 So, far we have discussed the application of IFF and in that we have discussed the stability of carbocation ok. Now, the second intermediate we have that is free daddy. What is free daddy? Can you see? Again if I write down this example, free bond, bond pair and it has only one electron present in this T-orbitor ok, means this electron is not taken away, it is there in the orbital group ok. So, now you see here how many electrons this carbon has here. If I talk about only carbon, carbon is four electron in the outer motion. So, these are the three electrons of carbon which is bonded with x, y and z ok and the fourth electron is present in this T-orbitor, it is not taken away, it is not withdrawn ok. So, it is there with the carbon atom, so carbon atom has all its electron present with it, it has no extra electron and not it has no extra electron present on this, now it has been and it did not lose its electron to the another molecule right. Four electron of carbon is there with the carbon itself and that is why this is neutral in nature, all the electrons are present with the carbon atom and it is neutral in nature. Now, you see few characteristics of this, this carbon atom which is you know free radical it is, it is also sp2 hybridized and it is sp2 hybridized, it is trigonal planet, trigonal planet how many electron it has see if I in the bonding state if I calculate all the electrons. So, we have pair of electron here, a pair of electron here, a pair of electron here, so it has total seven electrons. So, it has total seven electrons present, it is electron deficient also less than 8 electron, so it is electron deficient also, electron deficient but it cannot accept a pair of electron right because carbon cannot extend its octet, if it accepts a pair of electron the total number of electron becomes nine which is not possible, so it cannot extend its octet, so cannot accept a pair of electron, so can't be here as a Lewis acid, Lewis acid. Now, since it has one unpaired electron present with it, it is paramagnetic in nature, what is the magnetic behavior of carbocation, see in the carbocation we do not have present, we do not have this unpaired electron present over there, we have only these three bond pair with one positive charge on this, all the electrons are paired over there, so carbocation are diamagnetic in nature, carbocation are diamagnetic in nature, free radicals are paramagnetic in nature, like this we have two, three difference, different point or difference between these two you can keep in mind, it has no charge on it also but carbocation has one positive charge, its stability factor is also same as that of carbocation, so stability of free radical is directly proportional to plus I effect, inversely proportional to minus I effect, so this is what again you have to keep in mind here, that you also do plus will see that it is actually same as that of carbocation, so on the basis of this only we will discuss few examples, what example I will discuss then I write down the question, you can solve it on your own, then we will solve the questions, so first example you see I am taking this Nr3 and CS3, again you see the difference in the molecule is what, we have this group attach here and we have this group attach here, this group we know it shows minus I, strongly minus I effect it shows, very strong minus I group and this one is what plus I, stability of free radical is directly proportional to plus I, inversely proportional to minus I, order of stability will be, similarly you have to solve these questions I will just write down here, so these are the few examples I have written that you have to solve, you can pause it and solve and discuss in it, the first compound is this CH2, CH2 pH, CH2, CH2, CCL3, so which one has more minus I effect that is the question, now you can start solving this, I am starting this solution of this, so this group that is finite here, it has minus I nature but this CS3 has plus I nature, electron releasing electron withdrawing, stability of free radical increases by plus I nature, order will be this, again this CS3 has electron withdrawing nature, this CS3 has electron releasing nature, stability of second one is more than that of first, anode 2 and CN both has minus I nature, this is also minus I and this is also minus I, so both will reduce the stability of free radical, so both will reduce to the one which shows strong minus I effect that will reduce the stability more and hence the next one the other one is more stable, anode 2 shows stronger minus I than CN, just the relative strength you have to understand and the minus I effect or minus I effect of anode 2 is more than that of CN and hence the stability of this will reduce more, second one will be more stable, coming to this example anode 2 has minus I, CS3 has plus I and O has again minus I, so if it is A, B and C, plus I will increase the stability, so B is the most stable one, minus I will decrease the stability, anode 2, the minus I effect of anode 2 is more than that of anode because we have one more oxygen atom here, this will reduce the stability more, so after B the stability of C will be more and then we have A, B, C, A is the order, NF3, NF3 positive charge on this shows very strong minus I nature, this CS3 plus I and this also shows minus I, minus I of NF3 is a stronger than that of fluorine and this is plus I, so the order will be this is the first one most stable and then this will reduce, this will reduce maximum and then the second stability of order will be this and third one will be, this order will be first, second and third, okay, now you see this carbon atom is sp hybridized sp2 sorry, it is sp2 hybridized and this carbon atom if you compare it is sp3 hybridized, which one is more electronegative obviously sp2, this has electron withdrawing nature comparatively, it is more electron withdrawing nature and hence the stability of second one is more than to that of first one, okay, so this is the few examples we have discussed according to the, according to all the basis on the stability of free radical on the basis of I effect, okay, so far we have discussed stability of carbocation free radical with respect to I effect, okay, now the third type of intermediate we have and that is carb anion, okay, so the third intermediate we have, okay, so in this you see if the carbon atom if I write down this structure y and z, one pay on it, okay, so if you count the number of electrons for carbon here only carbon I am talking about, this carbon has one electron in this bond, one electron in this bond, one electron in this bond, right, one electron here with one extra electron, right, so total number of electrons if you calculate for carbon 2, 4, 6, 8, so it has 8 electrons basically, right, number of bond pair and lone pair if I write down, we have 3 bond pair and one, 3 bond pair and one lone pair, okay, so its steady number is 4, right, hibernation will be sp3, sp3 hybridization geometry is what is tetrahedral and the shape is it has how many electrons, 8 electrons, it means complete octane, it has one negative charge also present here, also present here, one lone pair we have, so we call it as electron rich as nucleophile since it has one in unbeared electron, so may act as nucleophile, may act as what, lewis place also, since the lone pair is present, so may be here as a lewis place, okay, one last thing here all the electron pairs are paired, all the electrons are paired, so it is also diamagnetic in nature, diamagnetic in nature, okay, tough stability due to IFI is exactly opposite to that of carbocation because in carbocation we have positive charge and here we have negative charge, okay, so now you see if I have, I will write down one example and then we will conclude the stability of this, so suppose if I write down this 3 CH2 negative charge, CH3 negative charge, 3 CH3 CH3, the three examples I have taken to compare its stability what we have to do, we have to again neutralize this negative charge, okay, like in case of Kato Katan we have discussed and we know the neutral species is more stable than the charged one, okay, so we have to neutralize this negative charge, to neutralize this negative charge we have to withdraw electron from this molecule, right, then only the negative charge density will decrease, so we require what, minus I group here, what kind of group this we have, it is plus I, so this will release electron, this will also release electron, 2 plus I group we have and here we have 3 plus I group present, so obviously the electron density here it will be maximum and here it will be minimum, so maximum electron density minimum stability, so order of stability will be what, if it is A, this is B and this is C, so order of stability will be A will be most stable and B will be C, this is the order of stability, this is 1 degree carbon ion because this carbon is attached with only one alkali, this carbon attached with 2 alkali group, so this is 2 degree or secondary carbon ion and this one is 3 degree or tertiary carbon ion, so the stability of carbon ion is what, it is 1 degree will have maximum stability, then we have 2 degree and then we have 3 degree, this also begins, 1 degree, 2 degree and 3 degree, okay. Now one more example I will write down here, which is CH3, CH2, CH2, negative molecule if you see, the difference is here only and bromine we have and we know these halosols has minus i effect, electron withdrawing, so more a minus i means lesser will be the electron density and more will be the stability, if we can withdraw more electron, the electron density will reduce maximum and here the stability will increase, so since fluorine has maximum minus i then chlorine and then bromine, so stability of this A, B and C if you compare, A will have maximum stability then B and then C, this is because of minus i, so what we can say with minus i if you compare this example and this example, right, this example and this example if you compare these two, I will write down here also, CH3, CH2, negative, FCH3, CH2, negative, if you compare these two, this has plus i nature and this has minus i nature, okay, so this will increase electron density, this will increase electron density, so if it is A, this is B, so stability of B is more than that of, okay and with this example what we can conclude here that the stability of the proportional to minus i and inversely proportional to plus i, this is what you have to keep in mind which is exactly reverse of plus i, right, reverse of cartoketan, in cartoketan it is simply of carbon and they have exactly reverse the order we have which is easily understandable because there we have positive sign and here we have negative sign, see I have written here louis phase, right, so this may behave as a louis phase because a lone pair present on this, okay, but these are not good louis phase, okay that's another thing but it may behave or not, yes it may be, okay, so we will see now some examples here, okay, we'll discuss again through examples on this and write down the examples, we solve it on your own and then we'll discuss these questions, you can pause the video and solve I am explaining it now, first question you see we have carbonyl, this carbonyl is 3 degree, this is 2 degree and this is 1 degree carbonyl and we know 1 degree carbonyl is most stable among the three, order we will do this, okay, this carbonyl is sp hybridized, this carbonyl atom is sp hybridized, sp 2, so which one is more like more electronegative sp, this can withdraw more electron since the stability will be, right, chlorine has electron with drawing nature and chlorine has electron with drawing nature, if you compare these two, the electron with drawing nature of this chlorine is more over here in comparison to this because it is closer, it is at the closest point than this carbonyl, from this carbonyl and we know the eye effect is a distance dependent effect, more close more will be the effect of that school, whether it is plus i or minus i, okay, so since this chlorine is closer to this carbonyl and effect of this chlorine is more here in comparison to this chlorine over here, right, hence this will withdraw more electron stability order will be, NF3 plus has strong what effect, NF2 has electron with drawing minus i and NF3 have what, minus i and which one has more minus i effect, which one has more minus i effect, tell me NF3 is a stronger minus i than NF2, right, so this will withdraw more electron stability order will be this, okay, here you see two methyl group electron releasing, increase the electron density, stability will decrease, here we have two electron releasing group, sorry withdrawing group to electron withdrawing group, okay, so if you compare these three, these are electron releasing so this has the least stability, so this is at the third point, third position, right, fluorine, chlorine both has electron withdrawing, chlorine, chlorine both has electron electron. But this chlorine this chlorine is same and one chlorine and chlorine we have, this chlorine has more electron withdrawing nature, So this will stabilize more, so this is the first position and this will be at the second position. CS3 has electron releasing nature, CT3 also has electron releasing nature, CT3 also has electron releasing nature. Which one has, suppose all are plus i, so which one has maximum plus i? This CT3 shows max plus i among the three and CS3 shows minimum plus i among the three. So the electron density will increase, but that increase will be least among the three, hence the stability of CS3 will be maximum and CT3 will be minimum. This carbon atom is again, it is sp hybridized, this is sp2 and this is sp3. Maximum electronegative pull more electron, maximum stability then this, so this is the stability order of these groups we have done. So with these examples you can easily understand how to do these questions on the base of i. Actually this is very easy to do all these questions, but when you get the questions in exam it is not like that in one compound we have only i effect possible. So it is also possible that in a given compound with i effect also there are other electronic effects like mesomeric effect resonance are taking place, then you have to be very careful at which effect is dominating and according to that of the effect only the nature of the product will decide. So this in coming lecture will compare like this also and we will decide, we will see at which effect is dominating when and according to that what should be the nature of the product. All these things we will discuss in the coming lecture. So this is all we have discussed three intermediates and their stability effect order, next we will see the other application of i effect which is nothing but acidity and basicity order of the molecule depending on i effect.