Question 1

What is the peroxide effect in hydrogen bromide addition to alkenes?

Accepted Answer

The peroxide effect reverses regioselectivity in HBr addition to alkenes, producing the anti-Markovnikov product instead of the expected Markovnikov product. In the presence of organic peroxides, bromine adds to the less-substituted carbon of the alkene, yielding the less-substituted alkyl halide. This occurs because the reaction proceeds through a free radical mechanism rather than an ionic pathway.

Question 2

How does the free radical mechanism explain anti-Markovnikov regioselectivity?

Accepted Answer

The free radical mechanism involves initiation, propagation, and termination stages. During propagation, a bromine radical attacks the less-branched carbon, forming a more-substituted alkyl radical. This intermediate is more stable due to hyperconjugation and inductive effects. The bromine radical approach encounters less steric hindrance at the less-substituted carbon, creating a lower-energy transition state that favors anti-Markovnikov addition.

Question 3

Why does the peroxide effect not occur with hydrogen iodide and hydrogen chloride?

Accepted Answer

The peroxide effect fails with HI and HCl because their propagation steps are thermodynamically unfavorable. For HI, the first propagation step—addition of iodine radical to the alkene—is endothermic. For HCl, the second propagation step—reaction of the alkyl radical with hydrogen chloride—is endothermic. These unfavorable energetics prevent the chain reaction from proceeding efficiently.

Question 4

What role do peroxides play in initiating the free radical reaction?

Accepted Answer

Peroxides, such as di-tert-butyl peroxide, serve as radical initiators. The weak oxygen-oxygen bond undergoes homolytic cleavage when heated or exposed to light, requiring only 159 kJ/mol of energy. This produces alkoxy radicals that abstract hydrogen from HBr exothermically (ΔH = -70 kJ/mol), generating bromine radicals that begin the propagation chain reaction.

Question 5

Why is a more-substituted alkyl radical formed during anti-Markovnikov addition?

Accepted Answer

When bromine radical attacks the less-substituted carbon of the alkene, it generates a more-substituted (tertiary or secondary) alkyl radical. Tertiary radicals are significantly more stable than primary radicals due to hyperconjugation and inductive stabilization from adjacent alkyl groups. This stability makes formation of the more-substituted radical thermodynamically favorable in the propagation step.

Question 6

What is the stereochemical outcome when a new chiral center forms during HBr addition?

Accepted Answer

When HBr addition to an alkene generates a new chiral center, a racemic mixture of products is obtained. This occurs because the bromine radical can approach the alkene from either face with equal probability. Since both faces are equally accessible and reactive, both enantiomers form in equal amounts, yielding a 50:50 mixture.

Question 7

How does steric hindrance influence regioselectivity in peroxide-mediated HBr addition?

Accepted Answer

Steric hindrance favors bromine radical attack at the less-branched carbon of the alkene. The less-substituted carbon presents less steric obstruction to the incoming bromine radical, resulting in a lower-energy transition state. This geometric advantage, combined with the greater stability of the resulting more-substituted alkyl radical, drives the regioselective formation of the anti-Markovnikov product.

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