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let's say we're given this molecular formula C 5 H 2 O and this proton NMR spectrum and we're asked to determine the structure of the molecule the first thing you could do is calculate the hydrogen deficiency index and so if we have five carbons here the maximum number of hydrogen's we could have is 2 n plus 2 where n is equal to 5 so 2 times 5 plus 2 is equal to 12 so 12 is the maximum number of hydrogen's here we have only 10 hydrogen's so we're missing two hydrogen's we're missing one pair of hydrogen's and so therefore the hydrogen deficiency index is equal to one immediately that makes me think about a double bond might be present in this molecule or one ring let's go down to the integration values so for this signal there's an integration value of 27 for this signal the integration value is 40 point 2 for this signal it's 28.4 and for this signal it's 42 point to remember what you do you divide all for integration values by the lowest one so the lowest one is of course 27 so 27 divided by 27 is of course one forty point two divided by 27 is pretty close to one point five twenty eight point four divided by 27 that's pretty close to one and forty two point two divided by 27 is once again pretty close to one point five remember these are just the relative numbers of protons giving you these signals but you can't have 1.5 protons giving you a signal we need a whole number and we need to account for 10 total protons here so if we multiply if it multiply 1 1 by 2 then we get 2 protons so this signal represents 2 protons if we multiply 1.5 by 2 we get 3 so this signal represents 3 protons if we multiply 1 by 2 we get 2 and if we multiply 1.5 by 2 we get 3 so if you add all those up 2 Plus 3 plus 2 plus 3 that's of course 10 protons so now we've accounted for all 10 protons using our integration value use next let's look at each signal one by one so we'll start with this signal so we have a ch2 since we have two protons so this signal represents a ch2 how many neighboring protons do those ch2 protons have all right so look we can figure that out by the number of Peaks on the signal so this signal here has one two three peaks and if you think about the n plus one rule if you have n neighbors you get n plus one Peaks so if we have three peaks all we have to do is subtract one to find out how many neighboring protons so three minus one is equal to two so this ch2 these ch2 protons have two neighboring protons let's think about the chemical shift for the signal so the chemical shift for this signal is between two parts per million and 2.5 and that's in the region for a proton next to a carbonyl and that would make a lot of sense because we calculate an HDI of one indicating there might be a double bond present and we need to account for an oxygen in in our molecular formula so let's go ahead and draw in a carbonyl and these ch2 protons are next to the carbonyl we know that because of the shift right the carbonyl the oxygen DeShields these two protons a little bit and gives us a higher value for the chemical shift compared to something in an alkane type region all right let's let's let's color these protons in here magenta so we're saying this signal is due to those two protons let's move on to the next signal so we have three protons so that would be a methyl group so ch3 how many neighboring protons for those methyl protons well there's only one peak here so one minus one is zero so zero neighboring protons what about the chemical shift so the chemical shift for this signal is once again passed to parts per million so this signal is D shielded right so these protons must be D shield a little bit and those must be next to our carbonyl so we'll draw in will draw in our methyl protons right being D shielded because they're next to the carbonyl and let's make these blue so these protons in blue here these three protons are giving us this signal and we should expect zero neighboring protons all right so those protons are on this carpet and the next door carbon has no protons on it so zero neighboring protons makes sense move on to the next signal so this signal represents two protons so that's a ch2 how many neighboring protons all right well it's count up how many Peaks we have one two three four five six so six Peaks so we subtract one from that so 6 minus 1 is 5 so we would expect 5 neighboring protons let's move on to the next signal 3 protons right so that's a methyl group so CH 3 how many neighbors well we have 1 2 3 Peaks so 3 minus 1 is 2 so we would expect two neighboring protons now these two signals right this signal here and this signal here we're talking about under 2 parts per million now so these must be further the furthest away from the carbonyl right so they're not they're not being D shielded as much as the two signals that as the two signals in this direction so these two neighboring protons for the methyl group must be these two protons right here so let's go ahead and draw that in so we have our methyl protons right here and let me let me make these protons red so these protons right here in red are giving us this signal and from the signal we know that these methyl protons are next to two neighbors and so we must have a ch2 we must have a ch2 next to that methyl and so let's go ahead and draw in our ch2 and this ch2 ch2 must be must be this signal right here okay so let's think about how many neighbors we expected five neighbors for these ch2 protons let's count them up one two three four five so five neighboring protons matches what we see on the NMR spectrum now those protons those magenta and red protons are actually in different environments and so the simplified version of the n plus one rule isn't quite true but it works for this example it works this example so we're going to go with it because all we care about is getting the structure of our molecule here all right so let's uh let's think about the red protons again right so the red protons proposed have two neighbors so how many neighbors are the red protons have here's one and here's two so this makes sense so that one makes sense this one makes sense right here we just talked about that let's think about the magenta protons so the magenta protons were supposed to have two neighbors so let's look at the carbon next door to this one so here's the carbon next door here we have a neighbor and here we have a neighbor so this makes sense and then we already talked about the blue protons right having a zero neighbors so everything everything seems to make sense here and to sum everything up make sure to count all of your atoms and you will get of course five carbons ten hydrogen's and one oxygen when you do that so this is the dot structure that we were trying to find let's take a look at another one here so we have a molecular formula of Ch H 10 and let's go ahead and calculate the hydrogen deficiency index so if we have eight carbons then we can have a maximum of two times eight plus two hydrogen's so two times 8 is 16 plus 2 is 18 so for 8 carbons 18 hydrogen's is the max here we have 10 hydrogen's so we're missing 8 hydrogen's or missing four pairs of hydrogen's so the hydrogen deficiency index is equal to 4 remember anytime your HDI is equal to 4 you should think about a benzene ring so I'm going to go ahead and draw a benzene ring in here and let's look at the integration alright so sometimes you'll you'll see the integration given like this so this this represents 5 protons for this very complex looking signal right here this signal over here represents 2 protons and this signal represents 3 protons so let's go back to these 5 protons with this with this complex signal right this is in the aromatic range right so approximately 6 point to H those must be five aromatic protons so we can go ahead and draw in five five protons off of our benzene ring so even though those protons are in slightly different environments right because this integration value is five here and we know a benzene ring is presence where we're done right we don't have to worry about the slightly different environments we know that these five protons are giving us this complicated signal over here all right next let's look at this signal so two protons so that must be a ch2 how many neighboring protons for the ch2 well there's one two three four Peaks so if there's four peaks 4 minus 1 is 3 so these these ch2 protons have 3 neighbors all right let's look at this next signal so 3 protons so this must be a ch3 how many neighboring protons do we have for the ch3 protons well we have 1 2 3 Peaks so 3 peaks 3 minus 1 is 2 so 2 neighbors and so two neighboring protons that must be these two neighboring protons and so we go ahead and draw an ethyl group right so this is an ethyl group pattern over here so there's a ch2 and there's your ch3 so let's draw in all of our protons so we have these protons and we have these protons let's color coordinate here just to make sure everything makes sense so these red protons right here they must be giving us this signal so these methyl protons right we'd expect two neighbors because we have three Peaks on our signal and here are the two neighbors one two so two neighboring protons let's look at these two protons next so these two protons are giving us this signal we would expect three neighbors because we have four peaks 1 2 3 4 so 3 neighbors so here's a next-door carbon so 1 2 3 3 protons and then this is also a next-door carbon but there are no protons on this carbon and so we see only these 3 neighbors and so everything everything seems to make sense and once again count up all of your atoms and make sure you've accounted for everything for example alright you have eight carbons you have to worry about so in your ring on your ring we know there are six one two three four five six and then we get seven and then we get H so the eight carbons is correct and that if you count all those up we have our 10 hydrogen's accounted for - so this is the this is the NMR spectrum for ethyl benzene alright this one was a little bit easier than the previous example