Suggest mechanisms that account for the outcome of the following reactions:

Name: Problem 9, Chapter 5: Advanced Organic Chemistry. Part A. Structure and Mechanisms
Uploaded: 2020-07-31T02:34:45-08:00
Duration: 13 min 47 s
Channel: Tom Rutherford
Description: Suggest mechanisms that account for the outcome of the following reactions:

Suggest mechanisms that account for the outcome of the...

Key Concepts

Regioselectivity and Stereoselectivity

Regioselectivity and stereoselectivity are principles that describe the directionality and outcome of chemical transformations in terms of the preferred positions (regiochemistry) and spatial arrangements (stereochemistry) of atoms in the products. They explain why certain products are favored over others when multiple reaction pathways are possible, and are essential for predicting and explaining the selectivity in chemical reactions.

Rate-Determining Step

The rate-determining step is the slowest step in a multi-step reaction mechanism, acting as a bottleneck that controls the overall reaction rate. Because it has the highest activation energy barrier, understanding this step is crucial for kinetic analysis and for designing strategies to accelerate the reaction. This concept is vital in both theoretical studies and practical applications, such as catalyst development and reaction optimization.

Transition States

Transition states are high-energy configurations that occur at the maximum energy point along the reaction pathway. They represent the point at which old bonds are partially broken and new bonds are partially formed, and they dictate the rate at which a reaction proceeds. While they cannot be isolated, studying transition states helps chemists determine the energy barriers associated with reactions, which is fundamental to understanding reaction kinetics and optimizing reaction conditions.

Reaction Mechanism

A reaction mechanism is the detailed, step-by-step description of how reactants transform into products. It involves the identification of all the elementary steps, including the sequence in which bonds are broken and formed, and explains how different intermediates and transition states lead to the final outcome. This concept is central to understanding and predicting the behavior of chemical reactions in organic synthesis and beyond.

Curved Arrow Notation (Electron Flow)

Curved arrow notation is a graphical tool used to depict the movement of electrons during a chemical reaction. It clearly shows how electron pairs are transferred, indicating bond formation and bond cleavage. This notation is essential for illustrating the detailed steps of a reaction mechanism as it helps in mapping out the path that electrons follow, thereby clarifying the transformation from reactants to products.

Reaction Intermediates

Reaction intermediates are transient species that form during the pathway of a multi-step mechanism. Although they are not usually isolated or observed directly, their formation and reactivity are critical to the overall mechanism. Understanding these intermediates provides insight into the progression and energetics of the reaction, and is key for rationalizing experimental observations and modifying reaction conditions.

Transcript

00:01 Okay, this problem is asking us to propose a mechanism for the following reactions.

00:03 Okay, so for the first one, we see that we have an alcohol, and we're going to react that with h2s of 4 in heat, in order to form this compound right here.

00:11 Okay, so very first thing, i'm going to do an acid -based reaction in which i have my very basic atom of this molecule, slash my oxygen, that's going to go ahead and get pronated by my h2s -o -4.

00:20 That's a super acidic acid, and i'm going to get the following intermediate, in which i have the same exact molecules before, the only difference being that i now have a protonated alcohol.

00:29 So i have water connected to my carbon.

00:33 Okay, so just like that.

00:34 Okay, so as we can see, that oxygen has a positive charge.

00:37 In organic chemistry, we want to avoid positive charges on oxygens because oxygens are electronegative, right? we want to have, if anything, negative charges associated with them.

00:47 But we can also have them just be neutral.

00:49 So i'm going to relieve this oxygen of its positive charge by simply moving electrons onto that oxygen.

00:54 Okay, and by doing that, i'll have that water behave as a leaving group, and then i'll end up with a carbocatine.

00:59 Right there, right? because i moved electrons away from that carbon, so that means that i have to have a positive charge on that carbon, or on that carbon.

01:06 Okay, so this is going to be my product, in which i have my secondary carbocadion right there associated with that, with this carbon, and then i have the remainder of my compound.

01:17 Okay, and then, of course, i have water as my leaving group, and that's not going to participate in the near future.

01:24 Okay, so next step, i see that i have that carbocadion, and i can see in my product, i want to form another six -member drain.

01:30 So right now i have a six member ring, this one, but i want to form this six member green.

01:35 How do i do that? well, i need to make sure that my electrophile is a certain number of carbons away from my alkyne, because my alkyne is going to participate as my nuclear file in this case.

01:47 So is this alken vulnerable to attack this carbon? and would that produce a six member drain? let's find out.

01:54 So if this is my nucleophile, we would have one, two, three, four, five carbons associated with my, cyclic structure.

02:02 So we know that a five member drain is not the same as a six member ring, so we have to move that electrophile elsewhere.

02:08 We have to move it onto the sixth carbon.

02:10 So we need to move that carbon onto that carbon.

02:12 Okay, and one way you can do that is by performing a methylide shift.

02:16 That's just moving a methyl group from one carbon to another.

02:20 So i'm going to move this methyl group onto that carbon, and by doing that, i will relieve that carbon of its positive charge, and relieving this carbon of its positive charge, and i relocate that positive charge onto this carbon.

02:30 Okay, and that will produce.

02:30 A tertiary carbacodyne, which is much more favorable than a secondary carbocadine.

02:35 Okay, so my product is going to be this, in which i have my alken, and then i have over here, i don't have that carbacadine any longer.

02:42 Instead, i have a methyl group, and then i have my two methyl groups over here with a carbacadine right there.

02:48 So that is a tertiary carbacodyne.

02:49 Again, that is much more favorable than a secondary carbokane.

02:52 Okay, so now is the point where i can have my nucleophile, in this case, my alken, that can behave as my nucleophile and attack this carbon, okay? and that's the electrophilic carbon.

03:01 Okay, so by doing that, i will produce the following compound.

03:04 I'll have this cyclohexane.

03:08 And then because i got rid of the alken, i don't have that alkyne any longer, but i will have a positive charge onto this carbon.

03:16 But again, i'm going to have that six -membered ring like this, and then on this carbon, that carbon corresponds to this carbon.

03:23 I'm going to have two methyl groups right there, and then one carbon over.

03:26 I'm going to have this single methyl group.

03:28 Okay, so that is my structure so far.

03:29 But again we need to make it into this structure.

03:31 We need to have this alkene eventually, and we also need to get rid of this positive charge.

03:36 So how do we do that? well, it's actually quite simple.

03:38 We just need to get the hydrogen right here.

03:41 We're going to go ahead and deprotonate that with my water.

03:43 So remember how we made water in one of the previous steps down here.

03:47 We're going to use that water to deprotonate this hydrogen.

03:50 And by doing that, i'm going to move the electrons from that carbon hydrogen bond onto the single bond, thereby making a double bond, which is needed, and we got rid of that positive charge in the process.

03:58 Okay, so that's my mechanism for this first one.

04:01 Okay, what about the second one? for the second one, we have kind of a similar situation in which we have h2s -o -4 and water, but we're going to go ahead and protonate my epoxide as opposed to an alcohol.

04:13 Okay, so epoxides, those are vulnerable to getting protonated because they have that basic oxygen on them.

04:19 Okay, so i'm just going to use the loan pairs on oxygen to get protonated by my h -2s -o -4.

04:24 Okay, so that's going to produce the following intermediate, in which i have simply a protonated epoxide.

04:31 Okay, but everything else is the exact same.

04:33 So, alky in there, and alky in there.

04:37 Okay, so next up, we are used to having potentially water come in, and then the water would potentially attack right here to produce basically transdials, right? but if we have potential for an intramolecular reaction, an intramolecular reaction is going to occur before an intermolecular reaction.

04:55 So by intermolecular, i mean that that would be the case in which water comes in, but by intram molecular, i mean that this alkyne is considered to be nucleophilic, right? so if we had water, for example, that water is considered to be nucleophilic, that would attack this carbon, et cetera, right? but we do have this alkyne associated with the same exact molecule, and that would be an intramolecular reaction.

05:16 And intramolecular reactions happen faster than intermolecular reactions.

05:20 So the first thing that's going to happen is i'm going to move the electrons from this alkene onto this carbon.

05:27 Okay, and the reason i'm picking this carbon is because if i have protonated my oxygen like this, then this oxygen has a very strong positive charge.

05:34 But as we heard before, oxygens do not like positive charges because they want to have as much electrons around it as possible, right, because they're very electron negative.

05:43 So as i have that positive charge on my oxygen, i'm going to be simultaneously moving electrons from this carbon single bond onto this oxygen.

05:52 So i'm not going to draw a concrete arrow because, well, actually, let's just erase this one for now.

05:58 I'm not going to draw a concrete arrow depicting the electron movement towards the oxygen, simply because it's not concrete, right? we only have the slight movement of electrons onto the oxygen, but when my alkene does attack it, then i can forcefully put the electrons onto the oxygen.

06:15 Okay, and that is going to behave in an s -n -1 slash s -n -2 reaction.

06:20 Okay, and that's kind of a unique reaction in which we have.

06:22 Have a little bit of both in which we have a carbocadine formation, but also the same exact nucleophile attacking my electrophile as the leaving group is leaving.

06:32 Okay, so we're going to get this.

06:34 If my nucleophile is my carbon number one, then i'm going to have my carbon number one either on this carbon or this carbon.

06:39 But as we can see over here, we're going to form two six member drains.

06:43 So we want it to be ideally a six member drain.

06:45 So let's see if it's going to be this one or this one.

06:47 So let's just consider it to be this one for now...

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