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Magazine Chapter 20: Q68P (page 1038)
The IR spectrum, 13C NMRspectrum, and 1H NMRspectrum of an unknown compound (C6H8O3) appear next.
Determine the structure, and show how it is consistent with the spectra.
Short Answer
Step by step solution
Spectroscopy of aromatic compounds
Several analytical techniques like mass, IR, and NMR can help identify a particular compound's structure. The functional groups in a specific molecule can be identified from infrared (IR) spectroscopy.
Determining the structure and showing how it is consistent with the spectra.
The formula C6H8O3 has 3 elements of unsaturation.
In the IR spectrum, the absence of the OH group shows that the compound is neither alcohol nor carboxylic acid.
There are two carbonyl absorptions: The one about 1770cm-1 is a strained cyclic ester, whereas the one at 1720 cm-1 is a ketone.
Anhydrides also comprise two peaks but possess a frequency higher than the ones in the spectrum.
In the H-NMR spectrum, the NMR has 4 types of protons.
The 2 H multiplet at 未=4.3is a CH2 group found next to the oxygen on one side.
The 1 H multiplet at 未=3.7 is also strongly deshielded, and a CH is also found next to the CH2.
The 3 H singlet at 未=2.45 is a CH3 on one of the carbonyls.
The remaining two hydrogens are highly coupled, a CH2 whereas the two hydrogens are not equivalent.
There are no vinyl hydrogens. Hence the remaining element of unsaturation must be a ring.
The proposed structure can be shown below.
In the13CNMR spectrum, the peaks are shown below.
Assembling the pieces to generate the structure
On each carbon, the top hydrogen is cis to the acetyl group and the bottom hydrogen is trans. Thus, the two hydrogens on each of these carbons are not equivalent, resulting in complex splitting.
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