For each compound, give the product(s) expected from (1) HgSO

Question

For each compound, give the product(s) expected from (1) HgSO₄/H₂SO₄ - catalyzed hydration and (2) hydroboration–oxidation.

b. hex-2-yne

Accepted Answer

Hey everyone and welcome back for the compound shown below predicts products that result when it undergoes the following a acid catalyzed hydration and be hydro operation oxidation. Let's recall that both A and B. They produce alcohols, right? So they produce alcohols. But with different regional chemistry, an acid catalyzed hydration follows more cosmic of regional chemistry. So for party we're going to analyze more cosmic of regional chemistry and for part B we're going to analyze and see more cosmic of regional chemistry. That's because an acid catalyzed hydration is different from hydration oxidation in the sense that there are mechanisms differ. So let's keep that in mind and let's begin solving part A. Okay, so now, for the part A we're just going to read through our AL kind, let's see that. So we're going to do our metal substations bonded to that carbon and then extend our bond to create that triple bond. And then we understand that we simply have a metal group over here, so that's our starting material and for part A it undergoes an acid catalyzed hydration. Let's draw a reaction arrow to illustrate what set of regions we need to use here. And acid catalyzed hydration requires us to use water in the presence of sulfuric acid because it capitalizes the reaction acid catalyzed and then essentially because we're dealing with an alka and we also want to make sure that we're using mercury to sulfate right now. Let's recall that because this reaction follows the Markov nick of regional chemistry, hydrogen is added to a less substituted carbon and our alcohol group is added to a more substituted carbon. However, when we think of this, al Qaeda is internal and that means the two carbon atoms are equally substituted. They both have one alcohol substitution. So we have two possibilities. There's no preference. It's more comic of school. But because there's equivalent substitution pattern, we can say that we have two options. Either simply add hygiene and disposition and the alcohol group at theater or swap them. We can also add our alcohol and this position and hygiene on the other position, basically that's because they're equally substituted. So more cosmic of school essential it sells us that there's no preference. There's no more or less substituted carbon. So we have just so we just have two possibilities. Let's draw the result on products. Now we think of the first product in blue, we are going to break one of our pi bonds and we are just getting the following. First of all, we can just redraw our metal substitue ints, monitor that carbon and now we can see that we have 12, three more carbon atoms. So let's show them that be 123. Now, because we have broken our pi bond, there's no necessity to join as linear. It's actually tribunal planer with respect to the double bond that we're going to indicate. Notice how this is our carbon number one and between carbons two and three. In our drawing we can say that there is a double bond. So if this is if this is carbon number one, then between two and three. Over here we're going to have one pi bond remaining. Now you may clearly see that we have added our wage group over here. Right, so let's throw it out. And now our second possibility is to add our alcohol group at the other carbon base load. We're here. So we're going to see it here, let's draw it. This will be our green pathway. So for the green pathway we can just read through our chain the position of the pi bond. This remaining pi bond will remain the same. But now wages added at this carbon. Okay, so we have to Eno's notice how each of these structures correspond to an N. O. And because Enel's are unstable, they course they basically undergo kit. So in all total tourism. So we will essentially see how that happens if we start with the blue structure shown, our first, you know, we are going to expand that oxygen hydrogen bond. That's because the lone pair on oxygen will be used in the formation of a pi bond. Then the adjacent pi bond will be broken. It will essentially participate in the proton transfer, it will attack that hydrogen and then these oxygen hydrogen electrons will be transferred onto oxygen in the form of a lone pair. And here we are illustrating our key to inal total memorization meaning equilibrium shifts towards the formation of a more stable key tone. Okay, so we're going to illustrate a longer arrow, a shorter one towards an unstable you know, and that would correspond to the first structure that we get. That's our found product. Now we are ready to throw our Carbonell group. We had our proton transfer. It has been added at this position and this is our first product. Right? Let's keep in mind that the disequilibrium corresponds to kit. So in all totem tourism. And now if we think about the total tourism of the other structure, we can expand that oxygen hydrogen bond once again and we can indicate that proton transfer. Okay, well done. And now we are ready to draw the result on key tone. Let's change the color to green. So we can clearly see that these are two different products. Now our Carbonell group will be over here where were destroyed and that will be our second product. So let's label these are products for part A this one and that one. And now we are ready to think of hydration oxidation. So let's move on to part B. Let's read through our starting material. Okay, now we clearly see our starting material and we want to recall that hydro operation oxidation implies that generally we're using these two reactions basically we want to use number one, boring in the presence of Tetra hydro veteran, which is our solvent followed by another reaction where we're using hydroxide and peroxide and a solution of sodium hydroxide for example. But this is generally whenever we're using al keane's, when we are dealing with al kinds, we want to make sure that we're using a more hysterically hindered version of boring, which is dice. I'm boring. So the first reaction would actually proceed with side to be H that corresponds to die siamo boring in the presence of tetra hydro Furin, which acts as our servant and number two addition of an aqueous solution of hydroxide and peroxide. As we already said, this reaction follows the anti more comic of regional chemistry. So the alcohol group would be added on a less substituted carbon. But once again, let's keep in mind that the two positions are equally substituted. So it's not only the more cosmic of school that produces cheap possible products, but also the anti more cognitive radio chemistry dictates that there will be two possibilities. Since the two carbon atoms are equal is substituted. Hi preparation oxidation will essentially allow us to add hydrogen and a wage in this arrangement. Or we can swap them and say that if we use another pathway in green, we can also add the alcohol group at the left carbon and hydrogen and the right most. So basically we can have two possibilities once again because similarly to the more comical school, the ante Markovic up reach your chemistry tells us that yes, these two carbon atoms are equally substituted. So you have two possibilities and we basically get the same products right Because essentially the whole goal of this reaction is to produce our emails first and each channel totem arises into more stable key. So forms. So we can say that if you look at the blue pathway, we already understand that the blue pathway produces an Enel which totem arises. We already saw how it looks like going back over here, we said that if we're adding alcohol and disposition, it undergoes total authorization to produce the same product. So we can simply redraw that product for this part because it's exactly the same. That's our keystone formed when our Enel totem arises and then the other product for the green pathway. It basically shows us that we're going to add O. H. At a different position. Or basically here, if you look at our previous analysis and that Enel totem arises into a key tone. So you can redraw that structure and that would be it. You may understand that we are getting exactly the same two products for part A and B. The reason for that is because we are dealing with an internal al kind. The two positions of an internal al Qaeda are equal is substituted. So it doesn't matter whether you follow the more Kafelnikov read your chemistry or the anti more cognitive. Read your chemistry due to equivalent substitution, you will always have two possible products and that's why the products for each reaction are the same based on our answers. We may conclude that the correct answer to the multiple choice question is B. However, if you have any questions, don't hesitate to leave a comment down below. Thank you for watching.

For each compound, give the product(s) expected from (1) HgSO₄/H₂SO₄ - catalyzed hydration and (2) hydroboration–oxidation.
b. hex-2-yne

Key Concepts

Catalytic Hydration

Hydroboration-Oxidation

Markovnikov's Rule

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