Question

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?

Answer 1

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.

Let's check the answer:

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