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1H NMR & IR : unknown structure determination

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Hi all,

 

I was struggling in solving this question which is a part of 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.

 

 

 

thanks all.

scan0004.jpg

scan0003.jpg

scan0005.jpg

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Im not sure how much help people can give seeing as its essentially homework.

 

As youve worked out that there is an aromatic ring from the saturation number (and the proton NMR confirms that). Take 6 carbons and 5 hydrogens off the given molecular formula and see what atoms you have left. That should help you a lot esp given the rest of the proton NMR

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1 doublet, 2 triplets, the doublet and one of the triplets are of the same length.

 

What if I had an aromatic ring with 1 substituent? :eyebrow:

 

Hint: a triplet is caused by 2 adjacent hydrogens, a doublet is caused by 1 adjacent hydrogen.

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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

methyl_benzoate.png

 

right?

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That would indeed make sense. Now can you correlate these to your IR spectrum?

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what are diagnostic bands used to interpret IR? try looking for strecthing frequencies of the functional groups that are in your proposed structure.

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That structure could be correct yes...but how can you twll which way the ester is (i.e. the way you've drawn it or having the carbonyl next to the methyl group). The chemical shift of the methyl group should tell you as well as the carbonyl stretch in the IR since one form will be in conjugation with aromatic ring

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