To propose a mechanism for an Î±-hydroxy amide is treated with Br 2 in aqueous NaOH under Hofmman rearrangement conditions.

Question

When an α-hydroxy amide is treated with Br2 in aqueous NaOH under Hofmann rearrangement conditions, loss of CO2 occurs and a chain-shortened aldehyde is formed. The mechanism involves the following steps: Base abstracts an acidic amide proton, yielding amide anion 1; The amide anion reacts with bromine in an α-substitution reaction to give N-bromoamide 2. Abstraction of the remaining amide proton by base gives a resonance-stabilized bromoamide anion 3; Rearrangement occurs to yield isocyanate 4; Water adds to the isocyanate to yield carbamic acid 5; Elimination of CO2 yields carbinolamine 6; Following proton transfer, expulsion of ammonia yields the final product aldehyde. Write out the mechanism and then draw the structures of bromoamide anion 3 and amide anion 1.

Answer 1

Question

Chapter 24.SE, Problem 43MP

Interpretation Introduction

Interpretation:

To propose a mechanism for an Î±-hydroxy amide is treated with Br₂ in aqueous NaOH under Hofmman rearrangement conditions.

Expert Solution & Answer

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Students have asked these similar questions

When an α-hydroxy amide is treated with Br2 in aqueous NaOH under Hofmann rearrangement conditions, loss of CO2 occurs and a chain-shortened aldehyde is formed. The mechanism involves the following steps: Base abstracts an acidic amide proton, yielding amide anion 1; The amide anion reacts with bromine in an α-substitution reaction to give N-bromoamide 2. Abstraction of the remaining amide proton by base gives a resonance-stabilized bromoamide anion 3; Rearrangement occurs to yield isocyanate 4; Water adds to the isocyanate to yield carbamic acid 5; Elimination of CO2 yields carbinolamine 6; Following proton transfer, expulsion of ammonia yields the final product aldehyde. Write out the mechanism and then draw the structures of bromoamide anion 3 and amide anion 1.

H₂C 0 NH₂ B₂, NaOH 98 H₂C OH When an a-hydroxy amide is treated with try in aqueous NaOH under Hofmann rearrangement conditions, loss of CO, occurs and a chain-shortened aldehyde is formed. The mechanhum involves the following steps: Base abstracts an acidic amide proton, yielding amide anion The amide anion reacts with bromine in an e-substitution reaction to give N-bromoamide 2. Abstraction of the remaining amide proton by base gives a resonance-stabilized bromoamide anion Rearangement occurs to yield isocyanate Water adds to the isocyanate to yield carbamic acid Elimination of CD, yields carbinolamine Following proton transfer, expubion of ammonia yields the final product aldehyde Write out the mechanism on a separate sheet of paper, and then draw the structures of carbamic acid 5 and isocyanate 4. H 400, ANH, You do not have to consider stereochemistry. You do not have to explicitly draw Hatoms De not include lone pairs in your answer. They will not be considered in the grading…

The reagent diisobutylaluminum hydride (DIBALH) reduces esters to aldehydes. When nitriles are treated with DIBALH followed by mild acid hydrolysis, the product is also an aldehyde. Propose a mechanism for this reduction.