Show how free-radical halogenation might be used to synthesize...

Question

Show how free-radical halogenation might be used to synthesize the following compounds. In each case, explain why we expect to get a single major product.

(a) 1-chloro-2,2-dimethylpropane (neopentyl chloride)

(b) 2-bromo-2-methylbutane

Accepted Answer

Hey everyone and welcome back in today's video, we are going to solve the following problem, propose a free radical hallucination reaction to synthesize each of the compounds given below. Also provide a reason why we get a single major product in each of these reactions. A one chloral cycle plantain and B. Two bromo two metal plantain. If we start with a, we want to propose an unknown substance treated with a halogen and the presence of light, right? Because this is the free radical halogen nation to form one chloral cycle of fencing, meaning we may just draw that five member grain and add a chloral substitution at position number one. We understand because chlorine is a halogen here. We're going to use cl two chlorine molecule in this reaction, we want to understand what the starting material is. Now let's think about the unique hydrogen in the ring itself because the hydrogen that has been substituted initially was at that position, right? We noticed that if we recall the principles of the free radical hydrogenation, one of the hydrogen atoms substituted with our halogen. So now we only have one hydrogen over here while the other one has been substituted with chlorine, meaning our starting material must be cycle pencil. Right, so this is the only possibility for us to proceed now, we will not worry about the reason yet. We will come back to this problem later, but let's use that exact same proposed logic to predict the starting material for B for B. There is another free radical hydrogenation reaction. So we need to add light to form two bromo two metal plantain. So if we draw a five member chain because we have been saying Position # two contains a metal group and a bromo substitutes. We are simply adding one metal group and one bro mean now once again because we're adding bro mean that means the reaction requires us to use bromine molecule in this free radical hydrogenation and similarly to what we did before we call that the hydrogen atom is simply replaced with broom in in this case. So initially we should have had we should have had hygienic disposition instead of roaming meaning our starting material must be to metal painting itself. So we are drawing A five member chain that's metal group. And now instead of roaming that we simply have that hydrogen atom which has been substituted with roaming. Now we have our reactions right? We understood that these must be our starting materials because we are simply replacing hydrogen atoms with chlorine and bromine respectively. So we simply got rid of chlorine and bromine and we added hydrogen to their positions to get our starting materials. So we're good with these reactions. And the only thing that we want to understand is just the reason why we're forming these mono chlorinated and mono brominated products as our major products and single products right? Major and single major and single thinking about our starting materials because that's much easier to think about if we start with A that's a cycle pension ring. And if you look at all of these five positions. These five carbons are all secondary carbon atoms because each of them has to al kill groups, bonnets to it. So each position is secondary and that means that each position is equivalent to another. Remember that radical stability increases in a similar fashion just as the car bo katan stability does. So primary radicals are less stable than secondary ones, while tertiary ones are the most stable. We can say that the radical stability increases from left to right. All of the possible radicals that we can form here are secondary ones, meaning there is only one possibility for the radicals to be formed in addition to that, all of these carbons are equivalent, right? Because you can always rotate your ring and you will notice that there is just one type of carbon atoms that we have, meaning it doesn't really matter which hydrogen will remove. We always produce that single product. Whenever we are thinking about that mono chlorinated product. So there is only one possibility to form that mono chlorinated product because all of these carbon atoms are equivalent and all of them are secondary ones. So that's the reason for a we can just say that only one product mono chlorinated product, right. Is possible. So here we're just thinking statistically there's only one possibility, right? We're not really worried about reactivity itself. All of these carbon atoms are equivalent and therefore there's just one possibility. But what about B and B? We have different carbon atoms, right. Why didn't we add roaming into this position? Why did we add broom into that position instead? And that's because this is the tertiary position which produces the most stable radical. Right? Blooming is highly selective. Remember that as opposed to chlorine bromine is much more selective and it will always be added to the position that forms the most stable radical. The most stable possible radical is at a tertiary position. And when we form a radical hear the booming molecule would approach and it would exclusively add broom into this position. The reactivity of the tertiary radical is much greater when booming reacts with it. And that's the reason for the formation of B as the major single mono brominated product because roaming is highly selective and the tertiary radical is the most stable radical that we can observe. This is our major and single product. And we may say that the proposed reactions in this problem for A and B. Are these reactions? In addition to that, we see that the reasons are for once again a only one product as possible and be roaming as highly selective and the tertiary radical is most stable. Right? Seeing these reasons, we can just conclude that the correct answer choice to this multiple choice question is C. However, if you have any questions, don't hesitate to leave your comment down below in the comments section and thank you for watching.

Show how free-radical halogenation might be used to synthesize the following compounds. In each case, explain why we expect to get a single major product.
(a) 1-chloro-2,2-dimethylpropane (neopentyl chloride)
(b) 2-bromo-2-methylbutane

Key Concepts

Free-Radical Halogenation

Selectivity in Free-Radical Reactions

Stereochemistry and Product Distribution

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