for today's problem we're gonna be looking at NMR spectroscopy. this is an NMR
spectrum that is going to correspond to a compound that has the empirical
formula C4H9Br
so we know that there are different structural isomers that can correspond to
C4H9Br
so we're going to want to draw, to be able to figure out exactly the structure of
the compound that corresponds to this NMR spectrum which has a doublet right
here that integrates to 2, a multiplet here that integrates to 1, a doublet here that
integrates to 6. so if this seems confusing go ahead and watch my tutorial on NMR
spectroscopy first and then go ahead and give this a shot
analyzing an NMR spectrum can be pretty tricky but in such a case as right now
we're trying to select a structure based on an empirical formula there's a couple
of approaches you can use. now one of them would be to simply write every
possible structural isomer that corresponds to that empirical formula
that's what I've done here, we know that if there needs to be C4H9
and a bromine atom we can either have the straight chain for the four carbons
and then we can have the bromine on a primary carbon or can be in a secondary
carbon or we could have the four carbons arranged like this and then therefore
either have the bromine on the tertiary carbon or on one of the primary carbons
on the edge so there's actually only four structural isomers for that empirical
formula. now you could have a problem where there's many more which would make
it more difficult to do this but in this case is very simple.
now once we have these we can just even look at which one corresponds to the
number of peaks that are in the spectrum so on this one
we'd probably expect four peaks, one for these two, one for these two protons, one for these two
protons
one for those three protons. there's only three peaks in the spectrum that's
probably not it. over here again we probably expect four peaks, these guys
this one, these two, and these three because we tend to have chemically
equivalent protons on the same carbon so usually different carbons will have
different, will give different peaks. over here we could have this guy this one
would only generate one peak because all nine of these methyl protons are
equivalent, they all see the exact same chemical environment, that one's also no
good. that leaves this, where we've got the two on here, the one on there. and
then these are all chemically equivalent so this ends up probably being the
structure. so let's draw this over here now if it's too complicated for you to
be able to draw every single structural isomer instead what you can do is go
ahead and begin to generate fragments of data from the NMR spectrum. we know that
if we have peaks that integrate to 3 that's oftentimes a methyl group. if it's 6
there might be two identical methyl groups if it's 2 that usually a CH2 of
some kind so there is information that you can begin to gather from the NMR
spectrum and start to piece together the structure that way, so that's another
way to do it but now that we've selected the structure let's go ahead and assign
all the peaks, so let's call these two A, call this B
and then these are identical to those will be C, so we know first of all
that these methyl groups are identical because as this bond rotates these methyls
spin around, they see the same chemical environment so all six of those
we know that if they're six protons that is gotta integrate to 6H, and then
we also know that these these three protons all see one proton next door
let's go ahead and draw all these in actually, so here are our hydrogens so
these three see this hydrogen next door
these three see that hydrogen next door so all of these six hydrogen see one hydrogen
next door, according to n + 1, that means the signal is going to be split
into a doublet so that takes care of our integration that takes care of our
splitting, in terms of chemical shift we know that these regions are the farthest
away from the deshielding group, the deshielding bromine, so we would also
expect for that resonance to be the furthest upfield and so we know all
three of these data are corroborating that this is peak C
because this is a doublet, its integrated to 6, and it is the furthest upfield, so
that's a very easy one to assign. now let's look at this proton, we know that there
first of all it's gonna be integrated to 1 because
there's only one of them, now in terms of splitting there are a lot of
neighboring protons we have 2 right here, 3 right there, 3 right there
for a total of 8, the splitting pattern is going to be very complicated, we would
simply call that a multiplet. once you get above five or so we end up just
calling it a multiplet instead of continuing with doublet, triplet, so we
know that that's going to be a multiplet and we know that it's not
going to be the most downfield, it's not going to be the most upfield
because it's closer to the bromine than those but it's further away from the
bromine than these so that one might be in the middle and so all of that
corroborates that this peak corresponds to the proton on position B so we
can assign that peak as well and then lastly just to go through
it we've got these two protons we know that is gonna integrate to 2 we know that
there is one adjacent proton so we know that that is going to be a doublet and
we know that these protons are the closest to the bromine so we would
expect those to be the furthest downfield so let's go ahead and say downfield and all
of that again corroborates that this is peak A. so once again when we're trying to
figure out the structure that corresponds to a particular NMR spectrum
we can do one of two things. we can either take this empirical formula, draw
all the possible structural isomers
which in this case ended up immediately telling us the answer or if it's too
complicated to do that we can start to piece together from the NMR spectrum
using integration using splitting and using chemical shift data to try to
start to put the pieces together
thanks for watching guys, subscribe to my channel for more tutorials and as always
feel free to email me