IR Spectra Answer Keys
Sheet labeled Introduction to Infrared Spectroscopy- 11: There are 5 problems. Detailed answers given that include how answer was obtained.
1. 1640 big peak- C=O of an amide
1600 (small shoulder barely sticking out of the C=O peak)- phenyl ring
absence of N-H bonds means amide is a tertiary amide (with 2 alkyl groups and no H's on the N)
2. 1600 is phenyl ring
1700 is C=O
the C=O must be part of an aldehyde because of the peaks at 2700 and 2800.
the "low" value of 1700 (vs. 1710) probably means that the aldehyde is conjugated with the phenyl ring
3. 1710 is C=O which is part of a carboxylic acid due to the very broad peaks around 3000
tiny shoulder on right side of C=O at 1650 is a C=C of an alkene.
"normal" value of C=O indicates isolated system (not conjugated)
4. 1740 peak is probably an ester (higher than "normal" C=O, verified by C-O at ~1200 in fingerprint region)
peak at 2200 is nitrile (C triple bonded to N), larger and a bit to the left of where an alkyne might be
5. 2100 shows alkyne (triple bond between carbons), verified as terminal alkyne by the sharp peak at 3300
broad peak at 3500 is O-H of an alcohol
Sheet labeled IR practice problems: 8 problems, only "answer" is given without specifying how answer was arrived at. These answers would NOT be acceptable on a test or exam.
1. aldehyde and alkene in conjugation
2. carboxylic acid
3. ketone and phenyl ring in conjugation
4. alcohol and nitrile
5. ketone and alkene in conjugation
6. ester
7. primary amine (2 H's on the N) and phenyl ring
8. primary amide and alkene, unclear whether conjugated or isolated