 can you see this hello can you see this now tell me yeah there i've shared just now gargi okay i don't know some problem is there this part and all some fluctuation is going on i don't know what it is and let them join just a second okay so uh whether they have seen the link guys you there can we start here right okay yeah so you see uh so we were talking about this mixture of these two product when when i passes through ppl from this because one product is is has tendency to rotate this ppl clockwise other one will rotate by the same angle anticlockwise right so finally we'll get the ppl undeviated ppl you'll get okay pain polarized right okay so when the mixture is not deviating the pale plane polarized light at all it means what this mixture is optically inactive this mixture as a whole is optically inactive correct what about this product a is optically active or not or b is optically active or not what about a and b whether it is optically active or not so individually when you see individually when you see the molecule is optically active but as a whole as a whole will get optically inactive product okay so this is very important you know properties of this s1 reaction we have to sum up all this okay we have discussed everything into this s1 reaction takes place in two step first step is the slowest step formation of carbocation rearrangement of carbocation also possible to get the more stable carbocation when we have when we have more stable carbocation but then the nucleophile will attack and will get the product okay now the studio aspect of this reaction is what you will get two different product into this one will retain the same configuration and other will have the opposite configuration exactly okay what is the doubt Sushant tell me Sushant tell me the doubt see why why the whole thing is optically inactive you understood this that a b are optically active right so we have equimoles of a and b suppose what happened happens a rotates the plane polarized light by 10 degree clockwise okay b rotates the plane polarized light by the same angle 10 degree anti-clockwise okay so as a whole if you miss this if it is plus 10 that this one is minus 10 so total it is zero so there will be no deflection into the plane polarized light what happens here you see if you allow this plane polarized light to pass through suppose a deflects like this plane polarized light by some angle with the same angle b also deflects this like this before leaving out this you know mixture and finally the mixture goes out by this direction direction finally it is not changing correct that is why it is optically inactive so the mixture is optically inactive but the individual product is optically active correct now in these two since it is an organic ms2 so ideal condition we are assuming then we get equal moles of this but if you if I ask you which one of these is predominant a and b which one of these is predominates in this reaction in this product yes yes shravan if there is no deflection plane polarized light it is optically inactive did you understand right which one is predominant a and b tell me why a gargi tell me the reason why a less crowding where is the less crowding the relative position you see vrat relative position is same only where is less crowding vrat tell me see the point is from this side the living group is going out right from this side and oh minus at the same time is trying to attack onto this positive charge carbon atom okay so obviously if you compare two different possibilities the side from which the x minus is going out from that side the attack of which minus is slightly difficult because of the hindrance between these two right so backside attack will be preferred that's why the option a is predominance so if they ask you you know this thing equi moles is practically it should be right but actually if you see you know the what if they ask you what is the predominance product then the one which the one which is the inverted product will be the more stable product we have because oh minus will attack from the opposite side attack will be preferred for OH minus correct so a will be predominant right if they ask you the major product into this but if they did not ask you the major product molecule as a whole we consider is optically inactive because you'll get equi moles of a and b correct did you understand this can we move on can we move on one more term in optical isomerism that we use the equimolar mixture of a and b this kind of mixture we call it as resmic mixture just you write down this resmic mixture we will discuss this again in optical isomerism the equimolar mixture of two optically active molecule right two optically active molecule remember one thing I'm giving you two optically active molecule if I replace this OH by chlorine also here okay if I replace this OH by chlorine also then this molecule also will be optically active but we cannot mix these two equally the point is the two molecule of opposite configuration with same molecular formula if mixed in equimolar mixture if mixed in equi moles then the mixture is said to be resmic mixture correct so this mixture is what this mixture is resmic mixture a plus b equimolar mixture of two opposite configuration having equal having same molecular formula are said to be resmic mixture so resmic mixture is optically inactive next the second reaction we have that is bimolecular nucleophilic substitution reaction bimolecular nucleophilic substitution reaction and in short we write it as sn2 two stands for that the order is two into this in the first one the order is one and here the order is two n stands for nucleophilic as is substitution and two stands for order is two second order it is now in this you see in sn2 the reaction proceeds write down sn2 the reaction proceeds in one step in one step the first one there are two step reaction here it is only one step in one step via a transition state transition step next line the nucleophile attacks the substrate carbon the nucleophile substrate carbon simultaneously pushing out simultaneously pushing out the leaving group pushing out the leaving group okay in this reaction if i write down rx plus nu minus so this nucleophile will attack the alkyl group from the backside backside attack will be preferred over here so we will get a transition state here where this nucleophile is trying to make bond with the alkyl group so they were partial bond here and at the same time it is pushing the leaving group out so we have x del minus and del minus right so this means what this bond is about to form and this bond is about to break so this is transition state ts okay and finally what happens this rx bond dissociates or you know it breaks down and the product we get is r nu plus x minus now you see that product is same we are getting here in the previous reaction also the product was this and here also the product was this but the mechanism is different what is different difference we have here first of all this reaction is single step reaction right it is single step reaction unlike uh sn1 reaction here we do not have any carbocation present okay so in this what we can write it is a single step reaction no carbocation formation no carbocation formation if carbocation does not form it means rearrangement also not possible no carbocation so no rearrangement rearrangement not possible why because carbocation itself is not forming if it is not forming how it will rearrange okay so rearrangement of carbocation is not possible the reaction rate depends upon both this substrate and reagent so rate of the reaction if i write rate will be equals to some constant k concentration of rx and concentration of nucleophile that's why it is second order reaction second order reaction understood now in this last point second order reaction see the rate again it is for chemical genetics we have rate is depending upon rx and nucleophile okay so that's why it is second order reaction we will not count we will not see the number of concentration term here that is not order but we'll see what we'll see in the rate determining step in the rate sorry in the rate determining step if you write it down the rate law expression here so whatever the power we have here that power will add so one plus one becomes two so two is a second order reaction okay so some of the power over here in the rate expression gives you the order of the reaction why bi-molecular because two molecules are involved over here and that remaining step we have two molecules but these two represents order not the molecularity see order and molecularity are two different term molecularity is the number of molecules involved in the rate determining step or number of molecules present in the rate law expression okay rate law expression is nothing but this rate is directly proportional to the concentration of this and this some constant k if included becomes equals to so number of molecules present in the rate law expression is nothing but the molecularity but in the rate law expression the sum of the power of the concentration term in rate law expression is nothing but the order of the reaction you don't worry with disaster you have to keep in mind so if somebody asks you what is two then two is the order of the reaction okay order and molecularity we'll discuss in chemical kinetics unimolecularly you see the last reaction the rate expression I have written this r is equals to k concentration of rx only one molecule however there we have reaction is this that's why I'm telling you don't bother with this the reaction was this only but the rate expression for sn1 was this here the rate expression is this okay so you don't worry with this just you keep in mind that what is this two and what is n in sn1 what is one in sn1 okay what is order why we have rate law this everything is there in chemical kinetics that we'll discuss in kinetics only clear done okay then in this you write down you must keep take care of this in mind that since carbocation does not form rearrangement is not possible okay but in this I'll write down one reaction here rearrangement is possible rearrangement is possible here but this rearrangement rearrangement so suppose we have this molecule ch2 double bond ch single bond ch2 br now when you have this nucleophile that will attack onto this carbon atom and this pi electron shift over here and this br comes goes out as the leaving group one step simultaneously okay so finally we get what we'll get ch2 double bond ch single bond ch2 and u this is the product we get one more mechanism I'll write down here this is first possibility otherwise what ch2 double bond ch single bond ch2 br and this we have nucleophile and this nucleophile suppose attacks over here and this br minus goes out so you'll get what ch2 double bond ch single bond ch2 and u suppose this reaction is sn2 dash mechanism we have and this is sn2 so we have two different types of reaction it goes however the alkene here is symmetrical so this won't make any difference in these two product but in unsymmetrical case but in unsymmetrical case I'm discussing this unsymmetrical case you may have different products just first you write down here one note then we'll discuss this write down sn2 dash mechanism the first one this one sn2 dash this mechanism sn2 dash mechanism leads to the same product as sn2 leads to the same product what happened need engine tell me yeah yeah that's what I'm telling you okay fine fine I understood I understood okay so we are getting same product here so sn2 dash mechanism leads to the same product as sn2 right however if the allylic halide is unsymmetrically substituted however if the allylic halide is unsymmetrically substituted substituted then the product formed then the product formed will be formed will be from the nucleophilic attack will be from the nucleophilic attack the less hindered end product formed will be from the nucleophilic attack at the less hindered end of the allylic system less hindered end Now you see two examples on this particular, you know, concept. Suppose I will give you this reaction. Here we have bromine, double bond, methyl. This reacts with, suppose any nucleophile we have NU minus, I will write down. And another product is this, again here we have nucleophile. What is the product we get? According to the point that I have given you, tell me the product here. Sn2 favored. Yes, we are talking about Sn2 only. Sn2 or Sn2 dash, that's what I want to ask. Sn2 first reaction, Sn2 dash, second reaction. Okay, you see here in this particular question, the answer will be same, right? Answer like the mechanism will be same. Here also it is Sn2. And here also it is Sn2. So that will be the nucleophile here, this one Sn2, right? So nucleophile here will attack on to this carbon atom and this bromine minus will go out and the product we will get here is this double bond and here we have CS3. Here also we will get the same thing. This nucleophile will attack over here and we will get the product, right? But when the base is something else, when the nucleophile is large base, in this case you see, this is the product we have. But when the nucleophile is this derivative of amine we are taking here ET ethyl whole twice. This is diethyl amine we have. So when you have large base like this or amines we have like this, then the product follows Sn2 dash mechanism in case of large base or amine related base. This you must remember, okay? So what happens here, this lone pair will attack on to this, sorry I will use other color. This lone pair will attack on to this double bonded carbon atom and this pi bond will shift here and this nucleophile will go out. So the product here we get is this and here this will attach with NH ET twice with positive charge onto this. Finally deep protonation takes place. Deep protonation means this H plus goes out and the product will be this ET whole twice double bonded. This is a double bond. This is the product we get. So what you have to keep in mind in presence of large nucleophile or large base like this amine thing you will have Sn2 dash allylic substitution like this it goes. Means the nucleophile will attack on to the what? On to less substituted carbon level bonded carbon atom. Correct this you must remember. Can we move on? Correct Bharat. That is the case of large base. You must have studied about Hoffman, Sedgef product and Hoffman product. So Hoffman product is also possible in case of large base if you remember. HSR guys, HSR Kurmangala guys you have done Sedgef and Hoffman. That if you remember there we have discussed that Hoffman product possible in case of large base. See you don't have to understand this. The point is you have to memorize this. If you have large base like this there's no logic it's experimental fact we have. In case of large base large base or large nucleophile will prefer to attack the less substituted double bonded carbon atom. Now less substituted means what? If I write down H2C double bond CH2 this is alkene it's not substituted alkene. But if I write down this CLCH double bond CH2. So this is substituted alkene. Now in this two if I ask you which carbon is more substituted this carbon is more substituted right. So here what happens you see CH2 double bond C this is the substitution we have. So when you have large base like this then the attack of this base will take on the less substituted double bonded carbon atom. Did you understand now Sushant Niranjan? Correct. Now the another question that you can answer I guess that like in the first case we have discussed the stereochemistry of SN1. So in this stereochemistry of SN2 what happens inversion or retention? What happens inversion or retention? Because the attack of nucleophile takes place from the backside always. Correct. Attack of nucleophile takes place from the backside always then it will always then it will always gives you complete inversion okay. There is no retention of configuration over here okay. It's not retention Sriram because nucleophile I'll write down this reaction you see. Suppose you have nucleophile here nu negative charge plus Cx with this we have 1 r double dash and then we have saline we have in the same plane r triple dash okay. Now in this what happens the nucleophile will try to attach with this carbon atom here and the same time this x is trying to go out. So we'll get what we'll get a transition state where we have nu del minus and this nu is trying to attach with this carbon and this bond is about to break x del minus right and here we have the carbon atom the all other alkyl group r double dash. So this bond is about to break and this bond is about to fall. So finally what happens the nucleophile attached with this carbon atom from the backside. So you see the nucleophile attacks from the backside. So if this has a particular configuration the configuration of this will be completely reversed okay. So this is what inversion right. There is no retention of configuration complete inversion takes place into this complete inversion takes place. This process we call it as Walden inversion. This is Walden inversion large base means Sn2 no it's not that large base means not Sn2 large nucleophile means what the attack of nucleophile takes place at the more substituted or bonded carbon atom. See like this if you think you will get confused okay. What I told you that if you have amines and derivative of amines will consider that as a large base that is the only thing you keep in mind okay. Amines and derivatives are considered as large base large nucleophile. Now you see for Sn2 reaction the reactivity order of methyl is maximum methyl is maximum methyl halide then we have 1 degree then we have 2 degree and then we have 3 degree. For Sn1 mechanism if you have tertiary alkyl halide it will be maximum then 2 and then 1 and then 6 3x. Why is it so because if you have tertiary halide then tertiary halide gives you tertiary carbocation and tertiary carbocation is more stable right. So if if the molecule has tendency to form more stable carbocation the reaction will go under will go under what will go under Sn1 mechanism because Sn1 mechanism favours formation of carbocation right but when we have methyl halide so methyl carbocation is not that much stable okay it's it's not stable actually. So when a stable carbocation possibility is not there then the reaction follows Sn2 mechanism order of reactivity according to Sn1 and Sn2 order of reactivity of alkyl halide according to Sn1 and Sn2 okay next one you see addition reaction the second one the second type substitution we have done right on this kind of reaction is possible addition reaction this kind of reaction is possible when the compound contains multiple bond this kind of reaction is possible when the compound contains multiple bond like in bracket it I don't in case of alkene or alkyl right down pi bond next line pi bond is comparatively weaker pi bond is comparatively weaker and hence it is easy to break and hence it is easy to break and shows hence it is easy to break sorry easy to break in the course of addition reaction okay generally the addition reaction shows like this you see if you have an alkene correct C double bond C with bromine Br2 the product we get here is what this pi bond dissociates and we get carbon single bond carbon bromine and bromine this is what the addition product we have if you have any alkyne C triple bond C plus Br2 you'll get C double bond C oh sorry we have Br Br and then plus again if you add one more bromine molecule you'll get carbon single bond carbon bromine bromine bromine and single bond right so this is how the reaction takes place but you know there are addition reaction this is the simple general thing I have given you but additional reaction are also of different types okay like we have addition of electrophile okay so addition of electrophiles are possible addition of nucleophiles also there right so an additional reaction also there are different different categories electrophilic addition nucleophilic addition and all right so three different types of reaction we have that you write down these are of three different types three different types the first one we have is electrophilic addition reaction addition reaction what do you understand by this electrophilic addition reaction addition of electrophiles addition of electrophiles yes so how this addition of electrophiles takes place okay so you write down into this the compound containing multiple bonds the compound containing multiple bonds write down yes electrophiles addition of electrophiles the compound containing multiple bond means alkene or alkyne decolorize a solution of bromine decolorize a solution of bromine in a suitable solvent in a suitable solvent like carbon tetrachloride CCl4 okay these properties also you have to keep in mind that's why I have given this right decolorize a solution of bromine in a suitable solvent like carbon tetrachloride this decolorization this decolorization is due to the formation of is due to the formation of addition product addition product the reaction and the reaction is known as addition reaction okay actually you see what happens in this kind of reaction when you have alkene and suppose I am taking the simplest one that is you know ethene when this combines with or when when this comes closer to a bromine molecule vr vr okay so what happens here because of this pi electron which is the you know it's not static right the electron cloud is like this above and down this carbon carbon single bond right so because of this pi electron this will polarize this bromine molecule this pi electron will polarize this bromine molecule and hence it forms like this CH2 and this will polarize the bromine molecule vr vr and finally a dipole moment exist start dipole moment is there in this molecule because of this polarization so here we have slightly positive charge and slightly negative charge what happens one of the bromine molecule when comes closer to this pi bond because of this pi bond the bond pair of electron is shifted towards the another bromine atom so this side it becomes partially positive and this side it becomes partially negative so because of this separation of charge will have dipole moment here also in bromine molecule and finally what happens the positive end combines with the negative end of this combines with the positive end of this bromine so we'll get CH2 CH2 positive charge onto this single bond CH2 vr right and the other bromine molecule behaves as a nucleophile now because this will taken up by this bromine molecule here so this molecule or this nucleophile finally attacks onto this positive charge carbon atom and we get this so this is nothing but the nucleophilic attack nucleophilic attack okay this compound we call it as pi complex pi complex and this pi complex converts into a what we say bromide ion vr minus and then one carbocation okay now you may have this doubt so when this nucleophile attacks onto this carbocation then why it is electrophilic addition right can you tell me why understood what is the answer shouldn't be both no actually you see this reaction is initiated by this reaction is initiated by positive end of bromine right this is the positive end of bromine we have right this behaves as an electrophile here and that's why it is electrophilic addition reaction okay since write down since the reaction is initiated with the positive end of bromine hence it is known as hence it is known as electrophilic addition reaction yes correct that's why the electrophilic addition reaction next you write down it has been observed next point this is important not on this point it has been observed that it has been observed that that the alkene undergoes alkene undergoes trans addition product means the trans trans addition product is the major product here trans addition product and this reaction is followed by see all these points and I'm giving you all these points are important okay and this reaction is followed by the formation of followed by the formation of a cyclic intermediate cyclic intermediate cyclic intermediate in bracket you write down bromonium ion bromonium ion how this cyclic intermediate forms see and how do we get the trans product also because till now the mechanism that I have given you from that cis or trans you cannot say you see the previous slide that we have in this molecule but randomly I have given this okay you do not know whether it is cis or trans I have written this like this you may say this is this product but this is the product I have given you whether it is cis or trans we haven't discussed okay from this mechanism you cannot say but when you look at this one how this you know cyclic intermediate forms so the intermediate carbocation we have here is this ch2 br and this bromine has three lone pairs and here we have ch2 positive charge unit sorry ch2 positive charge unit this carbon is what this carbon is electron deficient only six electron it has plus one positive charge on it correct first point which one this one what this one tell me shruti see what I said that this the alkene that you have pi electron when this bromine comes closer to this you know pi bond then because of this weak bond that you have since pi electron is delocalized like this it is continuously moving so because of this moving electrons this bromine molecule gets polarized okay so what happens one end of the bromine will get positive charge and it is closer to the pi bond and similarly the next end or the other end will get negative charge on it so like this the polarization takes place and we'll have a dipole moment over here because of the charge separation right finally this pi bond will attack onto this br positive ions and we'll get this intermediate okay the bond will dissociate one carbon will get positive charge and other carbon will get this bromine positive ion right so we'll get this ch2 positive ch2 br and one of the bromine ion br- will go out okay now this br- will behave as a nucleophile and attacks onto this positive charge carbon and get this okay this is pi complex we have right and since the reaction is initiated with the positive charge bromine atom that's why it is an electrophilic addition reaction correct so this positive charge bromine atom electrophilic addition reaction now you see here we are discussing what that like I said that the reaction proceeds by the formation of a cyclic intermediate and that we call it as bromonium ion okay before that I said what the major product is trans here means the addition of bromine takes place from the opposite side a two bromine atom comes from the opposite side to attach onto the carbon atom so it is a trans product which is dominating how this cyclic intermediate we get here this negative or this lone pair that we have this lone pair attacks onto the positive charge carbon atom since it is electron deficient this lone pair attacks onto this positive charge carbon atom and we'll get a cyclic excuse me and we get a cyclic intermediate like this which is ch2 ch2 bromine and bromine right this bromine atom now we'll have what charge we'll have on this on this bromine atom tell me what charge we have on this bromine atom plus one why it is plus one because one of the lone pair it donates to this carbon atom so we'll have positive charge onto this bromine atom so this is the cyclic intermediate now you see when we have this cyclic intermediate so again there are two possibilities what is that possibilities one possibility is what that this bromine nucleophile that you have br- this br- one possibility is what it may attack from this side onto this carbon atom and this bromine will take this bond pair of electron okay so in this case you'll get what you'll get this product ch2 br and here we have ch2 br this is cis product we have cis addition another possibility is what this bromine br- this nucleophile this nucleophile will attack from the backside onto this carbon atom and this bromine will take this bond pair of electron okay so in this case we'll get this which is ch2 ch2 br and br so this is the trans product we get the trans is more stable here so the question is what is the product we get then the answer will be this this will be the major product this will be the minor product understood backside attack and it is more stable also trans is always more stable okay in organic chemistry always we have whenever we have less in steric hindrance that path will follow okay always one note you write down we will not explain discuss this here again this is about optical isomerism we'll discuss it over there but just one note you copy it down here bromination of trans 2-butene write down bromination of trans 2-butene gives meso product bromination of trans 2-butene gives meso product bromination of cis 2-butene gives resmic mixture correct meso product will have the plane of symmetry bromination of cis 2-butene gives resmic mixture resmic mixture in bracket you write down pair of enantiomers enantiomers E N A N T I O M E R S these enantiomers and meso compounds will discuss later on in optical isomerism okay you just write it down in your notes yeah just a second meso compound is this M E S O meso enantiomers E N A N T I O M E R S we will discuss that synod anti breath next you write down addition to unsymmetrical alkene next you write down addition to unsymmetrical alkene write down see if you have this reaction if I write down this is the addition of HBR we are talking about this reaction is there in alkene chapter also okay hydrocarbon alkene chapter so we will not discuss in detail over there we will finish this addition of HBR this is the last part we are doing today so you see suppose if I give you one example of symmetrical first and then we will see unsymmetrical I think that would be better gives you you know better understanding CH3 CH double bond CH CH3 this is but 2 in plus HBR what is the product we get here in this H plus and BR minus will attach onto this double bonded carbon atom the product will be CH3 CH2 CH CH3 BR so this is a symmetrical alkene if you see symmetrical alkene and in this obviously we know we have H plus and BR minus right so whether you add this H plus on this carbon on this carbon that won't make any difference you can see I have added on this carbon but if you add this H plus over here also and BR minus on this carbon then also the product will be safe right the symmetrical alkene gives you only one product that's the point I'm trying to make okay now what happens if both reagent and the alkene are unsymmetrical what the topic we have unsymmetrical alkene I'm taking suppose CH3 CH double bond CH2 this one reacts with HBR so alkene is unsymmetrical reagent is also unsymmetrical when I take BR2 as the reagent this is symmetrical reagent HBR is unsymmetrical reagent so here the reagent and substrate both are unsymmetrical yes correct both are unsymmetrical so in this what happens there are two product possible one is what when H plus you add on this carbon and the other one when H plus you add on this terminal carbon the two product possible here I'll write down the both product here that is CH3 CH CH3 BR this is one possible product another one is CH3 CH2 CH2 BR this is the two product we get the point is which one is the major product here right this addition of HX or HBR right addition of write down this point note down this point addition of HX hydrogen halide general thing I'm writing down okay follows follows marconi cough rule marconi cough rule now what is marconi cough rule write it down marconi cough rule write down according to this rule according to this rule the negative end of the reagent the negative end of the reagent attached to the second one is minor it's correct pretty it's right second one is minor first one is major so the marconi cough rule write it down according to this rule the negative end of the reagent attached to the carbon atom containing lesser number of hydrogen lesser number of hydrogen okay less carbon atom what carbon atom carbon atom means doubly bonded carbon atom these two carbon atom these two carbon atom and negative part is BR minus this is H plus so negative part of the reagent that is BR minus BR minus will attach to the carbon atom which has lesser number of hydrogen atoms so here we have two and one so this product is the major product and this one is the minor product this is marconi cough rule second point to write down second point in presence of in presence of any peroxide in presence of any peroxide or light addition of addition of HBR follows anti marconi cough addition remember this point okay here I have written HX general thing here I have written HBR okay so this whenever you whenever you write down this reaction in presence of means this will be the major product when we have the peroxide present over here right so I'll just write down here if I this this particular color you just see if I write down here peroxide means this reaction is taking place in presence of peroxide so when the peroxide is there then the major product will be this one the lower one will be the major product and the upper one will be the minor product and this anti marconi cough will just reverse of marconi cough okay and this is only true in case of HBR okay so write down the peroxide effect or we also call it as peroxide effect or we also call it as kharash effect k-h-a-r-a-s-c-h okay so peroxide or kharash effect we have this is only true for HBR not for not for HCL, HI, HF for all these anti marconi cough rule is not true just for HBR yes anti marconi cough addition is true only in case of HBR that too when we have peroxide or light present peroxide or h-new is it clear did you understand this any doubt you have any one of you correct so we'll wind up the class here only fine on friday you will have the class the same class will continue from here nucleophilic addition we'll see okay so we'll continue the class from here on friday i will send you the schedule friday schedule i will send you okay tomorrow okay we'll continue from this fine okay so all of you take care bye bye