For each synthesis, start with bromocyclohexane and predict th...

Question

For each synthesis, start with bromocyclohexane and predict the products. Assume that an excess of each reactant is added so that all possible reactions that can happen will happen.

(a)

Accepted Answer

Hey everyone and welcome back in today's video, we are going to solve the following problem where the given synthesis predict the missing products. A through c assume that an access of each reagent is used. Now if we start solving our problem, we understand that the starting material is Roma cycle fencing is an L kill. Hey light. And then the first reaction, we're treating it with potassium hydroxide, ethanol in the presence of heat, ethanol here is a solvent. So first of all, let's analyze this reaction because potassium hydroxide associates in a solution. We are mainly interested in the presence of hydroxide. Hydroxide is a strong base strong nuclear file Because hydroxide is a strong base strong nuclear file. We have a competition between sn two NE two reaction but looking at the nature of the solvent that's ethanol and essentially it is a polar product solvent which favors elimination and in addition to that, we are also applying heat to our mixture, meaning it favors elimination as well. Due to these reasons the presence of a polar products solvent, strong base and heat. We may rule out that this reaction will follow an E two pathway. If we know that this will be an E2 pathway, we can easily predict the product. We control the starting material and we need to identify our beta positions. There are two beta positions. These two positions are equivalent because there's a line of symmetry, a vertical line of symmetry to be more accurate, which passes through our compound and we notice that eliminating this or the other hygiene will result in the same structure. So hydroxide acts as a strong base. We have the attack at the proton. We're forming a pi bond which is followed by the loss of the leaving group. And this is how we get our cyclops fencing. That will be our five member ring. And then we noticed that we have a double bond in this position. Now we have formed our structure A. We may label this for future reference and now we are ready to think about the synthesis of B. And whenever you're looking at this set of regions where you have ozone, followed by the addition of dimethyl sulfide and water, you may recall that this reaction is an ozone analysis reaction which corresponds to the oxidative cleavage of the carbon carbon double bond, meaning to draw our product as easily as possible. We are first of all going to break that carbon carbon double bond and we are going to label our carbon atoms. So let's suppose that we have 1234 and five. So let's break our double bond and starting with the first carbon, Right, this would be our carbon No. one Let's suppose that it's over here, then we have two three for in five. Now let's see how that works. So that would be once again our first carbon atom. Remember that we have broken our double bonds. So that would be the position at which the double bond starts. Right? Then we have carbon number 234. And this would be carbon number five, which itself has part of that fragment of the double bonds. We can just add another double bond over here. Now, essentially we want to understand that during the oxidative cleavage these two double bonds, they do not have any carbons here. We essentially want to incorporate an oxygen atom at each end of the resultant fragments. We are going to add one oxygen over here and another oxygen over here. Right. So these two oxygen's will correspond to the following oxygen's below. And now we can also add these implicit hydrogen is bonded to carbon number one and five. To clearly see that this structure is an AL to hide. Well then this is our structure B And now we are going to think about reaction to produce See and notice how in C. We are using elo magnesium bromide, followed by acidic hydraulic assist. So this corresponds to a Green Yard reaction. Right? Whenever we are using an L key chain bonded to magnesium bromide, followed by acidic hydrolyzed this this would correspond to the Green Yard reaction. So let's draw the green reagent. Let's keep in mind that as the problem suggests, every reaction takes place with an access of each reactant. So that means because we have two carbonell groups. We are going to repeat the same process for complete simplicity because we are not asked to draw the mechanism. We can just essentially show both steps at once. So that's our a little magnesium bromide. And as you recall, we have an attack at the electra Philip position, followed, followed by the cleavage of that pi bon right within our carbon in groups. So first of all, let's draw the arrow, the attack at the electrical position, followed by the cleavage of the pi bon. And similarly this is observed. My apologies. Let's add G for magnesium bromide. And now we may notice that exactly the same thing occurs for the other carbonell group because we're using an excess of each reactant. The two sides are al the height, so they're equivalently reactive. And we expect to see exactly the same attack at the electrolytic position at the carbondale group, followed by the cleavage of the pi bond. So now from here we are ready to think about our intermediate. We may essentially start by drawing our carbonell groups. But now you have to be extra careful because you only have a single bonded oxygen atom because it's single bonded. And because we have transferred these by electrons onto oxygen, we are producing an elk oxide and iron. It's not really important to show this implicit Haijun anymore because we do not have our carbon group. Instead, we're going to show how these three carbon atoms are attached to our chain. So if this is our carbon number one over here, let's label it clearly. Then we are attaching three more carbon atoms. Let's show them above. So that would be one, two and three, right, 123 and then we have a double bonded carbon atom at the end. So now we are ready to draw our chain. The original chain had carbons number 23, four and five. Right now this carbon number five will not have a double bond anymore. So we are ready destroy another single bond oxygen with a negative charge. That's our elk oxide. And we are essentially adding the same substitution at position number five. This is our position number five. Once again. So let's add another substitution that comes from the green yard region. And finally, because we have two elk oxide groups just as the reaction suggests we are going to protein ate them To form our alcohols. Right? We are going to form our alcohol products. So that would be a 30-plus acidic work up. And we are forming our final product which corresponds to the following structure. Let's first of all draw the left alcohol group that will be O. H. Then we have our five carbon atoms. So let's drop 1234 and five. We noticed that at position number five, we have another alcohol group because we're also protein eating it. And then let's draw our submission. So that would be our position, position number one. And then at position number five, we are drawing another one. Okay, well done. So this is our structure. See and for complete clarity, let's just list our final structures separately as we see our final products very easily. So first of all, let's start with structure A. Structure A. was our five member ring with a double bond. So that cycle fencing. Now structure B was our elder Hyde. We may essentially recall that we got this structure which looked like this. Let's crawl up. This is our structure B. Right, so we're going to try it, our carbonell group with hydrogen. Then we are going to add our carbon so that we carbon number 2345, hi Chen carbon group and finally for structure, see we are just going to rejoice that will be our alcohol group 12345. And then we are going to add another alcohol group as well as our substations added in the greener reaction. So let's label our final products A. B and see. Those are our final products based on the answers that we see here. We may simply conclude that the correct answer choice to the small cultures question is D However, if you have any questions, don't have States, leave a comment down below. Thank you for watching

For each synthesis, start with bromocyclohexane and predict the products. Assume that an excess of each reactant is added so that all possible reactions that can happen will happen.
(a)

Key Concepts

Elimination Reactions

Ozonolysis

Grignard Reagents

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