Suggest an arrow-pushing mechanism that accounts for the forma...

Question

Suggest an arrow-pushing mechanism that accounts for the formation of the following products.

(a) Chemical reaction diagram showing acid-catalyzed hydration with reactants and product structure labeled.

Accepted Answer

All right. Hi, everyone. So this question is asking us to propose a possible arrow pushing mechanism for the formation of the product in the following acid catalyzed hydration reaction. So here in our acid catalyzed hydration, we've got an alkene reacting with acid in the presence of water to yield the following alcohol. So here, right, recalled acid catalyzed hydration follows Markov Ko's rule. And what Markov Niko's rule states is that our substituent, in this case, our hydroxy group in the alcohol will add on to the more substituted carbon of the starting alkene. So let's go ahead and showcase the mechanism and recall that the first step is going to be the interaction of the pipe on here in our alkene with our acid. Now, when a strong acid dissociates in the presence of water, H 30 plus or hydro num is going to form. So in the first step, right, our pie bond is going to go ahead and take a proton from our hydro num ion. And as a result, we're going to end up with a Carbocaine. Now, here is the tricky part, at least to me because seven member rings always get me but there we go. But like I said previously, right, our pie bond from the All Keen has now been broken and we're going to end up with a hydrogen on one carbon of the starting All Keen and the other left as a car. So here recall that the more substituted a Carbocaine is the more stable it's going to be so not the pipeline has broken. We're left to choose between a primary carbon and a tertiary carbon and the tertiary carbon given that it's more substituted than our primary carbon would make a more stable carbo ion. So that is where our positive charge is going to go. However, we have another slight dilemma to consider because recall that whenever there is a carbo ion to consider, we have to also ask ourselves about rearrangement because recall that if a carbo ion can adopt a more stable configuration, it absolutely will via Carbocaine rearrangement. So it's worth remembering also that seven member rings aren't the most ideal. In fact, six member rings and five member rings are said to be the most ideal ring confirmations because smaller than that creates STAIC hindrance and anything larger than six carbons begins to lose its structural integrity, right? The ring itself becomes flopper, so to speak. So our car ion is part of a seven member member ring, but we can actually make a more ideal six member ring instead. And here's how the answer is via a ring opening because here if I go ahead and I label my carbo on number one and I go counterclockwise numbering my carbons here, I end up with six carbons and carbon number seven has my two methyl groups here. So the bond between carbon six and carbon seven of the ring can go ahead and open and carbon six can go ahead and connect with carbon one, leaving carbon seven on the outside. So as a result, right, we end up with a six member ring and always be sure to keep track of your substituent. So here are the eel groups and now, like we said previously, right, carbon seven is outside of the ring and carbon seven now bears the positive charge because a bond has been broken, right? So now it is electron deficient. And even though the carbo CAD ion in question, is still tertiary, the molecule overall is more stable. The reason being because it has a more ideal six member ring. So at this point, we can go ahead and introduce our nuclear file in this case, water. So our water molecule can go ahead and attack our carbo CAD ion now that the Carbocaine has adopted the most stable configuration possible. So here I'm going to go ahead and draw my carbon skeleton, my carbon backbone. And now I'm going to go ahead and add my water molecule accordingly. However, we have to be very careful because our water molecule now that it's attached to our carbon is actually going to have a positive charge on oxygen, which isn't always ideal or actually it's not ideal. So at this point, something else has to come in and behave as a base, removing a proton from our water molecule. And the species that will do that is actually another molecule of water. So here's a second molecule of water and it's going to go ahead and remove a proton from the first one because at that point, we end up with a neutral hydroxy group just like in our final product depicted, right. So here's my EO group or EO groups and now we have our hydroxy group neutral and there you have it right here is our completed mechanism of the acid catalyze hydration reaction. So the first step involved our pi bond being broken by taking a proton that resulted in the formation of a carbo ion which then rearranged to create a more ideal six member ring configuration. And after rearrangement, right, our nu file, in this case, water attacked or crumble on resulting in a positively charged water molecule attached to carbon. But after de protonation via a second molecule of water, we were able to generate a neutral hydroxy group at the very end of the process and there you have it. So with that being said, thank you so very much for watching if you stuck around to the end. And I hope you found this helpful

Suggest an arrow-pushing mechanism that accounts for the formation of the following products.
(a)

Key Concepts

Arrow-Pushing Mechanism

Nucleophiles and Electrophiles

Reaction Mechanisms

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