Show how you would synthesize the following compounds. As star...

Question

Show how you would synthesize the following compounds. As starting materials, you may use any alcohols containing four or fewer carbon atoms, cyclohexanol, and any necessary solvents and inorganic reagents.
(a)
(b)
(c)

(a)

(b)

(c)

Accepted Answer

Hey everyone and welcome back proposed the synthetic pathway to synthesize the following compounds starting from psycho pencil and any necessary regions and solvents. Starting with a. We understand that the starting material is cycle a pencil so we can draw a cycle a pencil our five member drain with an alcohol group at carbon # one. So that's our cycle of fentaNYL. We want to form an ether. Right? But in addition to that we also want to make sure that there's a methyl group. So that would be that essential implies there would be a green yard regent involved in this reaction. So to make use of the green yard region, let's first of all turn our alcohol into a carbon compound. So we can oxidize our alcohol. This is a secondary alcohol, right? Because the alcohol group is bonded to a carbon with two neighbors. So we can oxidize our secondary alcohol and say ketone very easily. If we use one of our oxidizing agents in this case we can use sodium hypochlorite in the presence of acetic acid even though we can use any other oxidizing agent that's perfectly fine. So you're going to turn cyclamen tunnel into the corresponding ketone. We are going to draw our carbonell group and we got it right now essentially we want to make sure that we add that metal group. So we can essentially use metal magnesium bromide because we need to add that metal group. Ch three mg B R. Right? This is what we are going to use here when we use metal magnesium bromide. This is our green your region essentially we can draw a partial attack at the the electra Philip position. And this is where we are forming our elk oxide with a metal group. So we control the intermediate. Here we are breaking the pi bond. Now we have auction with a negative charge and that metal group that we needed. Okay, so we got our elk oxide. Now the world last thing is we we noticed that we got something very similar but now we need to add a metal group at oxygen, meaning we can make use of an SN two reaction because we have a strong nuclear file. So why don't we use metal bromide, H three C. B. R. And basically the negative charge attacks the electra Philip position for by the loss of the leaving group. This is an essence to reaction. Right? And this is how we get our final product. We are essentially adding that metal group to our Alcock side. Okay, so we got our product. This is exactly what we wanted to synthesize. We can summarize the steps and we can say that if we start with psycho fentaNYL, we need several reactions to be performed to produce the final product. If you notice number one is addition of sodium hypochlorite in the presence of acidic asset number two treatment of the result on keystone. After we oxidize our secondary alcohol with the green region that be metal magnesium bromide. And number three, the third reaction, we performed an SNC substitution where we needed to add methyl bromide. So these are our reactant needed for the synthetic transformation. We can label them as our final answer. Now looking at reaction be, we went to synthesize a key tone and once again because we have additional carbon atoms, we are going to make use of the green your reaction and let's see how that works. So, first of all, what we're going to do here, we are going to use phosphorus try bromide to turn the alcohol into the corresponding al kill bromide. So what happens here is that your essential replacing your alcohol group with romaine? And you're getting como cycle of fencing. Right, that will be our blooming now because we have roaming attached, we can easily form the green region and to form the Green yard wage and we simply need to add magnesium in the presence of ether. As we already know, we are incorporating magnesium into our system. And we get our Green yard Rage and MGB R now that we got our Green yard region. We can easily use a carbon compound that will allow us to get our final key tone. Now the question is, what kind of carbon compound are we going to use? And we are going to use an elder Hyde right? Because we want to first of all get a secondary alcohol. So looking at the final product we are going to use and al to hide. And we of course need a metal group to make sure that we have that metal group in our product. Notice that the product is a key tone on one and it has a cycle Hexcel substitution on the other and it has a metal group. So we definitely want to use ethanol right? We want to use ethanol which has two carbon atoms within its al hide form. So what happens here is that when we treat our Griner regent with ethanol, we have an attack at the electrolytic position, breaking the pi bon and we are forming our elk oxide. So let's see how Elcock side looks like. That would be Our five member ring. Now we are essentially creating an additional bond that would be our carbon bonded to oxygen with a negative charge. And we have that metal group and that's why we needed to use ethanol because we want to make sure that there is this metal group over here, we got our Alcock side and now we are essentially ready to protein ate it to form an alcohol because when we get an alcohol we can oxidize it to a key tone later. So now the next step is treatment of Elk oxide with an asset acidic work up will give us our alcohol. So that would be our five member grain O. H. And all that we have to do. Now is just to oxidize it oxidizing secondary alcohols leads to the formation of ketones. so we can use our previously utilized oxidizing agents, sodium hypochlorite in the presence of acetic asset and now we are getting our final product. Okay, well done. So let's label this as our synthesis scheme because we got our key tone just as required. Once again we took our psycho pentagonal turned it into a bromo psycho pension using phosphorous tri bromide. Used magnesium to produce the Green Yard reagent treated it with ethanol to make sure that we get our secondary alcohol after acidic work up and then we oxidize our secondary alcohol into a ketone. And finally let's look at reaction. See for C we can once again troll cycle pence inal just as we did before. We are going to use the green dark reaction because we need to add two metal groups and actually carbon atom, right? We essentially need a source of carbon atoms. So what we're going to do, we are going to form the granular region first and to do that, we are going to turn it into an L kill bromide using phosphorous tri bromide, as you already know, we are essentially replacing our alcohol group with broom in That would be our five member ring roaming now because our goal is to form a green yard region out of it. We are going to add magnesium in the presence of ether and form our Green yard reagent, that's our granule region. And now when we got our Green yard wage and we want to make sure that we are adding the right carbonell compound because the carbon compound has two metal groups from the central carbon in the product. We are going to use acetone here. And if you look at the reaction scheme in particular, there's an attack at this position. We are breaking the pi bond and we're getting our Elcock side. So if we draw the result on the intermediate First of all, that would be our five member ring. And now we are adding it to this carbon atom. So this carbon atom and Saints oxygen with a negative charge. Right? We're breaking the pi bon and there are two methyl groups. Okay, so we got our Alcock side. And now when we get our Alcock side, the question is, do we really need to form an alcohol out of it? The answer is no, because we essentially want to form an ether at this position, we just need to add a metal group. So as we did before, we can simply use methyl bromide in an sN two manner. If we use methyl bromide as our regent, we can essentially notice that that strong nuclear file will attack the electrical position, followed by the loss of the leaving group and we will get our found product. Indeed. So that would be oxygen with a metal group and two metal group. So we're here. Okay, so that's our final product meaning this synthesis works for us, Let's discuss the steps once again, cycle pencil is treated with phosphorus. Try bro my to give us bromo cyclone fencing which is treated with magnesium and ether to give us our granite regent. This allows us to add cyclo pencil as a substitution to acid tone. The reason why we're using acid tone is because if you look at the final product, we have these two methyl groups. Right? So we definitely want to use acid tone. We're forming our elk oxide and iron, but instead of acidic work up, we're treating it in an essence to manner with metal bromide to add that metal group to oxygen because oxygen is a strong negatively charged oxygen or elk oxide is a strong nuclear file. So we're making use of this fact and applying an essence to reaction to get our final ether. And that would be it. We can conclude that the correct answer choice to the small cultural question as a B. However, if you have any questions, don't hesitate to leave your comment down below and thank you for watching

Show how you would synthesize the following compounds. As starting materials, you may use any alcohols containing four or fewer carbon atoms, cyclohexanol, and any necessary solvents and inorganic reagents.
(a)
(b)
(c)

Key Concepts

Alcohols and Their Reactivity

Synthesis of Ethers

Cyclohexanol and Its Derivatives

Watch next

Show how you would synthesize the following compounds. As starting materials, you may use any alcohols containing four or fewer carbon atoms, cyclohexanol, and any necessary solvents and inorganic reagents. (a) (b) (c)

Key Concepts

Alcohols and Their Reactivity

Synthesis of Ethers

Cyclohexanol and Its Derivatives

Watch next

Show how you would synthesize the following compounds. As starting materials, you may use any alcohols containing four or fewer carbon atoms, cyclohexanol, and any necessary solvents and inorganic reagents.
(a)
(b)
(c)