Leah here from leah4sci.com and then this video I'm going to show you how to solve for
the molecular structure when you're given a molecular formula and an H NMR graph. You
can find this entire video series by visiting my website, leah4sci.com/HNMR. The first thing
to do when you're given the molecular formula is to figure out your index of hydrogen deficiency
so that, you know, if to consider any PI bonds or rings. To do that, we'll use the equation
CnH2n + 2 which I teach in the index of hydrogen deficiency video link below. N represents
the number of carbon atoms, in this case, we have six. So if we set it up as 2n + 2
to solve four hydrogen, we have 2 x 6 + 2 or 2x2 is 12 + 2 is equal to 14. If we have
the maximum number of hydrogens, we know we have a fully saturated molecule with no PI
bonds and no rings. But in this case we only have thirteen hydrogens. So what does that
mean? It has to be an even number. Don't forget that halogens, because of how they bind to
carbon count just like hydrogen. So if we add up 13 hydrogen plus 1 hydrogen like halogen,
we actually have 14, 14 minus 14 is zero. That means we have no PI bonds and no rings
on this molecule. This helps us when we look at the structure because we know we don't
have to consider anything crazy. This also helps us with the oxygen because it immediately
rules out anything with a carbonyl. Remember, a carbonyl is a carbon double bound to oxygen,
which would require an index of hydrogen efficiency.
Now that we have an idea for the molecule, let's take a look at the graph. I'm going
to break this down and solve it the way I teach in my beginner NMR videos, which you
can find a linked below or by going to my website Leah4sci.com/NMR. I consider the right
side of the graph anything near zero. The boring region. This means that the carbons
and hydrogens on the right are further away from the exciting groups like halogen and
oxygen in our case, and the left is the exciting group. That means these two are likely sitting
close to the oxygen and the chlorine. I drew the peaks in different colors so that we can
refer to them by color and have a very clear idea of what we're looking at. Let's analyze
them starting from the right where we have three hydrogens. So three hydrogens is going
to be the CH3 with the peak of a triplet. How do we know it's a triplet? We can count
three distinct tops, right? So this is three points for a triplet we'll use and minus one
to find the number of neighbors. So three points, minus one tells us that we have two
neighbors nearby. This right here is a siege three group.
I'll draw the ghost atoms in gray so we know not to count them as part of our molecules.
But if we can match up a ghost to another piece, we know that we have a puzzle coming
together. For the green peak, we have six hydrogens. This has to be two carbons because
one carbon can have a maximum of four bonds, but six hydrogens have to be sitting on two
equivalent carbons, meaning there's some sort of symmetry there. And having a doublet, meaning
two points tells us that we have only one neighbor. 2 minus 1 is equal to 1. That means
there's one neighbor for these two different CH3 groups, which means we have a CH3, another
CH3 sitting next to a carbon that has one hydrogen splitting them both. I want to jump
over to the purple peak real quick, because this is a pattern that you have to learn to
recognize anytime you have a six hydrogen doublet and a one hydrogen multi plate, which
is this mess over here. You should recognize this as an isopropyl group. The purple group
is one hydrogen. Specifically, we have a septet, meaning we have seven points. But you'll often
see this just called multiplet because anything over five or six becomes too messy. So we
recognize. Oh, yeah it’s got a whole lot of neighbors in this case. Septet is seven.
Seven minus one is six neighbors. Well, what other six neighbors would it be if not for
our six hydrogen, two CH3 groups. So let's go ahead and put that right in here. No more
ghost CH. That right there is an isopropyl group. So we have one piece of the molecule.
Let's go back to the blue 2 hydrogen group. We have two hydrogens with quintets. Notice
we have 5 Points. A quintet is five minus 1 tells us that we have 4 neighbors and the
four neighbors could be one of two things. I can have a CH2 group on either side which
adds up to 4 or I can have a CH3 on one side and a CH on the other. In this case, we know
it has to be a CH3 on one side and a CH on the other because this is the only peak on
this graph that has two hydrogens. We have no other CH2s visible because nothing else
has two hydrogens. Last but not least, the Brown group on the left is one hydrogen in
a triplet. We've already said that a triplet is three minus one for two neighbors, and
that means this C has a CH2 nearby.
Now, notice we haven't even considered the oxygen on the chlorine yet. And the reason
for that is intentional. A lot of students will try to match everything up at the same
time. But if you let the carbons themselves match up like puzzle pieces that click, then
typically the rest of the groups will fall into place. And if not, you want to take extra
steps to figure out where they go. But why waste the energy if you don't have to? Before
we start putting everything together, let's make sure we accounted for all six carbon
atoms. Sometimes you'll have, say, a carbonyl that doesn't have any hydrogen and you have
to remember to create another puzzle piece on the side to take into consideration.
So what do we have? One carbon, two carbons. That's three. Another carbon is 4 another
is 5. Another 6, all of the carbons are accounted for. Now look at our ghost groups, for the
ghost groups, are there any ghost groups that match up a group that we have? For example,
we know we have a CH3, this is the only CH3. I don't even need to look at the shift to
see if they match up. I have no other option. So that tells me that the blue CH2 has a red
CH3 on the side. We accounted for this, let's cross it out, same here. And we also accounted
for this in the isopropyl group. The only thing that's missing is the brown CH. Well,
we know that we have to have another CH on the other side of the Blue Group. So let's
go ahead and put that CH right there. And that's it. We practically have the entire
molecule put together where we have two portions of the molecule. One portion is an isopropyl
group. The other one is a regular propyl group, Now what? This is where it gets exciting.
We have two pieces. We want to see how they connect. We also know we're still missing
an oxygen and a chlorine. If you count up the bonds that are here, you'll recognize
that the brown CH can have two more bonds because it needs to have a total of four.
And the purple CH already having three bonds, two to the methyl groups and one to the purple
hydrogen can have just one more bond.
I challenge you to figure out the rest of this without even looking at the graph. We
know we have to have an oxygen somewhere. Well, this one's got to have oxygen because
it's got two things. So it's actually got to have the oxygen and the chlorine. And the
only way to make this one molecule is to attach these two together. So without even looking
at where the shifts are happening, we figured out the molecule. But I do want to take you
back to the graph to prove it. How did we know that the oxygen was an ether rather than
an alcohol? Simple, alcohol is an OH and that hydrogen would show up as a broad, short singlet
on the graph. If you look at the graph, we don't have any singlets that tells us we don't
have any alcohols. We've already ruled out the possibility of a carbonyl. When we found
the index of hydrogen deficiency to be zero, leaving us in ether where an oxygen is bound
to two other carbon atoms as our only option. We know that halogens are considered exciting.
They're highly electro-negative and they tend to pull the nearby carbons and hydrogens towards
the four range a little bit higher, a little bit lower depending on what else is going
on and which halogen we have. We said that the Brown CH is holding the chlorine. Well,
guess what? The Brown C H is the highest shifted group. It is the most exciting. And doesn't
that make sense, given that it's next to both chlorine and oxygen? And the purple CH is
the second most exciting group. It's the second highest shifted. And doesn't that make sense
if it's directly attached to the oxygen, but not also holding a chlorine? So now that we've
matched everything up and we have all our puzzle pieces together, the final step is
to draw this into some sort of a coherent molecule.
Because right now it looks like a mess. I'll draw it out with all the atoms first. Then
we'll do it on line structure. So we have a CH3 down to a CH2 bound to a CH. That is
also holding chlorine. And we have an oxygen bound to the purple CH down to two green CH3
groups. In line structure that gives us one, two, three carbons where the third carbon
is holding a chlorine and bound to an oxygen, and attached to that we have an isopropyl
group. And there you have it. There is our molecule that came together by analyzing the
pieces. If you found this video helpful, first, give this video a thumbs up and make sure
you subscribe to my channel so that you don't miss out on any of the new videos to follow
this semester. I work through even more complex spectroscopy problems in the organic chemistry
study hall, not just H-NMR, but multi-spec where we're looking at C-HNMR, I.R. and masspec.
You have more information about the studyhall below or going to leah4sci.com/join.