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Exercises: Radical addition of hydrogen bromide to alkenes

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Exercise

The result of the addition of hydrogen bromide varies from experiment to experiment. Which of the following statements are correct?

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The addition of hydrogen bromide to 2-butene that has just been distilled yields the Markovnikov product 1-bromobutane.

The addition of hydrogen bromide to 1-butene that has freshly been distilled yields the Markovnikov product 2-bromobutane.

If 1-butene is previously exposed to air, the addition of HBr predominately results in the anti-Markovnikov product 1-bromobutane.

If 1-butene is previously exposed to air, the addition of HBr predominately results in the Markovnikov product 2-bromobutane.

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Exercise

Which reaction step initiates the radical chain reaction in the addition of hydrogen bromide to alkenes?

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The exothermic, homolytic cleavage of the O-O hydroxide bond.

The endothermic, heterolytic cleavage of the O-O suboxide bond.

The endothermic, homogenic cleavage of the C-H anhydride bond.

The endothermic, homolytic cleavage of the O-O peroxide bond.

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Exercise

In the first propagation step of the chain reaction of hydrogen bromide's radical addition to 1-butene, a bromine radical attacks the alkene. Which of the following statements are correct?

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The bromine radical is added to the μ bond of 1-butene, yielding a bromoalkyl radical.

The bromine radical is added to the π bond of 1-butene, yielding a bromoalkyl radical.

The attack of the bromine radical on 1-butene follows a regioselective course, which yields the less stable primary alkyl radical.

The attack of the bromine radical on 1-butene follows a regioselective course, which yields the more stable secondary alkyl radical.

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Exercise

Which of the following statements are correct?

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correct

wrong

partly correct

This is the right answer

In the radical mechanism, the less stable, primary alkyl radical is formed by the addition of a bromine radical to the double bond, while in the ionic mechanism the less stable, secondary carbenium ion is formed by the addition of a proton to the double bond.

In the ionic mechanism of HBr addition to 1-butene, a proton is first added to the double bond, yielding the more stable secondary carbenium ion.

When HBr is added to 1-butene, the radical mechanism yields the anti-Markovnikov product 1-bromobutane, while the ionic mechanism yields the Markovnikov product 2-bromobutane.

In the HBr addition to 1-butene, the radical mechanism yields the Markovnikov product 2-bromobutane, while the ionic mechanism yields the anti-Markovnikov product 1-bromobutane.

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