Fill in each box with the appropriate reagent:
a.

Question

Fill in each box with the appropriate reagent:

a. Chemical reaction diagram showing reagents needed to convert alkenes to alcohols and carbonyl compounds.

Accepted Answer

Hey, everyone and welcome back to another video for each reaction in the sequence below, right, the correct region, the starting material is a one methyl cyclo bete one in in reaction one, it is converted into two methyl cyclo butanol. Then we are performing our second reaction in which two methyl cyclo binal is converted into three methyl cyclo bet one in which is finally converted into two methyl sein aldehyde. So let's think about each reaction individually in the first reaction. The alkin that we are using attaches hydrogen at the more substituted position and the alcohol group at a less substituted position. So that means if you want to perform that conversion, we have to use hydra oxidation reaction because we have an cognitive radio chemistry, right, we are forming an alcohol which is a secondary alcohol instead of a tertiary one. So we are forming the ansar cognitive product. And therefore, we can say that for the first reaction, if we essentially want to convert our king into alcohol, we are going to use the following set of regions which corresponds to hydro operation oxidation. So that would be boring in the presence of aqueous hydroxide solution, hydrogen peroxide and water will essentially tell us that we have that aqueous solution of hydroxide. So this is our first reaction and just to state that we are using hydration oxidation reaction in which we are forming the anti Markov Kov alcohol. Now, in the next reaction, we want to turn that alcohol into a different alkin. And specifically what we're doing is that we noticed that we get our double bond within the four numbered ring. So we have to perform an elimination reaction. But the problem is that the alcohol group is a poor living group. And therefore, what we can actually do is that we can turn our alcohol into a good living group if we use chloride, our standard, right? So we are going to use Dozy chloride in the presence of pyridine. OK. So we are going to replace hydrogen with the tassel group. So we will get our tassel. This is a very common reaction. Pine is used to dero oxygen at the end so that it doesn't have a formal positive charge and it doesn't have that acidic proton. So in that step, we are, first of all going to get our good leaving group. So now we have our tale ester group and from there, we can actually get our elimination product. Let's notice that the double bond is obtained between two less substituted carbons, right. So we don't want to eliminate the hydrogen at the more substituted position, but we actually want to eliminate from the less substituted position. In other words, we have two beta hydrogens and we're going to form our Hoffman product, right. So if we form our Hoffman product, we want to use a hysterically hindered base. Specifically, we can use potassium turb oxide. We can actually write it as TB U O K O minus K plus, right? Because we have our ionic compound. So when we use potassium turb oxide, which is a historically inner base, we understand that if we draw it out, the negatively charged oxygen will take the proton, we are going to form our double bond followed by the loss of the leaving group. So we have a basic elimination reaction. And from there, when we get our Hoffman product, we can actually eliminate the arrow between our alcohol and Alcaine because we have a sequence of reactions and our final reaction if we carefully look at it is the oxidative cleavage of the carbon carbon double bond. Let's notice that there are two implicit hydrogens within our right. And specifically, if we use and also knows this reaction, we will be able to cleave our carbon carbon double bond. For that purpose. We have to recall that the first step is the addition of ozone at negative 78 degrees Celsius followed by reductive work up, which is dimethyl sulfide. In that case, we are just breaking our carbon carbon double bond. OK. And then we are going to open our ring as a result. So if we carefully label our carbon atoms, let's go back to the original sketch. And let's suppose that carbon number one starts at the double bond. If we go clockwise, that'll be 234. OK. And we're breaking the bond between carbons one and four. And what that means is we're going to get a four member chain. OK. So 1234, if we label our carbon atoms, we get 1234 at position number one, we actually have part of the double bond, right? Let's remember that during the oxidate of cleavage, the double bond, we are going to form two double bonds at carbon number one and carbon number four. And then we simply want to add oxygen atoms to each, right? Because we're breaking our double bond and then adding oxygen to each end. And now we have our implicit hydrogens at carbons one and four. So that's how we get to aldehyde groups. And carbon number number three has a metal group. So that's exactly what we're seeing. We're seeing our methyl group at carbon number three, followed by C group, right? Then we have C two and then we have another aldehyde group. So that makes sense. OK. So let's conclude our findings for the first reaction. We're using hydration oxidation for the second reaction. We have to perform two steps, right? So we're going to label them as number two and number two, right, specifically, we can also add number one as tazo chloride and Perine. Number two addition of potassium Urox side. So that'd be our reaction number two to convert our alcohol in. And finally, the third reaction requires us to use the oxidate of cleavage, which is our zinos reaction. We can use that ozone followed by the methyl sulfide. From here, we are ready to conclude our video. Thank you for watching.

Fill in each box with the appropriate reagent:
a.

Key Concepts

Oxidation of Alcohols

Common Oxidizing Agents

Reactivity of Carbonyl Compounds

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