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more complex splitting occurs when a proton has two different kinds of neighbors and a good example of this is the blue proton that I circled in cinnamaldehyde so the blue proton has a signal with a chemical shift about six point seven parts per million so down here is a zoomed in view of the signal for the blue proton let's look at neighboring protons so the blue proton is on this carbon and we have a carbon next door right here with one proton so there's one neighboring proton here's another carbon next door with one proton so we have two neighboring protons so let's try to apply the n plus one rule here so if n is equal to two we have two neighboring protons we would expect n plus one Peaks so two plus one is equal to three so a signal with three Peaks or a triplet but that's not what we see for the signal for the blue proton we see one two three four lines here so the n plus one rule doesn't work in this case and that's because the n plus one rule works when the neighboring protons are equivalent and here the two neighboring protons are not equivalent so we need a new way to explain the signal for the blue proton and we're going to use what's called a splitting tree so we're going to start with the the signal for the blue proton so here's the signal for the blue proton it's going to be split by the proton next to it this proton in red and the coupling constant between the red and the blue proton is 12 Hertz so let's go ahead and show let's show this signal is split into a doublet alright so let me go ahead and draw in the blue lines here so we get the signal split into a doublet the coupling constant is twelve Hertz so this distance here represents 12 Hertz so why can we think about it being split into a doublet well you could think about a variation of the n plus one rule here all right so we're talking about one neighboring proton so n is equal to 1 and is equal to 1 so 1 plus 1 is equal to 2 so we split the signal into a doublet with two lines all right now let's think about what happens with the other proton so this one okay well that's going to take each line of the doublet that we just drew and split it into another doublet so this this line gets split into a doublet and this line gets split into a doublet the coupling constant this time is six Hertz alright so this distance here represents six Hertz and this distance represents six Hertz so why was each line split into two well once again we can use a modification of the n plus one rule here alright so we have we're talking about one neighbor here so n is equal to one so one plus one is equal to two so each each line is split into two alright so each line is split into a doublet for this line and for this line this line is split into a doublet too so we get we get four lines for the signal of the blue proton if we go over here we see those four lines one two three and four and we call this a doublet of doublets or a double doublet alright what would happen if the coupling constants were the same so let's let's just pretend like they're both 12 Hertz let's go ahead and draw what we would see for the signal so we have our blue protons signal is is split in to a doublet by the red proton so let's go ahead and draw in our doublet here and let's say this this distance represents 12 Hertz alright so 12 Hertz here and let's say the coupling constant over here right was also 12 Hertz so instead of six Hertz so each line of the doublet we just drew is split into another doublet from our other neighbor and this time I'm going to I'm going to show a coupling constant of 12 Hertz alright so each line of the doublet in blue is split into another doublet so the line in blue on the left is split into a doublet and the line in blue on the right is split into eight doublet right because of our one neighboring proton all right well I changed the coupling constants and I said to pretend like now we're dealing with twelve here all right so the coupling constants are the same and notice what this gives us this gives us this gives us a triplet right so here here's one line and then we're going to get this peak and then we're going to get this peak here so if the coupling constants are the same you get a triplet you get what the n plus 1 rule predicted and so that's just something to think about with the origin of the n plus 1 rule let's do another example all right so let's let's go down here and let's look at this molecule so we're going to focus in on this proton here in blue all right and we have neighboring protons right so this neighboring proton the coupling constant between those two is 12 Hertz and then over here we have two neighboring protons and the coupling constant between the magenta protons and the blue is 7 Hertz so let's think about the signal for the blue proton here so here we have the signal for the blue proton which isn't split let's think about what the red proton is going to do we have one neighbor all right so one plus one is equal to two so we're going to split the signal for the blue proton in two so we get a doublet here so let me go ahead and draw in the doublet the coupling constant was twelve Hertz so this distance right here is 12 Hertz all right now we have now we have a situation where we're thinking about the magenta protons we have two of them all right so n is equal to 2 so 2 plus 1 is equal to 3 so the magenta protons are going to split each line of the doublet that we just drew into a triplet all right so we're going to get a triplet here let me go ahead and draw that in so for a triplet and our coupling constant is 7 Hertz so let me see if I can draw that here so that that distance is supposed to represent 7 Hertz let me draw that in all right so this distance is seven Hertz this distance corresponds to seven Hertz and one line of our doublet is split into a triplet now same thing happens for the other line right same thing happens for this line right here we need to split that into a triplets because of these magenta protons all right so we're going to split down into a triplet once again when you think about a distance of seven here seven Hertz so we have on both sides we have seven Hertz so let me go ahead and draw that in so this this is talking about seven Hertz and this is talking about seven Hertz so that one line is split into a triplet so let's draw that in here one two and three so finally how many how many Peaks would we expect for the signal for this blue proton so one two three four five six so we would expect six peaks six Peaks for the signal for this blue proton because of the neighboring protons which are on different types of environments so this is this is a complex splitting