The IR spectrum shows a peak at 1720 cm−1, and the 1H NMR shows a doublet near 1.10 ppm, a singlet at 2.10 ppm, and a septet at 2.50 ppm. What is the compound?

1 Answer
yumean1119
Dec 23, 2017

The compound is 3-methyl-2-butanone.

Explanation:

To begin with the IR-abosorption, 1720 cm−1 peak shows that this compound has a carbonyl group.
The molecule of C5H10O has a double bonding in the carbonyl group, and has no C=C double bonding.

Then, let's proceed to the NMR spectrum.
[1] A peak near 1.10 ppm corresponds to −CH3 group. This is a doublet, so there will be a single proton next to it.
[2] A peak at 2.10 ppm is for the −(C=O)−CH3. There is no proton in its neighbor.
[3] The fact that the peak at 2.50 ppm is a septet tells us there are six protons in the adjacent point. This might be for the −(C=O)−CH−(CH3)2.

Therefore, the structure will be
CH3−(CO)−CH−(CH3)2. The IUPAC name is 3-methyl-2-butanone.
![pubchem.ncbi.nlm.nih.gov)

Let's check the answer:
![https://www.chemicalbook.com/SpectrumEN_563-80-4_1HNMR.htm](useruploads.socratic.org)

Here is a chemical shift table.(PDF)
https://staff.aub.edu.lb/~tg02/nmrchart.pdf#search=%27proton+NMR+table%27

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