Understanding UV Spectroscopy Challenges

Good morning students we have completed our UV spectroscopy till now and today I will discuss some problems from your previous year papers from the UV spectroscopy. The questions of UV spectroscopy are covered in the examinations of the students. I have discussed all those types of questions in today's video lecture. After watching today's video lecture, no problem of UV will remain untouched. So, let's see... First problem we will take. UV spectrum of acetone shows two peaks at 279 nm and 189 nm. Ok, you have to see this lambda max value. given here 279 nanometer pay or 189 nanometer pay or up sick I identify the electronic transition for each peak Yannick II you do up kid or no transitions here 279 and 189 you can see transitions key values have key apps a push rain to a seat on me a cup of a high acid on up couta acid on is ch3 CH3CdO and two lone pairs are present on oxygen So these two types of transition will show up Carbonyl group is Nπ and ππ I have explained in detail in electronic transitions If this is π, then the π one will excite and go to π then it will be pi pi star transition and n pi star is the non-bonding electron if this non-bonding will shift here then it will be n pi star transition now we have seen that the different molecular orbitals present are present like this the lowest energy of the lowest molecular orbital is sigma and then pi non-bonding and then pi anti-bonding and then sigma anti-bonding so you can see that the energy gap between n pi star is the least and the energy gap between pi pi star is more so for n pi star transition you have to require less energy and I told you in the whole UV spectroscopy that this point will be used many times energy is inversely proportional to wavelength. So less energy means wavelength of lambda max value that will be high. Okay. So n pi star value will be more and pi pi star value will be less. So n pi star that is at 278 nanometer and pi pi star transition will occur at 189 nanometer. So, you just have to tell here that your peak is at 279 nanometer, that is due to n pi star transition and your peak is at 189 nanometer, that is due to pi pi star transition. Right? Then next question. Which absorbs UV radiation? Which compounds will absorb UV radiation? You can ask this question in this way. Which compounds will show UV spectrum? And which compounds are UV active? The UV radiation that we will absorb, that compounds will be UV active. That means the spectrum of those compounds will come in UV region. That means your band will come, peak will come from 200 nanometer to 800 nanometer. This is the UV region. Now you see, your heptane is heptane in this heptane only sigma sigma star transition is possible. And the energy gap between sigma and sigma star is the most. I have shown you in the last slide. So its wavelength will be very less, so much that it will collide approximately at 125 nanometer. And our UV region starts from 200 nanometer, so your heptane UV radiation is absorbed. Similarly, ethanol. Now, in ethanol, there is a sigma-sigma star transition and a n-sigma star transition. But these two transitions are at less than 200 nanometers. This means that it will not absorb UV radiation. Then you take cyclohexane, in cyclohexane also you will have only sigma sigma star transition, it will not absorb you. Then your benzene and butadiene, in both of them what will be your transition? Pi, Pi star transition is done and you know that both have conjugation present, here there are two double bonds in conjugation. So your value is 217 nanometer. So it will absorb 1,3-butadiene in UV region. Similarly, 3 double bonds are in the conjugation in benzene. And benzene also shows pi-pi star transition at 255 nm. Similarly, benzoic acid will also show n-pi star transition apart from pi-pi star transition. Similarly, both will be in acetone. n-pi star and pi-pi star. Nitrobenzene and aniline will have Npi and Pi So you will know which one will show Your heptane will not show It will not do heptane, ethanol, benzoic acid, cyclohexane, benzene, 1,3-butadiene, acetone, aniline, nitrobenzene. So this is your second question. This type of question can be asked from you. Which of the following will absorb UV radiations? The next question is, aniline absorbs at 280 nanometer. Aniline absorbs at 280 nanometer. But in acidic medium, it absorbs at 203 nanometer. Why this is so? Now what is aniline in aniline? This is your aniline, NH2 group is present and lone pair is present on nitrogen. So in aniline, the lone pair on nitrogen atom that undergoes resonance with the pi electrons of the benzene ring. This lone pair will participate in resonance with the pi electrons. And you know that I am not making all the structures like this, 4 of your resonating structures will become more of aniline. So, you can say that because of NH2 group being present, your conjugation has increased. And you know that wherever your conjugation increases, your value of lambda wax increases. So, what we had also studied, that this NH2 group acts as oxochrome. Okay? Like benzene, benzene absorbs at 255 nanometer, but you applied enyl on benzene, so now it will absorb at 280 nanometer. Okay? So why did this happen? Because of conjugation, due to presence of lone pair on nitrogen, that's why absorption shifts to longer wavelength. But now, Now what will happen in acidic medium? Aniline in acidic medium exists in form of anilinium ion. This is your anilinium ion. So your aniline will exist in the form of anilinium ion in acidic medium. And in anilinium ion there is no lone pair on nitrogen. On nitrogen... there is no lone pair available above which participates in conjugation with pi electrons of benzene ring. So here conjugation has decreased in acidic medium. Conjugation decrease means lambda max will also decrease. It means it absorbs at slight lower wavelength as compared to aniline which normally absorbs. 280 nanometer but acidic medium may aniline that will absorb it 203 nanometer clear the next what is main consideration in choosing solvent for UV spectroscopy UV spectroscopy you have to choose solvent for UV spectroscopy so what is your main criteria okay so what is your main criteria when you take solvent that is transparent Transparent in UV region. What does it mean? Transparent in UV region means that the compound should not have any peak between 200 nanometer and 800 nanometer. So you have to take a solvent which is transparent in UV region. And secondly you take a solvent which is less polar. Why? Because there should be no interaction with solute molecule. There should be no interaction with solute molecule of solvent. So this is your main criteria for choosing solvent. So the three commonly used solvents that you take, the most commonly used solvent that you take is your 95% ethanol. 95% ethanol is also cheap. and it is a good solvent, means it has good dissolving power. Almost all the solutes can be dissolved into 95% ethanol. Then you can take water, hexane, cyclohexane. These are some solvents which you take in UV spectroscopy. If someone asks you in UV spectroscopy if you have benzene, chloroform, carbon tetra chloride, can you take these solvents? So you can't take these solvents. You can't take benzene, chloroform and carbon tetra chloride because the bands of these compounds come in the region 242-280 nm. And that will interfere with the Bands of your solute. Okay? So, you have to take a solvent whose band does not come in its own UV region. Right? The next. The following dienes, you have some dienes given and their value of lambda max is given. And it is said, identify which one is which. Okay? Now see the values given are 176nm, 211nm and 215nm. Now see the conjugation and non conjugation. I am starting with C. See double, single, single, double. This is isolated diene. Then see here double, single, double, single. See the difference? conjugated. Similarly, see here double single double single. So these are both your conjugated dienes and C is isolated. So C is isolated then it is understood that its value will be the lowest i.e. it will absorb at 176 nanometer. Now talk about these two A and B. Now in A and B you can see that this is cis. Same group on same side or same group on opposite side. So this B is your trans. And trans absorbs at greater wavelength as compared to cis. So trans will absorb at 215 nanometer. It has more conjugation as compared to cis. In which steric factor is... Conjugation is slightly less because of this. Conjugation is not as effective as trans. Trans will absorb at 215 nm. Cis will absorb at 211 nm. And isolated will absorb at 176 nm. Then next, how will you distinguish between following isomeric dienes by UV spectroscopy? You have these three dienes which are isomeric. How will you distinguish between these two? So you can see that this is Homoannular Diene and this is Heteroannular Diene Both of them are conjugated but this is isolated Diene So if it is isolated Diene then naturally its lambda max will be least Now we will talk about A and B. A and B are conjugated dienes present. We will see lambda max of A and B. Calculate it. This is homoannular diene. Parent value will be 253 nanometer. And see the ring residue. 1, 2, 3. Three ring residues are there. So, for 1, you told that 5 is an increment, so for 3 it will be 15. Okay. Then there is one exocyclic double bond. Which one? This one. Okay. This bond is exocyclic to this ring X. Okay. So, its increment is 5. So, this compound will absorb at 273 nanometer. Then this is heteroannular. What will be your parent value for heteroannular? 2, 1, 4. Then bring residues. See. 1, 2, 3. Again 3 ring residues. 15. Okay. Then see. Exo is here. Yes. This bond is exo to this ring x. So 5. This value is added. So this is your 234 nanometer. So you will get to know the values of all three. If you see the values of all three, then you will see the lambda max value. So what will be your trend? The one whose lambda x will be the highest will be your A, that is the homoannular one. Then the one whose will be less than that will be your B, that is the heteroannular one. Then your C will be the isolated one. So by comparing these lambda max values of these three isomeric dienes you can distinguish between these three. Clear? The next question is Of the following pair of compounds, which one will absorb at longer wavelength? You have given pairs. So, first pair is ethyl benzene. I will draw it. This is your ethyl benzene. And styrene. Styrene is this. CH double bond CH2. This is styrene. So, by looking at it, you must have understood that styrene will absorb at greater wavelength as compared to ethyl benzene. Okay? Why this is so? Because in styrene, you can see that pi electrons are present that are in conjugation with pi electrons of the benzene ring. Okay? So, this is... double bond is extending conjugation conjugation co increase car Raha hapka is system may though extended conjugation kibajai say is key Joe lambda max said that will be more as compared to ethyl benzene ethyl benzene may is type key co interaction possible nahi ha there is no extending conjugation in ethyl benzene So this was our first, from ethyl benzene and styrene, styrene will absorb on greater. Then on second example, CIS and trans-1,3-pentadiene. We did this above also, trans will absorb at a greater wavelength as compared to CIS. Again, the effective overlap is more in trans, so the extent of conjugation is more. Then you have acetone and benzophenone. Now what happened in benzophenone? If I write benzophenone here, then this is acetophenone. So acetophenone will absorb more on you because you see what happened in acetophenone. There is benzene ring. So there are three types of acetophenone. pi bonds in conjugation with this carbonyl group which is acetone They were not there. Okay. So more conjugation. More conjugation means lambda max will be more. Okay. So ethyl benzene and styrene will be more than styrene. Cis and trans. 1,3-pentadiene will be more than trans. Acetone and acetophenone will be more than acetophenone. Acetophenone will be more than acetophenone. Acetophenone will be more than acetophenone. Acetophenone will be more than acetophenone. Acetophenone will be more than acetophenone. Acetophenone will be more than acetophenone. Acetophenone will be more than acetophenone. Acetophenone will be more than acetophenone. Acetophenone will be more than acetophenone. Clear? Then next. An organic compound having formula C5H8 on hydrogenation yields N-pentane. So, any compound if it is giving N-pentane on hydrogenation, then N-pentane is straight hydrocarbon. So, this is your unsaturated hydrocarbon. This should also be a straight chain. Unsaturated straight chain hydrocarbon. You have now understood this much. Why did you understand this? How did you understand? Because it has given N-pentane on hydrogenation, which is straight. There is no branching in it. So the UV spectrum of this compound is coming at 176 nanometers. So, it is coming at 176 nanometer. So, what do you predict? That it cannot be conjugated diene because conjugated diene value will be above 200 nanometer. So, it can be isolated diene. And what structure will be isolated diene? CH2 double bond CH, CH2, CH. Double bond CH2. Okay. Or your simple straight chain. Alkyne can also be. Pentine. C triple bond CH. Okay. So it can be your isolated diene. Which you can say. One. Two. three four four we know the one for pantodine up key host at the head yeah up given panting who's at the head ticket you could don't know key hydrogenation job up currently so you will get n pentane then indicate the increasing order of wavelength of lambda max increasing order may arrange Karna hey in the UV region So you have to do it in increasing order. In this, you have to look at the conjugation. So here three double bonds are in conjugation. Then single came, then double came, single came, double came, single came. And again, then your three bonds of benzene ring are in conjugation. So there is a lot of conjugation in this. Here conjugation is double bond, single bond, double bond. Here you have two pi bonds in conjugation. Here I will write how many were there? Three of these rings, four, this one, fifth this and third this. That means there are eight pi bonds in conjugation. Here two pi bonds in conjugation. Here you see double, single, single, double. So there is no conjugation. There is no conjugation. In C structure. Okay. So increasing order. Increasing means. First of all your lower one. So C's wavelength will be the lowest. Then your B. And then your A. A's will be the highest. Due to more conjugation in A. The more your conjugation. The more your pi bonds. The more you. I have told you that the energy gap between HOMO and LUMO decreases. Difference between HOMO and energy gap decreases means your lambda will be high. So C will be the lowest, B will be the highest and A will be the highest. Then Yn sigma star transition shift to longer wavelength in moving from methyl chloride to methyl bromide to methyl iodide. Now why this happens? If I tell you the values, this absorbs at 174 nm, this absorbs at 205 nm and methyl iodide absorbs at 258 nm. So you can see that if you move from chloro to bromo to iodo, then n pi star transition is shifting towards longer wavelength. Why this is so? So this depends upon Electronegativity of the halogen atom present. What happens is that the more your electronegativity is, means electrons are tightly bound to you. And to be as tightly bound, that is, if you want to remove them, then you have to excite more energy. So more energy is required for n sigma star transition. Okay. The more electronegativity, the more energy you need for n pi star transition. And the more energy you need means your wavelength is less. So, you know what is the order of electronegativity? Chlorine is more than bromine as compared to iodine. So, what will happen to you? Chloro key up. you can say that chloromethane's N-sigma star transition will come at lower wavelength i.e. 174 nm and when you move to methyl iodide, this will absorb at 258 nm because its iodine's lactonegativity is low, so it needs less energy for excitation and less energy means lambda max will be more. Then you can calculate lambda max of any compound. We have done woodward feasible rules so you can calculate lambda max of any compound. I will give you some examples. This is your system. This is your acyclic. There is no ring. This is acyclic conjugated diene. So, for acyclic conjugated diene, we have read the parent value of lambda max, that is 217 nanometer. Okay. Apart from that, you can see that there are two alkyl groups present in this. So, for 1, 5 and for 2, 10. So, this compound will absorb at 227 nanometer. Okay. The next, we will calculate this compound. So, first of all, you have to analyze the system. What is this? So, this is your alpha functional group. Your carbonyl, right? So, next carbon from carbonyl is your alpha and this carbon is your beta. So, this is alpha beta unsaturated. And this unsaturation is present in your five-membered ring. This ring is your five-membered ring. Okay? So, if your alpha, beta, unsaturated carbonyl compound is in 5-membered ring, then we had read the parent value, which is 202 nanometer. Then you see what else is present in it. There is one ring residue at alpha. If there is one ring residue at alpha, then its value is 10. Then you have one ring residue at beta. So, beta has a value of 12. And beta has an alkyl group. 1 CH3 group at beta. So, again its value will be 12. If you don't have anything else in this system, then sum up these. So, this comes out to be 236 nanometer. So, this compound will absorb at 236 nanometer. nanometer. Then calculate lambda max of the following. This is the structure given to you. And you have to find out this. So you can see that alpha beta is unsaturated ketone. It is present in six members. So the parent value of compound will be alpha, beta, this is your gamma, this is delta. So parent value is 215. nanometer. Demering residues, see, are at alpha. 1 delta per hai. So, ring residue at alpha, alpha ki value 10. Ring residue at delta, delta ki value 18. Okay. Aur dekhye, yeh jo aapka hai, this is exocyclic double bond to this ring A. Okay. To aapka, ek aapka exocyclic double bond ho gaya, uski value 5. Okay. Then this is. Double bond extending conjugation. Conjugation extending. Your parent system is only alpha, beta unsaturated. So this is double bond extending conjugation. And we have to add increment of 30 nanometer. Then you are seeing, in one ring only two double bond are present in conjugation. So this is a homoannular. conjugated diene or homoannular conjugated diene ke liye you have to add increment of 39 nanometer. Thik hai. Yeh saari cheeje aap isko add kar dijiye and this comes out to be 317 nanometer. To yaani this compound will absorb at 317 nanometer. Thik hai. So this was all about your problems on UV spectroscopy. I have discussed all types of your problems in this video lecture. Apart from these problems, nothing will come in your exams. So watch this video lecture very carefully. First watch the videos, then watch this video lecture in which I have discussed problems. So all your problems will be solved with UV spectroscopy. Thanks for watching the video.

After watching today's video lecture, no problem of UV will remain untouched. So, let's see... First problem we will take.

UV spectrum of acetone shows two peaks at 279 nm and 189 nm. Ok, you have to see this lambda max value. given here 279 nanometer pay or 189 nanometer pay or up sick I identify the electronic transition for each peak Yannick II you do up kid or no transitions here 279 and 189 you can see transitions key values have key apps a push rain to a seat on me a cup of a high acid on up couta acid on is ch3 CH3CdO and two lone pairs are present on oxygen So these two types of transition will show up Carbonyl group is Nπ and ππ I have explained in detail in electronic transitions If this is π, then the π one will excite and go to π then it will be pi pi star transition and n pi star is the non-bonding electron if this non-bonding will shift here then it will be n pi star transition now we have seen that the different molecular orbitals present are present like this the lowest energy of the lowest molecular orbital is sigma and then pi non-bonding and then pi anti-bonding and then sigma anti-bonding so you can see that the energy gap between n pi star is the least and the energy gap between pi pi star is more so for n pi star transition you have to require less energy and I told you in the whole UV spectroscopy that this point will be used many times energy is inversely proportional to wavelength. So less energy means wavelength of lambda max value that will be high. Okay.

So n pi star value will be more and pi pi star value will be less. So n pi star that is at 278 nanometer and pi pi star transition will occur at 189 nanometer. So, you just have to tell here that your peak is at 279 nanometer, that is due to n pi star transition and your peak is at 189 nanometer, that is due to pi pi star transition. Right?

Then next question. Which absorbs UV radiation? Which compounds will absorb UV radiation? You can ask this question in this way. Which compounds will show UV spectrum?

And which compounds are UV active? The UV radiation that we will absorb, that compounds will be UV active. That means the spectrum of those compounds will come in UV region.

That means your band will come, peak will come from 200 nanometer to 800 nanometer. This is the UV region. Now you see, your heptane is heptane in this heptane only sigma sigma star transition is possible.

And the energy gap between sigma and sigma star is the most. I have shown you in the last slide. So its wavelength will be very less, so much that it will collide approximately at 125 nanometer.

And our UV region starts from 200 nanometer, so your heptane UV radiation is absorbed. Similarly, ethanol. Now, in ethanol, there is a sigma-sigma star transition and a n-sigma star transition.

But these two transitions are at less than 200 nanometers. This means that it will not absorb UV radiation. Then you take cyclohexane, in cyclohexane also you will have only sigma sigma star transition, it will not absorb you.

Then your benzene and butadiene, in both of them what will be your transition? Pi, Pi star transition is done and you know that both have conjugation present, here there are two double bonds in conjugation. So your value is 217 nanometer.

So it will absorb 1,3-butadiene in UV region. Similarly, 3 double bonds are in the conjugation in benzene. And benzene also shows pi-pi star transition at 255 nm.

Similarly, benzoic acid will also show n-pi star transition apart from pi-pi star transition. Similarly, both will be in acetone. n-pi star and pi-pi star. Nitrobenzene and aniline will have Npi and Pi So you will know which one will show Your heptane will not show It will not do heptane, ethanol, benzoic acid, cyclohexane, benzene, 1,3-butadiene, acetone, aniline, nitrobenzene.

So this is your second question. This type of question can be asked from you. Which of the following will absorb UV radiations? The next question is, aniline absorbs at 280 nanometer. Aniline absorbs at 280 nanometer.

But in acidic medium, it absorbs at 203 nanometer. Why this is so? Now what is aniline in aniline? This is your aniline, NH2 group is present and lone pair is present on nitrogen.

So in aniline, the lone pair on nitrogen atom that undergoes resonance with the pi electrons of the benzene ring. This lone pair will participate in resonance with the pi electrons. And you know that I am not making all the structures like this, 4 of your resonating structures will become more of aniline. So, you can say that because of NH2 group being present, your conjugation has increased.

And you know that wherever your conjugation increases, your value of lambda wax increases. So, what we had also studied, that this NH2 group acts as oxochrome. Okay?

Like benzene, benzene absorbs at 255 nanometer, but you applied enyl on benzene, so now it will absorb at 280 nanometer. Okay? So why did this happen?

Because of conjugation, due to presence of lone pair on nitrogen, that's why absorption shifts to longer wavelength. But now, Now what will happen in acidic medium? Aniline in acidic medium exists in form of anilinium ion.

This is your anilinium ion. So your aniline will exist in the form of anilinium ion in acidic medium. And in anilinium ion there is no lone pair on nitrogen.

On nitrogen... there is no lone pair available above which participates in conjugation with pi electrons of benzene ring. So here conjugation has decreased in acidic medium.

Conjugation decrease means lambda max will also decrease. It means it absorbs at slight lower wavelength as compared to aniline which normally absorbs. 280 nanometer but acidic medium may aniline that will absorb it 203 nanometer clear the next what is main consideration in choosing solvent for UV spectroscopy UV spectroscopy you have to choose solvent for UV spectroscopy so what is your main criteria okay so what is your main criteria when you take solvent that is transparent Transparent in UV region. What does it mean? Transparent in UV region means that the compound should not have any peak between 200 nanometer and 800 nanometer.

So you have to take a solvent which is transparent in UV region. And secondly you take a solvent which is less polar. Why?

Because there should be no interaction with solute molecule. There should be no interaction with solute molecule of solvent. So this is your main criteria for choosing solvent. So the three commonly used solvents that you take, the most commonly used solvent that you take is your 95% ethanol.

95% ethanol is also cheap. and it is a good solvent, means it has good dissolving power. Almost all the solutes can be dissolved into 95% ethanol. Then you can take water, hexane, cyclohexane. These are some solvents which you take in UV spectroscopy.

If someone asks you in UV spectroscopy if you have benzene, chloroform, carbon tetra chloride, can you take these solvents? So you can't take these solvents. You can't take benzene, chloroform and carbon tetra chloride because the bands of these compounds come in the region 242-280 nm.

And that will interfere with the Bands of your solute. Okay? So, you have to take a solvent whose band does not come in its own UV region.

Right? The next. The following dienes, you have some dienes given and their value of lambda max is given. And it is said, identify which one is which.

Okay? Now see the values given are 176nm, 211nm and 215nm. Now see the conjugation and non conjugation. I am starting with C. See double, single, single, double.

This is isolated diene. Then see here double, single, double, single. See the difference? conjugated.

Similarly, see here double single double single. So these are both your conjugated dienes and C is isolated. So C is isolated then it is understood that its value will be the lowest i.e. it will absorb at 176 nanometer.

Now talk about these two A and B. Now in A and B you can see that this is cis. Same group on same side or same group on opposite side.

So this B is your trans. And trans absorbs at greater wavelength as compared to cis. So trans will absorb at 215 nanometer. It has more conjugation as compared to cis.

In which steric factor is... Conjugation is slightly less because of this. Conjugation is not as effective as trans. Trans will absorb at 215 nm. Cis will absorb at 211 nm.

And isolated will absorb at 176 nm. Then next, how will you distinguish between following isomeric dienes by UV spectroscopy? You have these three dienes which are isomeric.

How will you distinguish between these two? So you can see that this is Homoannular Diene and this is Heteroannular Diene Both of them are conjugated but this is isolated Diene So if it is isolated Diene then naturally its lambda max will be least Now we will talk about A and B. A and B are conjugated dienes present. We will see lambda max of A and B.

Calculate it. This is homoannular diene. Parent value will be 253 nanometer. And see the ring residue. 1, 2, 3. Three ring residues are there.

So, for 1, you told that 5 is an increment, so for 3 it will be 15. Okay. Then there is one exocyclic double bond. Which one?

This one. Okay. This bond is exocyclic to this ring X. Okay. So, its increment is 5. So, this compound will absorb at 273 nanometer.

Then this is heteroannular. What will be your parent value for heteroannular? 2, 1, 4. Then bring residues. See.

1, 2, 3. Again 3 ring residues. 15. Okay. Then see. Exo is here. Yes.

This bond is exo to this ring x. So 5. This value is added. So this is your 234 nanometer.

So you will get to know the values of all three. If you see the values of all three, then you will see the lambda max value. So what will be your trend? The one whose lambda x will be the highest will be your A, that is the homoannular one. Then the one whose will be less than that will be your B, that is the heteroannular one.

Then your C will be the isolated one. So by comparing these lambda max values of these three isomeric dienes you can distinguish between these three. Clear? The next question is Of the following pair of compounds, which one will absorb at longer wavelength? You have given pairs.

So, first pair is ethyl benzene. I will draw it. This is your ethyl benzene. And styrene. Styrene is this.

CH double bond CH2. This is styrene. So, by looking at it, you must have understood that styrene will absorb at greater wavelength as compared to ethyl benzene.

Okay? Why this is so? Because in styrene, you can see that pi electrons are present that are in conjugation with pi electrons of the benzene ring.

Okay? So, this is... double bond is extending conjugation conjugation co increase car Raha hapka is system may though extended conjugation kibajai say is key Joe lambda max said that will be more as compared to ethyl benzene ethyl benzene may is type key co interaction possible nahi ha there is no extending conjugation in ethyl benzene So this was our first, from ethyl benzene and styrene, styrene will absorb on greater. Then on second example, CIS and trans-1,3-pentadiene. We did this above also, trans will absorb at a greater wavelength as compared to CIS.

Again, the effective overlap is more in trans, so the extent of conjugation is more. Then you have acetone and benzophenone. Now what happened in benzophenone?

If I write benzophenone here, then this is acetophenone. So acetophenone will absorb more on you because you see what happened in acetophenone. There is benzene ring.

So there are three types of acetophenone. pi bonds in conjugation with this carbonyl group which is acetone They were not there. Okay.

So more conjugation. More conjugation means lambda max will be more. Okay. So ethyl benzene and styrene will be more than styrene.

Cis and trans. 1,3-pentadiene will be more than trans. Acetone and acetophenone will be more than acetophenone.

Acetophenone will be more than acetophenone. Acetophenone will be more than acetophenone. Acetophenone will be more than acetophenone.

Acetophenone will be more than acetophenone. Acetophenone will be more than acetophenone. Acetophenone will be more than acetophenone. Acetophenone will be more than acetophenone.

Acetophenone will be more than acetophenone. Clear? Then next. An organic compound having formula C5H8 on hydrogenation yields N-pentane. So, any compound if it is giving N-pentane on hydrogenation, then N-pentane is straight hydrocarbon.

So, this is your unsaturated hydrocarbon. This should also be a straight chain. Unsaturated straight chain hydrocarbon. You have now understood this much.

Why did you understand this? How did you understand? Because it has given N-pentane on hydrogenation, which is straight. There is no branching in it. So the UV spectrum of this compound is coming at 176 nanometers.

So, it is coming at 176 nanometer. So, what do you predict? That it cannot be conjugated diene because conjugated diene value will be above 200 nanometer.

So, it can be isolated diene. And what structure will be isolated diene? CH2 double bond CH, CH2, CH. Double bond CH2. Okay.

Or your simple straight chain. Alkyne can also be. Pentine. C triple bond CH. Okay.

So it can be your isolated diene. Which you can say. One. Two. three four four we know the one for pantodine up key host at the head yeah up given panting who's at the head ticket you could don't know key hydrogenation job up currently so you will get n pentane then indicate the increasing order of wavelength of lambda max increasing order may arrange Karna hey in the UV region So you have to do it in increasing order.

In this, you have to look at the conjugation. So here three double bonds are in conjugation. Then single came, then double came, single came, double came, single came.

And again, then your three bonds of benzene ring are in conjugation. So there is a lot of conjugation in this. Here conjugation is double bond, single bond, double bond.

Here you have two pi bonds in conjugation. Here I will write how many were there? Three of these rings, four, this one, fifth this and third this.

That means there are eight pi bonds in conjugation. Here two pi bonds in conjugation. Here you see double, single, single, double.

So there is no conjugation. There is no conjugation. In C structure. Okay. So increasing order.

Increasing means. First of all your lower one. So C's wavelength will be the lowest.

Then your B. And then your A. A's will be the highest.

Due to more conjugation in A. The more your conjugation. The more your pi bonds.

The more you. I have told you that the energy gap between HOMO and LUMO decreases. Difference between HOMO and energy gap decreases means your lambda will be high. So C will be the lowest, B will be the highest and A will be the highest.

Then Yn sigma star transition shift to longer wavelength in moving from methyl chloride to methyl bromide to methyl iodide. Now why this happens? If I tell you the values, this absorbs at 174 nm, this absorbs at 205 nm and methyl iodide absorbs at 258 nm.

So you can see that if you move from chloro to bromo to iodo, then n pi star transition is shifting towards longer wavelength. Why this is so? So this depends upon Electronegativity of the halogen atom present. What happens is that the more your electronegativity is, means electrons are tightly bound to you. And to be as tightly bound, that is, if you want to remove them, then you have to excite more energy.

So more energy is required for n sigma star transition. Okay. The more electronegativity, the more energy you need for n pi star transition. And the more energy you need means your wavelength is less. So, you know what is the order of electronegativity?

Chlorine is more than bromine as compared to iodine. So, what will happen to you? Chloro key up.

you can say that chloromethane's N-sigma star transition will come at lower wavelength i.e. 174 nm and when you move to methyl iodide, this will absorb at 258 nm because its iodine's lactonegativity is low, so it needs less energy for excitation and less energy means lambda max will be more. Then you can calculate lambda max of any compound. We have done woodward feasible rules so you can calculate lambda max of any compound.

I will give you some examples. This is your system. This is your acyclic. There is no ring.

This is acyclic conjugated diene. So, for acyclic conjugated diene, we have read the parent value of lambda max, that is 217 nanometer. Okay.

Apart from that, you can see that there are two alkyl groups present in this. So, for 1, 5 and for 2, 10. So, this compound will absorb at 227 nanometer. Okay. The next, we will calculate this compound.

So, first of all, you have to analyze the system. What is this? So, this is your alpha functional group.

Your carbonyl, right? So, next carbon from carbonyl is your alpha and this carbon is your beta. So, this is alpha beta unsaturated. And this unsaturation is present in your five-membered ring. This ring is your five-membered ring.

Okay? So, if your alpha, beta, unsaturated carbonyl compound is in 5-membered ring, then we had read the parent value, which is 202 nanometer. Then you see what else is present in it.

There is one ring residue at alpha. If there is one ring residue at alpha, then its value is 10. Then you have one ring residue at beta. So, beta has a value of 12. And beta has an alkyl group.

1 CH3 group at beta. So, again its value will be 12. If you don't have anything else in this system, then sum up these. So, this comes out to be 236 nanometer.

So, this compound will absorb at 236 nanometer. nanometer. Then calculate lambda max of the following. This is the structure given to you.

And you have to find out this. So you can see that alpha beta is unsaturated ketone. It is present in six members.

So the parent value of compound will be alpha, beta, this is your gamma, this is delta. So parent value is 215. nanometer. Demering residues, see, are at alpha.

1 delta per hai. So, ring residue at alpha, alpha ki value 10. Ring residue at delta, delta ki value 18. Okay. Aur dekhye, yeh jo aapka hai, this is exocyclic double bond to this ring A.

Okay. To aapka, ek aapka exocyclic double bond ho gaya, uski value 5. Okay. Then this is.

Double bond extending conjugation. Conjugation extending. Your parent system is only alpha, beta unsaturated.

So this is double bond extending conjugation. And we have to add increment of 30 nanometer. Then you are seeing, in one ring only two double bond are present in conjugation. So this is a homoannular. conjugated diene or homoannular conjugated diene ke liye you have to add increment of 39 nanometer.

Thik hai. Yeh saari cheeje aap isko add kar dijiye and this comes out to be 317 nanometer. To yaani this compound will absorb at 317 nanometer. Thik hai. So this was all about your problems on UV spectroscopy.

I have discussed all types of your problems in this video lecture. Apart from these problems, nothing will come in your exams. So watch this video lecture very carefully.

First watch the videos, then watch this video lecture in which I have discussed problems. So all your problems will be solved with UV spectroscopy. Thanks for watching the video.

Transcript for:Understanding UV Spectroscopy Challenges

Transcript for:
Understanding UV Spectroscopy Challenges