[bhambhadat]

My teacher assigned a spec problem for us to do as homework, and I've been having some trouble with it.  The data is attached in pdf.  We're supposed to propose a structure.  All I can tell is that it has an alkene/aromatic region and a carboxylic acid.  Can someone advise me as to how to derive the chemical formula?

I've also been assigned some spec problems from "spectrometric identification of organic compounds" by Silverstein.  The problems are 8.38 and 8.42.  If anyone is familiar with these problems, could they please offer some help? 

8.38

From the data, I speculated that the structure contains an aromatic ring, that it has a molecular formula of C10H12O2, an isopropyl ketone substituent, and an Ar-OH. 

8.42

I speculated that this structure has a formula of C10H12O2, an aromatic ring, an alcohol, and an ether. 

Thanks for the help--I'm not too experienced with this stuff and have been stuck for hours.

[Serenagreene]

For the PDF file

1)  I see 6 C (best case - none are overlapped)
2)  I see 8 H
3)  With a Mass spec of 176 I must then have 6 O.
4)  From the IR I see that there must be some OHs, a Ketone, and a ring.
5)  From the C NMR Dept - I see one C with 2H, and 2 with 1H
6)  The COSY is showing that the Hs between 3 - 4 and 4 - 5 are scalar coupled.

I would start drawing some compounds the fit the above and then calculate some of the C and H to see if they fit the spectras.

[bhambhadat]

Thanks, that was very helpful.  I came up with C6H8O6, and I have a good idea of the general structure now.  How were you able to know that there were 6 O's from the MS?  I see a couple of them, but would like to know how you saw all 6 in there. 

Also, when you said "ketone" do you mean generally a carbonyl group, or an actual ketone? 

Thanks again.

[agrobert]

There is no aromatic ring.  There are two carbonyls, an acid and an aldehyde.  The acidic proton is downfield at eleven and the aldehyde proton is at eight.  The other four protons are attached to coa conjugated carbon system.  To determine the geometry calculate the splitting of these alkenyl protons.    Trans has a higher J value than cis.  I don't understand the 6 Oxygen assumption either.

[bhambhadat]

I see what you mean with the h-nmr for the aldehyde, but shouldn't there be a c-nmr peak further downfield if an aldehyde is present?  Or is the 170 the aldehyde and the 160 the acid? 

My biggest remaining point of confusion is how the c-nmr gives two CH and one CH2, but the integration from the H-NMR gives that there are 8 (or a multiple of 8 ) hydrogens.  Since only four protons are covered by the carbons, does this mean that I have four alcohols?  If so, how would this fit in with the COSY?

[Serenagreene]

How I got the 6O

MS = 176 - (6C * 12) - (8H *1) = 96/16 = 6 O

There is a ring - not an aromatic ring, but it does have a double bound in it. 

I did mess up the COSY (just learned how to read them).  But I will tell you that the Hs for the OHs on the H NMR are the 3 downfield.

This is a common molecule that you probably consume every day.

[bhambhadat]

You're definitely right on the money--just realized it was ascorbic acid:  the ms matches perfectly.  Thanks for your help.

[bhambhadat]

Went ahead and scanned one of the problems I was talking about earlier (8.38).  The structure drawn next to the COSY is my prediction.  Any guidance with this problem would be appreciated.  Thanks.

Edit:  sorry about inverted image.  In adobe, go to view->rotate view to correct.