Anti-Markovnikov Hydrohalogenation

Anti-Markovnikov Hydrohalogenation

Assessment

Interactive Video

Chemistry, Science

11th Grade - University

Hard

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The video explains the anti-Markovnikov hydrohalogenation reaction, highlighting its regiospecificity, which results in the halogen adding to the less substituted carbon in the presence of peroxides. The mechanism involves a free radical process initiated by peroxides, leading to the formation of hydroxyl and bromine radicals. The bromine radical adds to the pi bond, forming a more stable alkyl radical intermediate, resulting in the anti-Markovnikov product. The video contrasts this with the Markovnikov mechanism, emphasizing the role of radical stability in determining the reaction outcome.

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5 questions

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1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What role do peroxides play in anti-Markovnikov hydrohalogenation?

They prevent the formation of the Markovnikov product.

They initiate the formation of free radicals.

They stabilize the carbocation intermediate.

They act as a catalyst for the reaction.

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

In the initiation step of the anti-Markovnikov reaction, what is formed from the homolysis of the oxygen-oxygen bond?

Water molecules

Bromine radicals

Carbocation intermediates

Hydroxyl radicals

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Why does the bromine radical add to the less substituted carbon in the anti-Markovnikov reaction?

To generate a more stable alkyl radical

To prevent the formation of water

To form a more stable carbocation

To increase the reaction rate

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the result of the propagation steps in the anti-Markovnikov mechanism?

Formation of a hydroxyl radical

Formation of a primary alkyl radical

Formation of a less substituted alkyl bromide

Formation of a Markovnikov product

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What happens if two radical species meet during the reaction?

They form a new radical.

They terminate the reaction.

They increase the reaction rate.

They form a carbocation.