The curved arrow notation and intermediate are to be shown to the radical and closed-shell mechanisms for the addition of HBr to propene. Each mechanism is to be completed by drawing the more stable intermediate in the box provided and adding the curved arrow notation. Realizing that both reactions proceed through the more stable intermediate, the opposite regiochemistry of the reactions is to be explained. Concept introduction: A hydrogen halide adds across the double bond of an alkene in a Markovnikov fashion under normal conditions. H + adds to the less alkyl-substituted C atom whereas the halide anion adds to the more alkyl-substituted C atom. The reason is that the first step of the mechanism is protonation of the C=C double bond, which proceeds to give the more stable carbocation intermediate. If a small amount of peroxide ( RO-OR ) is present, however, then the major product is produced according to anti-Markovnikov addition. The H atom adds to the more substituted C , and the halide atom adds to the less substituted one. Peroxides are radical initiators because they contain a weak O-O bond. Hence, the reaction in Equation 25-32 proceeds by radical mechanism. Radical addition to an alkene generally takes place so as to produce the more stable alkyl radical intermediate.

Question

Want to see the full answer?

Answer 1

Question

Chapter 25, Problem 25.23YT

Interpretation Introduction

Interpretation:

The curved arrow notation and intermediate are to be shown to the radical and closed-shell mechanisms for the addition of HBr to propene. Each mechanism is to be completed by drawing the more stable intermediate in the box provided and adding the curved arrow notation. Realizing that both reactions proceed through the more stable intermediate, the opposite regiochemistry of the reactions is to be explained.

Concept introduction:

A hydrogen halide adds across the double bond of an alkene in a Markovnikov fashion under normal conditions. H+ adds to the less alkyl-substituted C atom whereas the halide anion adds to the more alkyl-substituted C atom. The reason is that the first step of the mechanism is protonation of the C=C double bond, which proceeds to give the more stable carbocation intermediate. If a small amount of peroxide (RO-OR) is present, however, then the major product is produced according to anti-Markovnikov addition. The H atom adds to the more substituted C, and the halide atom adds to the less substituted one. Peroxides are radical initiators because they contain a weak O-O bond. Hence, the reaction in Equation 25-32 proceeds by radical mechanism. Radical addition to an alkene generally takes place so as to produce the more stable alkyl radical intermediate.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Alkene bromination. The two butane isomers react with Br2. Draw the key reaction inter- mediates and the product(s); use arrow pushing to explain the stereochemical outcomes of the reactions Final Products Intermediate Br2 Intermediate Final Products Br2

Draw the structures of the missing reactants, intermediates, or products, in the following mechanism. Use wedges and dashes to indicate stereochemistry when appropriate.

As we will learn, many antioxidants–compounds that prevent unwanted radical oxidation reactions from occurring–are phenols, compounds that contain an OH group bonded directly to a benzene ring.a.) Explain why homolysis of the O–H bond in phenol requiresconsiderably less energy than homolysis of the O–H bond in ethanol(362 kJ/mol vs. 438 kJ/mol).b.) Why is the C–O bond in phenol shorter than the C–O bond in ethanol?