# Sample Chemistry Paper on Interpretation of IR Spectra

## Interpretation of IR Spectra

CHE 240 IR Worksheet

For each of the spectra shown, (a) Calculate the degree of unsaturation (b) state  the functional groups that are indicated by the spectra and indicate which IR peak(s) match(es) each, and (c) propose a structure

Spectra 1

C6H12

Degree of unsaturation is used in the determination of number of Pi bonds (double bonds) and or the number of rings. Therefore the degree of unsaturation = 2(6) + 2 – 12 = 2/2 = 1. The one degree of unsaturation indicates that the compound contains one double bond or a ring. A peak at between 2950cm-1 to 3000cm-1 is associated with Sp3 C-H stretching vibration. The presence of sharp peaks at around 1600 cm-1 is due to C=C stretching vibration. Presence of peak at around 3100 is due to C-H stretch of an alkene. With the one degree of unsaturation, C-H, and C=C stretching vibration, it can be concluded that the compound is an alkene. Since the compound contains six carbon atoms, it has to be hexene and the structure is as indicated below

# Spectra 2

Degree of unsaturation = 2(6) +2 -14 = 0/2 = 0. This means that the compound has no double bond or a ring. It only possesses single bonds. Strong peak near 1000cm-1 is due to C-O stretching vibration.  A peak at around 1300 cm-1 is due to C-O-C stretch thus indication of an ether. A peak at around 845cm-1 is due to C-C-0 stretch. A peak at around 2900cm-1 is due to Sp3 C-H stretching vibrations. A Strong peak at around 1500 cm-1 is due to C-H bending. No broad peak at 3300-3500 cm-1 thus it is not an alcohol. Thus, the compound is an ether.

# Spectra 3

Degree of unsaturation = 2(6) +2 – 12 = 2/2 = 1.  This means that there is one double bond or a ring. A peak at around 2900cm-1 is due to Sp3 C-H stretching vibrations. A Strong peak at around 1500 cm-1 is due to C-H bending. The presence of sharp peak at around 1700cm-1 indicates C=O stretching vibration. No broad peak at 3300-3500 cm-1 thus it is not an alcohol. The presence of peak at around 2800cm-1 indicates that the compound is an aldehyde. With the one degree of unsaturation, H-C=O, Sp3 C-H stretching and six carbon atoms the compound can be concluded to be hexanal. The structure of the compound is as shown below

Spectra 4

Degree of unsaturation = 2(5) + 2 – 10 = 2/2 = 1. Since there is one degree of unsaturation, it means that there is only one double bond or a ring.  The presence of broad band at around 3400 cm-1 is due to OH stretching vibration. The presence of sharp peak at around 2900cm-1 is due to Sp3 C-H stretching vibration. Broad O-H band at around 3400cm-1 indicates that the compound is an alcohol. Since the degree of unsaturation is one, it means that the compound contains a ring structure. The compound therefore can be said to be a cyclopentanol since it contains five carbon atoms and a broad O-H band and the structure is as indicated below

# Spectra 5

Degree of unsaturation = 2(6) +2 -14 = 0/2 = 0. The zero degree of unsaturation indicates that there is no presence of double bonds or a ring structure. The compound is likely composed of single bonds. Strong peak near 1000cm-1 is due to C-O stretching vibration.  A peak at around 1300 cm-1 is due to C-O-C stretch thus indication of an ether. A peak at around 845cm-1 is due to C-C-0 stretch. A peak at around 2900cm-1 is due to Sp3 C-H stretching vibrations. A Strong peak at around 1500 cm-1 is due to C-H bending. No broad peak at 3300-3500 cm-1 thus it is not an alcohol. The degree of unsaturation being zero rules out the possibility of the compound being a ketone or an aldehyde. This is because the carbonyl group contains a double bond. Thus, the compound is an ether.

# Spectra 6

Degree of unsaturation  =     (2n + 2) –  x

2

(2 × 4 + 2) – 8         = 1

2

There is only one degree of unsaturation meaning there is one double bond or a ring. There is an absence of aromatic ring since aromatic ring has 4 degrees of unsaturation. Peak at around 2990 cm-1 means Sp3 C-H stretching vibration. Peak at around 1700 cm-1 means there is a C=O for carboxylic acid. The absence of peaks at around 2800 to 2700 cm-1 rules out the possibility of the compound being an aldehyde and also the fact that there is a O-H peak that is strong and broad around 3000 cm-1 which is consistent to carboxylic acids. For there to be one degree of unsaturation, Sp3 C-H, C=O, and O-H the compound has to be a carboxylic acid and in this case butanoic acid due to the four carbon atoms. The structure of the compound is as shown below

Spectra 7

# C5H10O

Degree of unsaturation = 2(5) +2 – 10 =2/2 = 1. This means that there is one double bond or a ring. The peak around 2964 cm-1 means Sp3 C-H stretch. The peak at around 2878 cm-1 is consistent with a C-H stretch for Aldehydes. The peak around 1716 cm-1 is also consistent with C=O stretch for Aldehydes. There are no peaks around 3300 cm-1 to 3500 cm-1 therefore it is not an alcohol. The rest of the peaks are in the finger print region and therefore cannot be identified reliably. Since the compound has 5 carbon atoms, Sp3 C-H stretching, and H-C=O it can be concluded that it is pentanal and the structure is as shown below

# C7H14O

Degree of unsaturation = 2(7) + 2 -14= 2/2 = 1. There is a likelihood of one double bond or a ring. The peak around 2959 cm-1 and 2934 cm-1 means Sp3 C-H stretching vibration. The peak at around 2862 cm-1 is consistent with the C-H stretching for Aldehydes. The peak around 1718 cm-1 is also consistent with the C=O group for Aldehydes. The absence of peaks around 3300 cm-1 to 3500 cm-1 indicates that the compound is not an alcohol. The compound therefore having one degree of unsaturation, C-H stretching, C=O, has to be a heptanal because of the seven carbon atoms. The compound is as shown below

# Spectra 9

C4H10O

Degree of unsaturation = 2(4) +2 -10 = 0/2 = 0. This means that the compound is saturated and therefore there are only presences of single bonds. The strong and broad peak at 3339 cm-1 is due to O-H stretching vibration for alcohols and it’s intermolecular bonded. The peak at 2957 cm-1 means Sp3 C-H stretching vibration. The peak at 1471 cm-1 is due to C-H bending. The strong peak around 1041 cm-1 is due to C-O stretching for primary alcohols. Since there is no peak around 1700 cm-1 it rules out the possibility of the compound being an aldehyde or a ketone. The compound is therefore an alcohol and since it has 4 carbon atoms it has to be butanol and the structure is as shown below

# Spectra 10

C4H8O2

Degree of unsaturation = 2(4) +2 – 8 = 2/2 = 1. This means that there is only one double bond or a ring. The peak around 2986 cm-1 and 2955 cm-1 means Sp3 C-H stretching vibration. Since there are absence of peaks around 2800 cm-1 to 2700 cm-1 this rules out the possibility of an aldehyde. The absence of peaks around 3300 cm-1 to 3500 cm-1 also shows that the compound is not an alcohol. The peak at around 1743 cm-1 is consistent with the C=O group for aliphatic esters. The peak around 1358 cm-1 means C-O for acyl which supports the possibility of the compound being an ester. The rest of the peaks are in the finger print region and therefore may not be used to identify the compound reliably. The one degree of unsaturation, Sp3 C-H, C=O, and C-O shows that the compound has to be ethyl ethanoate (ester) and the structure is as shown below

References

https://www.sigmaaldrich.com/KE/en/technicaldocuments/technicalarticle/analyticalchemistry/photometry-andreflectometry/ir-spectrum-table

Infrared spectroscopy absorption table. (2020, November 3). Retrieved June 7, 2021, from https://chem..libretexts.org/@go/page/22645