So in an earlier video we’ve seen the mechanism for the addition reaction of hydrogen halides to conjugated iines in this video we’re gonna look at how to control that reaction and to figure out how to control that reaction we need to remind ourselves about the mechanism so in this mechanism i said that the pi electrons in this bond right are going to act as a
Base and pick up a proton from the hydrogen halide and so those pi electrons in magenta became these electrons right here and we added the proton onto carbon one since that gave us a carbocation at carbon two which is an allylic carbo cation that is resonance stabilized and so once we formed our allylic carbo cation right there were there were some possibilities
Here one possibility is the halide anion that was formed in the first step of the mechanism is going to function as a nucleophile here and nucleophilic attack the positively charged carbo cation at the to position to give us the 1/2 addict as our product so this was the 1/2 addict down here so the 1/2 addict like that the other possibility is that once i form my
First a little it carbo cation i know that that allylic carbocation is resonance stabilized right so the electrons in this pi bond are going to move over here to the right in my resonance structure and that takes away a bond from carbon number 4 so carbon number 4 is right here and this carbon lost a bond so this becomes my positively charged allylic carbo cation
And my mechanism and so now my halide anion is free to attack carbon 4 and i would get the halogen adding on to 4 and the proton adding on to 1 so this is the one for adduct so let me go ahead and make that blue here so we get the 1 4 addict this way so i have the 1 2 addict in the 1 4 addict and i can actually control which one of these 2 products is is the major
Organic product in my reaction and to do that we have to think about the stability of these alkenes that are produced so if i look at the 1 2 addict and i want to know how stable it is i have to think about the degree of substitution across my double bond here so it helps to draw on your hydrogen’s so you can see what’s going on and so if i had drawn my hydrogen’s
On that alkene i can see there are three hydrogen’s and then i can easily see there’s one r group on that alkene so it’s mono substituted all right so this alkyne on the left is mono substituted it has one r group on the right i’m looking at the molecule on the right here and it’s easy if i draw in my hydrogen’s right across my double bond i can see there are two
Hydrogen’s and so there are two r groups on this double bond and so this is a die substituted double bond so i have a die substituted alkene which i know from earlier videos is the more stable one right so this is this is more stable then the mono substituted once this one is less stable so the mono sub suit is less stable the die substitute is more stable and so
Under certain conditions the one four addict is going to be the major product and we call those conditions thermodynamic control so if i go ahead and write thermo dynamic control here all i’m talking about is a little extra heat so if we add some heat to this reaction right this reaction will be under thermodynamic control and the most stable product is going to
Win so in the mechanism the additional heat is going to give enough energy for the carbo cation on the left to change into the carbo count on the right and therefore the halide anion is going to attack at the four position and we will get the one for addict as our major product so if you heat the reaction up a little bit everything’s at equilibrium and you’re
Going to form your more stable product under thermodynamic control which is your one four addict if you want to make the one two addict you have to do something a little bit different you have to cool it down a little bit so if you cool down your reaction conditions right so if you cooling down to say zero degrees you’re going to form the one two addict as your
Major product under what’s called kinetic control so kinetic control from general chemistry refers to the rate of reaction so how fast the reaction occurs and the one to addict is actually the fastest product form and we can see why if we look at the mechanism so if we go back up here right in the first step of the mechanism right we formed our carbo-cation our
Allylic carbocation and we also formed our halide anion and we can see that our halide anion is going to be on the right side of our carbo cation right here’s our halide anion the right side of my carbo cation and therefore the halide anion is very close to a carbo cation at the two position and so this halide anion does not have very far to go to nucleophilic
Attack a carbocation in the two position so it’s very close to that and so that’s why the 1/2 addict forms the fastest because if i think about the 1/4 addict the halide anion is still on the right side of my carbo cation but this time it has a much longer way to go right the halide anion has a long way to go over here to the right and if there’s not enough energy
In the system right the halide anion isn’t going to get there so it’s all about it’s all about the proximity of the nucleophile to the electrophile and since in the example on the left the nucleophile is very close to the electrophile it’s going to add the fastest and so the one two addict product would form the fastest and so if you keep things cold there’s not
Enough energy to form the 1/4 addict on your one two addicts would be the major product under kinetic control let’s go ahead and do an example of how we can control the product in the addition reaction to a conjugate a dying so i’m gonna start with a conjugated dying here and i’m going to make it symmetrical okay so there’s my symmetrical conjugated dying and it’s
Going to be let’s let’s react it with hydrochloric acid so i have hcl right here like that and if i number my my conjugated doni in here its symmetrical so there’s really matter which one i make number one i’m just gonna say that’s 1 2 3 & 4 so we’re going to react that symmetrical conjugate dyeing with hydrochloric acid and we’re going to do it two ways alright
So the first reaction we’re going to keep things cold so we’re gonna keep things at zero degrees here and so when i think about the mechanism i know the first the first step of the mechanism is for hi electrons to act as a base and pick up a proton since this is a symmetrical conjugate dyeing it doesn’t matter which pi electrons i use i’m just gonna say the ones
Between 1 & 2 alright so those electrons are gonna function as a base and pick up this proton which kicks these electrons off onto chlorine form the chloride anion here so let’s go ahead and draw the resulting carbo cations so once again i have to think to myself where is that proton going to go is that proton going to go to carbon one or is that proton going
To go to carbon two and i know that the proton is going to go to carbon one because if the proton goes to carbon one that gives me a carbo cation at the two position which is now an allylic carbo cation right so that positively charged carbo cation right here is right next to my double bond so i can see that i have an allylic carbo cation here in the next step of
The mechanism right our chloride anion functions as a nucleophile so i have my chloride anion here with a negative one formal charge functions in my nucleophile and since things are cold right the chloride anion is very close to this allylic carbo cation and then attacks at the two position and let’s go ahead and draw the product for this reaction all right so if
I if i have the chloride anion attack at the two position all right so now my chloride anion adds on the two position there’s still a methyl group of the two position and i still had my double bond here and a methyl group here so this would be the one to addict right so this would be the one to addict which is the major product under kinetic control right because
I we cooled things down this reaction so we’re gonna get the one two addict as the major one if i did this reaction and again but i changed the reaction conditions all right so we’re going to once again react this conjugated dying with hydrochloric acid but this time we’re gonna heat things up this time everything is under thermodynamic control so the first step
Of the mechanism is exactly the same right the the pi electrons function as a base pick up a proton and the carbo cation that results is going to be the allylic 1 so therefore the proton adds on carbon one and the carbo-cation is now at carbon two this time is that carbon two this time just like before we have our double bond and our methyl group like that so now
Since we know this under thermodynamic control we know that we’re going to get a resonance structure here all right so we’re going to go ahead and draw the resonance structure for this guy and we know that these pi electrons are going to move in here and i will get my resonance structure so if i move those pi electrons in there right now i have my methyl groups and
My alkyne and i took a bond away from carbon four right so if i number these again is b one two three and four so i took a bond away from carbon four so that’s where my carbo cation is going to go now so my carbo cation is at carbon four go ahead and finish my resonance brackets here and so my chloride anion is now going to function as a base right and it was it
Was actually up here to start with so let’s go ahead and put our chloride anion up here and it has a longer way to go this time but it doesn’t matter because now we’re under thermodynamic control and so the chloride anion is going to add on to the four position and therefore our major product right will be the one four addict all right so i’m gonna go ahead and
Draw my ring here and i know that i formed an alkene right here and i know that the chlorine is going to add on to carbon four like that and so this is going to be my major product and of course in the thermodynamic control your most substituted alkene is going to be your major product this is a tetra substituted alkene so let me just go ahead right here one four
Addict because we add a proton to one and our halogen to four for our one four addict which is more stable than the one two addict if you heat things up and so we’ve seen how you can control the addition reaction of conjugated iines simply by changing the temperature
Transcribed from video
01 Addition reactions of conjugated dienes 03 Addition reaction of conjugated dienes III Cont By L Hub
