The structure of A and B in the given reaction sequence is to be given. Concept introduction: In the Hoffman elimination, an amine with a poor leaving group is converted into a good leaving group, tetraalkyl ammonium ion, which is a moderately good leaving group, using excess of an alkyl halide (usually CH 3 I ). The bulkiness of the leaving group causes a anti-Zaitsev product or Hoffman product. It generates the least substituted alkene as the major product. The carbon atom to which the primary amine group is attached must be the secondary carbon with at least one alpha hydrogen atom. The index of hydrogen deficiency counts how many multiple bonds and rings are present in the structure.

Question

Want to see the full answer?

Answer 1

Question

Chapter 10, Problem 10.66P

Interpretation Introduction

Interpretation:

The structure of A and B in the given reaction sequence is to be given.

Concept introduction:

In the Hoffman elimination, an amine with a poor leaving group is converted into a good leaving group, tetraalkyl ammonium ion, which is a moderately good leaving group, using excess of an alkyl halide (usually CH3I). The bulkiness of the leaving group causes a anti-Zaitsev product or Hoffman product. It generates the least substituted alkene as the major product. The carbon atom to which the primary amine group is attached must be the secondary carbon with at least one alpha hydrogen atom. The index of hydrogen deficiency counts how many multiple bonds and rings are present in the structure.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Steps and explanation please

can you please draw out and list step-by-step the synthetic strategy for this rxn? thank you sm in advance

Answer 2

Question

Chapter 10, Problem 10.66P

Interpretation Introduction

Interpretation:

The structure of A and B in the given reaction sequence is to be given.

Concept introduction:

In the Hoffman elimination, an amine with a poor leaving group is converted into a good leaving group, tetraalkyl ammonium ion, which is a moderately good leaving group, using excess of an alkyl halide (usually CH3I). The bulkiness of the leaving group causes a anti-Zaitsev product or Hoffman product. It generates the least substituted alkene as the major product. The carbon atom to which the primary amine group is attached must be the secondary carbon with at least one alpha hydrogen atom. The index of hydrogen deficiency counts how many multiple bonds and rings are present in the structure.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 10, Problem 10.66P

Interpretation Introduction

Interpretation:

The structure of A and B in the given reaction sequence is to be given.

Concept introduction:

In the Hoffman elimination, an amine with a poor leaving group is converted into a good leaving group, tetraalkyl ammonium ion, which is a moderately good leaving group, using excess of an alkyl halide (usually CH3I). The bulkiness of the leaving group causes a anti-Zaitsev product or Hoffman product. It generates the least substituted alkene as the major product. The carbon atom to which the primary amine group is attached must be the secondary carbon with at least one alpha hydrogen atom. The index of hydrogen deficiency counts how many multiple bonds and rings are present in the structure.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 10, Problem 10.66P

Interpretation Introduction

Interpretation:

The structure of A and B in the given reaction sequence is to be given.

Concept introduction:

In the Hoffman elimination, an amine with a poor leaving group is converted into a good leaving group, tetraalkyl ammonium ion, which is a moderately good leaving group, using excess of an alkyl halide (usually CH3I). The bulkiness of the leaving group causes a anti-Zaitsev product or Hoffman product. It generates the least substituted alkene as the major product. The carbon atom to which the primary amine group is attached must be the secondary carbon with at least one alpha hydrogen atom. The index of hydrogen deficiency counts how many multiple bonds and rings are present in the structure.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Chapter 10, Problem 10.66P

Interpretation Introduction

Interpretation:

The structure of A and B in the given reaction sequence is to be given.

Concept introduction:

In the Hoffman elimination, an amine with a poor leaving group is converted into a good leaving group, tetraalkyl ammonium ion, which is a moderately good leaving group, using excess of an alkyl halide (usually CH3I). The bulkiness of the leaving group causes a anti-Zaitsev product or Hoffman product. It generates the least substituted alkene as the major product. The carbon atom to which the primary amine group is attached must be the secondary carbon with at least one alpha hydrogen atom. The index of hydrogen deficiency counts how many multiple bonds and rings are present in the structure.

Answer 6

Chapter 10, Problem 10.66P

Interpretation Introduction

Interpretation:

The structure of A and B in the given reaction sequence is to be given.

Concept introduction:

In the Hoffman elimination, an amine with a poor leaving group is converted into a good leaving group, tetraalkyl ammonium ion, which is a moderately good leaving group, using excess of an alkyl halide (usually CH3I). The bulkiness of the leaving group causes a anti-Zaitsev product or Hoffman product. It generates the least substituted alkene as the major product. The carbon atom to which the primary amine group is attached must be the secondary carbon with at least one alpha hydrogen atom. The index of hydrogen deficiency counts how many multiple bonds and rings are present in the structure.

Answer 7

Chapter 10, Problem 10.66P

Interpretation Introduction

Interpretation:

The structure of A and B in the given reaction sequence is to be given.

Concept introduction:

In the Hoffman elimination, an amine with a poor leaving group is converted into a good leaving group, tetraalkyl ammonium ion, which is a moderately good leaving group, using excess of an alkyl halide (usually CH3I). The bulkiness of the leaving group causes a anti-Zaitsev product or Hoffman product. It generates the least substituted alkene as the major product. The carbon atom to which the primary amine group is attached must be the secondary carbon with at least one alpha hydrogen atom. The index of hydrogen deficiency counts how many multiple bonds and rings are present in the structure.

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. 10 - Prob. 10.1P Ch. 10 - Prob. 10.2P Ch. 10 - Prob. 10.3P Ch. 10 - Prob. 10.4P Ch. 10 - Prob. 10.5P Ch. 10 - Prob. 10.6P Ch. 10 - Prob. 10.7P Ch. 10 - Prob. 10.8P Ch. 10 - Prob. 10.9P Ch. 10 - Prob. 10.10P

Concept explainers

Videos

Want to see the full answer?

Chapter 10 Solutions