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SolutionsHX Addition (X = halogen)

1. HBr with and without peroxides

Predict the result of the following reactions:

1-Methylcyclohexene with HBr and with HBr and peroxides (ROOR), and 3-bromo-2-methylcyclopentene with HBr
With HBr, 1-methylcyclohexene gives 1-bromo-1-methylcyclohexane; with HBr and peroxides, 1-bromo-2-methylcyclohexane as a mixture of isomers; 3-bromo-2-methylcyclopentene with HBr gives 1,2-dibromo-1-methylcyclopentane

Without peroxides the reaction goes through the more stable carbocation (Markovnikov rule); with peroxides, through the more stable radical, and bromine bonds to the less substituted carbon.

2. Rearrangement

3,3-Dimethylbut-1-ene with HCl gives two products: 3-chloro-2,2-dimethylbutane and 2-chloro-2,3-dimethylbutane. Explain how they are formed.

Protonation gives the secondary carbocation (CH3)3C–CH+–CH3, which gives the first product. A 1,2-methyl shift converts it into the more stable tertiary carbocation (CH3)2C+–CH(CH3)2, which gives the second.

3. Why only HBr?

Why is the peroxide effect observed with HBr but not with HCl or HI?

Both propagation steps of the radical chain must be exothermic. With HCl the step in which the radical abstracts the hydrogen is endothermic (the H–Cl bond is very strong); with HI it is the addition of I• to the double bond (the C–I bond is weak). Only with HBr are both favourable.

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