HNMR
Transcript
Let's walk through the details of HNMR of the number of peaks corresponds to the number of chemically different hydrogen nuclei. The integrated area underneath those peaks would represents the relative abundance of nuclei. And then the peak position or where they're locate on the spectrum reveals the chemical environment or the shielding of the nuclei.
Now if we have a compound with multiple carbons in it, the hydrogens that are attached to the carbons in this molecule will do something called coupling. Coupling is also called splitting because these chemically different nuclei that are on adjacent carbons will cause each others' peaks to split. So this hydrogen and this hydrogen are on adjacent carbons, meaning that they would interfere with each others' peaks and cause them to split.
On this column we have three hydrogens, here have two and here we have three. The two hydrogens in the middle would split the six chemically equivalent terminal methyl hydrogens peak into a triplet and the six methyl hydrogen would split the peak at the middle two and into a septal. Now perhaps this is easiest to understand if we work through a problem. So let's figure it out from an HNMR spectrum, the formula for a particular chloroethane.
First thing we'll do is draw our backbone, this is an ethane so there will be two carbons. If we look down here we have two peaks. The number of peaks corresponds to the number of chemically different hydrogen nuclei. So we're going to have two types of hydrogen on here.
Now, the area of these peaks represents the relative abundance of the nuclei. Here we have 1.95. Over here, we've got 5.89. So, what we can do in this case is divide 5.89 by 1.95. We could say this is approximately six to two or three. So we have a 3:1 ratio between peak one and peak two.
Where there's three times as many this peak one. Now the peak position reveals the chemical environment, so these hydrogens from two are further down field or less shielded. So we would expect to see chlorines or something else that would be in electron with drawing group near hydrogens from peak two. Now coupling tells us about the chemically different nuclei on adjacent carbons.
So if we look at peak two here we've got one, two, three splits. Three splits in the case that there are three hydrogens of the other carbon. On peak one here, we have one split. One split indicates that there is only one hydrogen on the adjacent carbon. So we're going to have one hydrogen and three hydrogens.
Since this is a chloral ethane, we can fill in the balance with chlorine. Now all of our data should point towards this. All our hydrogens in a 3:1 ratio. We have one, two, three to one, that matches up. And we expect that these three hydrogens to be more shielded than this one. Yes we would.
This one hydrogen has two electron-withdrawing groups in the forms of chlorine. These would de-shield this hydrogen and would expect to see it down field. Now, this one hydrogen will split the peak of these three into a doublet. These three hydrogens will split the peak of this one hydrogen into a quartet. So we have more than enough information to conclude that we have 1,1-dichloroethane.
Read full transcriptNow if we have a compound with multiple carbons in it, the hydrogens that are attached to the carbons in this molecule will do something called coupling. Coupling is also called splitting because these chemically different nuclei that are on adjacent carbons will cause each others' peaks to split. So this hydrogen and this hydrogen are on adjacent carbons, meaning that they would interfere with each others' peaks and cause them to split.
On this column we have three hydrogens, here have two and here we have three. The two hydrogens in the middle would split the six chemically equivalent terminal methyl hydrogens peak into a triplet and the six methyl hydrogen would split the peak at the middle two and into a septal. Now perhaps this is easiest to understand if we work through a problem. So let's figure it out from an HNMR spectrum, the formula for a particular chloroethane.
First thing we'll do is draw our backbone, this is an ethane so there will be two carbons. If we look down here we have two peaks. The number of peaks corresponds to the number of chemically different hydrogen nuclei. So we're going to have two types of hydrogen on here.
Now, the area of these peaks represents the relative abundance of the nuclei. Here we have 1.95. Over here, we've got 5.89. So, what we can do in this case is divide 5.89 by 1.95. We could say this is approximately six to two or three. So we have a 3:1 ratio between peak one and peak two.
Where there's three times as many this peak one. Now the peak position reveals the chemical environment, so these hydrogens from two are further down field or less shielded. So we would expect to see chlorines or something else that would be in electron with drawing group near hydrogens from peak two. Now coupling tells us about the chemically different nuclei on adjacent carbons.
So if we look at peak two here we've got one, two, three splits. Three splits in the case that there are three hydrogens of the other carbon. On peak one here, we have one split. One split indicates that there is only one hydrogen on the adjacent carbon. So we're going to have one hydrogen and three hydrogens.
Since this is a chloral ethane, we can fill in the balance with chlorine. Now all of our data should point towards this. All our hydrogens in a 3:1 ratio. We have one, two, three to one, that matches up. And we expect that these three hydrogens to be more shielded than this one. Yes we would.
This one hydrogen has two electron-withdrawing groups in the forms of chlorine. These would de-shield this hydrogen and would expect to see it down field. Now, this one hydrogen will split the peak of these three into a doublet. These three hydrogens will split the peak of this one hydrogen into a quartet. So we have more than enough information to conclude that we have 1,1-dichloroethane.
