Whether the product obtained by the bromination of optically active (R)-2-phenylpropanoic acid under Hell-Volhard-Zelinski conditions is optically active or racemic is to be explained. Concept introduction: Hell-Volhard-Zelinski reaction takes place through the formation of an acid bromide enol intermediate. If the chiral carbon of the optically active carboxylic acid is involved in the enol formation chirality will be lost during enolization, as the double bonds are planar. Bromination of this enol can take place from either face of the enol double bond leading to the formation of a racemic mixture. If the chiral carbon does not take part in enol formation, then the resulting product will be optically active.

Question

Want to see the full answer?

Answer 1

Question

Chapter 22.SE, Problem 53AP

Interpretation Introduction

Interpretation:

Whether the product obtained by the bromination of optically active (R)-2-phenylpropanoic acid under Hell-Volhard-Zelinski conditions is optically active or racemic is to be explained.

Concept introduction:

Hell-Volhard-Zelinski reaction takes place through the formation of an acid bromide enol intermediate. If the chiral carbon of the optically active carboxylic acid is involved in the enol formation chirality will be lost during enolization, as the double bonds are planar. Bromination of this enol can take place from either face of the enol double bond leading to the formation of a racemic mixture. If the chiral carbon does not take part in enol formation, then the resulting product will be optically active.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

reaction scheme for C39H4202 Hydrogenation of Alkyne (Alkyne to Alkene) show reaction (drawing) please

Give detailed mechanism Solution with explanation needed. Don't give Ai generated solution

Show work with explanation needed....don't give Ai generated solution

Answer 2

Question

Chapter 22.SE, Problem 53AP

Interpretation Introduction

Interpretation:

Whether the product obtained by the bromination of optically active (R)-2-phenylpropanoic acid under Hell-Volhard-Zelinski conditions is optically active or racemic is to be explained.

Concept introduction:

Hell-Volhard-Zelinski reaction takes place through the formation of an acid bromide enol intermediate. If the chiral carbon of the optically active carboxylic acid is involved in the enol formation chirality will be lost during enolization, as the double bonds are planar. Bromination of this enol can take place from either face of the enol double bond leading to the formation of a racemic mixture. If the chiral carbon does not take part in enol formation, then the resulting product will be optically active.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 22.SE, Problem 53AP

Interpretation Introduction

Interpretation:

Whether the product obtained by the bromination of optically active (R)-2-phenylpropanoic acid under Hell-Volhard-Zelinski conditions is optically active or racemic is to be explained.

Concept introduction:

Hell-Volhard-Zelinski reaction takes place through the formation of an acid bromide enol intermediate. If the chiral carbon of the optically active carboxylic acid is involved in the enol formation chirality will be lost during enolization, as the double bonds are planar. Bromination of this enol can take place from either face of the enol double bond leading to the formation of a racemic mixture. If the chiral carbon does not take part in enol formation, then the resulting product will be optically active.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 22.SE, Problem 53AP

Interpretation Introduction

Interpretation:

Whether the product obtained by the bromination of optically active (R)-2-phenylpropanoic acid under Hell-Volhard-Zelinski conditions is optically active or racemic is to be explained.

Concept introduction:

Hell-Volhard-Zelinski reaction takes place through the formation of an acid bromide enol intermediate. If the chiral carbon of the optically active carboxylic acid is involved in the enol formation chirality will be lost during enolization, as the double bonds are planar. Bromination of this enol can take place from either face of the enol double bond leading to the formation of a racemic mixture. If the chiral carbon does not take part in enol formation, then the resulting product will be optically active.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Chapter 22.SE, Problem 53AP

Interpretation Introduction

Interpretation:

Whether the product obtained by the bromination of optically active (R)-2-phenylpropanoic acid under Hell-Volhard-Zelinski conditions is optically active or racemic is to be explained.

Concept introduction:

Hell-Volhard-Zelinski reaction takes place through the formation of an acid bromide enol intermediate. If the chiral carbon of the optically active carboxylic acid is involved in the enol formation chirality will be lost during enolization, as the double bonds are planar. Bromination of this enol can take place from either face of the enol double bond leading to the formation of a racemic mixture. If the chiral carbon does not take part in enol formation, then the resulting product will be optically active.

Answer 6

Chapter 22.SE, Problem 53AP

Interpretation Introduction

Interpretation:

Whether the product obtained by the bromination of optically active (R)-2-phenylpropanoic acid under Hell-Volhard-Zelinski conditions is optically active or racemic is to be explained.

Concept introduction:

Hell-Volhard-Zelinski reaction takes place through the formation of an acid bromide enol intermediate. If the chiral carbon of the optically active carboxylic acid is involved in the enol formation chirality will be lost during enolization, as the double bonds are planar. Bromination of this enol can take place from either face of the enol double bond leading to the formation of a racemic mixture. If the chiral carbon does not take part in enol formation, then the resulting product will be optically active.

Answer 7

Chapter 22.SE, Problem 53AP

Interpretation Introduction

Interpretation:

Whether the product obtained by the bromination of optically active (R)-2-phenylpropanoic acid under Hell-Volhard-Zelinski conditions is optically active or racemic is to be explained.

Concept introduction:

Hell-Volhard-Zelinski reaction takes place through the formation of an acid bromide enol intermediate. If the chiral carbon of the optically active carboxylic acid is involved in the enol formation chirality will be lost during enolization, as the double bonds are planar. Bromination of this enol can take place from either face of the enol double bond leading to the formation of a racemic mixture. If the chiral carbon does not take part in enol formation, then the resulting product will be optically active.

Answer 8

Expert Solution & Answer

Answer 9

Expert Solution & Answer

Answer 10

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 11

Ch. 22.1 - Prob. 1P Ch. 22.1 - How many acidic hydrogens does each of the...Ch. 22.1 - Prob. 3P Ch. 22.3 - Write the complete mechanism for the deuteration...Ch. 22.3 - Prob. 5P Ch. 22.4 - If methanol rather than water is added at the end...Ch. 22.5 - Prob. 7P Ch. 22.5 - Draw a resonance structure of the acetonitrile...Ch. 22.6 - If methanol rather than water is added at the end...Ch. 22.7 - Prob. 10P

Want to see the full answer?

Chapter 22 Solutions