Determining Organic Structures From Ir/nmr

Determining Organic Structures From Ir/nmr

August 9, 2016 0 By Jeremy McCallum

Okay in this video we are going to use the following information ir and NMR to propose a structure that’s consistent with the analytical data so we need to use this data to draw a structure that is accurate to the IR NMR really we have three pieces of information we have our chemical formula our IR spectrum and then not.

Shown here at the bottom our NMR so we’re going to use those three pieces of information to solve a.

First thing that we need to do is use our chemical formula we’re going.

To calculate double bond equivalence DB es or also called the index of hydrogen deficiency using a formula to do that and the formula that we use is n plus 1 minus M divided by 2 plus T divided by 2 and that’s the formula that we use to determine the number of double bonds or triple bonds and rings we have in this formula N equals the number of tetravalent atoms carbon in our example M is equal to the number of mono valent atoms hydrogen’s and halogens and.

Of trivalent atoms which is nitrogen well you’ll notice here we don’t have to consider divalent atoms like oxygen so.

Oxygen divalent atoms are not in the formula so filling out our formula what we see.

Is we have seven carbons seven plus one and then we look at the number of hydrogen’s we have and halogens in this case that is 14 so minus 14 divided by 2.

And when we do this problem what we get is we have one DBE or double bond equivalent so our molecule either has one double bond or a ring associated with it now next what we’re.

Going to do is look at the IR spectrum so the first thing that we’re going to do here is analyze the spectrum and we’ll start in zone 1 so looking in zone 1 we don’t see any peaks at all so there’s nothing there but that is some info useful information because what we really know here is that we don’t have any we don’t have any OHS so there are no OHS in here knowing that we do have oxygen in the molecule we know that there are no o H bonds next we’re going to look.

At zone 2 and we see some peaks in zone 2 and what we.

Like to do in zone 2 what I described is drawing our 3000 line so I’m drawing.

This orange line here and you see it’s a little bit to the right of 3000 there are no Peaks immediately to the left of it but we do have peaks to the right of.

That oops in these Peaks are going to represent sp3 carbon to hydrogen bonds moving forward we get to zone 3 there’s an El Peaks here so no triple bonds and then that brings us to zone 4 so clearly there’s a peak here at about 1720 or 1730.

And that’s the biggest peak in the spectrum and what we know is that clearly is a carbonyl so there’s giant peak here is clearly a carbonyl peak and what’s important about this is we knew that we had one DBE and we’ve now seen that one DBE which is a carbonyl moving to zone 5 I don’t have any carbon-carbon double bonds and we can ignore everything in the fingerprint region below 15 so we’ve analyzed.

Our IR no o H Peaks sp3 carbon to hydrogen and our carbonyl peak which again we saw from our DBE so now let’s take a look and analyze our NMR and then finish solving our structure okay so we are going to scroll down and take a look at our NMR spectrum now what I want to do is keep in mind that we need to remember a couple things so I’m just going to on.

The side here kind of write down a couple things that we’ve learned and what we learned is we have a carbonyl somewhere okay so we will just.

Kind of put this information really over to the.

Side of things that we need to remember that was our one DBE and then obviously we have sp3 carbon to hydrogen bonds in here as well okay so looking at our NMR spectrum let’s count our number signals we have one signal here at five.8 and then we do see a little signal over here at zero so just one thing I kind of want to point out to remembers remember that we use an internal standard many times we’re running our NMR and that is TMS so this peak that we see at.

Zero what we want to do is remember that we’re not really going to look at that peak that is.

Our internal standard peak of TMS so we don’t have to worry about that peak there so let’s go ahead and label all of our peaks starting left.

To right and I like to use letters so the first peak we will call a okay we’ll move over our next peak is be followed by see this big doublet here.

Will be D and finally we have a triplet that is going to be e so we have five peaks one thing you’ll notice.

From our NMR spectrum is the splitting we can get by counting the number of Peaks and we’ll do that in a second but the integrals are not provided for us so what we have to do is measure them and if you remember we have this red line that goes throughout the spectrum that red line is really calculating the area under each of these curves and the area.

Under the curves represents how many hydrogen’s they are so what I’m going to do is literally take a ruler and measure the height of.

The line throughout the peak so I measure going to I’m going to measure the height from here to here using a ruler so if I take out a ruler and I measure this height I see the height.

From here to here is actually one centimeter so I’m going to write that down here so we’ll do our integrals in blue so our height here was one centimeter when I look at peak B I’m going to measure the height from here to here and that.

Height comes out to be two centimeters so we’ll write that down looking at peak C I’m going to measure now my height from here to here and again that comes out to be two centimeters so we’ll write that they’re looking at peak D I’m going to measure the height from the bottom of our line here way up to the top there so our height of the peak from here to here that’s representing the air yeah and using the ruler that measures out.

Centimeters so we’ll write that above the peak as six and then we’re going to move over to our last peak e and we’ll measure the height from here to here.

The height of e and when we measure that we get a value of 3 centimeters so what we’ve now done is actually measured out our relative integrals so let’s check these values here here I have a value of 1 to 2 2 to 2 6 to 3 and that’s going to represent a ratio of our peeps.

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