 In this video, we are going to see how the hyperconjugating effect of an alkyl group changes if I replace these hydrogen atoms with other heavier isotopes of hydrogen like deuterium and tritium. Isotopes, if you remember, are basically different atoms of the same element, right? Hydrogen has three isotopes, hydrogen 1, hydrogen 2 and hydrogen 3. All of these have the same number of protons, all of these have the same atomic number. And as you know, atomic number gives the chemical signature of a particular atom. So because all three of these atoms have the same atomic number, all of these are essentially hydrogen atoms. However, if you look at their mass numbers, which is nothing but the number of protons plus number of neutrons, their mass numbers are different, right? Hydrogen only has one proton and no neutrons, so it has a mass number of one, while this one, H2, has one proton as well as one extra neutron giving it a mass number of two, while H3 has one proton and two neutrons giving it a mass number of three. So these two isotopes, they have extra neutrons in the nucleus and these two are definitely going to be heavier compared to the hydrogen atom, right? So H3, which we commonly call tritium as it has a mass number of three and we represent it by the symbol T. Because tritium has the highest mass number, it's going to be the heaviest, followed by H2, which we call deuterium and represented by the symbol D, followed by the hydrogen atom, right? Hydrogen, in fact, is going to be the lightest of all of its isotopes. Now it turns out that these small differences in mass of these isotopes also affects their ability to show hyperconjugation to a surrounding cation or an alkane. So how does that happen? To understand, let's take a carbon-hydrogen bond. So let's say we have a carbon-hydrogen bond out here. Now we generally represent a bond by a straight line, but in reality the atoms in any particular molecule are not static, but these atoms keep vibrating. So a chemical bond is more like a spring. So you can think of this bond as vibrating like a spring, right? Now in a carbon-tritium bond, because tritium is heavier than hydrogen, so it turns out that it doesn't vibrate as much as a carbon-hydrogen bond. So therefore on an average, a carbon-hydrogen bond is longer compared to a carbon-tritium bond. Now because on an average these bonds are longer, this means that the electrons in this bond are further away from the nucleus of both of these carbon and hydrogen atoms. So therefore there will be a weaker electrostatic attraction out here between these electrons and the nucleus of each of these atoms compared to the electrons in a carbon-tritium bond. The electrons out here are much closer to the nucleus of both carbon and tritium, so there's going to be a stronger electrostatic attraction between these. So therefore because the electrostatic interactions are much weaker in the longer carbon-hydrogen bond, so it's easier to break a carbon-hydrogen bond compared to a carbon-tritium bond. In fact if you look at the bond dissociation energy data, if you look up these values on the internet, you'll find out that a carbon-hydrogen bond has a lower bond dissociation energy compared to a carbon-turtium bond. It's easier to break carbon-hydrogen bonds followed by carbon-turtium and the most difficult to break would be the carbon-tritium bond. Now coming back to hyperconjugation, because hyperconjugation involves breaking of these bonds, these carbon-hydrogen, carbon-turtium and carbon-tritium sigma bonds. It's a form of sigma pi resonance, so therefore because it's easiest to break these carbon-hydrogen bonds followed by carbon-turtium and carbon-turtium. So therefore the hyperconjugating effect of a carbon-hydrogen bond will be greater as it's easier to break compared to carbon-turtium and carbon-tritium. So if you look at the stability of these three cations, let's call this A, B and C. In all of this the cation can be stabilized via hyperconjugation. However this sigma pi resonance will be much easier in A as it's much easier to break these carbon-hydrogen bonds compared to carbon-turtium and carbon-turtium. So the stability imparted by the CHC group will be the highest followed by CD3 and CD3, right? So therefore stability wise A is going to be more stable than B followed by C, right? Now a quick note out here, from the point of view of hyperconjugation, a CHC group shows a greater plus H compared to CD3 and CD3. Now these alkyl groups, they can also show inductive effects, right? Because carbon is more electronegative than hydrogen, it can pull some of the electrons in this bond, it can pull some of the electron cloud towards itself, thereby making it partially negative, right? Now if you compare CH3 and CD3, even out here this carbon is more electronegative than the tritium atom, so even this can become partially negative. However because the carbon-turtium bond is shorter, the electrons in this bond can be much more influenced by the nucleus of the carbon atom compared to electrons in a carbon-hydrogen bond. Because this bond is longer, the electrons are further away, so the electron pull by the carbon atom is lower, so therefore the amount of charge that gets developed on the carbon atom of a CD3 group will be higher compared to that of CH3. So a CD3 group will have a greater charge density on the carbon atom, so it will push the electrons in surrounding sigma bonds much more compared to a CHC group, right? So from the point of view of induction, the trend is exactly the opposite, CD3 is a stronger plus I group followed by CD3 and CH3. So once again if you look at the stability of these three cations, on the basis of hyperconjugation, the stability of A is going to be greater than that of B which will be greater than that of C as a CHC group shows the highest plus I effect. But if we consider the stability purely on the basis of induction, so if we think about the stability purely based on the plus I effect, then because CD3 has the highest electron density, it will push the electrons in this sigma bond more towards this cation compared to CD3 and CH3. So therefore purely on the basis of induction, C should be more stable than B which should be more stable than A, right? So as you can see hyperconjugation and induction are working exactly in the opposite direction out here. But remember because hyperconjugation involves actual delocalization of this sigma electrons which will lead to an actual delocalization of this positive charge which is not the case in case of induction. So therefore these hyperconjugating effects are much stronger compared to induction and therefore the correct order of stability would be A greater than B greater than C. Now that you know how replacing hydrogen with its isotopes changes its hyperconjugative and inductive strands, can you figure out the stability order of these given cations? Pause the video and take as much time as you want and try and figure out what the correct stability order might be for these cations. Okay now that you're done let's see how each of these groups affect the carbocation stability. Now a CD3 group is an electron donating group, right? It can donate electrons to the benzene ring via hyperconjugation. Let's also not forget that this CD3 group is also an inductive group. It can also push electrons into the benzene ring via induction, right? So this is a plus H as well as a plus I group, right? In fact all of these are plus H and plus I groups. All of these are plus H and plus I, right? However as we have seen in our previous videos whenever we have an electron donating group attached to a benzene ring it only brings about negative charges at these positions, right? It brings about negative charges at ortho and para of the electron donating group. So when out here the negative charges will get developed out here, right? If you are not sure exactly how this is happening feel free to check out the previous video in which we talk about in detail about how these charges occur. Now as you can see in both A and C these negative charges will get developed right under the cation. So if you draw the hyperconjugating structures in case of CD3 if you keep pushing these electrons. If you keep doing that you will see that in one of the resonating structures will have a lone pair right next to the empty orbital of this cation. So both of these can overlap and stabilize the cation, right? So the CD3 group out here can stabilize this cation. Similarly if you come to see if you draw the hyperconjugating structures out here you will see that even out here this CD3 group can go and stabilize the cation. However if you come to B when we have a CHC group attached at the meta position if you draw the hyperconjugating structures out here if you keep moving these electrons. If you keep doing that you will see that in none of the resonating structures does this negative charge come right under this carbon atom. So therefore in both B and D because the electron donating groups are at meta so therefore they won't be able to stabilize the cation via hyperconjugation right? So in B and D because hyperconjugation is not leading to extra stability so hyperconjugating effects are going to be not that important right? Now that we have established how the attached groups are affecting the cation. Let's try and figure out the stability order. So A and C are clearly going to be more stable compared to B and D as they can stabilize the cation via hyperconjugation. And between A and C because CD bonds are weaker and easier to break compared to CD bonds. So there will be a greater hyperconjugating effect in case of CD3 compared to CD3 so A is definitely going to be more stable compared to C right? Now between B and D hyperconjugation doesn't play a role so the more important effect out here will be the induction and because CD3 shows a greater inductive effect compared to CH3. So therefore between B and D D is going to be more stable compared to B.