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Chapter 15: Q6. (page 570)

Question: Using Mechanism 15.1 as a guide, write the mechanism for the reaction of CH4with Br2 to form CH3 Br and HBr. Classify each step as initiation, propagation, or termination.

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01

Radical halogenation of alkanes

The first step is initiation that involves the hemolysis cleavage ofform two radicals.


The second step is propagation that involves the attack of the radical to the alkane to form alkyl and radical.


The third step is termination that involves the reaction of radicals together to stop the chain.

02

 Step 2: Explanation

The steps involved in the reaction are shown below:

Initiation:

Initiation reaction

Propagation:

Propagation reaction

Termination:

Termination reaction

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Most popular questions from this chapter

Questions: Compounds A and B are isomers having molecular formula C5H12. Heating A with Cl2 gives a single product of monohalogenation, whereas heating B under the same conditions forms three constitutional isomers. What are the structures of A and B?

Question: What products are formed from monochlorination of (R)-2-bromobutane at C1 and C4? Assign R and S designations to each stereogenic center.

Questions: Why is a benzylic C-H bond unusually weak?

Question: When HBr adds to (CH3)2C=CH2 under radical conditions, two radicals are possible products in the first step of chain propagation. Draw the structure of both radicals and indicate which one is formed. Then draw the preferred product from HBr addition under radical conditions.

Questions: Draw all constitutional isomers formed by monochlorination of each alkane.

a.

b.

c.
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