Organic Chemistry II
CHEM 2325
Packet #3
1.
The IR and 13C NMR of an unknown with a molecular formula of C7H12O4 is shown
below. What is the compound?
IR indicates that a carboxylic acid is present (see carbonyl peak in addition to broad peak
centered at ~3000 cm-1). Molecular formula, however, indicates the presence of 4 oxygen atoms.
The 13C NMR indicates that only one type of carbonyl is present and only 3 symmetrically
different carbons (with no other carbons attached to oxygen due to the lack of peaks in ether
positions). The compound must be a symmetrical dicarboxylic acid. The two possible structures
are shown below. The actual compound is 3,3-dimethylglutaric acid (compound on the right).
These two compounds could be distinguished by comparing the downfield shift of the aliphatic
carbons.
O
HO
O
O
OH
HO
O
OH
2.
Molecular formula is C6H10 and the IR and
structure?
13
C NMR are given below. What is the
The IR indicates a terminal alkyne is present. The 13C NMR indicates that only 2 other aliphatic
carbons are present in addition to the two sp hybridized carbons of the alkyne. With a molecular
formula of C6H10, there are 4 carbons in addition to the two alkyne carbons that account for only
2 symmetrically different signals. The only structurally possibility is to have a t-butyl group.
3.
The 1H NMR with relative integration values and IR with a peak at 1719 cm-1 for a
compound with a molecular formula of C9H10O2 are shown below. What is the structure?
3
3
2
2
The 1H NMR indicates the presence of an aromatic ring with peaks between 7.3-8.1 ppm. Due to
having 5 aromatic hydrogens, the most likely structure is to have a monosubstituted benzene ring
with an electron withdrawing substituent (due to the downfield shift of aromatic hydrogens).
The IR indicates a carbonyl is present. In addition an ester is suggested due to the presence of
two oxygens in the molecular formula, the slightly stronger energy carbonyl shift at 1719 cm-1,
and the presence of 2 hydrogens at ~4.3 ppm in the 1H NMR. The 2 hydrogens at ~4.3 ppm also
display a quartet splitting, thus in combination with the triplet at 1.4 ppm that integrates to 3 an
ethyl substituent must be present. This accounts for all atoms in the molecular formula.
O
O
4.
Following are 13C NMR of Isomers of C10H14. Each isomer displays similar 1H NMR
spectra with a quartet at ~2.6 ppm and a triplet at ~1.1 ppm (in addition to other peaks). What is
the structure of each?
The data indicates that these isomers are all disubstituted benzene rings with two ethyl
substituents. All isomers display peaks in the aromatic region and two aliphatic carbon signals.
Due to the fact that all isomers display a quartet at 2.6 ppm and a triplet at 1.1 ppm indicates that
the two substituents must be ethyl groups. The difference between the isomers is the symmetry
of the benzene ring with the two ethyl substituents, causing a different number of symmetrically
related carbons in the benzene ring. The isomers are thus the ortho, meta and para isomers.
A.
B.
C.
5.
Compound has formula C6H12O2. The 1H NMR shows two singlets at 3.6 and 1.2 ppm
and the IR shows a peak at 1740 cm-1. What is the structure?
The IR suggests an ester is present with the carbonyl peak at 1740 cm-1. The 13C NMR indicates
that in addition to the carbonyl carbon, three other symmetrically related aliphatic carbons are
present. The 1H NMR displays only two singlets at 3.6 ppm (suggests the carbon attached to the
ester oxygen) and at 1.2 ppm (carbon thus cannot be adjacent to either an oxygen or a carbonyl
due to the upfield shift). With a given molecular formula of C6H12O2, the symmetry indicates the
structure of the unknown.
O
O
6.
Given the MS, 1H and 13C NMR of an unknown compound, what is the structure?
170 172
2 peaks
3
2 2
The MS immediately indicates the presence of one bromine due to the M/M+2 ratio of 1/1. The
1
H NMR and 13C NMR suggests the presence of a benzene ring. In addition the benzene ring
must be para substituted due to 4 signals in the 13C NMR and two doublets in the 1H NMR. At
least one additional carbon is present due to the peaks at ~20 ppm in the 13C NMR and a singlet
at ~2.3 ppm in the 1H NMR. The integration to 3 in the 1H NMR also suggests a methyl
substituent. One bromine, 7 carbons (6 in the benzene and 1 methyl substituent), and 7
hydrogens as indicated in the 1H NMR would yield a molecular weight of 170. Since this is the
parent MW in the MS, the molecular formula is C7H7Br1 and the structure is the 4-bromotoluene.
Br