Starting from bromobenzene and any other reagents and solvents yo...

Question

Starting from bromobenzene and any other reagents and solvents you need, show how you would synthesize the following compounds. Any of these products may be used as starting materials in subsequent parts of this problem.

(a) 1-phenylpropan-1-ol

Accepted Answer

Hey everyone and welcome back in today's video, we are going to solve the following problem using Roma cycle plantain and any other suitable reagents and solvents as starting materials show how one cycle pencil athan one old can be synthesized. Now in particular, the problem suggests that our starting material is Roma cycle plantain. So we are going to draw a five member ring with bromo substituted at any position right? It doesn't really matter where we added. According to the problem, we went to form one cyclo pencil athlon one. Oh, so essentially that implies we have A five member dring bonded to two carbon atoms because it says ethanol and the alcohol group starts at position number one. So suppose that this is position number one. We have our alcohol group and the cycle pencil Supposition at the same position. This is our position number two. Right? So this is the alcohol that we want to synthesize and what we, what we may immediately notices that the main difference between the two is that we're adding these two additional carbon atoms and because we're adding two additional carbon atoms and forming an alcohol, it makes sense to make use of the greener reaction. So let's do that. And that's also perfect for us because we have roaming president already, we can easily form our grain yard region if we add magnesium in the presence of ether rights, we will form our our MGB R compound, which is the graveyard region. So to do that we need to add magnesium in the presence of ether and we will essentially incorporate magnesium over here to produce our Granier trajan. Now we get the green york region and now we need to think about which kind of carbon compound can we use? Because we are forming a secondary alcohol. We want to use an alga hide. This is very important to understand notice how we are going to form one carbon carbon bond. So due to the presence of that muscle group here and making this a secondary alcohol, we want to make sure that we choose the correct substrate. We need to carbon atoms right to be added to the chain. So that means we first of all start with our carbonell group because growing your reactions require carbon compounds and we understand that if this is our carbon number one, we will also want to have carbon number two bonnets. It so that be our carbon number two, that metal group. But we need to use an alga hide because we do not want to form a tertiary alcohol. Right? We are forming a secondary one. That's why we want to make sure to add hydrogen here, we don't see any additional carbon atoms. So just two carbon atoms, one within the carbonell group and the metal group bonded to it. In addition to that we have hydrogen. So that's perfect for us. What happens is that the election density is really shifted towards carbon because magnesium is highly electoral positive. Right? So that electron density essentially attacks the carbonell group. The carbon group. Right? We're breaking the pi bond and we are getting our elk oxide and iron in this step. We may say that our intermediate form intermediate form is the Alcock side. And I am and we may try it out. Now We are drawing our five member ring. There's no more magnesium bromide. Right? We're essentially getting rid of it. We're just transferring these bonding electrons into the formation of carbon carbon bond. So now this is what happens. We are creating that bond and this is our carbon carbon. But now it becomes an elk oxide oxygen with a negative charge because these two electrons had been transferred in the form of a lone pair onto oxygen. Now, in addition to that we have that metal group. Right, So this is our intermediate. We got our intermediate which is all cock side. If we want to get our alcohol where last step is a protein nation. Stop. So we understand that. To get our final product we are just going to protein eight Alcock side using a strong asset. So you can just say Our next regent is H30 plus in general. Just an asset that we can use for that proto nation in which one of the hydrogen atoms will be abstracted from hydro ni um we will have the loss of water and we will form our final alcohol. So we may say that if we conclude our steps number one was the formation of the graveyard trajan addition of magnesium in the presence of ether. Number two, we added ethanol. Right. We said that we want to use an alga hide here with two carbon atoms so that was ethanol that we chose. That's our al to hide. And number three we said that the final reaction was the protein nation step to make sure that the elk oxide and are informed is protein ated to give us our final alcohol. So this is our reaction scheme and the regions for this reaction are these three steps magnesium and ether, followed by the addition of ethanol and then protein eating our Alcock side intermediate with a strong asset. The correct answer to this? Multiple choice question is C However, if you have any questions, don't have seats leave your comment down below in the comment section and thank you for watching.

Starting from bromobenzene and any other reagents and solvents you need, show how you would synthesize the following compounds. Any of these products may be used as starting materials in subsequent parts of this problem. (a) 1-phenylpropan-1-ol

Key Concepts

Nucleophilic Substitution

Grignard Reagents

Reduction Reactions

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