Question

If the chemical formula is `"C"_6"H"_10"O"_3`, what is the degree of unsaturation? Then, given the information below, what is the compound?

The `""^(1) "H"` `"NMR"` has a peak at 2.5 ppm with 6 protons and 2 neighbors and a peak near 1.2 ppm with 4 protons and 3 neighbors.
The `""^(13) "C"` `"NMR"` has singlet peaks at 8 ppm, 28 ppm, and 170 ppm.

Answer 1

Your question might be "1.2 ppm with 6 protons and 2.5 ppm with 4 protons."

Explanation:

The spectra has only two peaks for `10` protons. It implies that the molecule has a symmetrical structure.

`1.2` ppm shift corresponds to a methyl group, and `2.5` ppm shift to a metylene group next to a carbonyl group. (See my previous post).

I surfed the Internet and found the "answer": Propionic anhydride.

However, the structure of propionic anhydride is inconsistent with "2.5 ppm with 6 protons and 2 neighbors and a peak near 1.2 ppm with 4 protons and 3 neighbors".

If the conditions are "`color(red)"1.2"` ppm with 6 protons and 2 neighbors" and "`color(blue)"2.5"` ppm with 4 protons and 3 neighbors", the answer would be perfect.

Answer 2

Warning! Long Answer. The compound is propionic anhydride.

Explanation:

Preliminary analysis

You know the formula is `"C"_6"H"_10"O"_3"`.

An alkane with six carbon atoms has the formula `"C"_6"H"_14"`.

The degree of unsaturation `U` is

`U = (14-10)/2 = 4/2 = 2`

Therefore, the compound contains two rings and/or double bonds.

`""^1"H NMR"`

The spectrum has 10 protons and only two peaks. The molecule must have a symmetrical structure.

A peak with 2 neighbours and aone with 3 neighbors corresponds to an ethyl group (`"C"_2"H"_5`, triplet-quartet pattern).

The `"6H:4H"` pattern tells us there are two ethyl groups.

However, I think you have the assignments reversed. The methyl group should have the smaller chemical shift.

A methyl group is normally at 0.9 ppm. Something is pulling it downfield
to 1.2 ppm.

A methylene group is normally at 1.3 ppm. Something is pulling it downfield
to 2.5 ppm.

We see from the table that a `"CH"_3` next to a `"C=O"` is shifted downfield to 2.2 ppm (a shift of 1.3 ppm).

We could expect a similar 1.3 ppm shift for a `"CH"_2` group; from 1.2 ppm to 2.5 ppm.

This is just where the `"CH"_2` group appears.

We now know that the partial structure is

`"CH"_3"CH"_2"(C=O)"` + `"(O=C)CH"_2"CH"_3`

These fragments add up to `"C"_6"H"_10"O"_2"`.

There is only one `"O"` atom left to insert. It must go in the middle:

`"CH"_3"CH"_2"(C=O)-O-(O=C)CH"_2"CH"_3`

The compound is propionic anhydride.

Confirmation:

1. The compound has two double bonds.

2. Three `""^13"C"` NMR signals tell us there are three different carbon environments,

We should expect to see

• `"CH"_3color(white)(mm)` at 10 ppm
• `"CH"_2 color(white)(mm)` at 30 ppm
• `"(C=O)O"` at 170 ppm

We see peaks at 8 ppm, 28 ppm, and 170 ppm. This is consistent with propionic anhydride.

3. The `""^13"C"` NMR spectrum of propionaldehyde is