1H NMR & IR : unknown structure determination

In summary, the author struggled to solve a question from a practice exam and needed help from the community. She attached two spectra to illustrate her point. The first spectrum showed the presence of an aromatic ring and double bond(s), while the second spectrum showed the integration of peaks corresponding to the different protons in the ring. Based on the spectra, the author predicted that the compound had an ester, but was not sure. Right-handed substitution of the CH3 group in the ester was confirmed by the NMR spectrum.
  • #1
dead actor
2
0
Hi all,

I was struggling in solving this question which is a part of a practice exam, and I need you to be patient with me and teach me step by step how to figure out the structure and how to read and analyze the spectra.

I attached the two 1H NMR & IR spectra so you will be able to know what structure I'm talking about.

I started with finding the saturation number which is 5 so I know that at least I have aromatic ring and double bond(s)
I predicted the compound to have an ester but I'm not sur.

please help me to know what is this structure.
The molecular formula is C8H8O2

and all the peaks and ppm readings are in the attached figures.

http://img257.imageshack...mg257/3670/scan0004t.jpg

http://img405.imageshack...g405/4821/scan0003aa.jpg

http://img213.imageshack...mg213/8636/scan0005j.jpg


thanks all.
 

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  • #2
dead actor said:
Hi all,

I was struggling in solving this question which is a part of a practice exam, and I need you to be patient with me and teach me step by step how to figure out the structure and how to read and analyze the spectra.

I attached the two 1H NMR & IR spectra so you will be able to know what structure I'm talking about.

I started with finding the saturation number which is 5 so I know that at least I have aromatic ring and double bond(s)
I predicted the compound to have an ester but I'm not sur.

Right. You note that there is the carbonyl stretch in the IR but no corresponding OH group in either the NMR or the IR spectrum.

What might the 'R' group be in that ester? Any hint from the NMR spectrum? What group gives a singlet at ~3.7-3.8ppm that integrates to 3H? (your spectrum is integrated to peak height... it should be by integrated by area in the future)
All of the aromatic protons are accounted for. What type of substitution do the integration and splitting patterns indicate? Mono-substitution? Di-substitution?
 
  • #3
chemisttree said:
Right. You note that there is the carbonyl stretch in the IR but no corresponding OH group in either the NMR or the IR spectrum.

What might the 'R' group be in that ester? Any hint from the NMR spectrum? What group gives a singlet at ~3.7-3.8ppm that integrates to 3H? (your spectrum is integrated to peak height... it should be by integrated by area in the future)
All of the aromatic protons are accounted for. What type of substitution do the integration and splitting patterns indicate? Mono-substitution? Di-substitution?
I have three signals in the aromatic region that accounts for 5 hydrogens. There is only one more signal of three hydrogens. It must be a CH3 group. so this gives me
C6H5 + CO2 + CH3?
so I'm only thinking of C6H5-COOCH3 which is a methyle benzoate
http://images.absoluteastronomy.com/images/encyclopediaimages/m/me/methyl_benzoate.png

right?

I don't know what type of subsitiution I have, may be a di ?
 
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  • #4
Right. It is a mono-substituted aromatic ring. There are three types of protons in that ring, 2 ortho, 2 meta and and a single para (think of the symmetry). The three signals in the NMR should integrate 2:2:1 (edit: or was that 2:1:2?) by area (not peak ht as they are split like crazy).
 
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  • #5
chemisttree said:
Right. It is a mono-substituted aromatic ring. There are three types of protons in that ring, 2 ortho, 2 meta and and a single para (think of the symmetry). The three signals in the NMR should integrate 2:2:1 (edit: or was that 2:1:2?) by area (not peak ht as they are split like crazy).

I was actually little bit confused about this but you cleared my confussion by letting this know to me as, I will search about this little more for my further knowledge.

Thanks!
 

FAQ: 1H NMR & IR : unknown structure determination

1. What is 1H NMR and IR?

1H NMR (Nuclear Magnetic Resonance) and IR (Infrared Spectroscopy) are analytical techniques used in chemistry to determine the structure of unknown molecules. They provide information about the functional groups and connectivity of atoms in a molecule.

2. How does 1H NMR work?

1H NMR works by exposing a sample of the unknown molecule to a strong magnetic field and then applying radio frequency pulses to excite the hydrogen nuclei in the molecule. The resulting signals are recorded and analyzed to determine the chemical environment and connectivity of the hydrogen atoms in the molecule.

3. How does IR work?

IR works by passing infrared light through a sample of the unknown molecule. The molecules in the sample will absorb specific wavelengths of the light, which can be used to identify functional groups present in the molecule. The resulting spectrum can be compared to databases to determine the structure of the molecule.

4. What are the limitations of 1H NMR and IR in structure determination?

1H NMR and IR are limited in their ability to determine the exact structure of an unknown molecule. They can provide information about the functional groups and connectivity of atoms, but they cannot determine the exact arrangement of atoms in a molecule. Other techniques, such as mass spectrometry and X-ray crystallography, may be needed to fully determine the structure.

5. How are 1H NMR and IR used together in structure determination?

1H NMR and IR are often used together in structure determination as they provide complementary information about the molecule. IR can identify functional groups, while 1H NMR can provide information about the connectivity of atoms within those functional groups. By combining the results from both techniques, a more accurate and complete structure can be determined.

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