What halides would undergo E2 dehydrohalogenation to give the ...

Question

What halides would undergo E2 dehydrohalogenation to give the following pure alkenes?

a. hex-1-ene

b. isobutylene

c. pent-2-ene

Accepted Answer

Hey everyone and welcome back in today's video we are going to tackle this problem which states identify the alcohol. Hey lights that would give the following pure Calkins through E two D. Hydro elimination for part A. We have beat one in. So we are going to start with that part. We are going to draw a beat one in which is a four member chain. And at position number one we have a double bond because we're dealing with E two D. Hydro halogen nation. We will think about this problem in reverse and we will get our starting material from the product meaning if we think about this problem we need to add a molecule of hx across the double bond where X is a halogen such as chlorine, bromine or iodine. Now if we neglect the more cosmic of school for a second, we essentially once understand what are the possibilities to add a checks across the double bond. Let's assume for the purpose of this problem that we are choosing bro. Mean as our as our halogen bonded to hydrogen. So the two possibilities to bond hydroponic asset across the double bond, those would be adding hydrogen a disposition and then adding broom in a disposition. So let's show what we get in such an instance. We get an al kill hey light which would look as follows. We would still have a four carbon chain but then broom in would be attached to position number one. So that's our first possibility and neglecting the more cosmic of school. We may also say that the second chance to get that L kill highlight is to reverse our thought for us and say what if we add hydrogen at position number one and boom in that position, number two. In such an instance we would get a broom in a position to So that would be our second possibility. So we have two possible alcohol. Hey lights to form an elk een Vis A and E two elimination because now we have to alcohol. Hey lights and we have beta protons. Now, if we identify the positions of those beta protons, we noticed that the leaving group is at the alpha carbon. So there is only one beta hygiene a disposition. And for the second structure we have two beta hydrogen surrounding the alpha carbon and disposition. So we have two beta hydrants over here. Now because the problem states that we want to synthesize a pure alkaline. We have to think about the mechanism of an E two reaction where we are using a strong base. So in this case we have two different beta hydrogen and we would form a mixture of two Elkins. We would remove this proton form a double bond at this position exactly what we need. But in addition to that product we would get another one which would be more substituted after removing this beta proton. So this structure doesn't work for us. But this one only has one beta hydrants. We may say this structure has one beta hydrogen and this structure has two different beta hydra jen's. So if we want to get a pure product, we are going with this one, we are going to choose one bromo butane is our first structure for a and that's perfectly fine. Let's see if that works for us. If we draw one bro mobile chain, we are going to d protein ate it at the beta position there is this hydrogen and we are using a strong base. Let's suppose. We just indicate as based with a negative charge and that base would attack the proton form the double bond and we would have the loss of the leaving group to form our final product, which is indeed but one in perfect. So now if we move on to part B and apply similar steps and similar logic, we want to synthesize two metal butte one in, we're going to draw a scene chain once again with a double bonded position number one In position # two has a method group. So whether the two possibilities to form al to form al kill highlights in this case and add H. B. R across the double bond. We may either go this way and add hygiene to position number one and brought into position number two or we may swap them and we may essentially get blooming at this position which is position number one and hae jin would be bonded to position number two. If we look at the blue pathway Essentially we get romaine at position one. So that's how the structure would look like There's also a methyl group at position two. So here it is. And now if we choose the green pathway we would synthesize another alcohol. Hey light. So that would be blooming at the same position as the methyl group. So it may indicate that there's brought me. Now let's find our beta hydrogen in the first structure. If we notice this is the alpha carbon, so we only have one beta hydrogen a disposition so one beta and in this case we have two equivalent metal groups. So they have exactly the same beta hydrogen, meaning that's one beta hydrogen. However, there's an adjacent one at this position. So there are two types of beta hydrogen. Again recall that these two methyl groups are equivalent to each other. So it doesn't matter which one you choose for your beta hydrogen, meaning there are only two different ones and we want to go with the structure that has only one unique beta hydrogen because that will give us a pure product. So this one works for us. We may choose this structure to be our substrate and now let's see if that works if we try to draw that is our structure, our starting alcohol highlight that would be Remain at position No one Position # two, We have a metal group and now a base would attack that beta hygiene. We already remember that. There it is this beta hygiene would be captured by the by the base. We would form a double bond at this position and simultaneously we would have a loss of the leaving group forming the alc in that we wanted to get. So that works for us. That's great. And now we may also solve part C. Applying the previously discussed steps. We want to get help three in, We made sure have three in part C. That would be a seven member chain and the double bond starts at position number three. So that's 12, three. So we have a double bonded disposition. What kind of alcohol hey lights can we actually form if we add hbr across the double bond, we have two pathways either add hydrogen broom in in this arrangement or swap them and add rooming here instead and hydrogen over here. So if you look at the blue pathway, We would get grooming at position number three. So that's blooming over here. And if we look at the green pathway instead, we would add roaming to position number four. So that's here. Now let's look at our beta hydra jin's. The first structure has two different data hydrants because that's the alpha position. So we may indicate the presence of these beta hydrogen. So there are two beta high genes in this position in the structure and for this one notice that there's actually a line of symmetry. This compound is symmetrical. So the two beta hydrants that we're seeing are actually equivalent. And there's only one type of beta hydrogen. That's very important to understand that. It doesn't really matter which one we removed. We will get the same compound in aipac naming you would just change the direction in which you label your carbon atoms for your parents but you will get the same product no matter which one you remove. So you may say that this one indeed has one beta hydrogen and it works for us. So we may now check if that structure undergoes the reaction to produce a pure product. If we redraw it brahman as at position number four and we may choose one of the beta hydrants to be removed. Suppose this one the base comes in attacks the protein is the structure and simultaneous that we have the loss of the leaving group to indeed produce our required product because notice the double bond will be at this position which is three heP team or heP three in well done. So now let's just summarize everything that we had to state. And this problem for a We have proposed a structure which is one bromo butane. We call that the molecule that we drew for it was one Bromo beating which looks like this for B. We proposed one bromo two methyl butane and the structure of it as you saw previously was this one We had obtained chain With bromine at position number one And a method group at position # two. And for C. The structure that we used was this one. So let's start with the name, that would be four bromo hep tane. We already see what the structure is here. It is. Now notice that we previously assumed that X. Was a halogen and we chose roaming for the purpose of this problem, but it doesn't matter. You may also choose chlorine or iodine. They're also good living groups. Just don't choose flooring because it's very bad leaving group actually. But here we get our three alcohol highlights for A B and C. Respectively. And the correct answer choice to this multiple choice question is C. If you have any questions, don't have states, leave your comment down below in the comment section and thank you for watching.

What halides would undergo E2 dehydrohalogenation to give the following pure alkenes?
a. hex-1-ene
b. isobutylene
c. pent-2-ene

Key Concepts

E2 Mechanism

Dehydrohalogenation

Alkene Stability and Substitution

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