Show how each of the following compounds could be prepared fro...

Question

Show how each of the following compounds could be prepared from the given starting material. Each requires a protecting group.

a.

Accepted Answer

Hey everyone. And welcome back, show the steps required to prepare the following compound from the provider's starting material. Hence, a protecting group is needed. Now, the conversion that we are observing is a conversion from one bromo to methyl pan 30 to 4 methyl heptane, 25 dial. And this conversion needs a protecting group just because we have an alcohol group and we are going to form another alcohol group which generally requires us to use or an organ, a metallic region. So specifically what we may notice is that the two structures look very similarly except from the fact that we are adding two extra carbon atoms to our structure by removing our bromine atom and adding an alcohol group because we have bromine in the starting material. And because we want to do two things, number one, Add two carbon atoms and number two synthesizing alcohol, a perfect solution for us is to use the Green Yard Regent. But before we use the Green Yard, and we need to protect our alcohol group in order to prevent the reaction between magnesium and our alcohol group, we want to make sure that the alcohol group stays So let's protect it using one of our common regions. First of all, let's redraw the starting material. And let's remember that one of the regions that we use as a protecting group is 3rd dimethyl chloride TB D MS C L. Now essentially what we have in that structure that silicon Bonded to Chlorine, which also has two male groups. So we're going to add one muffle group, another muscle group and then we also have a third beetle group. Now what happens in this reaction is that oxygen that has a lone pair can essentially attack the electro pic center which in this case, is silicon in an essence U reaction manner followed by the loss of the leaving group. So we are removing chlorine, right. It's a perfect living group for us. But then we want to also remove our acidic hydrogen and do that. We want to use a base, a very common base that we're using in this reaction is imidazole. So when we are forming our oxygen silicon bond, and then in the second step, we are essentially removing our proton and using imidazole as a base to make it shorter, we are going to use O TBD MS as a shorthand notation for us instead of throwing the whole fragment. So the structure that we get corresponds to an ether in which now the O H group becomes oh TBD MS which basically represents our oxygen bonded to TBD MS fragment or basically oxygen bonded to silicon and Silicon has two metal groups. Our 3rd be group and no chloral group anymore. OK. So we have successfully protected the alcohol group and now we are simply ready to form the grain out region because we do not have any alcohol. Now, the next step step number two requires us to form the green yard region which requires us to use magnesium in the presence of ether. In a general case, we can say that this is ether. But if we wish we can use a specific ether such as Diy ether, for example, say E TT O, right? And now we are forming our green yard region in which we are incorporating magnesium into our structure. Now that the greener region is formed, we need to think about the carbonyl compound that would allow us to add two carbon atoms. So let's keep in mind that we want to have the carbonyl compound. So one carbon is already at the carbon group and then we need another one. So that would be a methyl group. And the reason why we want to use an aldehyde as well is simply because we are adding two carbon atoms. And finally, we are going to get a secondary alcohol and to get a secondary alcohol, we want to use an aldehyde in the green yard reaction. So for that purpose, we are using ethanol. That's because we need two carbon atoms. We need to use an aldehyde. For a secondary alcohol, we have our green reaction in which our carbon magnesium bond, which is significantly polarized towards a more electron carbon. It basically acts as a nuclei and attacks the electrolytic position of the carbon followed by the cleavage of the C O pi bond. And this is how we get our al oxide and ion. So we are ready to illustrate that by simply removing magnesium bromide fragment from our structure. And that we understand that we are adding the two carbon atoms one and two. And now we get our oxide and iron, we know that we want to get an alcohol. So we want to proin our al oxide. And actually the protonation of the al oxide and ion using acidic work up will also detect our alcohol. So we will finally get our dial. So let's do that. We will get two steps at once to get our final product. In other words, using acidic word cup, we will both proin the alco side and we will dept protect the original alcohol group. So in this step, our oxide turns into an alcohol and we are deducting our original alcohol group which now returns H well done. So we got our final product. So now let's just summarize all of the steps for our synthesis. We will redraw the starting material, we will draw the final product and L V will include all of the steps needed for this transformation. Number one, we used our protecting group TB D MS C L in the presence of mizo. So that's the first step in which we are protecting our alcohol group. Number two, we are using magnesium in the presence of ethyl ether to form the green regent. Number three, we are performing the green reaction with SNL. And number four, we are using an acidic work up to do two things rotten, the resultant alco side and restore the original alcohol group. So we got our full synthesis scheme and we are ready to conclude that this census scheme corresponds to a successful conversion of the starting material to the final product. We have labeled our final answer in a green box and we are ready to conclude this video. Thank you for watching.

Show how each of the following compounds could be prepared from the given starting material. Each requires a protecting group.
a.

Key Concepts

Protecting Groups

Nucleophilic Substitution Reactions

Stereochemistry

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