Hydrogen deficiency index

when you get a spectroscopy problem on a test you're often given the molecular formula and asked to determine the structure of the molecule and sometimes the hydrogen deficiency index can be useful for figuring out the structure of your molecule let's start off by looking at hexane so here we have hexane with a molecular formula of c6h14 we say that hexane is completely saturated with hydrogen's so it has the maximum number of hydrogen atoms possible for the number of carbons so if we have six carbons 14 as the maximum number of hydrogen's that we can have and remember when you're talking about alkanes and you have n carbons you get 2 n plus 2 hydrogen's so here n is equal to 6 so we have 6 carbons so 2 times 6 plus 2 gives us 14 hydrogen's let's compare hexane with 1 hexene so now we have a double bond present and that changes the molecular formula of course so now the molecular formula is c6h12o6 but we have 12 hydrogen's and up here we had 14 so we're missing 2 hydrogen's and so we say that's one degree of unsaturation or a hydrogen deficiency index equal to one so an HDI equal to one so it's like we're missing one pair of hydrogen's here let's look at cyclohexane so here we have cyclohexane and the molecular formula is c6h12 once again we have 6 carbons but we have only 12 hydrogen's so we're missing 2 compared to hexane we're missing two hydrogen's we're missing one pair of hydrogen's so the hydrogen deficiency index the HDI is equal to one so if you were given this molecular formula c6 h-12 right here it's the same molecular formula right and you think about how many hydrogen's you're missing from the maximum number you're missing two hydrogen's that's a hydrogen deficiency index equal to one here we had one double bond and here we had one ring so with an HDI equal to one one possibility is you have one ring another possibility is you have one double bond all right let's look at benzene so here we have benzene a molecular formula of c6h6 so with six carbons the maximum number of hydrogen's you can have is fourteen and here we have six so 14 minus six is equal to eight so it's like we're missing eight hydrogen's so that's four pairs of hydrogen's so the HDI is equal to four so the hydrogen deficiency index is equal to four and I like to think about why the HDI is equal to four because right if this is four we have three double bonds here one two three and and we have one ring right we have one ring here so three plus one is equal to four and we get an HDI equal to four right up here we had one ring and we had an HDI equal to one up here we had one double bond and an HDI equal to one we could also we could also calculate the HDI using a formula so we could write that the hydrogen deficiency index is equal to one half times what's in the parentheses which is two times number of carbons plus two that's the whole 2n plus two idea minus the number of hydrogen's so let's go ahead and do it for this molecular formula with six six carbons and six hydrogen's let's plug in those numbers so we have one half this would be two times number of carbons that's two times six plus two minus number of hydrogen's that's six so let's do the math so this be 1/2 2 times 6 plus 2 is 14 minus six 14 minus six is eight times one half we get an HDI equal to four so if you have a spectroscopy problem and you calculate the HDI equal to four think about the possibility of a benzene ring being present in the structure of your molecule let's compare let's compare benzene 2 to another molecule we're adding in a nitrogen so here's benzene and we know there are six hydrogen's 1 2 3 four five six six hydrogen's let's add in a nitrogen alright so we are adding in a nitrogen so now we have annalen as our compound here how many hydrogen's all right so this would be one two three four five six seven so we have seven hydrogen's here so we've added one nitrogen right and we get one more hydrogen so we need to modify the formula for HDI's let's go ahead and do that so we would write the hydrogen deficiency index is equal to 1/2 times what's in our parentheses which would be two carbons plus two times number of carbons plus two and now we're going to add the number of nitrogen's that we have so add the number of nitrogen's and then subtract the number of hydrogen's so let's go ahead and do this calculation so 1/2 times once again we have 6 carbons 1 2 3 4 5 & 6 so 2 times 6 plus 2 plus the number of nitrogen's we have one nitrogen so we add a 1 in here minus the number of hydrogen's we said there were 7 so minus 7 and let's do this this be 1/2 times this be 14 plus 1 is 15 so 15 minus 7 is equal to 8 times 1/2 is equal to 4 so an HDI of 4 all right remember an HDI of 4 think about a benzene ring being present in the structure of your molecule all right let's do another one let's let's compare benzene to chlorobenzene so over here on the Left we have benzene once again 6 hydrogen's and here we have chlorobenzene so we've added in a halogen and how many hydrogen's do we have in glory benzene 1 2 3 4 5 so only five hydrogen's this time so we've added a halogen and we've lost a hydrogen we went from 6 to 5 so we need to modify our formula again so HDI is equal to 1/2 times was in the parenthesis 2 times number of carbons plus 2 plus the number of nitrogens so plus number of nitrogen's minus the number of hydrogen's and we would also need to subtract the number of halogens so minus the number of halogens that you have alright let's let's do it for chlorobenzene so 1/2 times how many carbons we have still 6 1 2 3 4 5 & 6 so 2 times 6 plus 2 0 nitrogen's alright so 0 nitrogen's how many hydrogen's 5 4 chlorobenzene so minus 5 and we have 1 halogen all right we have one halogen here it's the chlorine so minus 1 so let's do that math so we have 1/2 times 2 times 6 plus 2 is 14 right so we have 14 minus 6 over here so 14 minus 6 is 8 times 1/2 is equal to 4 so we get an HDI equal to 4 and once again we think benzene ring so HDI is equal to 4 we think about a benzene ring being present all right let's think about one more one more example so if we look on the left once again we have benzene so here's benzene with the six hydrogen's and we've added in an oxygen alright so for this molecule we've added in one oxygen what happened to the number of hydrogen's so we have 5 on the ring 1 2 3 4 5 and then this one 4 6 so even though we've added in a hydrogen there's no change in the number even though we've added an oxygen I should say there's no change in the number of hydrogen's and since there's no change in the number of hydrogen's we don't need to modify our formula here for the hydrogen deficiency index so I'm going to go right back up here and box this so this is our final version for calculating the HDI and you don't have to use this formula every time sometimes you can just think about how many hydrogen's are missing but sometimes sometimes using this formula allows you to calculate the HDI which helps you to think about what's present in the structure of your molecule and so I'll show you several examples of how to use the HDI to figure out your dot structure