Question

Given the following spectra, can someone help me identify the unknown compound:

Photo 1: IR spectra

Photo 2: Proton NMR

Photo 3: Proton NMR

Photo 4: Carbon NMR

Photo 5: Close up image of the Carbon NMR values.

I previously asked this question and was requested to provide more information, but I was unsure what the individual needed. If you need additional information, please be descriptive with what you need! Thank you so much :)

45 3500 3000 2500 2000 Wavenumbers (cm-1) 1500 1000 500

Answer 1

Since there are five types of peaks in CNMR, so we can say that the compound has five different types of carbon atoms. But the peak around 77 ppm corresponds to that of solvent (CDCl₃), so there are only FOUR different types of carbons are present.

From CNMR, we can say that a highly deshielded small peak is present which must be of C=O, also from IR it can be further proved which has a sharp strong band at 1704 cm^-1 that there is a C=O group present, but since there is no peak in HNMR around 9-10 ppm which corresponds to aldehydic proton, so we can say that C=O is present as ketonic group.

In IR since many peaks are below 3000 cm^-1 (NOT above 3000 cm^-1), thus it can be concluded that there is no =C-H stretching, only -C-H stretching is present. Also there are peaks around 1100 cm^-1 which corresponds to C-O stretch.

From HNMR, it can be concluded that three different kinds of protons are present as there are three peaks are shown. The sharp triplet at 0-1 ppm crresponds to -CH₂CH₃. The peak around 1.6 ppm which is a quartet corresponds to -CH₂CH₃ but here proton is deshielded, so it is -CO-CH₂CH₃. The peak around 2.4 ppm is highly deshielded so it must be O-CH_3.

In CNMR, the peak around 13.7 corersponds to -CH₂-CH₃, the peak around 17.4 also corresponds to CO- CH₂-, the peak around 44.7 ppm corresponds to CH₃-CO-.

The compound will be: CH₃-O-CO-CH₂-CH₃.