 I once again welcome you all to MSB lecture series on interpretive spectroscopy. You must be very happy now because I will be solving from here onwards problems. One thing you should remember when we are learning science subjects and if we learn those subjects through problem and those subjects are not a problem at all, okay. For example, chemistry is not a problem if it is learned through problems. Physics is full of mathematics, full of problems and if we solve problems understanding mathematical logic everything would be very easy. Similarly, physics is full of problems and if we learn physics through problems and physics is not a problem. Similarly, spectroscopy and if we learn spectroscopy through problems using our basics it is not a problem. So it is enjoyable. So let us start problems and next 8 to 9 lectures I will be focusing on solving very very different type of problems. Having data from all these spectroscopic methods that I had discussed in my last 50 lectures or so. So let me begin with some simple questions, simple problems and then go on to little bit more complex problems. Let us look into mass spectrometer related problem here. The question is what are the masses of the charged fragments produced in the following cleavage pathways? Two reactions are there. What you have to do is you have to identify the masses of the charged fragments produced in the following pathways. And also the what kind of fragmentation happens is also given here. The first one is alpha cleavage of triethylamine. The second one is maclaferty rearrangement of 4-methyl 2-pentanone. Let us look into one at a time. In order to know about the fragment and its mass we should know the reaction. For example, what is the molecule we are considering and what kind of breakage happens and when the breakage happens what kind of fragments comes out. So let us begin first with alpha cleavage of triethylamine. Let me write here we have lone pair. To have a better picture of cleavage I have expanded something like this. Now the cleavage occurs here of course we are considering a radical cation. When it proceeds through alpha cleavage what we get is this one here. So if you just look into this one 1, 2, 3, 4, 5 are there C5, H12 and N. And if you look into M by Z value this will be 86. So that means 60 plus 12 plus 14 is 86. So this is the one they have asked for what are the mass of the charged fragment. In case of alpha cleavage of triethylamine the mass of the charged fragment produced is 86. Now let us look into the second one here. So this one is 4-methyl 2-pentanone. Now for convenience of showing this rearrangement I would write in a little different way. Of course this is where it is going to fragment. And now if you just see here now once we know that one the job is done for this one molecular weight is 100. Then this one is C6, H12, O. So this would give this rearrangement to this portion I am writing here. You just look here this is C3, H6, O. This is 58 and then this one is MZ equals to 36 plus 642. This is the one the question is asking for. So here this is one in case of this one the equals 58 and in this case equals 86. So this is the answer for this question. What are the masses of the charged fragments produced in the following cleavage pathways in case of alpha cleavage of triethylamine? So that means how to do this one while discussing mass spectrometry. I showed you the fragmentation of different functional groups in organic molecules. And if you are familiar with those things if you look into amides, amines, alcohols, ketones, aldehydes you will be knowing. Once you know those things you should be able to split accordingly to give the right kind of fragments the question is seeking from you people. So this is how you can do it. I hope you have understood this one. Now let us go to another very simple problem. What is the frequency of a wave with a wavelength of 200 centimeters? Of course we are familiar with energy how it is related to frequency equals h nu and nu equals C by lambda. Where C is the velocity of light it is 3 into 10 raised to 10 centimeters per second or 3 into 10 raised to 8 meters per second. So we can use here and 200 centimeters the wavelength given here. If you calculate what we get is of course here 1.5 it comes it will be 1.5 into 10 raised to 8 hertz. This is the answer very simple here. Once we know the equation from this one we can calculate wavelength if the frequency is given. If the frequency is given wavelength can be considered or if wavelength is given frequency can be calculated or you can also calculate the energy corresponding to that frequency of electromagnetic radiation. Now another one a radio transmits a frequency of 100 hertz. What is the wavelength of this one? Again you can use the same analogy here. So here 100 nu is given here. Nu equals 100 hertz is given here. We have to calculate the wavelength. So nu equals C by lambda in the same way lambda equals C by nu. So this will corresponds to C will be 3 into 10 raised to 10 and then here we have 100. So it gives so centimeters and it can convert this into 10 raised to 6 meters. So this is the answer for this one. Because 10 goes it becomes minus 2 it becomes minus 8 minus 2 it becomes 3 into 10 raised to 8. So one can convert some of those things very nicely provided. We remember this one very easy to remember. So now without calculations simply by reading you should be able to answer which molecule 25 haptadiene or 24 haptadiene would you predict to have a lower heat of hydrogenation. So just look into that one and probably try to write the molecular formula once if you write it you can notice that in 24 haptadiene what we have is we have conjugation. If the conjugation is there what happens pi pi star gap lowers as a result what happens reactivity increases. That means if you want to do hydrogenation it needs low heat of hydrogenation. So that means basically it is 24 haptadiene. So I would predict 24 haptadiene to have lower heat of hydrogenation than 25 haptadiene. This is due to the conjugation between the double bonds in 24 haptadiene which is stabilizing. So very simple answer and also very simple analogy. And also while discussing about UV spectroscopy we did mention that conjugation extensive conjugation brings the homo-lomo gap. One should remember that one. And what is the energy range for 400 nanometer and 500 nanometer in the UV spectrum where a compound absorbs. So here again we should use energy we have to calculate. Now earlier we are converting wavelength into frequency, frequency into wavelength. In this one what we should do is we should calculate the energy associated with 400 nanometer as well as 500 nanometer radiation. So we should use E equals h nu we should use that is 6 point converted into meter. So if you simplify this one what it gives is 4.965 into 10 raise to minus 19 joules. So what we have done is everything is in given in meter. For example velocity of light is 3 into 10 raise to 10 centimeters that should be made in 3 into 10 raise to 8. So here also we have made it 10 raise to 7. And then if you just take this one to numerator and we get this value here. So this is the one for 400 how about 500 in the same way we can calculate. So this will give a value of 3 point j that means what is the energy range? The energy range will be from 3.972 into 10 raise to minus 19 to 4.965 into 10 raise to minus 19 joules. So this is the range that is asked in this question. Now let us look into which one would absorb a longer wavelength 1, 3 cyclohexadiene or 1, 4 cyclohexadiene. Similar question I discussed about heptardiene. So now you can see here in this one 1, 3 we have conjugation and here no conjugation is there. So obviously this would absorb longer wavelength. So 1, 3 cyclohexadiene would absorb longer wavelength. More conjugation in the molecule longer the wavelength absorbed because the gap is lower. If the gap increases wavelength decreases and frequency increases. This also again is a simple question. Now where is radio frequency radiation as the energy scale of the electromagnetic spectrum? So radio frequency radiation includes radio waves that is used in NMR and micro waves that is used in EPR. Is at the low energy end of the electromagnetic spectrum? It is a type of non ionizing radiation. So this information you should know. Also in the beginning I have showed you the electromagnetic spectrum starting from UV to the higher energy or lower wavelength to higher wavelength. Try to be familiar with those things so that you know UV, visible, x-ray all those things where exactly the energy range falls in that electromagnetic radiation spectrum. So now there is a question. 1H requires 200 MHz of energy to maintain resonance in a magnetic field of 4.7 Tesla. If autumn x requires 150 MHz calculate the amount of energy required to flip the nuclear spin of x. Is it greater than energy required for hydrogen? So that means 1H nucleus requires 200 MHz of energy to undergo nuclear transition in a magnetic field of 4.7 Tesla. And if autumn x requires 150 MHz, so how much energy is needed for this x-nuclear to undergo nuclear transition? That is the question. That means is it greater than the energy required for hydrogen? One has to calculate. So here simply we should use the equation E equals H nu. And of course H value we know and then this value is given for H 150 MHz, 150 into 10 raise to 6. So this would come around 993 10 raise to minus 28 is equal to 9.93 into 10 raise to minus 26 joule. So this is for that means the energy required is 9.93 into 10 raise to minus 26 joule to cause resonance for x. Next we calculate for hydrogen. Similarly we can calculate for hydrogen here to know the difference. This is for x and this is for H. So the value is smaller here. If you see here the value is smaller. So you calculate this and you should be able to compare the values here. That is it. Now let us look into one more example here. Calculate the energy required to flip the spin of a nucleus at 400 MHz. How about the changing the frequency to 500 and 300? That means now you should calculate the energy required to perform transition of a spin at 400 MHz field strength. And then we have to see what would happen to the same energy would decrease or increase if you use frequency of 500 or 300 MHz. First let us calculate for 400 MHz. We will consider 400 MHz for that one. So what we get is 2.648 into 10 raise to minus 25 joules. And then if you consider the same for 500 what we get is 3.310 joules. And then we calculate for 300 MHz. 6 to 10 raise to minus 25 joules. So that means for 500 MHz the energy increases for 300 relative to 400 energy decreases. And this we know if you recall as magnetic field strength increases the gap whatever the gap is there that increases. So this is true 500 MHz we need high energy and 300 MHz we need low energy and 400 is in between. So for CHCl3 chloroform there is a peak at 2904 hertz in a 1HNMR spectra taken from a 400 MHz spectrometer. Convert this value to PPM. So that means hertz to PPM PPM to hertz we should be familiar. And how to calculate the chemical shift in PPM we have to divide frequency by spectrometer frequency that is it. For example here we know 2904 divided by 400 MHz so that would be 2904 divided by 400 into 10 raise to 6. So this would give 7.26 by 10 raise to 6 means it is 7.26 PPM parts per million. Of course if you look into 1HNMR spectrum of chloroform it shows a sharp peak at 7.26 PPM. So this is how we can calculate we can convert chemical shift into PPM. Why we represent chemical shift in PPM is when we represent chemical shift in PPM it is independent of field strength. That also we looked in some problems in my next lecture. Until then have an excellent time reading and solving more problems on spectroscopy to make yourself familiar with interpretation, characterization, elucidation of structures of new molecules that you see. Thank you.