 Welcome back to Centrum Academy again, today in this session we are going to discuss the last part of the general organic chemistry, which is nothing but tautomerism. Tautomerism you must have seen that this is a part of isomerism, it is given in isomerism. But since this particular isomerism type of isomerism requires the concept of requires the concept of this particular type of isomerism requires the concept of resonance. That is why we are discussing it here only. So, here in this tautomerism, what is tautomerism? Tautomerism is a special type of functional isomerism, special type of functional isomerism. In which the two isomers exist in dynamic equilibrium. The next point you have to understand here, these isomers inter convertible are inter convertible and called tautomers and called tautomers. And that is why and this phenomenon is tautomerism. For example, you see if I take this molecule, this is example. This is keto form, ketone functional group. So, keto form and this is enol form. Why enol? Because this ene stands for double bond alkene and this all stands for this functional group. And this kind of conversion we call it as keto enol tautomerism. So, basically this particular type of isomerism which is tautomerism. In this what happens like suppose I have taken this example. One more example if I write down here for keto enol tautomerism. C H 3 C double bond O C H 3 C H 2 double bond C single bond OH C H 3. Now, in this example you see the second one which is nothing but this C H 2 H here C double bond O C H 3. So, in this hydrogen which is adjacent to this keto group whether this hydrogen or this hydrogen. This hydrogen is acidic hydrogen or reactive hydrogen since this is present adjacent to the keto group. Because of the electron withdrawing nature of the C double bond O, this hydrogen can come out as H plus. Now, what happens this hydrogen because of this electron withdrawing group comes out and attach on to this oxygen. And we get C H 2 double bond C single bond O H and C H 3. This is what this is how the reaction happens right. And why it happens since you have taken this molecule you know that is acetone. If you take this molecule acetone into the into any bottle or any vessel closed vessel right. Then what happens after sometime at time t is equals to 0 right there is only acetone present. So, if I write down here at t is equals to 0 this is 100 percent acetone right and there is no in all form. So, 0 percent it is initially when you have taken this into a vessel closed vessel. At time t is equals to t any other time if you see the content of this will less than 100 percent. And this content will be greater than 0 percent right means few amount of this in all has been formed right. And we are not doing anything into this this happens on its own right. This is an spontaneous process right it is an spontaneous process we are not doing anything into it. You take this acetone into a bottle after sometime you will get a few amount of in all is also there right and few amount of acetone is also there right. The one which is more in the composition that is more stable right means whatever the molecule is more stable the composition of that will be more into that vessel or bottle whatever you are taking right. So, stability also we will discuss in this ok. And this kind of conversion see actually what happens when this H plus comes out right then this carbon has one lone pair negative charge on it ok. And then this H plus has again the tendency to attach with this molecule again right. So, this H plus which is coming out can attach onto this carbon or on this oxygen also why on this oxygen because this oxygen has lone pair on it right. So, it can easily accept this H plus ion from this ok. So, basically we have three two points where this H plus can attach on this carbon or on this oxygen right. So, when this hydrogen comes onto this carbon will get the same molecule or when this hydrogen comes onto this oxygen will get this molecule ok. So, this is right where this hydrogen will attach that depends on the stability of this molecule or this molecule whichever molecule is more stable there only H plus will attach and that particular compound will be more in major more in composition right. So, now few more examples if I write down suppose if I take this example oxygen we have here and double bond O when you write down the enol form of this will get OH here and this oxygen will be here as it is ok. Now, you see in all these examples alpha hydrogen is must required ok. So, this hydrogen is alpha hydrogen right this is alpha this is alpha position right. Similarly, here also this is alpha hydrogen and here also we have alpha position right. So, alpha hydrogen shifts onto this oxygen that is what is happening here in all these example ok. So, condition is what for tautomerism we must have we must have at least one sp3 hybridized alpha hydrogen that is the necessary condition. You must have at least one sp3 hybridized alpha hydrogen correct that is the condition we have now you see this tautomerism can takes place in acidic medium also and basic medium also right. So, what is the mechanism is of those reactions in acidic and basic medium that we are going to discuss now ok. The first one we are looking at is base catalyzed mechanism ok. Suppose the molecule I am taking is CH 3 write down like this CH 2 hydrogen here C double bond O CH 3 both are inter convertible that is why we have taken the reversible sign here. So, base is what base is suppose we have OH minus right. So, what happens this OH minus will take this alpha hydrogen this OH minus will take this alpha hydrogen this bond pair comes over here and this bond pair will go on to this oxygen ok. So, the thing here CH 2 double bond C single bond O with 3 lone pair CH 3 ok. This is step is the rate determining step with the oxygen will have here the negative charge ok. Now, here what happens the H 2 O from this H 2 O this will take this H plus ion and OH minus will come out. So, finally this will convert into CH 2 double bond C single bond OH CH 3 plus OH minus ok. This is how the reaction proceeds for base catalyzed mechanism. One point is very important here for you to understand is suppose since both side we have 3 alpha hydrogen right. If one side we have like this suppose we have R CH 2 C double bond O CH 3 then what is the product here we get because both side we have 2 different number of alpha hydrogen. So, in base catalyzed mechanism hydrogen comes out from the carbon alpha carbon which has more number of hydrogen. So, what happens this is hydrogen and this hydrogen will come over here. So, I will write down R CH 2 single bond COH double bond CH 2. So, in base catalyzed mechanism in base catalyzed mechanism hydrogen comes out from the carbon which has more number of hydrogen carbon means alpha carbon ok. This is what you have to keep in mind. So, write down here note H plus comes out from the carbon which has more number of hydrogen. Carbon means alpha carbon ok. In base catalyzed mechanism we have to take care of this thing. The next one is acid catalyzed mechanism acid catalyzed mechanism. In this one I am taking the same example H plus we have and the lone pair of this oxygen will attack on to this H plus here right. We get CH 2 single bond COH CH 3 positive charge on oxygen and here we have H right. Next what happens this pi electron will go here and this pi electron go here H plus will come out ok. So, we will get CH 2 double bond COH CH 3. This is how the mechanism takes place in acid catalyzed reaction ok. Here you have to keep this point in mind that H plus comes out from the alpha carbon which has less number of less number of hydrogen ok. This is the two mechanism we have ok. Now in all this mechanism because you do not have to draw this mechanism again and again in the exam ok. So, what you have to keep in mind suppose this is first position this is second and this is third right or here this is again first position second and third right. Randomly I am giving some number I am assigning some number over here right 1 2 and 3. So, hydrogen comes out from the third atom attach on the first and there will be a double bond between second and three ok. Hydrogen comes out from the second third atom attach on the first and there is a double bond between second and three that is what the product we have here ok or 1 2 and 3 I will write this side 2 and 3 I will write this side ok. Here it will not make any difference because both side we have CH 3 ok. Hydrogen comes out from the third atom attach on the first right and there will be a double bond between second and third this is what we have here this is what you have to keep in mind ok. Hydrogen comes out from the second atom attach on the third and then there is a double bond between second and third atom ok. Now this is what the theory part we have of tautomerism right depending on the stability of these two compound keto and enol form the one which is more stable the composition of that will be more in the mixture in the equilibrium mixture ok. So, which one is more stable and for stability what are factors we have to consider that we are going to discuss now right. One thing I will tell you generally keto form is more stable than enol form and why is it so because the C double bond O keto form is more stable enol form ok. So, reason is what the bond energy of C double bond O is more than the bond energy of C double bond C or the bond energy of C single bond O right. So, this bond energy and this bond energy the bond energy of C double bond O is more than this that is why the keto form is generally more stable than enol form. But in few cases for example if I write down here like this example there are few factors which actually affects the stability of keto enol form for example this one the enol form of this will be what the oxygen is one this is 2 and this is 3 this is sp3 hybridized this is alpha carbon right. So, hydrogen comes out from this 3 attach on the first. So, we will get here what hydrogen comes out from 3 attach on the first and there is a double bond between second and third these two pi bond already we have right. So, in few cases like in this case the enol form is more stable why this one is more stable here because of aromaticity this is more stable right. So, this kind of factor if you have then may be stability will be different, but otherwise or generally keto form is more stable than enol form ok. We will see now few examples into it and then we will discuss the stability comparison ok. So, the next we are going to discuss the stability of stability comparison of keto enol form keto enol form case 1 there are few cases into it will the first discuss the case 1 when we have in case of mono carbonyl compound mono carbonyl compound one carbonyl group will be here for example you see the product here it will be right this is keto and this is enol the equilibrium mixture if you see the composition of this enol form here it will be 10 to the power minus 4 percent this is very less at equilibrium because this molecule is more stable and why it is more stable because of same reason I have told you that higher bond energy of C double bond O right and here the bond energy is less ok. So, that is the reason this is more stable form here now in second case you see this one double bond O. So, its enol form will be this the composition of this will be 1.2 percent in this section this composition is also important it is the experimental value I am giving you ok you have to memorize this ok again this is more stable at equilibrium more stable and this is more stable at equilibrium ok. This is the first case when we have mono carbonyl compound there is no exception it is straight forward only ok keto form is more stable ok. Case 1 case 2 you see when we have di carbonyl compound two examples I will take here suppose we have C H 3 C double bond O C double bond O C H 3 and the another one we have is this two ok. So, here you see what happens in the first case ok these two examples you see here we have lone pair lone pair repulsion lone pair on this oxygen right and lone pair on this oxygen. So, here we have lone pair lone pair repulsion right and to minimize this repulsion the two one of this part like the half part of this will rotate itself. So, this double bond O will come to this side to minimize the repulsion. So, we get here H 2 C H C double bond O C double bond O C H 3 and when you write down the enol form of this is hydrogen will come over here. So, H 2 C double bond C O H C double bond O C H 3 right in this case if you write down the product will have O H here will have double bond and here we have double bond O correct ok. This is alpha hydrogen comes over here and we get double bond here this molecule is stable because of hydrogen bonding. Now, because of hydrogen bonding this molecule is highly stable and its composition is around 100 percent here because of hydrogen bonding. So, this is one factor we have which you know which influence the composition of this mixture because of hydrogen bonding it is getting stability. Hence the composition of this will be more in fact it is 100 percent here, but here since we have this repulsion and there is no such you know hydrogen bonding possible here. So, this two if you compare this one is more stable here obviously and the composition of this will be 0 percent means this reaction will not go in forward direction at all ok. This won't form actually this is highly unstable molecule because of this C double bond O group here di carbonyl compound ok. If you have di carbonyl compound like this no case 3 we have which is 1, 3 di ketone 1, 3 di ketone for example you see what is 1, 3 di ketone C H 3 C double bond O C H 2 C double bond O C H 3 1, 3 di ketone ok. So, what happens here you see when you write down the tautomeric product here this is alpha hydrogen this is alpha hydrogen for this molecule this is alpha this is alpha right. So, what happens here this alpha hydrogen will not take part in the reaction this is methylene group and when this methylene group is separated by two carbonyl group right then it is highly acidic ok. It is highly acidic which is in between two carbonyl group right because both are electron withdrawing nature both have electron withdrawing nature. So, effect of these two group will be maximum here right in comparison to these two and that is why this hydrogen is acidic here. So, the product of this if you write that will be C H 3 C single bond OH double bond C H single bond C double bond O C H 3 the composition of this will be here around 76 percent this composition you must remember and why it is 76 percent that also you can understand this is active methylene group here and if I write down this molecule as this this carbon is here C double bond C single bond OH right this C is here double bond O C H 3 and C H 3 understood if I write down random if I do it on numbering here suppose 1 2 3 4 5 ok. So, this carbon is 3 right this is 4 and this is fifth carbon right this is 1 and this is 2 this is what it is now in this is what happens this is a stabilized by hydrogen bonding again because of this hydrogen bonding this molecule is again more stable than the previous one and that is why its composition is 76 percent ok this composition is important you must memorize this value ok. So, whatever the composition I am writing it down it is important you must write it down sometimes in the test you will get in this in the matching column ok the composition you will get in the matching column I have seen few questions in J advance ok. So, this composition you must remember experimental value I am giving you you can compare logically which one is more stable, but composition you have to memorize ok. Another example we will see next example you see P H C double bond O C H 2 C double bond O C H P H here now the product here we will get is P H C OH double bond C H single bond C double bond O P H ok the composition of this will be 96 percent you see here we have extended conjugation conjugation is here also you see pi sigma pi lone pair sigma pi. So, this is we have conjugation here right here also we have conjugation, but because of phenyl group present both side we have extended conjugation here and that is why its composition is more extended conjugation gives more stability. So, this is another example 96 percent next example you see if instead of this phenyl if I put hydrogen H C double bond O C H 2 C double bond O H the product here it will be H C OH double bond C H single bond C double bond O H right. So, because of this minus m nature this is also more stable and its composition will be 84 percent this is 84 percent. So, this 76 84 and 96 this composition you must remember third example we will see C H 3 C double bond O C H 2 C double bond O O C 2 H 5 this you see it shows minus m effect electron withdrawing minus m minus I E ok this group is electron withdrawing. Because of this electron withdrawing nature the tendency to lose this hydrogen atom will be comparatively less we are comparing these molecules with this. So, the product here we will get is C H 3 C single bond OH double bond C H single bond C double bond O O C 2 H 5. So, because of this electron withdrawing group the composition of this will be lesser than the previous one and it is found to be 44 percent. So, the value will be this product is same similarly the product will right because of this group electron withdrawing nature the tendency to lose this hydrogen as H plus will be less and hence it is plus m sorry plus m and minus I right plus m and minus I electron releasing tendency tendency to lose H plus will be less over here electron because of this lone pair you see it has electron releasing tendency and hence tendency to lose H plus here it will be less hence composition will also be less it shows minus I also, but plus m will be dominant here. Another example you see highly bonding is possible is not possible here in this case because of the steric repulsion the repulsion in this methyl group ok and the composition of this is found to be only 7 percent. Because of steric hindrance again another factor because of steric hindrance the composition decreases stability decreases composition decreases even if you further increase the size of this group the composition will be less over there. So, when the size of this group is more than the methyl group the composition will be even less than from 7 percent right. So, that is the another case the steric hindrance is the another factor I will write down here because of steric hindrance few more examples will discuss these are the 3 examples you see. Now, from this if I write down the product here the product will be this double bond double bond OH here the product will be OH double bond here the product will be this ok. Now, you see it has 4 pi electron and hence it is anti aromatic and anti aromatic is highly unstable composition will be 0 percent right. It has 6 pi electron aromatic and this composition will be 100 percent since aromatic compounds are highly stable right it has 8 pi electron anti aromatic highly unstable composition will be 0 percent ok. So, this is it for this tautomerism part we have seen how to compare the stability of various you know various compounds various ketoenol form right. Those factors are hydrogen bonding steric hindrance aromaticity right and then resonance also is one of the factor extended resonance we also have seen right. These 3 4 factors you must keep in mind whenever you have to you know compare the stability of tautomeric structures all those examples and the composition that I have written you must remember those composition it is important again I am telling you in few questions I have seen they have given the composition value also the percentage value also right. So, that is it for this particular session we are done with tautomerism GeoC is finished now this is the last portion we will see you next into the another chapter thank you very much.