Show how you would synthesize octan-2-one from each compound. ...

Question

Show how you would synthesize octan-2-one from each compound. You may use any necessary reagents.

(c) 2,3-dimethylnon-2-ene

Accepted Answer

Hey everyone and welcome back to another video. In today's video, we are going to solve another problem which reads show how to three that metal hex gene can be converted to pence and Sione provide any necessary regions. Let's start by drawing our starting MTI hex two in So that'd be six carbon atoms. We are going to throw 123456. Right. And we understand that that position to let's suppose that this is our carbon number one, then this is carbon number two. We have our pi bond. So that's hex to in and our carbons to and three can same muscle group speakers, it says C three die metal. So this is our starting material. And we understand that we are producing plants and she owns so that'd be five carbon atoms 12345, right. And then at position number two, we have our carbonell group, we got tents and noticed that the starting material is an elk. Een the final product is a key zone. Recall that one of the ways to easily convert Elkins into carbon compounds is via. And arsenal's reaction or basically oxidative cleavage and arsenals is is one of the ways to perform that conversion. So we can say that we may achieve this conversion via. Also knows this. This really makes sense because if you think of the structure contains six carbon atoms in the parent, the product has five. So we're shortening the chain and in addition to that we are forming a carbon compound. Now let's recall the steps for our noses noses reaction reaction number one would be addition of ozone in literature, you will often see this value negative 78 degrees Celsius. So this forms our Morelos oh night intermediate, which then rearranges into ozone ID. And then we need to use reductive work up the metal sulfide in the presence of water. And this allows us to form carbon compounds afterwards in a general case, so that we don't have to worry about the mechanism. We can just recall that whenever you have an L. Keen with hypothetical substitutions such as R one R two. And then let's suppose that here you have your L. Kill group R. Three and hydrogen. We whenever we perform an osteoporosis reaction. Using these two steps for simplicity, let's say those analysis, we are going to break our pi bonds were basically our double bond, right and separated chief fragments apart. If we separate the two fragments, we get the following, we have hydrogen bonded to carbon and then we have our three. That's one half of our fragment and then the other fragment with essentially correspond to a double bond R. T. R. One. So we are just visually breaking apart. Right notice that we can easily combine them into the original structure and have the reason why we're doing so, it's just because if you add two oxygen atoms to each end of each double bond, you will get the right products. One key tone. One al to hide and indeed we get a key tone. So oxidative cleavage might work for us. Let's see if it works. So if you read through the starting material, We may essentially read through our 6th member chain. So that would be 123456. We are going to add our pi bond metal groups. And now if we think about those analysis we're essentially going to break this double bond, right? We identify the position of the double bond, we break it. We may essentially state that we're performing an analysis reaction and reductive work up. Right. And essentially we can redraw the two fragments on the left. We have 1234 carbon atoms. That will be the fifth one. Right? Because we're essentially breaking this double bond, we can draw five carbon atoms for the first part. So every 12345. Right? And I would notice how position number four. Or basically if we go the other direction, position number two can sense that fragment of the double bond. So if that's our number one, that's our number two. We're going to add a double bond and the other fragment would correspond to a double bond carbon atom bonded to T methyl groups. Now, all that we have to do is just add those oxygen atoms. And these are our products to different key tones. Now isn't this the product that we want to form, that's 12345 carbon atoms at position number two we have our carbondale group. So this is indeed 10, 10 to own. It's not a problem that we get that by product. We are interested in the production of pence entity own and then we are going to isolate it. So we see that we indeed get our products. So close analysis will work for us, meaning if we go back to the reaction scheme, we can say that to perform this conversion, we need to follow two steps number one edition of ozone at negative 78 degrees Celsius and step number two would be a reductive work up addition of dimethyl sulfide in the presence of water. We have proposed our regions so that would correspond to our final answer and based on the regions proposed, let's conclude that the correct answer to the multiple choice question is C thank you for thank you for watching and I'll see you in the next video

Show how you would synthesize octan-2-one from each compound. You may use any necessary reagents.
(c) 2,3-dimethylnon-2-ene

Key Concepts

Carbonyl Chemistry

Alkene Reactions

Reagent Selection

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