 It is with pleasure I welcome you all once again to MSP lecture series on interpretive spectroscopy. In my last lecture I started discussion on IR spectroscopy and I was discussing about some of the fundamentals concerned with IR spectroscopy. Let me continue from where I had stopped. I am repeating again. So the energies of photon associated with the infrared region are not large enough to exert electron from the one level to another level, but the energy associated is quite adequate to induce vibrational excitations of covalently bonded atoms and groups. And if you recall the covalent bonds present in the molecules are not rigid sticks or rods holding two atoms together, they are more like stiff springs that can be rotated, provided they are bonded with single bond or they can be stretched or they can be bent or they can be the scissor. So all kind of things can happen when the energy is applied in the infrared region for these molecules. So these are different types of vibrational motions are characteristic to a molecules component atoms. So that means all organic and disorganic compounds will absorb infrared radiation that corresponds in energy to these vibrations and infrared spectrometers allow chemists to obtain a production spectra of components that are a unique reflections of their molecular structure. So coming or showing all kind of these vibration motions. In my last lecture we also discussed in about three triatomic bent molecules like water and also we looked into triatomic linear molecules like carbon dioxide and also we looked into possible vibrational motions for CH2 in a molecule. So now let us continue imagine a diatomic molecule to two balls of mass M connected by a spring with a force constant F that can be calculated by the following equation Hooke's law. If you recall that can give you vital information about what we are talking about. So here that can be given in the form of nu equals 1 over pi into square root of F over 2 M the vibrational motions and frequencies of a structure containing several balls of different masses connected by springs with different force constants can be studied using classical mechanics which can be correlated with the motion of a molecule. If you understand the classical mechanics studied from physics understanding and extrapolating that to chemical molecules would be rather simple. So that we can understand the relationship of the stretching frequency with respect to force constant and the mass of the two constituent atoms in the form of reduced mass. The simplest way to study polyatomic molecules is by treating various parts as diatomic species. At a given time if several atoms are there and several bonds are there understanding would be rather difficult. So what we should do is instead of making an attempt to study a polyatomic molecule you take a component as a diatomic species every segment and then we can try to understand so that interpretation would be rather easy. So this work well with one of the two one of the two atoms is not bonded to any other atom in the molecule. So that means we have a terminal atom other one is bonded to another. So in that case it helps very well for example if you take CH what have C may be connected to something else but CH is a terminal and similarly NH in case of methyl I mean we can consider CH or NH both have a terminal atom and similarly we can consider CH and OH in case of acetone. So something like this so we can always simplify for better understanding of all these facts that are governed by quantum mechanical rules. So now using Hooke's law the stretching frequency of a diatomic molecule can be readily calculated. So here the equation is simplified and more useful it is not very complicated it is very easy to remember nu equals bar that is wave number equals 130.3 is a constant we obtained after simplifying into square root of f over u where f is the force constant and mu is the reduced mass. Of course reduced mass how to determine we know if a diatomic molecule having two atoms of mass M1 and M2 then mu will be M1 into M2 divided by M1 plus M2. And at the end I would show you lot of examples and how to determine for a given molecule whether force constant or stretching frequency so one of those things can be conveniently measured and by simply determining the reduced mass. Reduced mass we know that if you know the atomic weight determining or calculating reduced mass would be very easy. So now I have given some here you can just look into it different groups I have considered and also the corresponding mass reduced mass also in atomic mass units I have given here and now force constant in Newton per meter is also given here for all these bonds I have shown on the left side and also the corresponding frequency also shown here stretching frequencies for these things are shown here. So by simply using the equation I showed you nu bar equals 130.3 into square root of f over mu so you can use that one and you can verify whether all these things are not it says you can consider as a practicing exercise and you can calculate for all for example reduced mass is already given 700 is already given force constant you should be able to determine this one or you assume this is given and this is given this not given and then you can calculate. So you practice with all those things to make yourself familiar with determining force constant later I will show you from force constant you can also estimate the bond distance that is very interesting before you get data from x-ray bond angles and bond distances at least bond strength you can determine simply by looking into some of these parameters. So f for elements on the periodic table increase from the left to right across first two rows so as the vibrational frequencies so the trends also you can see here the stretching frequency for elements on the periodic table increase from left to right across first two rows so as the vibrational frequencies so that you can clearly see here. So some of these trends are also very useful in remembering or understanding or comparing the data when we have two or three molecules having different groups on adjacent atoms. So now the knowledge of symmetry is very helpful to understand the factors influence the intensities of IR bands symmetry and point groups and characteristics table all these things comes in group theory I am sure there are good courses and books are there look into it make yourself familiar if you want to dig deep into understanding IR spectroscopy but as far as interpretation is concerned whatever I am dealing in this course is quite adequate. To observe vibrations in IR it must be anti-symmetric with respect to the molecular center of symmetry that is important to observe vibrations in IR infrared spectroscopy it must be anti-symmetric with respect to the molecular center of symmetry because this produces an oscillating dipole moment that can interact with the electric field of the radiation so observe light if the vibrations are symmetric with respect to the molecular symmetry then the vibrations are inactive IR inactive so one should remember if the vibrations are symmetric with respect to the molecular symmetry then the vibrations are IR inactive and you do not see any bands corresponding to that one for example if you consider CC double bond stretching vibration is not seen in IR and only the asymmetric modes are observed at 895 and 1200 centimeter minus one for example if you consider this example here dichloroethylene center of symmetry is there for this one and then if you consider symmetric and asymmetric stretching so this is not seen this is symmetric stretching it is not observed on the other hand asymmetric is observed and then hence we are seeing two bands at 895 and 1200 centimeter minus one so certain things one should remember now let us consider two molecules having acetylene all kind groups in it one is methyl acetylene and dimethyl acetylene so here observed at in case of methyl acetylene C triple bond C is observed at 2150 centimeter minus one whereas this one is inactive for the same reason I said and then if you consider CH2 vibrations wagging is there and twisting is there and then the twisting mode produces no change in dipole moment and hence IR inactive in symmetric molecules again in symmetric molecules you do not see these things because the twisting mode produces no change in the dipole moment and hence IR inactive in symmetrical molecules so few things one should remember so you should not worry why I am not getting the reason is given here to make you familiar with the what is symmetric stretching how it looks like I have put here these cartoons radial the symmetric in symmetric stretching you can see here it is moving in this direction and then anti-symmetric you can understand very nicely simultaneous moving one is coming in one is going out and then whereas in symmetric both are projecting out and then scissoring you can clearly see the direction this is a scissoring and then this is rocking in the moving in the same direction so this is rocking and this is wagging and then this is twisting here twisting would be something like this wagging is something like this and twisting is something like this a screw action so these are the fundamental stretch one can see symmetric and anti-symmetric symmetric stretching anti-symmetric in radial motion and in latitudinal you can see scissoring and rocking and longitudinal it is wagging and twisting so the six type of vibrational modes we come across in many molecules most of the molecules so now let us look into few more aspects let us consider coupled interactions what is coupled interactions consider two independent but identical diatomic molecules that vibrate with identical frequencies consider two independent but identical diatomic molecules that vibrate with identical frequencies so when they are part of a molecule they are mechanically coupled since the vibrations of any of these groups affect the other group for example you consider acetylene and then you consider diacetylene I have shown here but when they are sufficiently separated by distance little or no coupling are observed and the two frequencies will be similar if they are farther from each other so they would be looking identical in acetylene they are strongly coupled compared to diacetylene so for example if you consider here nu s 33 30 centimeter minus one and nu asymmetric is 32 95 centimeter minus one and delta nu the separation is about 35 centimeter minus one here 33 30 and 32 95 30 the separation between symmetric and asymmetric the difference between symmetric and asymmetric stretch is 35 centimeter minus one then when you look into here with a acetylene here symmetric one is 33 75 and asymmetric is 32 80 and the difference between symmetric and asymmetric is 95 centimeter minus one so now you can understand clearly but when they are sufficiently separated by distance little or no coupling are observed and the two frequencies will be identical 33 30 30 the difference is very marginal but when they are farther what happens they are independent look identical but when they are coupled they have an influence here that can be clearly seen here now let us consider another example to evaluate the coupled interaction first we consider acetylene and diacetylene now we consider two more molecules to have a clear idea about whatever we discussed again we consider here they are coupled but they are separated by a single cc bond now if you look into stretching frequency symmetric stretching is 1600 centimeter minus one and asymmetric stretching is 1640 always asymmetric will be having larger value compared to symmetric one that you should remember and the difference between symmetric and asymmetric is 40 centimeter minus one in case of acetylene besides about 35 centimeter minus one so now if you consider this allene here and nu s is 1070 centimeter minus one this is symmetric stretching and asymmetric stretching it is 1960 centimeter minus one and the separation is larger 890 centimeter minus one this confirms what we concluded about coupled interactions with no strong coupling the IRCC band is expected to be around 1600 centimeter minus one so another point to remember is a symmetric vibrational mode is higher in frequency and the symmetrical one is lower in frequency always irrespective of whether they are coupled or not what one should remember is symmetric vibrational mode is lower in frequency compared to the asymmetric vibrational mode the values are reflected here you can see clearly here so now little bit more information about coupled interactions so vibrations of two atomic groups are not coupled unless the individual frequencies are identical one should remember for example we have a double bond and a triple bond they are not going to be coupled so vibrations of two molecules are not coupled unless the individual frequencies are identical the coupling is stronger when the frequencies are the same the coupling is stronger when the frequencies are the same the strong coupling requires a common order between the groups so when the vibrations are orthogonal to each other coupling is negligible and the two groups are orthogonal to each other so in that case what happens this coupling is not observed even if it is observed it is very weak and it can be neglected the bending and stretching vibrations can couple provided the stretching bond form one side of the changing angle there is another point so bending and stretching vibrations can couple provided the stretching bond forms one side of the changing angle now let us look into few more terms that is fermi resonance over tone and combination what are those let us look into one more term called fermi resonance fermi resonance is a special case of mechanical coupling which results from a fundamental vibration with an over tone or combination now I would tell you what is over tone and what is combination fundamental IR transitions or absorptions take place between the ground state vibration level E naught and the first excited levels for example I can say E 1 equals delta E star equals E 1 star minus E naught this equals H nu naught and here E 1 equals delta E double star I have given equals E 1 double star minus E star equals H nu double star what are those new star and new double star new star and new are different vibrational frequencies that means a transition from ground state E naught to a higher energy level E 2 is known as over tone whereas a transition from E 1 to E once double star is known as combination one should know the difference between over tone and combination a transition from ground state E naught to a higher energy level E 2 is known as over tone whereas a transition from E 1 star to E 1 double star is known as combination within. So overtones and combinations are forbidden by simple harmonic oscillatory theory of molecular vibrations, but they become weakly allowed when unharmonicity is taken into consideration. So now let's look into this example here. I have displayed IR spectrum of benzyl chloride. So normally an overtone or combination band is very weak, but when fermi resonance occurs, sharing of intensity takes place and as a result it can become quite strong. So one such example I have shown here is for benzyl chloride. You focus your attention to CO stretching frequency since it has one CO. Now one can anticipate only one single stretching frequency for CO group here, but in case of benzyl chloride, two bands are observed nearly at 720 and 760 centimeter minus one. So they are called fermi doublets. The spectrum may suggest the presence of two kernel groups in the molecule is misnomer. So one should not confuse that they have two compounds in it. The lower frequency is due to the overtone of the CH out of plane bending mode at 865 centimeter minus one in fermi resonance with the CO mode. So the frequencies 1760 and 1865 into two 1730 are very close in frequency. So that means basically if you multiply by this one into two you get 1730. So that means they are very close in frequency. 1720 what we are getting and when you take the combination we get 1730. So this how sometime why that happens because CH out of plane, overtone of the CH out of plane. Here we get two frequencies. So now let's look into hydrogen bonding. How one can gauge hydrogen bonding using IR spectroscopy as a tool to what extra hydrogen bonding is there, whether weak hydrogen bonding is there, whether strong hydrogen bonding is there. That information of course it can come very nicely from IR spectroscopy or NMR spectroscopy but nevertheless IR can also give you some hint about the possible hydrogen bonding in molecules. Hydrogen bonding occurs between hydrogen atom bonded to an electronegative element such as oxygen or nitrogen to another atom via its bonding or non-bonding electrons that can overlap the S orbital of hydrogen or oxygen or nitrogen lone pairs or electron pi bonds. That means basically if you consider O here, consider water and if you take this, this can interact with so like this. So this is a typical hydrogen bonding. This can happen between hydrogen atom attached to an electronegative element and also it has in the within the molecule or in neighboring molecules we have another electronegative atom. So they can nicely overlap through this one S orbital. Hydrogen bonding will be stronger when the bond formed are linear. So it should be if it is linear like this 180 degree or close to 180 degree, you can assume that the bonds are very stronger that can also be reflected from the stretching frequency. How to assess the presence of hydrogen bonding using IR? So one thing is broadening of the band can be seen and also increase in intensity can be observed and also shift to lower wave numbers. So these three if you gauge you should be able to tell yes there is a hydrogen bonding in the given molecule. This can be directly assess it and understood from IR data. So one is broadening of the band and increase in intensity and shift to lower wave numbers. So hydrogen bond involved OH and NH bonds show stretching bands between 2500 to 3000 centimeter minus one lower than those without hydrogen bonding. The change in the stretching frequency is a measure of the strength of the hydrogen bonding which is in the order of four to six kilo calories per mole. Hydrogen bond involved OH and NH bonds show stretching bands between 2500 to 3500 centimeter minus one lower than those without hydrogen bonding. The change in the stretching frequency is a measure of the strength of hydrogen bonding which is in the order of four to six kilo calories. So let us learn more and also start looking into data and incidation or interpretation in my next lecture until then have an excellent time. Thank you.