr addition in the peroxide route?
Which hydrogen halide shows the standard peroxide effect here?
What forms on hydration of 2-methylpropene with acid?
Practice Problems
- Write ethene products with hydrogen, bromine, HBr, and cold concentrated sulphuric acid.
- Compare both possible carbocations from protonation of propene.
- Give HBr addition products of hex-1-ene with and without peroxide.
- Write the two radical propagation steps for the peroxide effect on propene.
- Explain why the same peroxide effect is not observed with HCl and HI.
- Compare an alkyl hydrogen sulphate product with an alcohol formed through acid-catalysed hydration.
Key Takeaways
• Addition consumes alkene pi bonding while forming new sigma bonds. • Ordinary hydrogen halide orientation follows the favoured carbocation pathway. • The HBr/peroxide route uses radicals, with bromine adding first. • State reagents and conditions before predicting the major product.