Propose a mechanism to show how 3,3-dimethylbut-1-ene reacts w...

Question

Propose a mechanism to show how 3,3-dimethylbut-1-ene reacts with dilute aqueous H₂SO₄ to give 2,3-dimethylbutan-2-ol and a small amount of 2,3-dimethylbut-2-ene.

HINT: When predicting products for electrophilic additions, first draw the structure of the carbocation (or other intermediate) that results from electrophilic attack.

Accepted Answer

All right. Hello everyone. So this question is asking us to draw the mechanism that demonstrates the reaction of 33 dimethyl hex one, E with a diluted aqueous H two. So four which produces 23 dimethyl hexane 30 as the major product and 23 dimethyl hex to E as the minor product. So, before we do anything involving the mechanism, let's go ahead and draw out our starting material and our products starting off here with 33 dimethyl hex one E, that's a six carbon parent chain with a double bond in between carbon one and two and two methyl substituents branching off of carbon number three. And this all keen is reacting with H two. So four, that's diluted in water. As for our major product, we have 23 dimethyl hexane 30, so that's the same six carbon parent chain with a hydroxy group branching off of carbon. Number three and two methyl substituents one on carbon number two and another on carbon. Number three. This means that our major product is an alcohol and our minor product is another all keen. We have 23 dimethyl hex two in thats six carbon parent chain with a double bond in between carbons, two and three and one methyl substituent on carbons two and three of the parent chain. All right. So now, with respect to the formation of the major product, we're going from an alkene to an alcohol in the presence of dilute aqueous acid recall that this combination of reagents is referred to as the acid catalyzed hydration of an all king. And when we say hydration, we're describing the addition of water across both carbons of the alkene itself. So one carbon gains a hydroxy group or oh group and the other carbon gains a hydrogen. Now, with respect to the addition itself, acid catalyzed hydration does in fact follow Markov Niko's rule, meaning that the hydroxy group would be installed onto the more substituted carbon of the starting alkene. And we can understand why by following the mechanism, the mechanism itself is characterized by the formation of a carboy ion intermediate. And it's this carbo cion that helps us understand why we end up with the minor product, which is another all keen stemming from a beta elimination process. But we will get to that for now, let's go ahead and start with the mechanism and I'm going to go ahead and copy paste our starting material on the bottom of the screen here. All right. So let me scroll down once more just to open up some space. And now let's go ahead and begin with our mechanism. Now, because the mechanism or the reaction itself is catalyzed by that strong acid. Our first step is going to involve the electro pilic addition of a proton to the ke. So here I have my all key and here I have the structure of sulfuric acid. So first, the pi electrons of the alkane itself are going to take a proton from sulfuric acid. And as a result, the least substituted carbon of the starting all keen in this case, carbon, one of the parent chain is going to take that proton. The reason for this is because it allows for the formation of a more stable carbo cion intermediate because now we end up with a secondary carbo Kion on position number two. So now I'm going to pause here to recall one very important aspect of mechanisms involving carbo C on intermediates. And that's rearrangement every time you come across a carbo cion in a given reaction mechanism, you must make a habit of checking for rearrangement because if a reion is ever able to gain higher stability by changing positions, it's going to do precisely that. And here that is precisely what's going to happen because we have a secondary carbo cion at the moment that is adjacent to a ordinary carbon that contains methyl substituents recall that one of the many ways that rearrangement can occur is via a methyl or cul shift in general. So now we're going to see a methyl shift because methyl groups are relatively small. One of these attached to the ordinary carbon can migrate over to the secondary carbon that has a positive charge at the moment. This is referred to as a metal shift. And so what that does is change the positioning of the methyl substituents, which we did note in the structures of our products because now one metal group has migrated over to carbon. Number two, leaving a tertiary carbo cion on position three of the parrot chain. So at this point, now that we have our major carbo cion intermediate, we can go ahead and showcase the formation of both the major and the minor product to form the major product, which was an alcohol, water is going to behave as a Nile and it's going to attack the carbon with a positive charge. On the other hand, to form the elimination product that was shown water can also behave as a base and eliminate this beta proton that's attached to carbon two of the parent chain. Recall that a beta proton is a proton attached to a carbon atom adjacent to the carbo cion in this case. So let's go ahead and showcase those possibilities. I'm going to start with the formation of our alcohol product. So when showcasing how water behaves as a Nile, I'm going to draw those arrows in red, those mechanism arrows. So now in red, I'm showcasing a molecule of water attacking the tertiary crumble cion directly and now oxygen is attached to carbon number three of the parent chain. However, because this nucleic attack took place when oxygen was part of a molecule of water at this point, oxygen has a positive charge because it has three covalent bonds at this point and only one lone pair of electrons. So because of this one more step is necessary in which an additional molecule of water removes a proton from the oxygen that has a positive charge. This gives back a loan pair of electrons to oxygen and produces a neutral hydroxy group in the product. So now we have our major product and because we're done with this portion of the mechanism, let's go ahead and return to our carbo c on intermediate after rearrangement because now it's time to showcase how water behaves as a base during beta elimination. Now do keep in mind that there are multiple different beta positions that can be or rather that have a proton that can be eliminated. But here we're only going to focus on the beta proton attached to carbon two of the parent chain because that is going to form the minor product that we saw in the beginning of this video. So now in purple water behaving as a base here removes that beta proton and it subsequently breaks the bond between carbon two of the parent chain. And that beta proton, those electrons then travel in between carbons two and three to create a pie bond in our all keen product. And that's how we get our all keen pot and there you have it. Now, we've completed the mechanism showcasing the formation of both the alcohol and allen products from our starting material. And so with that being said, if you watch this video all the way through, thank you so very much. And I hope you found this helpful.

Key Concepts

Electrophilic Addition Mechanism

Carbocation Stability

Markovnikov's Rule

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