indepth explanation of ANTI ADDITION for RE - NEET 2026 180 / 180

Added: 2026-05-15

[00:00:01]Hello all my dear students, warm welcome to H and S in world of chemistry. As I've told you the mnemonic this mnemonic and its in-depth explanation I will let you know in the next video as I have asked in the comment section. First of all, this mnemonic was tam tushar car What is this?

[00:00:26]Something is wrong written over here.

[00:00:28]Let me write down.

[00:00:29]See some m m should be there. m It's m.

[00:00:35]Or you can go like mat roast race MSC.

[00:00:42]T means trans, A means anti, M means meso, R means racemic, S means syn.

[00:00:49]So for example, if any alkene is cis, any alkene is cis and syn addition is going to take place, the product will be meso.

[00:00:59]If any alkene is trans anti addition is going to take place, then it will be meso. If any alkene is cis, anti addition is going to take place, racemic will be the mixture. For example, Br2 Whenever you are going to add Br2 to alkene, always remember anti addition will take place. Now question arises, why anti? Why not syn?

[00:01:26]That mechanism I'm going to tell you.

[00:01:29]Now see.

[00:01:30]This is your alkene.

[00:01:32]It's very clearly visible that this is cis alkene. Why cis?

[00:01:37]Because both the methyls are on same side.

[00:01:41]Now this is nucleophile. This alkene is nucleophile. Why it is nucleophile?

[00:01:46]Because of the pi electron. It is electron rich, so it is an example of nucleophile.

[00:01:54]Br2 is electrophile. Now some people may ask why we have to as electrophile. This bromine is having lone pair of electrons, three lone pair of electron.

[00:02:05]This is also having three lone pair of electron. So, it is an example of electrophile. And to understand why it is an example of electrophile, we should know molecular orbital theory MOT approach. From there, you will get answer why bromine is it is electrophile. Now, first bromine first attack of bromine can be at this place, at this place, at this place, or at this place. There are four places.

[00:02:36]There are total how many places, kids?

[00:02:39]There are total four places where bromine can attack. My dear student, just understand there are total four places where bromine can attack. Bromine can attack at this place, at this place, at this place, or at this place. There are four places.

[00:02:56]Now, at any place, let me attack this bromine. For example, I have added this bromine from the top of this lobe.

[00:03:09]Pure P orbital, pure P orbital. One orbital one pure P orbital is made up of two lobes. One pure P orbital is made up of two lobes. So, there are total four lobes. And suppose, for example, bromine has attacked over there. From upside, something is getting attacked.

[00:03:25]This methyl was coming towards your side, wedge formula. That hydrogen is going away from you, dash formula. So, this will remain towards your side only. This will remain away from you only. But, they will come little bit downside. If somebody is adding box on my head, bromine box, so my nose was coming towards your side, so it will remain towards your side only, but I will go a little bit down.

[00:03:53]Because it has also to occupy space.

[00:03:56]Now, 109.

[00:03:58]Now, bromine was having lone pair.

[00:04:00]Remember, my dear students, whenever any electrophile is having lone pair, it will donate its lone pair to another adjacent carbon, and three-membered ring formation will take place.

[00:04:14]Three-membered ring formation will take place. This is known as non-classical carbocation. What? Non-classical carbocation. This non-classical carbocation is highly unstable.

[00:04:28]But, this bromine has donated its electron to this carbon, so this carbon's magnitude of charge got decreased.

[00:04:38]Otherwise, pure carbocation is having positive charge, one positive charge, but it is having partial positive charge, so non-classical carbocation, charge on carbon is very less, so no no rearrangement no reso no rearrangement will take place.

[00:04:59]No rearrangement will take place.

[00:05:01]Remember that. In this case, no rearrangement will take place.

[00:05:06]Remember that what's written over here.

[00:05:09]So, let me write down. No rearrangement will take place. What I'm saying?

[00:05:15]No rearrangement will take place. Let me write down.

[00:05:20]Due to this, no rearrangement no rearrangement will take place because the charge on carbon is very less, so never the rearrangement will take place.

[00:05:33]If five-membered ring is there, it will remain five-membered only. This is no rearrangement case. Remember that. Now, the another bromine will attack from the opposite of that because there is no space. So, this bromine will go and attack from opposite side of that bromine. And that's why it is known as anti addition. Now, you got what is the meaning of anti addition?

[00:06:03]Because this is three-membered ring.

[00:06:05]This bromide cannot attack from the same side as that bromide bromine is there.

[00:06:10]It has to go opposite side. If this would have been pure carbocation, then the pure carbocation is always having two lobes of pure p orbital. So, above the plane, below the plane, both side attack equal amount is going to take place. But here, backside attack, anti addition. And anti addition, you will get this bromine. Now, my dear students, according to my trick, this is cis alkene, cis Br2 is doing anti addition. Br2 will always do anti addition and Baeyer's reagent. What is Baeyer's reagent? 1% cold dilute alkaline KMnO4. Baeyer's reagent is 1% cold dilute alkaline KMnO4. What is the function of this? It will add two OH. And the previous video already I have mentioned. Now, according to the trick, cis anti, the product has to be racemic. This product has to be racemic. And for racemic, the one compound should be having RR configuration, another one should have SS. So, let's see. It has to come either RR or SS. So, according to CIP rule, let's apply CIP rule.

[00:07:28]Number this carbon is chiral. Let's see.

[00:07:31]Number one to this carbon, number four LPG least priority to this. And this carbon is attached with bromine, so this will get number two. This will get number three. So, the arrow is moving arrow is moving like this and this is your anti-clockwise direction, but LPG is away from you. So, this is your S configuration.

[00:07:56]You have to apply RS system according to CIP rule that you should know. Now, my dear student, let's see this.

[00:08:02]This has to come S. Let's see and let's check it out.

[00:08:07]One that carbon two, three, and four. Arrow is going anti-clockwise.

[00:08:16]And your LPG is away from you, so S configuration. If this is having SS configuration, so if this bromine would have attacked from there, then the anti addition would have taken from here. So, its mirror image will be RR. So, RR and SS, they are known as racemic mixture.

[00:08:38]Now, the same thing we can do for trans addition. The same thing could be done for trans addition. For trans, you have to make trans alkene. One bromine will come from upside, another bromine will come from downside. And the same thing you are supposed to do and you will get one carbon as R configuration and another carbon as S configuration. One is rotating the light in the clockwise direction, another will rotate with the same magnitude in anti-clockwise direction. So, meso because there will be plane of symmetry. See, there is no electricity, but still your Baba is making video for you kids.

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[00:10:12]Take care. And after this video the hydrocarbon long question sequence reactions question will come just after 1 hour.