A mechanism for the final step in Edmund degradation- the acid catalyzed rearrangement of the ATZ derivative to the PTH derivative is to be proposed. Concept introduction: The various steps in the acid catalyzed rearrangement of the ATZ derivative to the PTH derivative in Edmund degradation are i) Protonation of lactone oxygen ii) Nucleophilic attack of water on the protonated lactone iii) A proton transfer from O to N iv) Opening of the five member ring v) Rotation of the bond and attack by N to form a new five member ring vi) Proton transfer vii) Loss of water viii) Deprotonation. To propose: A mechanism for the final step in Edmund degradation- the acid catalyzed rearrangement of the ATZ derivative to the PTH derivative.

Question

Want to see the full answer?

Answer 1

Question

Chapter 26.SE, Problem 59AP

Interpretation Introduction

Interpretation:

A mechanism for the final step in Edmund degradation- the acid catalyzed rearrangement of the ATZ derivative to the PTH derivative is to be proposed.

Concept introduction:

The various steps in the acid catalyzed rearrangement of the ATZ derivative to the PTH derivative in Edmund degradation are i) Protonation of lactone oxygen ii) Nucleophilic attack of water on the protonated lactone iii) A proton transfer from O to N iv) Opening of the five member ring v) Rotation of the bond and attack by N to form a new five member ring vi) Proton transfer vii) Loss of water viii) Deprotonation.

To propose:

A mechanism for the final step in Edmund degradation- the acid catalyzed rearrangement of the ATZ derivative to the PTH derivative.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

5. A solution of sucrose is fermented in a vessel until the evolution of CO2 ceases. Then, the product solution is analyzed and found to contain, 45% ethanol; 5% acetic acid; and 15% glycerin by weight. If the original charge is 500 kg, evaluate; e. The ratio of sucrose to water in the original charge (wt/wt). f. Moles of CO2 evolved. g. Maximum possible amount of ethanol that could be formed. h. Conversion efficiency. i. Per cent excess of excess reactant. Reactions: Inversion reaction: C12H22O11 + H2O →2C6H12O6 Fermentation reaction: C6H12O6 →→2C2H5OH + 2CO2 Formation of acetic acid and glycerin: C6H12O6 + C2H5OH + H₂O→ CH3COOH + 2C3H8O3

Show work. don't give Ai generated solution. How many carbons and hydrogens are in the structure?

13. (11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the molecule depicted below. Bond B 2°C. +2°C. cleavage Bond A •CH3 + 26.← Cleavage 2°C. + Bond C +3°C• CH3 2C Cleavage E 2°C. 26. weakest bond Intact molecule Strongest 3°C 20. Gund Largest argest a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in appropriate boxes. C Weakest bond A Produces Most Bond Strongest Bond Strongest Gund produces least stable radicals Weakest Stable radical b. (4pts) Consider the relative stability of all cleavage products that form when bonds A, B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B, and C are all carbon radicals. i. Which ONE cleavage product is the most stable? A condensed or bond line representation is fine. 13°C. formed in bound C cleavage ii. Which ONE cleavage product is the least stable? A condensed or bond line representation is fine. • CH3 methyl radical Formed in Gund A Cleavage c.…