Answer Key for IR assignment
Spectrum 1: 1,6-Heptadiyne
| Frequency |
Assignment |
| 3299 |
v(CH, alkyne) |
| 2850-2940 |
v(CH, alkane) |
| 2118 |
v(CC, alkyne) |
| 1454, 1434 |
CH2-bending |
Spectrum 2: Benzonitrile
| Frequency |
Assignment |
| 3066 |
v(CH, aromatic) |
| 2229 |
v(CN, triple bond) |
| 1598, 1490, 1447 |
v(C=C, aromatic) |
| 688, 738 |
d(C=C-H, oop-bending) |
Spectrum 3: Acetophenone
| Frequency |
Assignment |
| 3085, 3065, 3005 |
v(CH, aromatic) |
| 2923 |
v(CH, saturated) |
| 1685 |
v(C=), ketone) |
| 690, 750 |
d(C=C-H, oop-benbing |
Spectrum 4: Diethyl phthalate
| Frequency |
Assignment |
| 3071 |
v(CH, aromatic) |
| 2906-2983 |
v(CH, saturated) |
| 1726 |
v(C=O, ester) |
| 1226, 1124 |
v(C-O, ester) |
Spectrum 5: Glycerol
| Frequency |
Assignment |
| 3351 (br) |
v(OH, alcohol) |
| 2937 |
v(CH, saturated) |
| 1650 (br) |
d(OH, alcohol) |
| 1043 |
v(C-O, alcohol) |
Remark: The spectrum has too many peaks for a simple molecule like methanol. It also can not be assigned to benzyl alcohol because it doesn't possess any v(CH, aromatic).
Spectrum 6: Propionitrile
| Frequency |
Assignment |
| 2882-2995 |
v(CH, saturated) |
| 2247 |
v(CN, triple bond) |
| 1462, 1431 |
d(CH2, bending) |
Spectrum 7: Bromoform
| Frequency |
Assignment |
| 3028 |
v(CH, saturated) |
| 1142 |
d(CH, bending) |
| 654 |
v(C-Br) |
Spectrum 8: 3-Chloro-2-methylpropene
| Frequency |
Assignment |
| 3085 |
v(CH, unsaturated) |
| 2980 |
v(CH, saturated) |
| 1655 |
v(C=C, alkene) |
| ~760 |
v(C-Cl) |
Spectrum 9: Benzyl alcohol
| Frequency |
Assignment |
| 3169 (br) |
v(OH, alcohol) |
| 3032 |
v(CH, unsaturated) |
| 2874 |
v(CH, saturated) |
| 1602, 1496, 1454 |
v(C=C, aromatics) |
Remark: The peak at 1700 cm-1 is too weak for a 'real' carbonyl peak. In addition, the v(OH) peak is too narrow for an carboxylic acid peak.
Spectrum 10: Di(n-butyl)amine
| Frequency |
Assignment |
| 3294 (w, br) |
v(NH) |
| 2810-2959 |
v(CH, saturated) |
| 1631 |
d(NH) |
| 1463, 1376 |
d(CH2, CH3) |
Spectrum 11: 2-Adamantanone (solid!)
| Frequency |
Assignment |
| 2916, 2854 |
v(CH, saturated) |
| 1720 |
v(C=O, ketone) |
Remark: The IR spectrum shows too many peaks for a simple compound like acetone (Z=3N-6!).
Spectrum 12: p-Nitrobenzylbromide (solid!)
| Frequency |
Assignment |
| 3038-3107 |
v(CH, unsaturated) |
| 2861 |
v(CH, saturated) |
| 1611 |
v(C=C, aromatic) |
| 1540, 1346 |
v(NO2) |
Spectrum 13: Octadecane (solid!)
| Frequency |
Assignment |
| 2854-2958 |
v(CH, saturated) |
| 1467, 1378 |
d(CH2, CH3) |
| 721 |
d((CH2)4) |
Remark: The peak at 721cm-1 is too weak for a v(C-Cl). The compound is a hydrocarbon and has to have at least 17-18 carbon atoms to be a solid! Decane, pentane, etc are liquids at room temperature.
General comments:
1. The average grade for the assignemnt is 34 points (out of 40). If you scored significantly less than 30 points, I would advise you to see your TA or the instructor to get a better insight on IR spectroscopy.
2. One of the most common mistakes was not to analyze the v(C-H) region correctly to determine what type of compound (alkane, alkene, aromatic, or mixed) is present.
3. The v(OH) peaks for acids and alcohols look very different and can be easily identified by looking at them.