A mechanism for the reversible equilibrium isomerization of 2-substituted-2-cyclopentanones to 5-substituted-2-cyclopentanones is to be proposed. Concept introduction: The steps involved in the reversible equilibrium isomerization of 2-substituted-2-cyclopentanones to 5-substituted-2-cyclopentanones are i) abstraction of a proton from the Î³-carbon to form a resonance stabilized enolate ion ii) Protonation at the Î±- carbon iii) Abstraction of a proton from other Î±- carbon to form a different resonance stabilized enolate ion iv) Protonation of the Î³- carbon. To propose A mechanism for the reversible equilibrium isomerization of 2-substituted-2-cyclopentanones to 5-substituted-2-cyclopentanones.

Question

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

Answer 1

Question

Chapter 22.SE, Problem 27MP

Interpretation Introduction

Interpretation:

A mechanism for the reversible equilibrium isomerization of 2-substituted-2-cyclopentanones to 5-substituted-2-cyclopentanones is to be proposed.

Concept introduction:

The steps involved in the reversible equilibrium isomerization of 2-substituted-2-cyclopentanones to 5-substituted-2-cyclopentanones are i) abstraction of a proton from the Î³-carbon to form a resonance stabilized enolate ion ii) Protonation at the Î±- carbon iii) Abstraction of a proton from other Î±- carbon to form a different resonance stabilized enolate ion iv) Protonation of the Î³- carbon.

To propose

A mechanism for the reversible equilibrium isomerization of 2-substituted-2-cyclopentanones to 5-substituted-2-cyclopentanones.

Expert Solution & Answer

Trending nowThis is a popular solution!

See solution

Check out sample textbook 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.…