 once again welcome you all to MSP lecture series on interpretive spectroscopy. So now we are discussing about IR spectroscopy. While it is not necessary to remember all of the peaks for each functional group, learning a few key characteristics will be enough to answer 99 percent of all IR related questions. I showed you a few IR spectra in my last lecture. If you just look into it, you can identify certainly functional groups presence, and also the fingerprint region. So fingerprint region would not really tell you much information, except for the fact that that can be compared with the known samples pure spectrum. But if you want to see whether you have performed some reaction and product have changed to something else, that information would come only when you focus your attention to functional groups. And then that's good enough to answer 99 percent of all IR related questions. For example, if you have a formaldehyde spectrum is that I will show it is not there. One can see that there are few different possible ways that bonds can behave, which leads to many different peaks. So I would come back to that one again before that let's look into the all kinds here. In case of all kinds, we look into aliphatic hydrocarbons and also we look into cyclohexane. Now we shall look into all kinds here. Here in this case, the stretch would come around CH around 33, 10 centimeter minus 1. And alkyl CH stretch would come around 28, 57 to 29, 41 centimeter minus 1. And C triple bond C stretch would be around 2, 1, 1, 9 centimeter minus 1. Then when you look into CH bend, which is attached to a triple bond, the war tone will be around 12, 50 centimeter minus 1. And then CH bend, fundamental vibration would appear at 6, 30 centimeter minus 1. Strong and broad absorptions around 700 to 6, 110. And war tones in the range of 13, 70 to 12, 20 centimeter minus 1 are characteristic of all kinds. If you have an alkanic group, you should focus your attention to this range. In case of aromatic hydrocarbons, CH stretch would appear around 3,008 centimeter minus 1. And CH 3, CH stretch would be around 28, 50 to 29, 40. And then war tone or combination bands would be in the range of 2000 to 16, 67 centimeter minus 1. And C double bond C present in the ring would be having stretching frequency around 16, 0, 5. And also 14, 95 and 14, 66 centimeter minus 1. In-plane CH bend would be 1, 0, 5, 2 and 1, 0, 2, 2 centimeter minus 1. And out-of-plane CH aromatic bend would be around 7, 41 centimeter minus 1. And out-of-plane C double bond C bend would be 4, 38 centimeter minus 1. So these are the few things for your information. There's no need to remember all those things. So now what would happen if you have halogen bonds are there? So carbon to halogen bonds such as carbon to fluorine, carbon to chlorine, carbon to bromine. In that case, what happens? CH stage would be in the range of 1110 to 1350 centimeter minus 1. Whereas CCL bond stage would be 550 to 8,500 centimeter minus 1. 850 centimeter minus 1, not 8,500. And then CBR stage would be 550 to 690 centimeter minus 1. The strong CH to wagging bond for CH to X will be around 1150 to 1300 centimeter minus 1. So these are the important regions as far as CX bond where X is a halogen is concerned. The spectrum of propanamide is given here. And you can identify NH stretching vibration is around 3,500 to 3,200 centimeter minus 1. And also we have a carbonyl group. So it comes around 1650 to 1690 centimeter minus 1. In the same region, NH and CN combined vibrations also come, 1590 to 1650 centimeter minus 1. And then CH vibration, one can identify here. And then this 1500, if you take 1500 to this range, this is called fingerprint region. Then let us look into the structure of anizole now. So anizole is methaxybenzene. Its molecular formula is C7H8O. Anizole has 5 ESP to CH links. So here 1, 2, 3, 4, 5 ESP to CH links are there. And they have between 2960 to 2838 centimeter minus 1. And 3 C double bond C aromatic linkages are there between 1600 to 1500 centimeter minus 1. And 1 CO linkage is there here at 1249 centimeter minus 1. And 1 aliphatic CO bond, this one, this aliphatic CO bond is at 1100 centimeter minus 1. And also 3 ESP3 CH linkages here. So these are the bonds you can anticipate some bands in IR spectrum. So now we should look for the associated frequencies in its IR spectrum for its identification. So now we have identified from the prior information we have about the different searching frequencies and other things. And we have identified whether we have ESP, ESP3, ESP3 carbon is there. And when we look into spectrum, we can see whether all these things are there or not. So 2960 to 2988. So we have here. And then 3 C double bond C around 1600 to 1500 we have here. And then CO linkages at 1249. So here 1249 it should come here. And then 3 ESP3 CH linkages. The CO bonds have two pinnacles you can see here. One because of the asymmetrical COC at 1250. So 1250 somewhere here this is. And the other at 1040, this is the one, this is the one 1040. Because of symmetrical stretch, asymmetric bending is stronger since the double bond character increases due to the resonance. The double bond feature gives increased bonding strength and improves the elongation frequency. So this is the spectrum of any sort. Whatever the bands we identified from prior knowledge could be seen here with little or no difference. So now IR spectroscopy typically has a spectrum reading organized as shown below. So we didn't discuss about those things. For example, say transmittance versus the frequency in wave numbers. So if you just consider this one. This spectrum is for formaldehyde and is typical of most infrared spectra. IR readings are inverted compared with UV visible spectra. Thus a sample that did not absorb at all would record a horizontal line at 100% transmittance while one that absorbs a significant amount would be towards the bottom. So this is in contrast to the spectrum that is recorded and plotted in case of UV visible or NMR. The advantage is, as it is clearly mentioned, the horizontal line at 100% transmittance shows there's no absorption at all. But only wherever absorption is there, that will be coming towards the bottom. Each peak dips on the graph represent various bond characteristics and because most functional groups have specific bonds, they can be identified due to this reason. A typical IR spectrum will be plotted like transmittance versus frequency. Or, you see, one can look like this, then this will be absorbance. So this would look rather odd. That's the reason the convenient ways to plot in the other way I showed. So something like this. So the absorption spectrum here, the complexity of infrared spectra in the range 1450 to 600 centimeter minus 1, makes it difficult to assign all the absorption bands. And because of the unique pattern, it is often called the fingerprint region. That's what I told you in my earlier lecture. It is not necessary to remember any peak in this region as they do not help you determine any significant functional groups. This is only to identify, yes, this pattern is there or something is there. Yes, fingerprint region. There's no need to worry about interpreting all the bands. It is a little bit more complex and it's unnecessary. Absorption bands in the region, 4,000 to 1450 region are usually due to stretching vibrations of diatomic units and this is sometimes called the group frequency region. This is also quite important. In the IR spectrum shown below, so here, focus on 3750 areas. There is a sharp peak at 700 here. This corresponds to a CO group and it is most likely a ketone or an aldehyde. So that means you see here, you should be able to think that this has to be due to a ketone or an aldehyde and if you see double bond O is there. No broad or sharp groups above 3000 mark. So that means there is no OH or NH. Why we decided it's ketone or aldehyde is because no OH group is there. So carbide acid is ruled out. So there is a sharp peak just below 3000 maybe due to CH, CH2 under CH3 bonds but it could also be part of aldehyde CH bond. So that means it gives you some idea about the what type of molecule this IR is representing. If the choices are between 2-meter cyclobutanol, 3-meter butanol and 2-chloropentonic acid, certainly it should be clear to recognize that this is only due to 3-meter 2-butanol has both necessary CO bond and there is no OH bond at all. So that means when you have option of identifying the spectrum with 3 or more choice by simply analyzing we should be able to arrive at the right kind of molecule this IR spectrum is representing. So now I have given little bit elaborated and extensive table here for different groups and also the range in which absorption bands are seen and also what kind of assignments and comments can be given for those things. When you look into BOH the range is 3300 to 3200. It is sharp, strong, broad band due to OH states, OH states here. When we have BH or BH2 it appears around 2650 to 2350 sharp a doublet for BH2 here and then also we can see 1200 to 1150 medium strong or 1980 to 920 medium and again 1430 medium strong. So they are respectively assigned to BH2 deformation of BH, BH2 wag and benzene ring vibration. That means if there is a benzene ring is there and then if we look into BN it shows the band in the region of 1460 to 1330 very strong stretch borosins and aminoborans. If you have borosins and aminoborans possibly you can identify using this range and BO 1380 to 1310 very, very strong. This is a stretch boronates and boronic acids and then if you have CBC 1280 to 1250 and anti-symmetric stretch we can identify. Similarly if you look into silicon compounds if you have silicon hydroxide SiOH bond we can see in the range 3700 to 3200 and OH stretch this is very similar to what we come across in case of alcohols or OH and then if you have SiH or SiH2 or SiH3 that means primary saline secondary saline or tertiary saline primary saline tertiary saline or tertiary saline is not there here yes tertiary saline is the SiH, SiH2 or SiH3 we can see they all appear in the range of 900 to 820 centimeter minus 1 and SiOH stretch for the other one and then this is SiH stretch here and then SiH deformation or WAG will appear in 950 to 800 and SiA or if it is connected to a phenyl group or aromatic group it comes around 1430 as a medium and strong and 1100 very strong and this indicates this ring mode here and then SiOC aliphatic 1100 to 1050 very, very strong and this is due to anti-symmetric stretch and then SiOAR will show around 970 to 920 very strong it is again SiO stretch and if you have SiOSI we can see in the range of 1100 to 1000 very strong and it is again due to anti-symmetric stretch. So, now we about phosphorus compounds if we have a direct pH bond phosphorus acids and esters and phosphines we come across 24, 25 to 23, 25 centimeter minus 1 this is essentially due to new pH and then we can also see bands in the region 23, 20 to 20 to 40 this is again pH stretch this is very sharp and if you see 10, 90 to 1180 medium strong pH to deformation and then 990 to 910 medium it is for pH bag for POH in phosphoric acid or phosphorus acids and esters and salts they appear in the range 2700 to 2110 this is OH stretch 1 or 2 broad and often they are weak and then if you see something around 1040 to 920 strong absorption band that is again due to POH stretch and POC in aliphatic compounds will be in this range 1050 to 950 very strong it is POC anti-symmetric and then 830 to 750 POC symmetric stretch here OME and OET compounds of P1D. So, trimethyl phosphate or triethyl phosphates to show this kind of range for POC symmetric stretch and then aromatic compounds if you have PPH3 or something like that RPOPH thrice triphenyl phosphate we come across 1250 to 1160 very strong aromatic CO stretch and then 1050 to 870 we have PO stretch. So, this should be C double bond C not CO stretch and then PC aromatic compounds we have 1450 to 1430 P directly on a ring sharp band and then quaternary aromatic 1110 to 1090 P directly on a ring sharp band is observed and in many compounds pentavalent tetra coordinated compounds we come across P double bond O and also we can identify whether the PO double bond O is from the aliphatic compounds or aromatic compounds. They have distinct range again for aliphatic ones P double bond O appears in the range of 1260 to 1240 they are usually strong and sharp. If you see in the range of 1350 to 1300 indicates that we have aryl oxide groups are present on phosphorus aryl oxide phosphates in that case P double bond O would appear around 1350 to 1300 a lower frequency pH stretch and then 1250 to 1180 when OH is on P and in phosphine oxides typically it will be in the range of 1140 to 1200 this is for new PO. So, sometime we have phosphines and after performing catalysis or something if they are converted into corresponding phosphine oxides immediately one can look into IR and get some idea why catalytic activity has brought down the reduced maybe catalyst phosphine present is decomposed or it has come out of the metal in the form of a phosphine oxide this information also comes from IR spectroscopy and then if you look into SH in thiols like mercaptans you can see 2580 to 2500 this is due to SH stretch strong in the Raman and then CS 720 to 600 weak CS stretch strong in the Raman and then SS in disulfides 550 to 450 very weak or absent again SS stretch yes in active in the Raman and then SH double bond wasulfoxides they have this characteristic region 1060 to 1020 and dialkyl sulfides in the range 1220 to 1190 both are due to SH double bond work stretch and then SO2 sulfone sulfonamide sulfonic acid sulfonate sulfonyl chloride dialkyl sulfates sulfonyl fluoride all come in the range 1300 to 1290 and this is for SO2 antisymmetric stretch and SO2 for symmetric stretch and of course in case of 1420 to 1390 SO2 antisymmetric again 1220 to 1190 SO2 symmetric stretch and in case of SOC when we have SOC C is from dialkyl we have 1050 to 850 this is SOC stretching two bands are observed usually and for SOC sulfates what we have is 1050 to 750 we can also and spitier two or more bands so these are all just for information in case if you come across some compounds and if you have a year on hand you can readily analyze and look into it whether such groups are present or not so few facts we should remember while analyzing one such important factor is inductive effect what it does is increase in the inductive effect increases the bond order and hence the force constant in carbonyl compounds therefore the wave number of C double bond was stretching vibration increases with the strength of the inductive effect of X which is shown by the following examples for example I have given here for example if the X is there when X equals H 1740 when X equals CL it becomes 1800 it is considerably shifted by about 60 centimeter minus 1 and then the second one is mesomeric effect so first one is inductive effect in mesomeric effect mesomerism diminishes the bond order and hence the force constant resulting in lowering wave number of the stretching vibration so that means it diminishes bond order and hence stretching frequency drops considerably for example if you take this compound here X equals H 1740 is there but when you replace X by RL group it comes to 1685 and then when you take here X equals H 1645 to 1640 now when you replace X with RL it drops to 1600 to 1145 centimeter minus 1 so this effect one should know inductive effect and mesomeric effect what influence they have on carbonyl carbon groups if inductive effect is there stretching frequency increases and if mesomeric effect is stretching frequency decreases so then if one were wishing to use IR to identify successful conversion of 3 ethylbutanic acid to 3 ethylbutanamide which band change would prove the most useful there are four options are given I read again if one were wishing to use IR to identify successful conversion of 3 ethylbutanic acid to 3 ethylbutanamide which band change would prove the most useful or which band change is important in identifying this reaction so four options are there the gain of a broad band at 3200 and a narrow band at 1750 the gain of a narrow band at 3400 and the loss of a broad band at 3200 the loss of a narrow band at 3400 and the gain of narrow band at 1750 the loss of a broad band at 3200 and the gain of a narrow band at 1750 so that means in all we have CO group is there so CO group is almost left unchanged and in that case what happens initially we had OH so that is disappearing and in that place we are getting INH that means basically the 3200 due to OH is disappearing probably option B is the correct one here you can also see I have given a yes spectrum for both alcohol and amide you can see here this is for amide and this is for alcohol alcohol it comes around 12 OH is here and then it comes to 7 here you see NH1 otherwise this portion remains almost looks like unchanged here so that means here the catch is this one but however one can also use NMR conveniently to see whether the reaction is completed or not in both of them products are very pure that means the OH group is converted converted into amide here that can also be confirmed from NMR spectroscopy so we shall look into more examples maybe combined examples having both IR NMR and UV data in my next lectures until then have an excellent time thank you