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Chapter 15: Q20. (page 570)

Question: Which compounds can be prepared in good yield by allylic halogenation of an alkene?

a.

b.

c.

Short Answer

Expert verified

Answer

a. Heptane

b. Cannot be made

c. Cannot be made

Step by step solution

01

Halogenation

It involves the treatment of alkene with NBS in the presence of sunlight. The reaction yields the product via allylic substitution.

02

 Step 2: Explanation

a.

The allylic halogenation of cycloheptane forms 3-bromocyclohept-1-ene in a good yield. The reaction is shown below:

Formation of product in a

b.

The given compound can not be made via allylic halogenation method as the starting material (alkene) would produce the mixture of allylic halides.

c.

The given compound can not be made via allylic halogenation method as the starting material (alkene) would produce the mixture of allylic halides.

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Most popular questions from this chapter

Question: a) Draw all constitutional isomers formed by monochlorination of each alkane with Cl2and h谓. (b) Draw the major monobromination product formed by heating each alkane with Br2.

Question: Which alkyl halides can be prepared in good yield by radical halogenation of an alkane?

Question: Draw the products formed when each alkene is treated with NBS + hv .

a.

b.

c.

Question. Ethers are oxidized with to form hydroperoxides that decompose violently when heated. Draw a stepwise mechanism for this reaction.

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.
See all solutions

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