Determining Organic Structures From Ir/nmr

Determining Organic Structures From Ir/nmr

August 9, 2016 0 By Jeremy McCallum

Sure that corresponds to the number of hydrogen’s do we need to multiply that by this by.

A value do we need to divide it by something so let’s add these values up 1 hydrogen plus 2 plus 2 gives.

Us 5 plus 6 is 11 plus 3 is 14 and if we remember from our.

Formula we had 14 hydrogen’s so these values actually correspond to the number of hydrogen’s we have the signal for a is represented of one hydrogen B is to C is to 6 D is 6 and E is 3 okay so let’s start assigning some structures here so let’s first start with a so if I count the number of Peaks I.

Have I have one two three four five six seven so looking at peak a what we can do is talk about the number of Peaks we have seven peaks makes this a.

Septet so let’s write that information down here we have a septet and let’s go ahead and go through and do this for all of them before we actually assign our structures looking at B B clearly has three peaks so that makes be a triplet okay so.

2 comma T to say it’s a triplet if we count our peaks at C we have 1 2 3 4 5 6 so peak C has six Peaks so we can label that a sextet okay so peak series a sextet.

Looking at D that’s easy to.

See that’s a doublet so we can just abbreviate that with a D and finally looking at our peak II here this is clearly a triplet so let’s.

Come over to e and label that as such e is a triplet so again looking at peak a integral of 1 sub 10 B integral of 2 and a triplet see has an integral of 2 and as a sextet D integrates to 6 and.

Is a doublet and E integrates to 3 and as a triplet so now we have our full spectrum that we can work with and we can go ahead and start assigning pieces so I’d like to start from the thing that is furthest downfield alright that means the peak with the highest number on the left so we’ll start our analysis really looking at a ok so because a.

Integrates to 1 we can say that this is a single hydrogen that is connected to a carbon atom because there’s one hydrogen and that is a septet meaning that.

Is next to six hydrogen’s now.

Obviously we can’t put six hydrogens on one carbon okay so if it’s a septet the n plus one rule says it’s next to six hydrogen’s we can’t put that on one.

Carbon but we can put that on.

Means is there are two carbons connecting.

To our singlet and each of these carbons has three hydrogen’s on it so connected.

To that peak a I have two methyl groups so let’s draw those in here and now clearly.

We see we have for our peak a I have six neighbor three neighbors there are three neighbors here and a total of six neighbors and seven peeps so what I like to do here is draw a little squiggly line and that tells me that this is a piece I just made so we’re really done looking at pk alright so looking at our spectrum I’m just going.

To kind of cross this off we’ve evaluated that peak now notice I just drew in two methyl groups here and here so let’s find those in the spectrum and this is an isopropyl group right what we’re looking at here is an isopropyl group so which of our Peaks has.

An integral of six easy to see that’s peak D so the two hydrogen’s that I drew ORS are the two methyl groups that I do he drew here correspond to beak peak D so we can label that that as peak D if we double check this has an integral of six because we do six hydrogen’s and it is a doublet and that makes complete sense six hydrogens here there are adjacent to one hydrogen they have one neighbor so.

A and D is definitely a piece.

We have okay so let’s move on now we’ll take a.

Be okay peak be here in a role of two and a triplet so let’s.

Have two hydrogen’s connected to a carbon atom and that is a triplet so it has two neighbors so let’s draw those two neighbors in here and that correspond so that corresponds.

To B so now B is done and what we need to note here is I’ve just drawn two hydrogen’s so B is done and sorry I didn’t cross out D let’s.

Cross D out so now I have a choice so the two hydrogen’s that I do.

In here can either be our triplet at E or C now it’s very clear to see because I just drew in two hydrogen’s here that it clearly cannot be peaky because II integrates to three so what we drew next to.

Be this peak here has to be peak C because C has an integral of two so let’s take a look at peak C now integral of two it’s a sextet which means it has five neighbors so looking at C to.

The left of it are two neighbors so I need to.

Draw three more neighbors so let’s draw those neighbors in I’m going to draw another carbon and that carbon is going to be nekked connected to three hydrogen’s okay so see is now done we’ve analyzed see and conveniently we only have one peek left e e integrates two three and is a triplet.

And that’s exactly what I just drew so looking here three hydrogen’s so an integral of three and it’s next to two neighbors making it a triplet so this final piece is BC e and that corresponds to a propyl group so let’s just draw in our connector here to show that that is a piece that must be connected to something and now looking at our NMR we’ve analyzed all of.

Our peeps and by doing that what we’ve done here is we really have two pieces from our NMR we have our isopropyl group consists of a and D and we have our propyl group which consists of B C and E so what we need to do now is look at our chemical formula and find out what is missing what atoms are we missing okay so we know our formula C 7h 1402 so let’s first count our carbons.

So over here we have three carbons plus.

We’re actually missing one carbon atom so we’re missing a carbon atom okay the other thing we’re missing is we have two O’s so we are missing two oxygens and now what we need to do is take these missing pieces or.

Take these missing atoms and turn them into pieces and here’s where.

We remember our IR our our IR clearly showed that we have a carbonyl so clearly these pieces are going to correspond to a give ourselves a little more room here these pieces are going to correspond to a carbonyl and that carbonyl has two bonds associated with it and.

Then we still have a leftover oxygen so let’s draw that oxygen in and of course that oxygen draws two bonds or has two bonds as well so here are two other pieces that we have so now if we count our number of carbon atoms we have three carbon atoms.

Over here another three carbon atoms and then there’s another one so seven carbon atoms count our hydrogen’s here we have six seven plus another two four seven.

Gives us 14 and two oxygens so every atom is now accounted for now that every atom is accounted for we can try to put.

Our pieces together and how do we do that well what we have to do.

Is now see how these are connected and there’s really two possibilities our isopropyl group could be connected to our oxygen which is.

Connected to our carbonyl and our propyl group or the isopropyl group could be connected to the carbonyl and the oxygen and then the propyl group the two carbon pieces here are terminal pieces they only form one.

To form at the end the pieces we derived here have two connections so.

Those have to be connected together so let’s do that let’s cross.

That out and show that clearly we have an.

O connected to a C double bond.

O and both of these are now connected to something else so now what we’ve done is shown that we only have three pieces so now our decisions even easier is the isopropyl group connected to my oxygen or is.

It connected to the carbonyl and same for the propyl group now how do we decide what pieces are connected what we have to do is look at the chemical shift of this peak a and the chemical.

Shift of peak B and see what makes sense looking at the chemical shift of peak a this has a value of five okay so that’s pretty downfield that’s a high ppm number.

Look at the value of B that comes at about 2.3 between the range of two point three okay so what we know is that carbon to hydrogen to oc2 h2o normally comes between three and four and we don’t have anything there this peak at five is further downfield it’s a higher ppm number the carbonyl an H so in H to a C to.

A carbonyl that normally comes between two and three and we do have a peak there so what.

This clearly indicates is B is not going to be connected to our Oh it’s going to be connected to our carbonyl so what we can do is draw our final structure what I’m going to do here is draw my isopropyl group that isopropyl group is connected to our oxygen that oxygen is of course connected to our carbonyl and then we finally have a propyl group attached okay.

So from our analysis this is clearly in the top our final structure.

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