Provide stepwise mechanisms for the following reactions: |...

Name: Problem 19, Chapter 11: Advanced Organic Chemistry. Part A. Structure and Mechanisms
Uploaded: 2020-07-23T22:22:37-08:00
Duration: 7 min 44 s
Channel: Tom Rutherford
Description: Provide stepwise mechanisms for the following reactions:

Key Concepts

Intermediates

Intermediates are transient species that form during a reaction mechanism that exist between the reactants and products. They often cannot be isolated but play a crucial role in the formation of the final product by providing insight into the pathway of the transformation.

Reaction Mechanism

A reaction mechanism is a detailed, step-by-step description of the process by which reactants are converted into products. It explains the sequence of elementary steps, including the making and breaking of bonds, and helps in understanding how the overall transformation occurs at the molecular level.

Stepwise Mechanism

A stepwise mechanism breaks down a complex reaction into individual stages, with each step involving an elementary reaction. These steps can involve the formation of short-lived intermediates and distinct transition states, allowing for a clearer understanding of the reaction’s progression.

Transition States

Transition states represent high-energy, unstable configurations that occur momentarily as reactants are converted to intermediates or products. Understanding transition states helps in elucidating the energy barriers involved and in identifying the rate-determining steps in a mechanism.

Arrow Pushing

Arrow pushing is a visual representation technique used to depict the movement of electrons during chemical reactions. It is essential for illustrating how bonds break and form during each step of a mechanism, thereby clarifying the electron flow throughout the reaction process.

Transcript

00:01 Okay, this problem is asking us to propose a mechanism for each of these reactions.

00:04 So let's go to the first one.

00:05 This one we have a secondary alcohol.

00:07 We know it's a secondary alcohol because the carbon in which my alcohol is attached to is connected to one, two other carbons.

00:12 Okay, we're transforming that secondary alcohol into this tetra substituted alken.

00:17 Okay, and we know it's tetra substituted because on each of these carbons of the alken on this particular carbon, for example, we have one, two, and on this carbon we have one, two, two plus two equals four.

00:28 So tetra, tetra substituted.

00:30 Alken.

00:31 Okay, so let's start off with this mechanism.

00:33 So i'm going to go ahead and because i recognize i have a super strong acid, i'm going to go ahead and protonate my most basic atom of this molecule, which is my oxygen.

00:42 So let's go ahead and protonate that, resulting in water instead of an alcohol.

00:48 So now i have h2o connected to a carbon.

00:52 Okay, so in these types of problems when we have a, first of all, an acid and then heat, the heat is indicative of an e slash elimination reaction.

01:04 So what we're going to do is recognize that we're going to try to make an alken as apparent by the product and also recognize that we're on a secondary carbon, right, because our alcohol is attached to that secondary carbon.

01:14 If we're on a secondary carbon, it's going to prefer an e1, right, because secondary can prefer either sn1 or e1, but because it's heat, we know it's going to be e1.

01:25 So because it's e1, that's going to occur in a unimolecular transition state in which i have two sequential steps instead of simultaneous steps.

01:34 Okay, so that means i'm going to have to kick off my water and that's going to be its own separate step.

01:39 So that's going to result in this intermediate.

01:42 So i have a positive charge on that carbon and then my h2o right here.

01:46 Okay, so right now we have a secondary carbocadion.

01:50 We recognize that that is decent, but it has the potential to be even better.

01:53 And it has the potential to be even better because we have this methyl group right there.

01:57 If we were to shift this methyl group over onto that carbon, we'd result in, first of all, the relief of this secondary carbon caton and the resulting tertiary carbocadone right there.

02:09 Okay, so let's go ahead and do that.

02:10 I'm going to shift this methyl group, this one right here, onto this carbon.

02:15 Okay, and that will result in the following.

02:17 So i'll get my cyclohexane.

02:20 I have that methyl group still there, but then i moved the methyl group over here.

02:24 So now this carbon, this former carbocodon is now covered.

02:28 It doesn't have a charge, but this carbon does.

02:31 So that is a tertiary carbon caton.

02:32 And the driving force for that particular step is that we went from a secondary carbonateon, which isn't particularly stable, into a tertiary carbon ion, which is much more stable.

02:41 Okay, so now we're going to go ahead and do my elimination reaction.

02:45 So my elimination reaction, as i said before, this is done in heat, so that's why we prefer to do elimination reactions.

02:50 My oh2 -h2, that's going to come in, and it's going to try to relieve this carbon of its positive charge by making an alkyne.

02:57 And in in order to make an alken, we must deprotonate a hydrogen and then kick off our leaving group, which we already kicked off our leaving group and now we just have to deprotonate a hydrogen.

03:07 So in order to make that tetrasubstited alken, right here, i want to make it so that i can make this carbon have one less proton.

03:16 So that means i'm going to have the deprotonation of that hydrogen, moving the electrons from that carbon hydrogen bond onto this bond, and in the process, relieving this carbon of its positive charge.

03:25 So that will result in my tetra -substituted alkyne.

03:30 Oops, that is not a circle.

03:33 Let's see.

03:33 Oops, i can't do that, so i'll just go like this.

03:37 Okay, tetra -substitated al -keen with h -3 -0 -plus.

03:41 Okay, next up.

03:43 Oops, that was h -3 -0 -plus.

03:45 Okay, next up, we have this one.

03:47 So, again, propose the mechanism for this problem...