Where does the bromine go?
"Where does the bromine go?" is a free, interactive Class 11 Chemistry simulation for NEET and JEE, built on the lesson "Properties and reactions of alkenes" in Hydrocarbons. Pick an alkene and whether benzoyl peroxide is present. The first species to attack joins one double-bond carbon; the cards compare the two intermediates it could leave behind. Both routes pass through the more stable secondary intermediate on C2. With H⁺ first, Br lands on C2; with Br• first, Br is already on C1. For symmetrical but-2-ene the two routes give one product. JEE Main asked 34 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
From the lesson Properties and reactions of alkenes in Hydrocarbons. Change the values and watch what happens.
In JEE Main
42 shifts analysed
34 questions from Hydrocarbons across the 42 JEE Main shifts Lumi has analysed (2017–2026).
The idea behind it
NCERT §9.3.5
- Alkenes C₂ to C₄ are gases, then fourteen members are liquids, and higher ones are solid. Ethene has a faintly sweet smell; other alkenes have none. None of them dissolves in water, but all dissolve fairly well in non-polar solvents such as benzene and petroleum ether. Each added CH₂ raises the boiling point by 20–30 K.
- Bromine or chlorine adds to give a vicinal dihalide (through a cyclic halonium ion); iodine does not add under normal conditions. Loss of the reddish orange colour of bromine in CCl₄ is a test for unsaturation.
- Hydrogen halides add to give alkyl halides, with reactivity HI > HBr > HCl. A symmetrical alkene gives one product: ethene + HBr gives bromoethane.
- Markovnikov rule (1869): when an unsymmetrical reagent adds to an unsymmetrical alkene, its negative part joins the double-bond carbon that carries fewer hydrogens. Propene + HBr gives mainly 2-bromopropane.
- Mechanism: H⁺ attacks first, forming the more stable carbocation (secondary rather than primary), which forms faster; Br⁻ then bonds to that positive carbon.
- Peroxide (Kharash) effect: with benzoyl peroxide, HBr adds to propene against Markovnikov's rule, giving 1-bromopropane, by a free-radical chain through the more stable secondary radical. Hex-1-ene gives 2-bromohexane without peroxide and 1-bromohexane with it.
- Only HBr shows the peroxide effect. H–Cl (430.5 kJ mol⁻¹) is too strong for the radical to break, and I• radicals join to form I₂ rather than add, even though H–I (296.8 kJ mol⁻¹) is weaker than H–Br (363.7 kJ mol⁻¹).
- Cold concentrated H₂SO₄ adds by Markovnikov's rule to give alkyl hydrogen sulphates; water with a few drops of concentrated H₂SO₄ adds by the same rule to give alcohols.
- Cold dilute aqueous KMnO₄ (Baeyer's reagent) turns alkenes into vicinal glycols, and its loss of colour is a test for unsaturation. Acidic KMnO₄ or K₂Cr₂O₇ cleaves them to ketones and/or acids; but-2-ene gives two molecules of ethanoic acid.
- Ozonolysis adds O₃ to form an ozonide, which Zn–H₂O splits into smaller carbonyl compounds; the fragments reveal where the double bond was. Polymerisation at high temperature and pressure with a catalyst turns ethene into polythene and propene into polypropene.
Questions
3 answers
- What does the "Where does the bromine go?" simulation show?
- Pick an alkene and whether benzoyl peroxide is present. The first species to attack joins one double-bond carbon; the cards compare the two intermediates it could leave behind.
- Does JEE Main ask questions from Hydrocarbons?
- Yes. JEE Main asked 34 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
- Is this simulation free?
- Yes. It is free and runs in the browser on a phone or a laptop. It belongs to the lesson "Properties and reactions of alkenes" in Hydrocarbons (Class 11 Chemistry).
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