[flora_932]

hello,

i have a NMR for an unknown, with an unknown empirical formula. All the peaks i have are multiplets.
The chemical shifts are:

0.7-1.0
1.2-1.5
1.8-2.3
2.3-2.5

i have attached the NMR, but i didn't see it in the preview, i can email it to you

[Markovnikov]

So you have an NMR of an unknown compound, where you yourself did not perform a reaction. And you don't have any idea on empircal formula nor do you know the integrals?

Do you know if the sample is pure - as in no solvent traces?

[flora_932]

it should be pure...
these are the knowns that we can pick from:

[Markovnikov]

Well, then what you could do is list what functional groups it cannot be? Can it be aldehydes - no, can it be alkenes - no, etc.

[flora_932]

sorry, i didn't include more information..

from my IR, it can be concluded
that it is not alkynes, alkenes, alcohol, secondary amide, nitro

it can be ester, ketone, aldehylde, tertiary amide, aromatic, alkane, alkyl halide

and then my next step was to look at this nmr, which is about deciding how many hydrogens there is, but our teacher did go in depth with the multiplet multiplicity. i understand the single, double, triple, quartet multiplicity, but not the multiplet one.

it can be concluded that it is not a C=OOH, since it doesnt go beyond 10,
nor a O=C-H (not sure the proper name for this),
nor aromatic,
nor C=C-H (alkene),
nor RCH2O, RCH2X, RCH2N (not sure of the names of these)


it could be
C (triple bond)CH,
=CCH2R,
cyclohexane-CH2R,
RCH2C=O,
hydrocarbons of R-CH3, R-CH2-R, R3CH


and then i would proceed to my mass spec, which i can only come to the conclusion of it being 154g/mol as the molecualr weight

[orgopete]

How about the NMR integral? I find that is very helpful, especially when identifying an unknown.

[flora_932]

what is the NMR integral?

[orgopete]

That is the horizontal line that steps across a spectrum which indicates the area of a signal. The area is proportional to the number of hydrogens in it. Alternately, you may have a digital integral, just a number for each signal rather than that line.

If you had a compound with a MF C4H9Br, you could virtually identify the compound with only the integral. There are four possibilities; 9H, 2:1:6H, 2:2:2:3H, or 1:2:3:3H; t-butyl bromide, isobutyl bromide, sec-butyl bromide, and n-butyl bromide.

[flora_932]

i dont think i learned that...
so im not sure what the answer is

[Med90]

do you have carbon data? From there you can identify whether your compound has carbonyl group?

[flora_932]

nope, we were only provided with the IR, NMR and mass spec to figure out our unknown (which from the list is 3 pages of unknowns)

[orgopete]

Okay. Let's look at what we have. From the NMR, there appear to be three singlets from CH₃ groups. I think it is an isopropyl group and a methyl group without adjacent hydrogens. The IR shows a C=O group. The MS shows MW=154. An isopropyl group is MW=43, a CH₃C=27, and C=O = 28.

What is missing?
154 - (43 + 27 + 28) = 56.

How can we get 56? C_xH_y = 56?
x can't be 5 (60 + 0H) or 3 (36 + 20H).

x = 4, y = 8

MF of unknown would then be C₁₀H₁₈O = 154

There appears to be symmetry present, therefore I suggest a ring in order to make it symmetrical. That would make several hydrogens of the CH₂ groups overlap.

[bromidewind]

That sharp peak on the IR spec at 1725 cm^-1 is a dead dead dead giveaway for a ketone. An aldehyde would have two peaks around 2700 cm^-1. Based on your mass spec, there is no aromatic ring involved. The molecular ion number of your mass spec is 154, and I don't see any indication of a nitrogen, bromine, sulfur, or chlorine. So I'm guessing that your chemical formula is probably C₁₀H₁₈O. This gives you two elements of unsaturation.

The mass spec indicates that a methyl group fragments first, then an oxygen. The rest is just a lot of hydrocarbons breaking off, I didn't really bother looking at them too much. If you count the difference in the peaks, then you will be able to tell how many carbons and hydrogens are breaking off.

I don't think you have any alkenes or alkynes in there, as indicated by your IR spec. The two elements of unsaturation are probably due to some serious branching. The fact that almost all of your hydrogens on the proton NMR are at the very bottom indicates that they are highly shielded, so there's probably a few methyl groups near the oxygen, probably branching off from the C in the C=O.

Sorry I couldn't be of much more help in arriving at an answer, but I hope that this helps set you in the right direction.

Edit: Sorry that I repeated most of what orgopete said, I should have paid more attention to it. Also, if you think you've gotten the right structure, you can compare your results with submitted standards via sites like http://webbook.nist.gov/chemistry/ and http://www.ebi.ac.uk/nmrshiftdb/.

[orgopete]

Should I identify the compound at this point?