How could you distinguish between the following pair of compounds from their IR spectra?
#"HOCH"_2"CH"_2"CHO"# (3-hydroxypropanal) and #"CH"_3"CH"_2"COOH"# (propanoic acid)

Question

Sarah Angotti · Accepted Answer

The main differences between these molecules' IR spectra are in the #OH# stretches and in the #C=O# stretches.

While the alcohol #OH# stretch is broader, the carboxylic #OH# stretch is less broad. While the aldehyde #C=O# stretch is sharp and "well-defined", the carboxylic #C=O# stretch is broader and more "smeared".

COMPARING THE STRUCTURES

Follow the structural formulas from left to right to determine that the structures are:

Notice how the functional groups are clearly different; the left compound has an alcohol hydroxyl and an aldehyde carbonyl group, while the right compound has a carboxylic hydroxyl and a carboxylic carbonyl group.

SIGNIFICANT VIBRATIONAL MOTIONS

To break it down further, consider the ways these compounds can stretch and bend.

It's a good rule of thumb to assume that single bonds can stretch and bend, and double bonds can stretch, but don't bend very well. Also, as an aside, alcohol bends aren't really necessary to consider because their stretches are so distinctive.

Here's what I see. I've bolded the unique motions, and put the frequencies in wavenumbers (Techniques in Organic Chemistry, Mohrig).

3-hydroxypropanal -

#sp^3# #"C"-"H"# bend (alkane) --- #1480~1430# and #1395~"1340 cm"^(-1)#; medium to weak; sharp

#sp^3# #"C"-"H"# stretch (alkane) --- #2990~"2850 cm"^(-1)#; medium to strong; sharp

#\mathbf("C"="O")# stretch (aldehyde) --- #1740~1720# (saturated) and #1715~"1680 cm"^(-1)# (conjugated); medium; sharp

#\mathbf("C"-"H")# stretch (aldehyde) --- #2900~2800# and #2800~"2700 cm"^(-1)#; weak; sharp Fermi doublet

#\mathbf("C"-"O")# stretch (alcohol) --- #1300~"1000 cm"^(-1)#; strong; sharp

#\mathbf("O"-"H")# stretch (alcohol) --- #3650~3550# and #3550~"3200 cm"^(-1)#; medium to strong; broad
propanoic acid -

#sp^3# #"C"-"H"# bend (alkane) --- #1480~1430# and #1395~"1340 cm"^(-1)#; medium to weak; sharp

#sp^3# #"C"-"H"# stretch (alkane) --- #2990~"2850 cm"^(-1)#; medium to strong; sharp

#\mathbf("C"="O")# stretch (carboxylic acid) --- #1725~1700# (saturated) and #1715~"1680 cm"^(-1)# (conjugated); strong; sharp

#\mathbf("C"-"O")# stretch (carboxylic acid) --- #1300~"1000 cm"^(-1)#; strong; sharp

#\mathbf("O"-"H")# stretch (carboxylic acid) --- #3200~"2500 cm"^(-1)#; medium to weak; broad
You should eventually see that the significant differences in the frequencies given above are:

The carboxylic #"C"="O"# stretch is generally stronger, slightly more upfield, and broader.

The carboxylic #"C"="O"# stretch is less broad (spans a smaller range of frequencies near #"3000 cm"^(-1)#)
COMPARING THE SPECTRA

This will become clearer once we examine comparable spectra on this nice website called the SDBS.

Although I couldn't find 3-hydroxypropanal's spectrum, I did find propionaldehyde/propanal's spectrum, which still shows a major difference with the #"C"="O"# stretch when comparing to propionic/propanoic acid's spectrum.

Pay attention to the circled peak; it's sharp and "well-defined".

The circled peak here is broader and more "smeared". Therein lies the difference!

(Ignore the #"O"-"H"# stretch for this comparison because it's an invalid comparison in that regard.)

Maya Allen · Answer

undefined The IR spectrum of 3-hydroxypropanal (HOCH2CH2CHO) would show a broad peak around 3200-3600 cm^-1 due to the O-H stretching vibration of the alcohol group, whereas propanoic acid (CH3CH2COOH) would exhibit a sharp peak around 2500-3000 cm^-1 due to the O-H stretching vibration of the carboxylic acid group. Additionally, propanoic acid would also show a peak around 1700-1750 cm^-1 due to the C=O stretching vibration of the carboxylic acid group, which would not be present in 3-hydroxypropanal.

How could you distinguish between the following pair of compounds from their IR spectra? #"HOCH"_2"CH"_2"CHO"# (3-hydroxypropanal) and #"CH"_3"CH"_2"COOH"# (propanoic acid)