Write a stepwise mechanism for each of the radical rearrangement reactions listed in Scheme 11.6

Write a stepwise mechanism for each of the radical...

Key Concepts

Radical Reaction Mechanisms

This concept involves reactions where radicals—species with unpaired electrons—serve as key reactive intermediates. Such mechanisms generally proceed through initiation, where radicals are generated; propagation, where radicals react with stable molecules to produce new radicals; and termination, where radicals recombine to form stable products. Understanding these mechanistic steps is crucial for analyzing how radical intermediates are formed, rearranged, and eventually quenched in complex reactions.

Stepwise Mechanism

A stepwise mechanism explains chemical transformations as a series of discrete steps, each featuring its own intermediate species and transition state. In the context of radical rearrangements, this means detailing each elementary reaction, such as bond breaking, formation, and radical shifts, leading to an overall conversion. This approach helps in predicting the path of the reaction and understanding the energy profile of the process.

Radical Rearrangement

Radical rearrangement refers to the structural reorganization of a radical intermediate, typically involving the migration of an atom or group within the molecule. This movement often leads to a more stable radical center or a configuration that is better suited for subsequent reaction steps. Recognizing these rearrangements is essential for deciphering reaction pathways, especially when analyzing complex sequences illustrated in reaction schemes.

Reaction Schemes

Reaction schemes are visual representations that outline the sequence of steps in a chemical mechanism. They summarize the formation and interconversion of intermediates and the flow of electrons during a reaction. In the study of radical rearrangements, these schemes help in tracking the movement of radical centers and the successive steps that lead to the final product, making it easier to visualize and rationalize the mechanism.

Transcript

00:02 This is the answer to chapter 20, problem number 55 from the smith organic chemistry textbook.

00:09 This problem asks us to draw a stepwise mechanism for each of these two reactions.

00:15 So in a, we have this lactone that we're treating with methyl magnesium bromide.

00:22 And we end up with a non -cyclic product.

00:27 And if you count carbons, you can see that we've added two methylgene bromide.

00:32 Groups.

00:33 And so we need to draw a mechanism that's going to account for that.

00:37 Okay.

00:39 And so the first thing that's going to happen here is methyl magnesium bromide is going to attack the carbonyl carbon in the typical fashion of a granite reagent.

00:56 So like this.

01:04 And so after that first step, we have this intermediate.

01:08 So we have this intermediate.

01:12 So we we've added our methyl group here.

01:16 We have the alcoxide here.

01:20 I forgot to draw the oxygen right here.

01:23 So let me fix that.

01:24 Okay.

01:25 So we have this alcoxide with a negative charge.

01:33 We've added our methyl group.

01:35 And so the next thing that can happen is the alcoxide can go to reform the double bond there.

01:45 That's going to cause this bond to break, this other carbon -oxygen bond.

01:52 And when that happens, that is going to open up our ring.

01:58 And so now, three, four, five, six.

02:08 So now we'll have this with a ketone here and the alk -oxide over here.

02:20 Okay.

02:23 And so from here, and let me just double check.

02:30 Right.

02:30 Yeah.

02:30 So the problem specifies that we're using excess methamagnosium bromide.

02:35 And the reason it has to be excess is because we can now have a second equivalent of methyl magnesium bromide come in and attack this ketone.

02:49 And so again, this will proceed just like a regular greenyard.

02:54 And so we now have another methyl group here and an alkoxide here.

03:08 And so we have two alcoxides now.

03:17 And when we go to do our water workup, we can take care of both of these at the same time.

03:27 So we can draw both of these being protonated at the same time as a single step.

03:37 I shouldn't omit negative charges, so each of these has a negative charge.

03:42 And we can just draw each of them grabbing a proton from different water molecules simultaneously.

03:50 And so that is going to get us to our final product, which would be this dial...

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