The IR spectrum shows a peak at "1720 cm"^(-1), and the ""^(1) "H" "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 C_5H_10O 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 -CH_3 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)-CH_3. 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)-Ccolor(red)H-(CH_3)_2.

Therefore, the structure will be
CH_3-(CO)-CH-(CH_3)_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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