[Spilogale]

Hi

I'm a senior taking organic chemistry. We were given an IR/ NMR assignment and while I completed identifying two other spectra on my own, I'm a bit stumped by the weird peaks on the NMR spectra. There are four short peaks which do not match the patterns given to use for a quartet, they appear to either by random or increasing in height from left to right and the peaks are in the 7-6 range.

If it is a quartet (next to 3 H), then there's 1H next to 3H.

The given formula is C5H8O2 with a given molecular weight of 100.12.

The IR spectra suggests an OH group, that it is a carboxylic acid, etc.

What other information can I provide to help try to clarify this so I can understand the problem better?

[phth]

What is the degree of unsaturation?  that will tell you how many ring/double bond/carbonyls are in the structure. 

[Spilogale]

The degree of unsaturation is 2.

[Spilogale]

The complete calculation of integrations that I have are:

Sum of all integrations is 50

A 6/50 * 8 = 0.96 1 H next to 3H

B 4/50 * 8 = 0.64 1 H to 0 H

C 2/50 * 8 = 0.32 1 H to 0 H

D 23/50 * 8 = 3.68 3 H next to 0H

E 15/50 * 8 = 2.4 3 H next to 1 H

I have attached the IR/NMR spectra.

[phth]

looks like an ester (1080 peak is C=O-O-C ether stretch), that is corroborated by the 3.5ppm 1H shift.  Methyl ester (C2O2H3).  Left over is C3H5.  The peak at 1724 is the carbonyl stretch, but it should be 1740;  that means it is next to conjugation i.e. double bond.  This is corroborated because of the 5.5-7 ppm peaks.  There are 2 CH's and one more methyl which is next to one. so it can be trans or cis which is determined by the J coupling size, but this is a really bad NMR spectrum, and it looks like the spectra are a mixture of both.  Your teacher is bad and shouldn't give you better questions IMHO

Mod Edit: Swearing modified. Do not swear. Dan

[phth]

Trans is probably the answer I would go with because it is more stable and easier to have laying around

[Spilogale]

So regardless of trans/cis, you're suggesting

CH3-CH=CH-C=O-O-CH3

It has the right DU, right formula.

3 next to 1 CH3-CH=

1 next to 3 for =CH-CH3

3 next to 0 -O-CH3

1 next to 0 for the CH-C=O-O-CH3

Aren't I missing a 1 next to 0 somewhere or did I mess up the integrations? or does it count twice because of the double bond (is this what is going on with the 5.5-5.7 peaks?)?

[phth]

U got 9 hydrogens in the integration.  This is probably from error (the machine is ±10% [today's standards]) on top of you having to measure by hand, which could be inaccurate in addition to it being an old spectrum.  The peak is has to be a doublet.  It is a spread out doublet because it has a very small allylic coupling (called W coupling) to the methyl group on the end.  This makes it a doublet of quartets; the spectrum looks like it was taken in 1960 and the instruments were not good enough to see the small lines, so it just looks like a blob broadening the peaks increasing the integration which is why it is off.

[Spilogale]

Wow, you're %@#% right about those H's!

Our spectra come from software that required using a virtual machine for XP, so I imagine they are pretty old.

[Dan]

The complete calculation of integrations that I have are:

Sum of all integrations is 50

A 6/50 * 8 = 0.96 1 H next to 3H

B 4/50 * 8 = 0.64 1 H to 0 H

C 2/50 * 8 = 0.32 1 H to 0 H

D 23/50 * 8 = 3.68 3 H next to 0H

E 15/50 * 8 = 2.4 3 H next to 1 H

I think B/C is one signal with a large coupling constant (which is a clue as to the double bond geometry).

It does look as though this was recorded on a very low frequency spectrometer.

[Spilogale]

I just wanted to say thanks, it was indeed CH3-CH=CH-C=O-O-CH3

[Babcock_Hall]

I would like to amplify something that Dan said.  When one takes NMR data at low frequency, the spectra do not always strictly obey first-order rules with respect to nuclei which are coupled and are also relatively close in chemical shift.  As one moves to higher frequency spectrometers, the spectra become more nearly first-order.