gas chromatography can be used for qualitative analysis so identifying what species you have present and to do this we're going to use a retention index i which is a pro proposed as a parameter for identifying solutes so we'll look at two different scenarios the first one on the left which is a non-polar stationary phase and so thinking back to like dissolves like that things that are non-polar are going to be retained longer than things that are polar so anything polar going through is going to come off rather quickly because it's not retained onto the stationary phase and so here if you're looking at all of your possibilities here we can look at where they come off on this column you see the longest one that's retained is going to be your octane but again even after four minutes you have very little retention so for the nonpolar stationary phase the determinant of retention will be the volatility of the solutes anytime you're looking at nonpolar stationary phase your compounds will always elute or come off the column in the order of their boiling point in the nonpolar stationary phase so for non-polar stationary phase you're looking at the order of the boiling point so as blind point increases it stays on a little bit longer and so here we see again 10 is that octane pentanol so still going to be some differences in regards to polar nonpolar but one of our longest ones is going to be the octane the strongly polar stationary phase obviously things that are polar are going to be retained longer than things that are nonpolar so any of your alkanes are always going to be um your alkanes such as octane pentane hexane all of your alkanes are nonpolar which is why over here this one compared to petanol pentanol is polar so the polar came off before the nonpolar and then in regards to the boiling point is how they elute so over here we have the determinant of retention is going to be the hydrogen bonding or the strongest intermolecular forces so the stronger the intermolecular force the longer it gets retained another way to look at that is that it's going to retain strongly polar solutes we have to think back to polar nonpolar in regards to how this is going to elute so the first looking at the different groups the first groups here are alkanes so you notice the alkanes came off first it has 2 5 8 and 10. so pentane hexane peptane and octane and then they came off the column in order of chain obviously the longer the chain the little bit longer it stayed on next we have our ketones so our ketones are going to be polar and so next we have one four and seven so they stay on a little bit longer because they're polar compared to the nonpolars so these are non-polar these are polar and polar alcohols having that oh group have the hydrogen bonding so those stand even longer and so then here we have our hydrogen bonding so we have three which is propanol you have six which is butanol and then nine which is pentanol so in that case the longer chain the longer it stays on as well so we're looking at for non-polar stationary phase looking at the volatility of the solutes polar stationary phase looking at non-polar come off first and then polar it's going to be in regards to the strength of the intermolecular force that we have going on so the kovitz retention index i is for any given solute can be derived from a chromatogram of a mixture of that solute with at least two normal alkanes having retention times that bracket that of the solute so here we're looking at i equals 100 little n in this case represents the number of carbon atoms in the smaller alkane so let's talk through a hypothetical so what if we're looking at octane and no name so octane i equals 800 and the n would be eight for octane and then we'll bracket that with no name so no name i will equal 900 and n will be nine so you're always going to be bracketing your sample with two alkanes so this would be if we want something that falls in between the two capital n is going to be the number of carbon atoms in the larger alkane so here actually this should be capital n so it's n little n plus big n minus n then it's going to be log t r prime so that's the adjusted retention time of the smaller alkane so n would be the smaller alkane big n will be the larger alkane and then you have your unknown so this is assuming that on the chromatogram that your retention time of your unknown fell between the retention times of octane and nonin the retention index for a normal alkane is always equal to the number of carbons times 100 so this is why we get octane n equals eight so it's 800 900 whatever so this is always true regardless of column packing temperature or any other chromatographic column conditions so the retention index for normal alkane is always going to be equal to the number of carbons times 100. the retention index for all other compounds will vary with column variability so depending on your column packing your temperature any other conditions the retention index of the other compounds may vary where they fall and their eye value will vary but alkanes will not so here's a plot of the logarithm of adjusted retention times versus the number of carbons in a homologous series so here we have uh series here so it's five four so these are gonna be your alkanes here's six this year sevens here eight nine and then this is going to be adjusted based on where they fall so in this case benzene is going to fall right between the 6 and the 7 at 6.44 dimethylbutane will fall between 5 and 6 at 5.37 the retention index for a normal alkane is independent of temperature and column packing so again i for heptane is always 700 octane is always 800 nonin is always 900 and the retention index of all solutes may vary from one column to another so again those conditions are going to vary depending on what you're looking at so we can look at this data um and look at predicting some orders of evolution going into calculating some eye values so starting with orders of evolution so here we have benzene butanol pentanone nitro one nitro penta uh propane and purity and then we have different phases here and these phases are listed in order of polarity with the greatest polarity down here meaning these are polar this up here is going to be nonpolar so predict the order of elution in a column containing polyethylene glycol as your stationary phase so polyethylene glycol is right here and here's the information about those so these are going to be your retention index so we want to look at what order they're gonna elute based on what we know about them so starting with one pentanol the first thing you want to identify is do you have something polar or do you have something nonpolar so one petanol is going to look like this which is an alcohol and it's going to be polar two hexanone the hexagon part tells us it's a ketone and so it's going to look like that and that's going to be polar as well heptane so hepta7 and it's just an alkane and then we have octane which is also just an alkane one two three four five six seven eight okay so we have these two over here are going to be nonpolar and nonpolar all right so we're looking at a polar column so we're looking at the polyethylene glycol which is a polar column and for a polar column the nonpolar will elute first in order of boiling point so between three and four haptain has a boiling point of 98 degrees so it's gonna be heptane first so three and then next will be octane so your nonpolar always allude first and order boiling point then next will be your polar will elute and your polar will elute going from weakest intermolecular force which is dipole dipole well technically london dispersion but dipole dipole if we have a polar cast that has to be it's potentially strongest all the way to hydrogen bonding so looking at these two between an alcohol and a ketone your ketone is going to be dipole dipolar but not hydrogen bonding so next will come the two and then last will be your alcohol one so that's looking at the order of those coming off the column so you always look at what type of column do you have polar nonpolar and then decide which is going to come off first keeping in mind like dissolves like so if the retention times of ch4 is 0.5 minutes and we give more information find the retention index for the unknown so here is 0.5 minutes octane is 14.3 minutes the unknown is 15.7 minutes and then the no name is 18.5 minutes well the first thing you should be looking at is the fact that in our equation of i we need tr prime so we need the adjusted um retention index for the dust retention time so over here i'm gonna do that for each of mine so for the unknown the adjusted is going to be 15.7 minus the 0.5 you subtract off the dead time of your solvent so 15.7 minus 0.5 gives you 15.2 minutes the octane is going to be 14.3 minus the 0.5 which is going to be 13.8 minutes and then we have last would be our known um yet no name and the no name is going to be 18.5 minus 0.5 which is 18. now we can look at our equation so i equals 100 times n plus n minus m and then we have log of tr prime unknown minus log t r prime n divided by log of tr prime big n minus log tr prime alola so in this case our little n is going to be our octane and our big n will be our no name so i equals 100 so little n is going to represent octane so it's eight plus the no name is nine minus eight and then we have log of tr prime of our unknown just 15.2 minus the log of tr prime of our little n which is the octane at 13.8 divided by log of tr prime big n which is 18 minus the log of tr prime of little n which is 13.8 so take a moment pause the video and calculate this and you should be getting 836. now you want to make sure your number falls where it should so octane is going to have an i value so remember the i i value of octane is going to be 800. the i value of no name is going to be 900 so your number should be somewhere between 800 and 900. you can also look at this and say well kind of halfway between those is right here you're a little bit below half so does 836 make sense yes it does so where would i'm just going to erase this because it's in the way so where would an unknown with a retention index of 936 a loot in the figure above so would it aleut ah let me change this so basically would it come off the column before octane between octane and unknown between unknown and no name or after no name while looking at our numbers 800 836 900 well 936 would be somewhere over here so it's going to be after the no name