[mirandamix]

hey everyone,

I have this question, that I've been trying to solve like forever.
so here's the thing:

there's a subsance that consists of C, H, N and O.
one molecule of this substance has 12 H atoms and a mass fraction of 26,06% O.
we got experimentally using a proton-nmr the following signals:
* 11,50 ppm   Singulett  (Integral 1)
* 1,80 ppm     Quartet   (Integral 2)
* 0,75 ppm     Triplett    (Integral 3)

Q: Determine the molecular formula of this Compound.

thank you in advance )

[Hybrid]

it is so hard give me a while, i tried symmetry rings acids ethers but it is difficult

[mirandamix]

thank you for replying )
i've been trying hard to solve it... but nothing came out so far...

[Hybrid]

it is difficult i constructed 2 symmetric molecules but with O either 25.3 or 28.2 %, i never been able to adjust O to 26.06 % . are u sure there is no other information missed ?

[ATMyller]

The singlet is the key. Hardly anything goes that far downfield if it's only C, H, N and O. Carboxylic acids, hydroxyl next to carbon double bond and amines next to double bonded oxygen like phthalimide or barbituric acid.

[mirandamix]

it is difficult i constructed 2 symmetric molecules but with O either 25.3 or 28.2 %, i never been able to adjust O to 26.06 % . are u sure there is no other information missed ?

hey, it's actually a long question and this one is like part (a)... in the other parts there are 2 spectrums... One is the IR-Spectrum of the compound and there's another one which is the 13C-nmr-spectrum of the compound.

I'll upload the two spectrums as soon as possible, I guess they might help.

[mirandamix]

in the IR-spectrum there are these 5 marked distinct absorption peaks:
A: 3209 - 3077 cm¯¹
B: 2961 - 2855 cm¯¹
C: 1720 - 1679 cm¯¹
D: 1521 - 1502 cm¯¹
E: 1461 cm¯¹

so A is the group N-H or O-H ??
B is guess C-H
C should be C=C or C=N or C=O
D and E same as C

[mirandamix]

yeah and there's a tip in the question that it is not an aromatic compound...

[Hybrid]

sure it is not aromatic, it will be difficult to construct symmetric aromatic cpd with all aromatic H substituted , give as C 13 may it help
i can't get benefit of IR except excluding presence of CN

[mirandamix]

so in the 3C-nmr-spctrum there are 5 signals and here's a pic attached...

[Hybrid]

this question remember me when one of dr. give the postgraduate students spectroscopy interpretation question and no one answer it  and they try to ask other dr.s of our department and also no one of the dr.s " even emeritus one" answered it .  this is hard question and need time and one to be mentally cooool

[mirandamix]

oh nooo.. so i won't get the answer?? i need it and as soon as possible...
if you can't get the wright answer, can you at least show me the way to get to something.. like anything??

[mirandamix]

right answer**

[Hybrid]

1. having 12 H but only 6 appear mean it is symmetric cpd
6 H at left = 6 H at right

2. quarette & dublet set refer to CH2CH3 gp. so now we have 2 CH2CH3

3. singlet at high field correspond to COOH proton. so must have 2 COOH

now we have 6H with main 6C & 4O

but %O make it very difficult , try to think u may get it but till now i couldn't get the right answer

N.B no CN " from IR peaks" so N must be fully substituted " to assign no H peak in NMR"

[mirandamix]

thank you so much for the explanation,

but i thought quatret refers to R-NH2 or R-OH ?? and triplet to R-OH ...
maybe i'm wrong...

concerning %O I really don't get it.. how do we calculate it??

and how did you know that there is no CN from IR peaks?

so until now we have C6H12O4 what about N??