Skip to content

Addition of X2 to Symmetrical Conjugated Dienes

  • by

Hi all in this video i want to talk about the addition of x2 to symmetrical conjugated dienes x2 could be either br2 or cl2 in the previous video i talked about adding hx hbr or hcl i also discussed how to figure out the major product by analyzing whether we want the kinetic versus the thermodynamic but also determining which one is the kinetic is it the one

Two or the one four and which one is the thermodynamic is it the one two or the one four product that video seen here uh that’s where i discuss those terms i wasn’t too pleased with how i presented them so here in this video i just wrote it out and we could just read off of the slide what we want to remember or what our little voice wants to be saying to us

While we take the exam let me show you actually on the document camera these three very important statements the first thing to know or to remember whether you’re adding hx or x2 is that below zero degrees celsius the kinetic product is favored at higher temperatures the thermodynamic product is favored now the kinetic is the one that’s made faster so this is

For my classes it’s going to be always true the 1 2 product is always the kinetic it’s always the one that’s made faster now the tricky part is the thermodynamic product the one that’s actually more stable it could still be the one two product but it could also be the one four product so again the one two product is always kinetic so kinetic is easy to find it’s

The one two product the thermodynamic you wanna analyze the stability of the products double bond and it could either be the one two or the one four now obviously if the one two and the one four are identical then you know it’s both um you don’t really have a one two or one four product you have uh both pathways giving one product and we saw that actually in a

Previous video where uh two the potential products were actually identical but let’s focus on this and let’s take our first reaction oh actually let me remind you what happens if we only have one double bond so for instance if i have this alkene and i’ll make it yeah i’ll make it cyclic when you add b or two so this is from organic chemistry one when we learned

About alkenes uh we give a chlorine chlorinated solvent ccl4 okay that’s because if we run it under water we get the halohydrin and normally there was no temperature when we talked about this in organic chemistry one but you get the dibromide and it’s not necessary for uh what we’re discussing now but if you recall it was anti-edition anti-edition but now we

Have a diene okay so we are going to add two bromines just like we did here on one of the double bonds but we also have to think about the resonance uh of the intermediate now i’m not going to go through the mechanism in this particular video i want to show you how quickly to assess what the products are and how to pick out the major product oh i didn’t write

It here but the problem here is draw all the products and select the one major product so let me recopy this we have br2 carbon tetrachloride as a solvent and this is important we’re running this under very cold conditions negative 78 degrees c so right off the bat we’re going to say we’re going to choose the kinetic product so the temperature tells me right

Whether we want the kinetic or the thermodynamic as the major product but the second statement says oh we’re done we know at least we’re done determining whether it’s the 1 2 or the 1 4 the 1 2 is always the kinetic so if i know i’m going for the kinetic the one that’s made faster then for sure i’m going to pick the 1 2 product now these dienes both of them i said

They were symmetrical and that means whether you choose double bond a or double bond b you will get the same set of products so i’ll just choose double bond a and the one two right is one and two one two and you’re adding the bromine on those two carbons you know this is a little bit easier to visualize because uh more easy the visualize than adding hbr right

Because with hbr you have to remember that you’re making the more stable carbocation by adding the h in this case on the end to put the carbocation here because you’re adding two things you’re adding an h and the br but in this case we’re adding two brs so um we don’t have uh regio isomers right and this is our first product and yes this is the one two product

And it is definitely the kinetic okay now how do you draw the one four product okay i don’t know if it’s a thermodynamic or kinetic but i do know that i have 1 2 and i should have a 1 4 product as well if you think about it and label this is the one two three four what happens is again i’m not doing the mechanism but you shift the double bond okay over and

You have this we’re still going to get the one having a bromine and now it’s the four right it’s the four that gets the other bromine so this is one two three four this is the one four product and now we have to assess whether this is uh the thermodynamic or not look at the double bond this is disubstituted versus mono so look at the third statement the thermodynamic

Or more stable product could be the one two or the one four we are determining this by the stability of the double bond okay you could add that if you want for these type of reactions with the conjugated dienes this thermodynamic is determined by the stability of the double bond so this is the thermodynamic okay these two are not identical so we’ll draw both

The thermodynamic and our major product is going to be oh low temperature below zero is going to be the kinetic so this is the major okay so again the temperature is telling you whether to go for the kinetic or the thermodynamic as the major product you typically are drawing two products if you want to draw everything you’re drawing the one two and the one

Four it doesn’t matter in this case whether we choose double bond a or double bond b okay because they would give the same set of products and for my class i don’t care whether you draw the double bond as sister trans i’m more interested whether you could draw the 1 2 and the 1 4 and determine which one is the thermodynamic there’s no decision for the kinetic

It’s always the one two okay for the second problem we’re above zero degrees so here we’re aiming for the thermodynamic let’s see what we get when we react this dying again it is symmetrical chlorine uh the chlorinated solvent here it’s just random but i’ll choose chloroform what’s important is the temperature and our initial analysis or our first thing that

We should determine is because of the temperature we want the thermodynamic product now i don’t know right i don’t know if the thermodynamic is the one two or the one four it could be either i have to wait till i draw the products it doesn’t matter whether i use double bond a or b so i’m going to use double bond a and if i use double bond a we’re adding two

Chlorines here and here that is the one two product and by a default or automatically this is the kinetic now the 1 4 product is again if you need to label uh your carbons you could label it as such one two three four okay we’re still working with double bond a but that double bond a could have resonance and remember how we did it up here you shift the

Double bond towards um the two like right here we we shifted it two to two so we have a new double bond here carbon number one still gets something and then carbon over four right gets the other bromine so we we are doing or we are drawing the one four product this is two in both cases right the one two and one four carbon number one gets something it gets a

Cl here and carbon number four gets the cl so we have that product right there my goodness this is the one for product and now we have to determine the thermodynamic product because you are making both of these you know we made both of them appear but we have a major and a minor do we have a major and minor over here i didn’t quite plan this out do you see

The issue here the thermodynamic is based off of the stability of the double bond but both of them are disubstituted so if i had to choose oh so first of all if i gave this on an exam um i would give them an option to say well both of them are equally stable so both of them could be considered the thermodynamic and they would circle i i would allow students

To circle both okay so uh we could do it that way um but if i’m gonna force myself to figure out which one’s more stable i might probably just go with um steric hinderance and because these two chlorines are bumping into each other right they’re bigger than hydrogens i would rather have them more spread out both double bonds are disubstituted so i can’t use the

Degree of substitution as a decider i would say this is the thermodynamic so it does happen to be the 1 4 and i would maybe say slight major or i’ll just put major here okay now like i said because they’re both disubstituted i would write here i would also accept that both are the major product or you can even say that both are the thermodynamic product and

Ultimately if you say it that way you’re claiming that they’re equally stable and based off of our knowledge of our rules of stability of double bonds they are equally stable they’re both what they’re both substituted but they’re also both cis right this is a cis double bond same as that so again um either circle both or right down somewhere they’re equally

Stable hence there is no one thermodynamic product or like i did here i use steric hindrance as sort of a tiebreaker now i think i’m happier now uh i think i just had to write it out so i got the wording the way i liked it but again these are the three phrases or the three things that the little voice should be saying in your head when you’re working out not

Only you know addition of x2 but also the addition of hx these three points

Transcribed from video
Addition of X2 to Symmetrical Conjugated Dienes By fieldguide2chemistry