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Chapter 8: Q19Q (page 395)

Propose a mechanism for the addition of bromine water to cyclopentene, being carefulto show why the trans product results and how both enantiomers are formed.

Short Answer

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first step is addition of a bromine molecule to cyclopentene ring to form bromonium ion.

Step by step solution

01

Step 1:

first step is addition of a bromine molecule to cyclopentene ring to form bromonium ion.

02

 

Now water molecule is added in the second step. Here two possibilities occur due to the presence of bromonium ion. first if water attack from the back side (dash bond) then bromine come upward (wedge bond) while if water molecule attack from the front side(wedge bond) then bromine go backward(dash bond). The two products formed are enantiomers as the only difference in orientation of the bromine atom and hydroxyl group

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Most popular questions from this chapter

The chiral BINAP ligand shown in figure 8-8 contains no asymmetric carbon atoms. Explain how this ligand is chiral.

Predict the carbenoid addition products of the following reactions.

(a) trans - hex - 3 - ene + CH2I2, Zn(Cu)

(b) cis - hept - 2 - ene + CH2I2, Zn(Cu)

The chiral BINAP ligand shown in figure 8-8 contains no asymmetric carbon atoms. Explain how this ligand is chiral.

In the hydroboration of 1-methylcyclopentene shown in Solved Problem 8-3, the reagents are achiral, and the products are chiral. The product is a racemic mixture of trans-2-methylcyclopentanol, but only one enantiomer is shown. Show how the other enantiomer is formed.

Show how you would accomplish each of the following synthetic conversions.

a) Trans-but-2-ene→trans-1,2-dimethylcyclopropane

b) Cyclopentene→

c) Cyclohexanol→
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