Alkene halogenation

let's take a look at the halogenation of alkenes so on the Left I have my alkene and I'm going to add a halogen to it so something like bromine or chlorine and I can see that those two halogen atoms are going to add anti to each other and so they'll add on opposite sides of where the double bond used to be let's take a look at the mechanism so we can figure out why we get an anti addition so I start with my alkene down here and I'm going to show the halogen approaching that alkene so it's going to approach this way and I put in my lone pairs of electrons like that if I think about that halogen molecule I know that it's nonpolar because if I think about the electrons in the bond between my two halogens here both halogen atoms of course have the exact same electronegativity so neither one is pulling is pulling more strongly and so overall the molecule is nonpolar however if the pi electrons in my alkene so I'm going to say that these electrons right here are my PI electrons if those electrons get too close to the electrons in blue they would of course repel them since electrons repel each other since they're like charged and if the electrons in magenta repel the electrons in blue the electrons in blue would be forced closer to the top halogen like that giving the top halogen a partial negative charge and leaving the bottom halogen with a partial positive charge that's losing a little bit of electron density now I could think about that bottom halogen as acting like an electrophile because it wants electrons and so in this mechanism the PI electrons are going to function as a nucleophile and the PI electrons are going to attack my electrophile like that at the same time these electrons over here is electron pair on the left side of the halogen is going to attack this carbon and the electrons in blue are also going to kick off on to the top halogen like that so let's go ahead and draw the result of all those electrons moving around so now I have carbon singly bonded to another carbon like that the electrons in magenta formed a new bond between the carbon on the right and my halogen like that and these electrons over here I'm going to mark in red so this lone pair of electrons of my halogen are going to form a bond with the carbon on the left like that and that halogen still has two lone pairs of electrons on it so I'm gonna put in those two lone pairs of electrons and that halogen is a plus 1 formal charge this is called a cyclic halonium ion and it's been proven to occur in this mechanism if I think about that positively charged halogen hodgins are very electronegative and they want electron so the electrons in magenta let's say are going to be pulled a little bit closer to that halogen which would leave which would leave this carbon down here losing a little bit of electron density giving it a partial positive charge and so that's going to function as our electrophile on the next step and our nucleophile is going to be the halide anion created in the previous step right so we had a halogen that had three lone pairs of electrons around it it picked up the electrons in blue right so now it has four lone pairs of electrons eight total electrons giving a negative 1 formal charge meaning it can now function as a nucleophile so if I if I think about this cyclic halonium ion here the halogen on top is going to prevent the nucleophile from attacking from the top it's going to have to attack from below here so this negatively charged halide anion is going to nucleophilic attack this electrophile here this carbon and that's going to kick these electrons in magenta off on to this halogen here so let's go ahead and draw the results of that nucleophilic attack alright so now I'm going to have my two carbon still bonded to each other like that and the top halogen has has swung over here to the carbon on the left it used to have two lone pairs of electrons it picked up the electrons in magenta so that's what the carbon on the Left will look like the carbon on the right is still bonded to two other things and the halide anion had to add from below so now we're going to have this halogen down here like that and so now we understand why it's an anti addition of my two halogen atoms let's go ahead and do a react so we're going to start with cyclohexene as our reactant here and you're going to react cyclohexene with bromine so br-2 now if i think about if i think about the first step of the mechanism i know i'm going to form a cyclic halonium ion so i'm going to draw that ring and i'm going to show the formation of my cyclic halonium ion it's called a bromonium ion alright so I'm going to form a ring like this and the bromine is going to have two lone pairs of electrons it's going to have a plus 1 formal charge like that and also I know from my mechanism I'm going to form a bromide anion at the same time so I'm going to have a negatively charged bromide anion like that when I think about where that bromide anion is going to attack I know that it's going to attack one of these two carbons here so it could attack the one on the left it could attack the one on the right let's go ahead and start with the carbon on the right and draw the product so if a lone pair of electrons the bromide anion attack this carbon right here that would kick these electrons off onto the bromine and we could go ahead and draw the result of that so I would have my ring and the bromine on top is going to swing over to the to the carbon on the left here so now this bromine is going to go look like that and the bromide anion has added from below the plane of the ring like that so that's one possible product the bromide anion could also attack the the bromonium ion from the left side so this lone pair of electrons could attack this carbon which would kick these electrons off onto the bromine and so we could go ahead and draw the result of that nucleophilic attack so in this case the top bromine would swing over to the carbon on the right and it would pick up an extra lone pair of electrons and the bromide anion again added from below the plane of the ring like that now these two molecules are actually different molecules let's go ahead and redraw them so it's a little bit easier to see and we're going to stare down this way at the molecule this is the top of your head so let's go ahead and redraw that molecule so this is the same molecule as if we're looking down on it if we're looking down on it this this carbon right here would be this carbon and I can see there's a bromine coming out at me in space so I'm going to put a bromine coming out at me in space at that carbon I move to this carbon over here that's this carbon so there must be a bromine going away from me in space at that carbon so that's one possible product on the right we do the same thing all right so we're going to put our eye right here we're going to look down this is the top so I can look at this carbon first so I can see there's a bromine down at that carbon so I go ahead and draw my cyclohexane ring and at this carbon there's now a bromine down and of course at this carbon over here there's a bromine coming out at me so I represent that with a wedge and these are my products all right and if you've already had stereochemistry you know that these two products are enantiomers to each other there are actually different molecules there are non-superimposable mirror images so we can see that the absolute configurations have been reversed so if I think about this carbon right here bromine coming out at me bromine going away from me this one down here bromine going away from me and for this one a bromine coming out at me so that's the halogenation reaction