Draw the structure of a compound of molecular formula C4H8O that has a signal in its 13C NMR spectrum at ≥ 160 ppm. Then draw the structure of an isomer with no rings of molecular formula C4H8O that has all of its 13C NMR signals at < 160 ppm.

1. Butyraldehyde; 2. but-3-en-1-ol

Explanation:

1. Peak ≥ 160 ppm

The formula C₄H₈O shows that the Index of Hydrogen Deficiency = 1.

It could be caused by either a ring or a double bond.

A peak at ≥ 160 ppm strongly indicates a C=O group, so the rest of the molecule can contain no rings or double bonds.

There are no other heteroatoms, so the compound most be an aldehyde or a ketone.

One compound that meets these criteria is butyraldehyde, CH₃CH₂CH₂CH=O (see Fig. 1.)

2. No rings; all peaks < 160 ppm

If all peaks are < 160 ppm, there can no C=O group.

There is no ring, so there must be a C=C double bond.

There is no other unsaturation, so the O atom must be present as an alcohol or an ether.

One compound that meets these criteria is but-3-en-1-ol, CH₂=CHCH₂CH₂OH (see Fig. 2).