Haloalkanes and Haloarenes · Lesson 5 of 16
Preparing Haloalkanes from Hydrocarbons
“Distinguish radical replacement from alkene addition and predict products from symmetry and reaction conditions.”
• Enumerate radical monohalogenation products using equivalent hydrogen positions. • Predict ordinary HX addition and peroxide-dependent HBr addition. • Distinguish addition across a double bond from allylic or benzylic substitution. • Use conditions and plausible intermediates to explain the principal products.
Light can change the reaction pathway
An alkane has no π bond available for simple addition. Under UV light or suitable heat, chlorine or bromine can react through a radical chain in which a hydrogen is replaced by halogen. In contrast, an alkene can add HX or X₂ across its C=C bond. Seeing “bromine” in both descriptions does not mean the same bonds change. Specify light, radical conditions, solvent and the substrate before predicting a product.
For methane chlorination, initiation is Cl₂→2Cl• under light. Propagation includes Cl•+CH₄→HCl+CH₃• and CH₃•+Cl₂→CH₃Cl+Cl•. A radical is regenerated, allowing the chain to continue; radical–radical combination can terminate it. The net mono-substitution is CH₄+Cl₂→CH₃Cl+HCl. The dots denote unpaired electrons. Ordinary full-headed curved arrows describe electron pairs, so they must not be used indiscriminately for one-electron radical steps.
Further chlorination and different hydrogen environments can produce mixtures. For butane, replacement at an end or an internal carbon gives 1-chlorobutane or 2-chlorobutane, and further substitution can also occur. Product ratios depend on both the number of available hydrogens and their relative abstraction rates. Symmetry determines the number of different constitutional products, not their proportions.
Problem
Enumerate monochloro constitutional products of 2-methylbutane.
- 1.Write CH₃CH(CH₃)CH₂CH₃. The two methyl groups attached to the branched CH are equivalent, while the other terminal CH₃, CH₂ and branched CH are distinct.
- 2.Replacing H in these four environments gives ClCH₂CH(CH₃)CH₂CH₃, CH₃CH(CH₃)CH₂CH₂Cl, CH₃CH(CH₃)CHClCH₃ and CH₃C(Cl)(CH₃)CH₂CH₃.
- 3.These are 1-chloro-2-methylbutane, 1-chloro-3-methylbutane, 2-chloro-3-methylbutane and 2-chloro-2-methylbutane. Equivalent positions are counted once; stereoisomers are a separate issue.
Problem
Which C₅H₁₂ isomers give one, three and four distinct monochloro connectivities?
- 1.2,2-Dimethylpropane gives one: all its methyl groups and their hydrogen positions are constitutionally equivalent.
- 2.Pentane gives three: end C1/C5, C2/C4 and middle C3 environments.
- 3.2-Methylbutane gives four as enumerated above. No relative product yield is implied by the counts.
Ordinary HX addition: choose the better cation pathway
In ordinary ionic addition to an unsymmetrical alkene, proton addition can generate different carbocation-like pathways. The pathway leading to the better stabilised cation is usually favoured, followed by halide capture. Propene with HBr, without peroxide, predominantly gives 2-bromopropane: CH₃CH=CH₂+HBr→CH₃CHBrCH₃. Forming a secondary rather than primary cation helps explain this Markovnikov orientation. It is a regioselectivity rule under specified conditions, not a guarantee of one product for every complicated alkene.
Problem
Predict HBr addition to styrene, HCl addition to but-1-ene, and HBr/peroxide addition to allylbenzene.
- 1.Styrene is C₆H₅CH=CH₂. Ordinary HBr addition favours the benzylic cation pathway and gives C₆H₅CHBrCH₃, 1-bromoethylbenzene.
- 2.But-1-ene plus HCl under ordinary addition conditions predominantly gives CH₃CH₂CHClCH₃, 2-chlorobutane, through the secondary-cation orientation.
- 3.Allylbenzene is C₆H₅CH₂CH=CH₂. With HBr and peroxide, radical anti-Markovnikov addition gives mainly C₆H₅CH₂CH₂CH₂Br. The extra side-chain CH₂ must not be lost when reading the starting drawing.
Peroxide changes HBr addition through a radical chain
Peroxide-derived radicals can generate Br• from HBr. Br• adds to the terminal carbon of propene so that the new radical is secondary: Br•+CH₂=CHCH₃→BrCH₂CH•CH₃. That carbon radical abstracts H from HBr to give BrCH₂CH₂CH₃ and regenerate Br•. Thus bromine ends on the less substituted carbon, and H on the more substituted one. The orientation is explained by the more favourable radical intermediate rather than by a reversed carbocation rule.
The familiar peroxide effect applies usefully to HBr; the corresponding propagation energy balances do not support an equivalent general anti-Markovnikov addition rule for HCl or HI. Adding “peroxide” to an HI arrow does not provide a reliable route to a primary iodide. To make 1-iodobutane from but-1-ene, use HBr/peroxide to obtain 1-bromobutane and then NaI/dry acetone exchange, rather than claiming direct anti-Markovnikov HI addition.
Adding X₂ makes a vicinal dihalide
Bromine can add across C=C, breaking the π component and forming two C–Br bonds. Ethene gives BrCH₂CH₂Br: CH₂=CH₂+Br₂→BrCH₂CH₂Br. Loss of bromine colour can indicate reaction with unsaturation, but it is not unique proof that an unknown is an alkene because other reactive compounds can also consume bromine. In ordinary non-radical bromination, a bridged halonium intermediate helps explain anti addition to many alkenes. A dark, non-radical addition is different from a light-driven allylic substitution.
Carbon tetrachloride has historically been used as a non-nucleophilic medium in bromine-addition descriptions. Its presence on an equation is a solvent label, not an atom source in the addition product. The chemistry can be explained without treating historical solvent use as a recommendation for routine handling. For cyclohexene under the specifically stated Br₂/heat or light radical conditions, the intended allylic product is 3-bromocyclohex-1-ene; addition can compete, so this condition set should not be advertised as a universally selective preparation.
Ring halogenation and side-chain halogenation need different conditions
Toluene or another alkylbenzene with halogen and a Lewis-acid ring-halogenation catalyst can react on the ring. Under light-driven radical conditions, a benzylic C–H bond can instead be replaced because the benzylic radical is resonance-stabilised. For p-nitroethylbenzene with Br₂ under the stated radical conditions, substitution is expected at its benzylic CH₂, giving p-O₂NC₆H₄CHBrCH₃. The nitro group does not turn this into bromination of the ring under the same radical scheme.
Problem
Identify C₅H₁₀ that resists ordinary chlorine addition in the dark but gives one C₅H₉Cl connectivity in sunlight.
- 1.C₅H₁₀ can represent an alkene or a cycloalkane. Lack of ordinary addition in the dark supports a saturated ring rather than a normal alkene.
- 2.Cyclopentane has equivalent ring positions for constitutional monochlorination, giving chlorocyclopentane under radical conditions.
- 3.The inference uses both observations and the formula. “One product” refers to constitutional positions and assumes the stated comparison, not a guarantee of complete selectivity against further chlorination.
Quick Quiz
Quiz
What does symmetry determine in monochlorination?
Explanation: Equivalent sites give the same connectivity; rates need additional information.
What is the major ordinary HBr product from propene?
Explanation: The usual ionic pathway favours the secondary-cation orientation.
Which reagent change supports anti-Markovnikov addition to propene?
Explanation: A viable radical propagation chain gives terminal bromide with HBr.
What is formed when ethene adds Br₂ under ordinary addition conditions?
Explanation: Two bromines add across the original double bond.
Which C₅H₁₂ skeleton gives only one constitutional monochloride?
Explanation: The four methyl groups of neopentane are constitutionally equivalent.
Practice Problems
- Give a route from but-1-ene to 1-iodobutane and explain both stages.
- Predict the principal product of 1-methylcyclohexene with HI under ordinary addition conditions.
- Explain why CH₃CH₂CH₂CH₃ with Cl₂/light need not give only one chloride.
- Distinguish styrene from allylbenzene in HBr product prediction.
- Why can bromine decolourisation be useful without being a unique alkene test?
- Why are the products of ring halogenation and benzylic halogenation different?
Use HBr/peroxide to give CH₃CH₂CH₂CH₂Br through radical anti-Markovnikov addition. Then use NaI in dry acetone to replace Br by I with NaBr precipitation. Direct HI/peroxide is not the analogous general solution because its radical chain is not comparably favourable.
Key Takeaways
• Radical halogenation replaces H, whereas alkene addition consumes a π bond. • Equivalent hydrogen positions control constitutional product counts. • Ordinary HX addition often follows the more stable carbocation pathway. • Peroxide-dependent anti-Markovnikov addition is a useful HBr-specific pattern. • Halogen addition to an alkene produces a vicinal dihalide under suitable conditions. • Allylic and benzylic radical substitution must be distinguished from double-bond or ring addition. • A reaction arrow needs conditions to communicate which pathway is intended.