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?

Question

Ellie Adams · Accepted Answer

The compound is 3-methyl-2-butanone.

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://tutor.hix.ai

Zoe Alexander · Answer

undefined The compound is likely an ester with the formula RCOOR'. The IR peak at 1720 cm^(-1) indicates the presence of a carbonyl group (C=O), while the ^(1)H NMR spectrum shows characteristic signals for the methyl (singlet at 2.10 ppm), methylene (doublet near 1.10 ppm), and methine (septet at 2.50 ppm) groups in the ester structure.