If the force constant is approximately the same for #C-C#, #C-N#, and #C-O# bonds, how can I predict the relative positions of
their stretching vibrations in an IR spectrum?

Question

Skylar Alexander · Accepted Answer

You can't.

The #"C-C"#, #"C-N"#, and #"C-O"# groups all appear in the region from #1300# to #"1000 cm"^-1#.

The region from about #1500# to #"500 cm"^-1# is called the fingerprint region.

It usually contains a complicated series of absorptions that are caused by many different vibrations within the molecule.

This region is often useful only for confirming the presence of functional groups that you have identified in other areas of the spectrum.

group, so one of the strong bands between #1200# and #"1100 cm"^-1# is probably a #"C-O"# stretching vibration.

Nicholas Antonuccio · Answer

undefined The relative positions of stretching vibrations in an IR spectrum are determined by the reduced mass of the atoms involved in the bond and the force constant. The reduced mass is inversely proportional to the masses of the atoms forming the bond. Therefore, in bonds with lighter atoms, such as C-O, the stretching frequency will be higher compared to bonds with heavier atoms, such as C-C or C-N. However, the force constant also plays a role, with stronger bonds resulting in higher stretching frequencies. Overall, the C-O bond is typically stronger and lighter, so it will appear at a higher frequency compared to C-N and C-C bonds in an IR spectrum.

If the force constant is approximately the same for #C-C#, #C-N#, and #C-O# bonds, how can I predict the relative positions of their stretching vibrations in an IR spectrum?