Hydration of alkynes (via oxymercuration) gives good yields of...

Question

Hydration of alkynes (via oxymercuration) gives good yields of single compounds only with symmetrical or terminal alkynes. Show what the products would be from hydration of each compound.

b. hex-2-yne

Accepted Answer

Hey everyone and welcome back in today's video, we are going to solve the following problem. Oxmoor curation reduction is an excellent method for obtaining single compounds from the hydration of symmetrical internal or terminal AL kinds draw the products of the oximeter curation reduction reaction for the following compound to Pantene. Now based on the identity of this structure, we understand that this is an AL kind because it says pen sign, we are going to have a five member parent chain. Most importantly, this is an AL kind which must have a triple bond. And based on the name, we understand that the triple bond starts at position number two. So let's suppose that this is our position to meaning we need to add an additional metal group over here. So we have a total of three carbon atoms. We need two more to make sure that we get a five member chain. Now this is our starting material. And according to the problem, we want to draw the oxy mercure ation reduction product here. Now let's recall the reagents needed for an oxy mercure ation reduction reaction. And those would be number one. It consists of two steps that would be mercury to a sedate and water. And the second step would be our reduction step where we are using sodium world hydrate and sodium hydroxide. Now, most importantly, to solve this problem efficiently, we have to recall that oximeter creation reduction follows the more cosmic of regional selectivity M R. Let's use the shorthand notation for this one, which essentially implies that we are going to form alcohols at the very beginning by following the more cosmic of school. If we look at the given al kind, we understand that the substitution pattern is exactly the same at each carbon atom. The left carbon atom within the triple bond has one L. Kill substitute print. And the other carbon atom within the triple bond has one L. Kill substitution as well. So because we have two equally substituted carbon atoms, we have two possibilities for our products to be formed. So let's look at the first product in the first product we have this possibility in green we call that the ox mercury emission reduction reaction allows us to add hydrogen to one end which is less substituted. But in this case it doesn't matter. So let's suppose that we're adding hydrogen to this carbon atom and then the O. Wage group at the other end of the triple bond. So this is our first possibility. Again. The more comic of school is not applicable because the two carbon atoms are equally substituted. And the second possibility is to use the black pathway where we are actually adding hydrogen to a different carbon atom and then the alcohol group ad position number two. So if we look at the product that we get for the black pathway, let's look at it at the very bottom if we add hydrogen and the alcohol group across that high bond, we are going to break it and therefore we are going to get a double bond and stud. So we may draw a five member chain that we want 2345. We understand that at position number two, We're now happy double bond. There's no more triple bond because the pi bond has been broken and we are adding an alcohol group at position two and hydrogen at position three. There's no necessity to show that implicit hydrogen, that's perfectly fine. But now we have to recall the keto in also summarization, which essentially tells you that if you expand this hydrogen, the lone pair on oxygen can reform or actually form. That would be more correct to say because this is an N. All right, it's an alcohol with an adjacent double bond. So we can actually use the lone pair and oxygen to form a double bond at this position, meaning the adjacent pi bond will be protein ated to add a protein at this position. And as a result we are going to transfer these electrons in the form of a lone pair on oxygen to form a more stable keto form. So we are going to draw an equilibrium. That's our keto total memorization. And now you may notice that we're getting a key tone our five member chain. Now we will have a double bond in this position and oxygen. So this is our first product. We may label it in black because it corresponds to our color. Use for the idea of how the submissions are added. This is our first product and then we are ready to look at the Green one. So for the green one, we understand that we are adding hygienic, position C. And a wage at position three. So we're getting another email which looks a bit differently because even though the double bond still starts at position C. Now hydrogen is added over here, it's implicit and the alcohol group is added at position number three. Now, because this is another email, right? This is an Enel as well. We can essentially show that keto totem authorization to form a more stable zone. So the lone parent oxygen forms a double bond. We get the protein nation stop and then let's actually fix this arrow to make sure that we clearly see that substation over here. So that would be an attack at the proton and then we have a transfer of oxygen, hydrogen electrons in the form of a lone parent oxygen. And this is how we get our second product. We have keto, you know, summarization and then we have our five member chain and our carbon yield group. So this is our second product. We can say that the second product corresponds to the green product where the carbon group is over here at position three, position two and position three. So these are our final products and based on the identity of the products, we may conclude that the correct answer choice. This multiple choice question is B. However, if you have any questions, don't hesitate to leave your comment down below in the comments section, Thank you for watching.

Hydration of alkynes (via oxymercuration) gives good yields of single compounds only with symmetrical or terminal alkynes. Show what the products would be from hydration of each compound.
b. hex-2-yne

Key Concepts

Hydration of Alkynes

Oxymercuration-Demercuration

Markovnikov's Rule

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