Propose a mechanism for the following reaction. (Hint: Number the carbons to help you see where

Propose a mechanism for the following reaction. (Hint:...

Key Concepts

Reaction Mechanism

A reaction mechanism is a step-by-step description of how reactants are transformed into products. It involves identifying the sequence of bond-making and bond-breaking events, intermediate species, and any rearrangements that occur during the reaction. Understanding the mechanism helps predict the reaction pathway, energy barriers, and possible side reactions.

Curved Arrow Notation

Curved arrow notation is a symbolic tool used to represent the movement of electrons during chemical reactions. It shows the flow of electron pairs from one atom or bond to another, allowing chemists to visualize and rationalize the steps of the reaction mechanism.

Intermediate Formation

Intermediate formation refers to the transient species that form during a reaction mechanism. These species, such as carbocations, carbanions, radicals, or other stabilized intermediates, exist briefly before transforming into the final product. Recognizing these intermediates is critical for understanding complex reaction pathways and predicting product distribution.

Carbon Numbering

Carbon numbering is a strategy used to keep track of the position of carbon atoms throughout a reaction mechanism. By assigning numbers to the carbon atoms in a molecule, chemists can accurately follow the movement and rearrangement of these atoms, ensuring the mechanism’s clarity and correctness. This technique is especially useful in reactions where structural changes or rearrangements occur.

Transcript

00:01 Okay, this problem is asking us to propose a mechanism for the reaction shown here.

00:04 Okay, so i'm reacting a dyeing with two alkenes with my carboxylic acid and acidic medium.

00:10 Okay, and that is all in the goal to produce this ester right here.

00:13 Okay, so let's go ahead and analyze this mechanism to explain how that ester forms.

00:18 So i recognize immediately that i have h2s -o -4, sulfuric acid.

00:22 Sulfuric acid is one of the strongest acids that we use in organic chemistry.

00:25 So i'm probably going to use an acid -based reaction to begin with.

00:29 Okay, the question is, am i going to use my acid to proteinate this compound, my dine, or am i going to proteinate this carboxylic acid? okay, and that answer lies in the product.

00:39 Okay, so as we can see here, we have the formation of an ester, right? i have this cyclic structure connected to this carboxal oxygen, right? there's just a lot going on there, but basically it's just an ester.

00:49 Okay, if i were to protonate this carboxelic acid, i would end up with a proton attached to this oxygen.

00:55 Okay, that would mean that the oxygen would have a positive charge.

00:57 We know that positive charges on oxygens are not necessarily the most favorable thing in the world.

01:01 So what i'm going to do is move the electrons from this carbon -oxygen bond directly onto the oxygen.

01:06 Okay, that would relieve the oxygen of its positive charge and instead place a positive charge on the carbonyl carbon.

01:11 Okay, then that carbon would be vulnerable to nucleophilic attack.

01:15 And as i can see here, i'm forming an ester.

01:17 The most likely sequence of events would be that in alcohol were to attack this carbon to eventually form my ester.

01:24 Okay, but as we can see here, we do not have an ester.

01:26 Sorry, we do not have an alcohol, we have a dine, right? i do not have any alcohols to eventually form an ester.

01:32 So the other option is to protonate my dine.

01:36 Okay, so i'm not going to protonate my carboxylac acid.

01:39 Instead, i'm going to protonate my dine.

01:40 The question is, which carbon am i going to protonate? am i going to protonate this carbon from this alkyne? or am i going to protein this carbon from this alkyne? okay, so a question that might arise is, why did i choose those carbons as opposed to their opposite ends of the alkyne? so this one and this one.

01:56 That is because if i are to protonate this carbon or this one, the resulting carbocadion, so the carbocadine formed on this carbon and this carbon, would be more stable than if we are to protonate those carbons.

02:08 So that might be a little bit confusing, but hopefully by drawing out the resulting carbocadines, it would make sense.

02:15 So again, i'm going to protonate, for example, this carbon right here.

02:19 So that means i'm going to have ch3.

02:20 I'm going to get rid of that double bond.

02:22 I'm just going to have a single bond, and i'm going to have a carbocadine right there.

02:25 Okay, so this carbocatin is secondary, right? it's a secondary carburekodine.

02:29 If i were to protonate the other carbon, so this one right here, i would end up with a ch2, and then i would have this single bond right there, and a positive charge right here.

02:37 So this is a primary carbocadine.

02:39 We know that the more substituted that carbocyan is, the more stable it is.

02:44 So a primary carbonyne is not very favorable, but a secondary carbokadine is more favorable.

02:49 So that's why we were to hypothetically protonate that leftmost carbon.

02:53 Okay, but what about over here? so for this one, again, same concept.

02:56 I would end up with a tertiary carbocadine over here versus a secondary carburekodonide over there.

03:00 That's where i'm going to protonate this carbon right there.

03:02 Okay, so i'm going to end up with ch2, single bond, carbocadion, tertiary carbocadine.

03:08 So the question is, which carbon am i going to protonate? am i going to protonate this one? or i'm going to protonate this one.

03:13 I'm going to protonate this one.

03:15 And that is simply because the resulting carbocadion, in this case, a tertiary carbocadine, is more favorable than the resulting secondary carbocadine if it are to protonate this carbon.

03:24 Okay, so just to kind of summarize, i'm protonating this carbon because i'm forming a tertiary carbocatin which is the most stable carbocatin.

03:33 Okay, so with that in mind, let's go ahead and proceed with this mechanism.

03:36 I'm going to use the electrons from this pie bond to get protonated by the acid, a sulfuric acid, right? just like this, after i do that, i'm going to have to move the electrons onto this oxygen of my sulfuric acid.

03:48 So as far as that immediate product goes, i'm going to have the same number of carbons as before.

03:52 In fact, i'll just highlight this and duplicate it.

03:54 So same number of carbons.

03:57 The only difference is that instead of a ch double bond, i'm going to have ch2 single bond with a carbocadion on this carbon right there.

04:08 Okay, so now we have, we're getting closer, right? we're eventually going to form this cyclic structure, right? but how do we do that? how do we form a cyclic structure from a basically linear molecule? i can do that by recognizing that i have electrons in this pie bond, right those are an electron source that can behave as a nucleophile so what i'm going to do is i'm basically going to form my cyclic structure right away i'm going to recognize that this is an electrophilic site and this is a nucleophilic site so i'm going to go ahead and use these electrons to go ahead and attack this carbon okay and the book recommends to number the carbon so i'm going to do exactly that remember i'm using this carbon to behave as my nucleophilic atom in this case because if i were to draw a resonance form of this i would end up with the electron going here, resulting in a carbocetyone right there, which is more favorable than the carboketion being right here.

04:59 Okay, so that is why i'm using this carbon as my number one, as my nucleophile.

05:03 So this is going to be my number one.

05:05 I always label my nucleophile as atom number one.

05:07 So one, following along each of the carbons, here's my two, here's my three, here's my four, here's my five, and then my electrophile is always the last number in my sequence of numbering.

05:17 So this is number six...

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