For each of the following target molecules, design a multistep...

Question

For each of the following target molecules, design a multistep synthesis to show how it can be prepared from the given starting material:

c. Chemical reaction diagram showing the conversion of an alkene to a ketone with a specific structure and reaction arrow.

Accepted Answer

Hey, everyone. And welcome back to another video designing multis stepp synthesis to show how the following target molecule can be produced using the given starting material. The starting material is a vinyl cyclo pens and we are producing two cyclo pencil acid aldehyde. Now, you may notice that this synthesis can be easily performed if we use hydration oxidation, right? We can use Bora in the presence of tetra hydro Furin followed by hydroxide hydrogen peroxide and water. So that would give us our primary alcohol and then we could oxidize it into an aldehyde using a mild oxidizing agent such as PC C. However, we are going to analyze the different synthesis in today's video using an alk kind as an intermediate. So let's see how it works. First, our idea is to get an alkin out of an alkin and do that. The very first thing we want to form is our dibromm alkin. So we're going to use bromine and the common so solvent that we are using in the electrolytic addition is carbon tetrachloride. And we are now going to draw our first product in this synthesis. So we are essentially taking our side chain we have two carbon atoms. And now our goal is to add two bromine atoms to our double bonded carbon atoms so that we are breaking our pi bond. And we have our electrolytic addition to form our di bromo ALCA. Now, as we know, we can easily form an alk kind if we perform a double de hydrogenation. So you can state that one of our reactions is a double dehydrogenating in that reaction. we're using an axis of a strong base. And in particular, it would be a good idea to use a classical base which is amide and ion. And we're adding sodium to represent that it's a sold compound, right? But since sodium as an iron spectator, we can just use amide and iron on its own. And if we use an excess of it, we perform our elimination reaction. So first of all, we noticed that we have our beta hydrogens with respect to each bromine, right? And specifically, when we remove one of them, we are forming a double bond followed by the loss of the living group, then we remove the other one forming a triple bond, followed by the loss of the living group. And finally, you have to be very cautious because sodium amide or just amide itself is a strong base, right? So that means when we form our triple bond, the terminal alky will essentially be dep proton. Let's remember that hydrogen bonded to an S P carbon is significantly acidic, right. So we are actually going to get an ayite an ion because the hydrogen atom bonds and S P carbon is significantly acidic and it can also be removed by an excess of AM. So we're using three equivalents two for our elimination and one more to form a satellite. So essentially, if you want to get an alk kind, instead we're going to add an additional step. So the first step is addition of sodium ami access. And the second step will be the protonation step where we are going to use water, right? Our mild proton donor. So in that case, a satellite will get proton during the second step and we will get our alkin. And finally that we have our alkin, just as we mentioned, we can use hydration oxidation. The reason why we're using hydration oxidation is because according to the anti Markov Kov Radio chemistry, hydra oxidation will essentially allow us to add alcohol grew up to the less substituted carbon and hygiene is the more substituted carbon. So we are going to add the set of regions needed for that reaction. Number one, that would be dialy borane, right? So we can use R two B H in the presence of tetra hydro, that's our solvent. OK. And now the second step, just as we said, we are going to use hydroxide hydrogen peroxide and water. So those would be our two steps and we will readily produce our aldehyde Now, just in case you're wondering why we're getting an aldehyde. You may recall that the first step in this reaction is the production of an eno. So we are going to show how it's formed. If we essentially go down a bit. Let's remember that. As we said, we are going to add our alcohol group to the less substituted carbon, right? So we're going to get our two member chain and then we have our alcohol group. But in addition to that, we have our pie bond, one of the pie bonds remains. And because enos are not so stable, they undergo ization or basically keen to. So if we expand our oxygen hydrogen bond, the lone parent oxygen forms a pi bond, we're breaking the edges and pi bond transferring the electrons to oxygen. And that's how we get the final aldehyde. You may notice that we can just redraw that structure neatly. We don't have any pyon anymore. And instead we have 12 carbon atoms. And at that position, at the very end of the side chain, we are forming our carbonyl group. And of course, we have our implicit hydrogen atom which corresponds to an aldehyde now. So that's why hydration oxidation out of a terminal alkin forms that required aldehyde. So let's label our synthesis scheme and let's discuss it in detail. The first step was the electrolytic addition of bro me into an to form our di bromo ALCA. Then we used our double D hydro halogen in which we used an excess of sodium amide. And we needed to proin the ace and to isolate our alkin. And finally, we used Hidra oxidation to produce an aldehyde out of a terminal alkin. That will be it. Thank you for watching.

For each of the following target molecules, design a multistep synthesis to show how it can be prepared from the given starting material:
c.

Key Concepts

Alkynes and Alkylation

Stereochemistry

Functional Group Transformations

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