Predict the major products of the following reactions, and giv...

Question

Predict the major products of the following reactions, and give the structures of any intermediates. Include stereochemistry where appropriate.

(i)

Accepted Answer

Hey everyone and welcome back in today's video, we are going to draw the reaction intermediates and major products for the following reaction. Show this theory of chemistry where applicable, the starting material is an AL keen because it contains a double bond and looking at the set of four agents we are using. But as in permanganate, which is a cold dilute solution, followed by another reaction where we are adding an acquittal solution of sodium hydroxide. If you look at the set of two regions, you may recall that they correspond to a reaction which produces a sin di hydroxy elation. Meaning we are expecting to see an addition product where we are adding two alcohol groups in sin arrangement. And that means they're both wedges or dishes. They're both pointing up or down with respect to the plane of the ring. So let's see how that works because we also need to include an intermediate. So we may just draw our starting material. We are going to draw our cyclops, benzene ring with an ethyl substitue in and now in the world. First step of the reaction we call that potassium permanganate as an ionic substance. So, but as Minister aren't spectator in a solution, we essentially may only think about permanganate on because that's the most important part for our problem. So, permanganate essentially has four auction atoms. Let's draw manganese and then see oxygen atoms here would be double bonded to manganese. There'll be an extra oxygen with a double bond and the remaining one would be single bonded, We would have a negative charge. Right. Camera four amount of work corresponds to a negative charge. So now because manganese is a metal and metal's correspond to a low electro negativity. This oxygen manganese bond is strongly polar and it's strongly polarized towards oxygen, which essentially allows the dense the high electron density within this pipe bond to attack the carbon atom within the double bond which attacks back or basically it attacks the adjacent oxygen atom. And then from here we are going to break this pi bond because we're forming a new bond and transfer these electrons in the form of a lone pair on manganese. And this would correspond to our first intermediate in this first stop we are adding But that simple manganese so that we can, you know, for cold and dilute. Let's draw the result and intermediate. So once again we may start by drawing our five member dring our ethyl group. And now we understand that at this position. Let's recall that we have Cindy Hyde rocks elation. So because first of all, the molecule has a line of symmetry, there is just one substation at this position. So the molecule is symmetrical with respect to this horizontal line. This means it doesn't really matter if we add two wedges or two dashes because the compound would be a Cairo that would be miso structure. Right? So we can add to wedges or two dashes. It doesn't really matter. So let's go with two, which is indicating that now this intermediate would have a group pointing up and we have two oxygen atoms, you may notice that the two oxygen atoms are now single bonded to our carbon atoms because we have broken two pi bonds. So those t oxygen atoms bonded to manganese, which now can send a lone pair of electrons. And then we have our double bonded oxygen and our single button oxygen with a negative charge. Well done. So this is our first intermediate. We can label this our intermediate because it's part of our final answer. And then the next step we are essentially treating this intermediate with sodium hydroxide. We already know what the final product is because this is sin di hydroxy elation. We are simply going to replace all of this part with two hydrogen atoms. Right? We are going to make this an alcohol group and the following one an alcohol group. So this is how we get our dial. Let's draw a reaction arrow. Remember that? Our region in the second step is sodium hydroxide solution and now we are ready to produce our final products. So that will be our five member ring. Let's draw an ethyl group as well. And now here we are forming our first alcohol group. Nothing changes regarding the configuration. And similarly here we have an alcohol group which indeed corresponds to a sin dial. Right? The two alcohol groups are pointing up Once again remember that it doesn't matter if you try to register two dashes. The compound is a Cairo, it has a line of symmetry. So if it's a missile compound, you don't need to worry whether you're adding permanganate from the top face or the bottom phase. That would be exactly the same structure. It doesn't have a pair of finance commerce. It only has one product to the fact that this is a missile structure. So this will be our one and only product. And we are ready to summarize our conclusions. So first of all, in the first step we added, but as in per magnate called dilute solution, we understood that we have formed our intermediate. That's why we labeled it in green and then this intermediate. In the second step upon hide releases and strongly basic conditions or treatment of that intermediate with an aqueous solution of sodium hydroxide. We are producing our dial. We are replacing all of this fragment with two hydrogen atoms to give us two alcohol groups. This is a meso structure, so we do not have and in antimatter pairs we only end up with one possible product. Now, looking at our answers, we may simply conclude that the correct answer to this multiple choice question is d However, if you have any questions don't hesitate to leave a comment down below. Thank you for watching

Predict the major products of the following reactions, and give the structures of any intermediates. Include stereochemistry where appropriate.
(i)

Key Concepts

Dihydroxylation

Stereochemistry

Oxidation States and Mechanism

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