 Hello students, good morning. Today in this session, we are going to discuss the chapter called Phenol. So, without any delay, let us start the session today. So, all of us know the formula or the structure of Phenol, which is nothing but we have a benzene ring and on this ring, we will have one hydroxy group present OH. This is Phenol, this particular molecule, we call it as Phenol. Similarly, like this, there are other molecules also, which formula we should know, because sometimes formula-based questions also Jay asked, like if I tell you, once they have asked about how many lone pairs are present in a structure called melamine. So, when they give you the name of the structure, without knowing the, when they give you the name of the molecule, so without knowing the structure of that molecule, you cannot find out how many lone pairs are present in that. So, that is why for few molecules, the structure is important. And for melamine that is given in polymer chapter, so as soon as we will cover the various chapters, we will discuss all those different molecules whose structure we should know. So, like I have given you the structure of Phenol, similarly, if I have OH, two OH groups present on the benzene ring at different, different positions, suppose we have OH here and OH here, this is one structure, the another structure is this one, one OH on the top and then OH on the bottom. Another structure is this OH here and OH at ortho position. Next structure, we have this one, when three OH group is present on the benzene ring, OH and OH. The name of this structure, when OH is present at the metaposition, we call it as resorcinol, resorcinol. This one is hydroquinol, when you have double bond O here and double bond O here, it is hydroquinol and no any in that case. Now, when OH present on this ortho position, we call it as catechol, and this one is pyrogallon. So, these are the few structures whose name are important, whose name is important in this chapter, you must know the structure of all these molecules. Now, the next thing is two properties of phenol we will see first. So, basically in all the organic chapters, if you see, we have chemical, physical properties and the preparation method. So, chemical, physical properties, again we have physical properties like the color density and all, melting point, boiling point, solubility, all these things we have and chemical reaction again involves chemical reactions only. The mainly chemical reactions and preparation methods are important. However, in this chapter, the preparation method of phenol is not that much important, it is very 2, 3, 4 methods we have that we will see and it is the direct reaction. But the chemical reaction of phenol is important, there are few name reactions given in this chapter that we should, but definitely know, right. So, they done first the few properties of these molecule, you see phenols, like I said, it is a crystalline solid which is colorless in nature, right, which is colorless actually. So, we will write down it is colorless, colorless crystalline solid, okay. It is colorless crystalline solid, solid. It is soluble in water, soluble in water, why? Due to hydrogen bonding, it shows hydrogen bonding with water and hence it is soluble. The boiling point of phenol is much higher of phenol is much higher than the corresponding, then the corresponding chromatic hydrocarbon, hydrocarbon and hellerines. Corresponding means what? The molecular mass must be comparable, okay, then only we can do the comparison, okay. Now, the next thing is what? Thus, this phenol shows acidic behavior also, it shows acidic properties, it shows acidic properties, okay. Acidic property means what? The tendency to release H plus ion, okay. Now, you see if the molecule or the phenol structure if I draw here, it is this, when it releases H plus ion, it forms O minus and H plus. This ion, we call it as phenoxide ion, okay. O minus on the top, this one, we call it as phenoxide ion, which is the conjugate base of phenol. Okay, phenoxide ion is what? Phenoxide ion is the conjugate base, okay. So, we know if the conjugate base is stable, right, the tendency for this molecule, for this reaction to go in forward direction will be more, right. If it is a stable, then only this has tendency to convert into this particular ion, okay. And then the reaction goes in forward direction, more amount of H plus will get and hence more will be the acidic nature, right. This phenoxide ion, however, is resonance stabilized, okay. Phenoxide ion is what? Phenoxide ion is ion is resonance stabilized. Now, since it is stable through resonance, hence the phenol has tendency to convert into this and shows acidic behavior, right. So, this is the important part we have, okay. However, if I see, phenols are weak acids, okay. These phenols are weak acids. So, these are not a strong acid, but yes, they have acidic behavior, tendency to lose H plus ion, okay. Now, you see this OH group or if I, this negative charge if I consider, when you draw the resonating structure, you will get negative sign at ortho and para position, okay. So, if I take this phenoxide ion, you see, this is the phenoxide ion, suppose we have O negative. When we draw the resonating structure, what we get, you see, this negative charge comes over here and this pi electron comes over here, right. So, we get this and double bond O. Now, again, if you draw the resonating structure of this, then what happens? This negative charge comes over here and this pi electron comes over here. So, the structure will be this double bond, double bond, negative charge, double bond. Further, we can draw the resonating structure of this, okay. And we will get a negative charge here then and then again on the oxygen. I will draw all the structures here. This is the five resonating structure possible here, five resonating structures possible, right. This is the resonating structure here. Now, in this molecule, what we see, the negative charge we are getting at ortho and para position and hence this group, which is OH present on the ring, we call it as ortho para directing group, ortho para directing group. This is one thing, okay. Now, any group present at ortho or para position or even at meta position that affects the acetic behavior of phenoxide, okay. That affects the acetic behavior of phenoxide, okay. The basic thing is what? More stable phenoxide ion, more will be the acidity of phenoxide, right. Any group present at this position, this position and this position affects the acidity of phenoxide, because this group, these groups affects the stability of phenoxide ion, okay. So, what we can write in the next step that the acidity of phenols, acidity of phenols is directly proportional to the stability of phenoxide ion, stability of phenoxide ion, okay. More stable phenoxide ion, more will be the acidity of phenoxide, okay. Now, you see the phenoxide ion is this, O minus, okay. And if any group attached ortho para position, this group, this affects the stability of phenoxide ion. How the stability of phenoxide ion decreases? If we decrease the electron density here, right, means G, this group should be electron withdrawing group, okay. So, electron withdrawing group increases the stability of phenoxide ion and hence the acidity of phenoxide, okay. So, in the next line, what we can write? If this G is electron withdrawing group, electron withdrawing group, which in short, we write it as EWG, electron withdrawing group, hence the stability phenoxide ion increases. And hence the acidity of phenols increases, acidity of phenol increases. So, and the next thing is what? What? If G is electron releasing group, electron, electron releasing group, which is ERG, then what happens? Stability of phenoxide ion decreases. It means the acidity of phenol decreases, right. So, basically one thing you have to keep in mind and what is that? That acidity of phenol is directly proportional to electron withdrawing group and inversely proportional to electron releasing group. This is the basic thing we have, okay. The acidity of phenols is very important for exam point of view also, whether you are going to write JEE exam or board exam or any other exam, okay. Even in board exam also, they ask this orders, okay, right. Second thing is what? If the number of withdrawing group is more, then more will be the acidity of phenol, okay. More number of electron withdrawing group, more acidity. More number of electron releasing group, less acidity. For example, we will see some examples on this, which is this one, suppose. We need to compare the acidity of acidity of given molecules, given molecules, okay. First question is this. Next question is this. Try these questions. You need to assign the order of acidity. This example is also very important. Now, we will do this one by one. So, first of all, what we need to do, you see. An O2 group, you should know this. An O2 group is electron releasing, sorry withdrawing group, electron withdrawing group. How do we know that electron withdrawing group? The condition for electron withdrawing group we have is between the first and second atom, we must have unsaturation. We can have double bond, we can have triple bond also. Condition is what? The electronegativity of y must be greater than the electronegativity of x. This is the condition. If you have unsaturation present between the first and second atom, provided, like providing this electronegativity of second atom is more than to that of x, this condition. Then it shows electro-electron withdrawing nature, okay. Electron releasing nature is what? ERG. On the first atom, we must have atom, we must have at least one lone pair, right. At least one lone pair present. Like for example, if you write here OCH3 is electron releasing group OH, electron releasing group, okay. Then what we have NH2, electron releasing group, derivative of amines, all these are electron releasing group because you see oxygen has two lone pair here, two lone pair here and one lone pair on nitrogen. Electron withdrawing group, the example is what? Cynide group, C triple bond N, aldehyde, C double bond OH, right, COOR. That is also an electron withdrawing group because between first and second atom, we have unsaturation present and second atom is more electronegative than the third atom, okay. So, as you see here, all these are electron withdrawing group, NO2, NO2, NO2. More electron withdrawing group means what? More stability of phenoxide ion and hence more will be the acidity. The acidic order of this molecule will be, will be, this is the first most acidic second, third and fourth. Why? Because three anode group, 2, 1, 0, right, acidity order will be this. In the second one and there is only one anode group present, okay and the position of anode group is different, okay. So, at ortho position we know we have minus R effect possible. At meta position, we have only minus I possible. At para, we have minus R possible. Here we have no effect. All these minus effect, obviously, they stabilize the phenoxide ion. So, this one is obviously the least acidic, okay. So, the fourth one is, this one is the least acidic. So, number is, rank is 4, okay. Now, minus I and minus R. Obviously, minus R will stabilize more the phenoxide ion in comparison to minus I, okay. So, this minus I will show the less acidic, will be less acidic than these two molecules because minus R and minus R we have here. So, this one is obviously third then. Fourth, third, okay. In this two, we have minus R effect, first and third. We have minus R effect and minus R does not depend upon distance, distance, okay. It only, the only thing is required is conjugation, okay. Minus R minus R we cannot compare with minus R here. Then we consider the next effect which is minus I and here also we have minus I, okay. Now, the minus I effect, however, minus R dominates minus I always, okay. Minus R dominates minus I always, right. But since we cannot do this with minus R effect, hence we are considering minus I. The effect of minus R will be same on this molecule and this molecule. Now, we will see minus I then. Now, minus I will be more effective over here because it is more close to the hydroxy group. Since we know the minus I effect is distance dependent effect, okay. More close, more will be the effect of minus I or plus I, whatever it is, okay. Right. So, fourth, third, minus I is more effective here. Hence, this one is the most acidic and then second, third, fourth rank will be this, okay. Now, in the next example you see chlorine attached at ortho-parameter position. Chlorine shows always minus I effect. Helogen group always shows minus I. It can also show plus R since chlorine has lone pair on it, okay. Plus R it can also show, but its minus I effect is more dominating. See, in all the cases, plus R or minus R dominates plus I or minus, plus I or minus I always, plus R or minus R, plus I or minus I always. This is the order we have. Except in case of halogens, in case of halogens. In case of halogens, the order will be reverse, minus I will be effective. Why? Because you see, this chlorine, I will write down here, this chlorine Cl has the lone pair present in 3P orbital and this carbon has 2P orbital. So, the size of orbital is different, right. 2P is this, 2P is this. So, when the size is different, the overlap is also difficult, okay. Same size of orbital can overlap easily. That is why the plus R nature of this is not dominating here, okay. So, here we have minus I, minus I and minus I. It is distance dependent. So, this is more close to this. This is most acidic than this one and then this one and then acidic nature is this. Now, fluorine, chlorine, bromine, iodine, again we have halogen. So, obviously we have minus I here, minus I here, minus I here and minus I here, okay. Why? Because in case of halogen, minus I is more dominating because the difference in size of the orbital 2P and 3P. Now, you must have this question, if you think, this carbon is 2P orbital and this fluorine also has 2P orbital. So, this, the size of the orbital is exactly same. So, why not this fluorine shows plus R nature because this fluorine also has 3 lone pair on it. Why it does not show plus R nature? The reason of this is what? Because fluorine is the most electronegative element. That is why the minus I effect is more dominating. Here, the case is what? The electronegativity is less but the size of the orbital is not same, right? Size of the orbital is different. That is why overlap is not possible. Here, size is same but since fluorine is the most electronegative element, so its minus I effect again dominates. So, basically overall the reason is what? Whatever the reason I said that you must keep in mind but what you have to keep in mind plus R or minus R always dominates plus I or minus I in except in one case, that is in the case of halogens. In case of halogen, we consider what? I effect dominates R effect. So, now we have maximum minus I here because it is the most electronegative element. So, order of acidity will be 1, 2, 3 and 4 electronegativity. So, like this we can compare the acidity of phenols. Two more examples we will see which is the important example we have as far as this J is concerned. Suppose we have to compare the acidity of these molecules. One more example we will see. These two examples you see. So, you see NO2 has minus R effect, right? It shows minus R. Here the minus R effect of NO2 is not possible, right? This minus R effect is not possible in this case. Why it is not possible? Because see here, here in this two molecule CS3 and NO2 and CS3 and NO2 we have steric repulsion, right? And to minimize the steric repulsion, this NO2 group will change its plane and it becomes what you see here? To minimize the steric repulsion, this NO2 will change its plane, right? And hence like why it changes plane to minimize the steric repulsion, okay? So, this change in plane of NO2 to minimize is the repulsion, okay? Plane changes. And when the change in plane is there, that's why this NO2 will not show, hence does not show plus minus R effect or resonance effect simply. Because for resonance the molecule must be planed, okay? So, minus R is not possible here, but here the minus R is possible, right? Minus R possible. Now because of minus R effect, this phenoxide ion is more stable, hence this molecule is more acidic than the second molecule, okay? It's a very good question, important also. Now the same thing if you see here, but here we do not have a steric repulsion. Why? Because this Cn and this Cn, all these Cn's are linear group C triple bond N. So, in linear group there is no steric repulsion, okay? So, I'll write down here in linear molecule no repulsion because it is linear, right? So, this group also shows this shows minus R, this also shows minus R, okay? Apart from minus R, this CS3 and CS3 shows plus i and plus i here also. Plus i is more dominating in this case because it is closer to H. And because of plus i effect, the stability of phenoxide ion decreases. Why? Because these are electron releasing effect, it releases electron. This is also electron releasing effect, but the effect of these two group on here will be lesser than these two group on here because it is closed, right? Hence, the acidic order again, it is more in the case of this molecule, right? So, here we are considering steric repulsion because NO2 is not linear, right? But here since C triple bond N we have, cyanide group is nothing but this, right? So, it is linear group, sp hybridized, sp hybridized. Hence, we are not considering repulsion, then minus R, minus R is same in both the molecule, then we consider plus i of it and hence the acidic order, okay? So, like this, they ask questions on acidic order of phenol, okay? Which is important. Apart from this, there are three, two, three, four reactions of preparation method of phenol we have. That reaction is not that much important. But like I said, there are certain name reactions in this chapter like rhymer time reaction, cumane phenol process, okay? Clayson rearrangement, those reactions are important. So, we will see that in next session. Thank you.