MS,IR, and NMR help

[Keaver]

I have a MS, IR, NMR problem and I don't know where to go from here. I have everything that I know for sure written on the doc and I'm pretty much stuck trying to figure out this structure. I've never done any NMR, IR, or MS before so this is pretty confusing right now. Any direction on the C13 spectral data would be fantastic and does what I have written so far seem correct?

 

 

 

 

So far the closest thing I've got to the answer is a tert-butyl group attached to a carbon with a double bond O an OH coming off of it. This comes out to the relative formula mass of 102 but there's only 2 types of H where as the H NMR says there should be different kinds of H.

[hypervalent_iodine]

I'll go through each of these one step at a time for you.

 

Mass Spec

 

• You've identified the base peak fine, but the more important piece of information you have there is the molecular ion peak, which for your purposes we can assume is probaly the peak with m/z of 102.
• Because you're only dealing with a fairly simple molecule, you can take a guess at what the molecular formula might be by using the rule of 13. Given that you have some sort of carbonyl peak in your IR, it's safe to assume you have at least one oxygen in your molecule, so figuring out the formula should be fairly straight forward (especially since you also have the integration from your ¹H NMR handy).
• Once you have that, you will want to work out your double bond equivalents, so you know how unsaturated your compound is.

 

 

IR

 

• You need to look a bit more at how you've identified your carbonyl peak. If you look at your ¹H NMR, you'll see that a carboxylic acid doesn't match the data you have.
• The tiny peaks >3000 cm^-1 are not alkene peaks. This will be more apparent once you've worked out your molecular formula and your DBE. Most likely these peaks are noise or an echo from another functional group.
• You seem to have ignored the peaks just below 3000 cm^-1, which though they aren't terribly interesting, may be worth a mention for the sake of completion.

 

 

¹H NMR

 

• This is where you start building up an idea of the connectivity of your molecule. Have a look at each of your peaks and first identify what kind of functional groups they might be and what the splitting pattern tells you about their neighboring environment. So, what does it mean if a peak is split into a septet (note: not a heptet) or a doublet?

 

 

¹³C NMR

 

• Same as with ¹H NMR, you will want to look at a correlation table and see what kinds of functional groups your peaks may correspond to.
• Typically a carbon NMR would be proton decoupled, which means you don't get the splitting pattern that your exercise sheet reports. Sometimes, though, it is useful to turn the coupling on and in these types of experiments, what you get is an idea of how many hydrogens are attached to each carbon. This is essentially the same idea as proton-proton coupling in your ¹H NMR, in that a multiplicity of n will mean that the carbon is bonded to n-1 hydrogens. So, a quartet would correspond to a -CH₃ group, a doublet to a -CH group and so on.
• If you add up the number of hydrogens you get by looking at the multiplicity in this NMR versus the number of hydrogens you get by looking at the integration of the proton NMR. These won't be equivalent, which should tell you or confirm something about your compound.

 

 

Hope that's of some help.

 

 

 

 

 

 

[Keaver]

Ok for the MS the 102 is the fomula mass

 

for the IR I believe the carbonyl group is actually O=CR2

the peaks just below 3000 are then just C-H

 

for H NMR I find that the ratio of hydrogens is 3:1:6

When split into a septet means there are 6 adjacent Hydrogens and a dublet indicates 1 adjacent hydrogen.

 

 

I counted up the hydrogens which would then be 7 hydrogens from the

 

This is as far as I can see right now I think I have the right carbonyl group now though.

 

The or it would be C8H18 if it was just a normal alkane

Edited December 3, 2012 by JamesM22

[hypervalent_iodine]

Ok for the MS the 102 is the fomula mass

 

Yes, so using the rule of 13, what is your molecular formula?

 

for the IR I believe the carbonyl group is actually O=CR2

the peaks just below 3000 are then just C-H

 

Go back to your proton NMR and see if this makes sense. If you had a ketone group, where would the RC=OCH₂R' or RC=OCH₃ group occur? Do you see this in your proton or your carbon NMR?

 

for H NMR I find that the ratio of hydrogens is 3:1:6

When split into a septet means there are 6 adjacent Hydrogens and a dublet indicates 1 adjacent hydrogen.

 

Excellent. So start building up your molecule. Your septet is some kind of CH group and it's next to 6 hydrogens. What kind of functional group(s) is your doublet? If it's next to one proton, which peak does that one proton occur at in your NMR? You can at least build up half of your compound from that information alone.

 

 

I counted up the hydrogens which would then be 7 hydrogens from the

 

Yes, 7 hydrogens from the carbon NMR, but what about the proton NMR? What does this tell you about the peaks in the carbon NMR?

 

The or it would be C8H18 if it was just a normal alkane

 

Nope. C₈H₁₈ has a molar mass of 114 g/mol. You're looking for something with a molar mass of 102 g/mol. If it were a plain old hydrocarbon with no other heteroatoms, then it would be C₇H₁₈. It's obviously not a simple hydrocarbon, so what would the formula be with 1 oxygen? What about 2? Do any of these make sense with your proton NMR, which suggests you have some multiple of 10 hydrogens?

[Keaver]

Yes, so using the rule of 13, what is your molecular formula?

 

 

 

Go back to your proton NMR and see if this makes sense. If you had a ketone group, where would the RC=OCH₂R' or RC=OCH₃ group occur? Do you see this in your proton or your carbon NMR?

 

 

 

Excellent. So start building up your molecule. Your septet is some kind of CH group and it's next to 6 hydrogens. What kind of functional group(s) is your doublet? If it's next to one proton, which peak does that one proton occur at in your NMR? You can at least build up half of your compound from that information alone.

 

 

 

 

Yes, 7 hydrogens from the carbon NMR, but what about the proton NMR? What does this tell you about the peaks in the carbon NMR?

 

 

 

Nope. C₈H₁₈ has a molar mass of 114 g/mol. You're looking for something with a molar mass of 102 g/mol. If it were a plain old hydrocarbon with no other heteroatoms, then it would be C₇H₁₈. It's obviously not a simple hydrocarbon, so what would the formula be with 1 oxygen? What about 2? Do any of these make sense with your proton NMR, which suggests you have some multiple of 10 hydrogens?

Ok I finally ended up getting it and this post helped a ton there were 2 oxygens. One was double bonded the other was not there were 2 CH3 on the end attached to a CH. It all checked out with the NMR and it fit.

[hypervalent_iodine]

Fantastic! I'm glad I could help. Also, the functional group you had is what's called an ester, which has a general formula of -COOR, where R is some alkyl group. The name of your compound, if you're interested, is methyl isobutyrate.