Test 4
Dr. Bachrach4/25/2011
1)Provide the IUPAC name for the
following compounds.
a)
(5S)-1-methyl-5-propyl-1,3-cyclohexadiene
b)
(2Z,4Z,6S)-2-chloro-6-methyl-5-propyl-2,4-nonadiene
5) Predict the product(s) of the following
reactions
a)
b)
c)
d)
3) The
IR and 13C NMR spectra below are for a molecule with molecular
formula C7H14O. Determine the structure of this compound,
match each carbon with its peak, and identify all important IR signals.
From the IR: signal at 3000cm-1 is the C-H bond stretch,
1710 cm-1 is the
C=O stretch, and the signals around 1400 cm-1 are the C-C stretches.
So most important for solving the structure is the compound has a C=O group.
The degree of unsaturation is
(16-14)/2 = 1. Since we know there has to be a C=O grouop,
that accounts for the one degree of unsaturation>. So
the rest of the molecule must be alkyl only!
For the 13C NMR, we see only 4 peaks, indicating
only four unique carbons. The signal at 210ppm is the C=O carbon. The three
other peaks are in the typical alkyl region, with the lowest one typical for a
methyl group.
Since we have 7 carbons, one of which is the C=O carbon, and
only 3 more unique carbons, and there cannot be a ring (because of the DOS) the
molecule must have the C=O in the middle of a symmetric system. This can be
done in only 2 ways:
The compound on the left (A) has 4 unique carbons,
while the compound on the right has only 3 unique carbons (the four methyl groups
are identical!). So the correct answer is compound A.
4) How would you distinguish the following two isomer using 1H
and 13C NMR?
13C NMR: compound A has
5 peaks. Compound B has 5 peaks. The chemical shifts in A and B are likely to
be quite similar. So 13C NMR is not by itself really solve this
problem for us.
1H NMR: compound A has 3
unique Hydrogens and compound B also has three unique
hydrogens.
For compound A, we would expect a singlet at ~5 (2H), a
singlet at ~1.5 (6H), and a singlet at about 2.0ppm (2H).
For compound B, we would expect a triplet at ~5 (2H), a
triple at ~2 (2H) and a singlet around 1.8 (6H)
So the most obvious thing is the difference in the splitting.
A would have three singlets, while B would have one
singlet and two triplets.
2)Provide the appropriate reagent to
affect the following reactions.
a)
<
b)
6) The 1H NMR below is for a
compound with molecular formula C11H16. Determine the
structure of the molecule and justify it by matching each proton with its
corresponding peak.
Degree of unsaturation = (24-16)/2
= 4
The peaks at 7.2 indicate a phenyl ring, which accounts for
all 4 degrees of unsaturation.
Let’s look at the signals at 0.6 and 1.6 – a triplet and a quartet. This looks
to be an ethyl group with
the neighboring carbon having NO hydrogens.
Let’s look at the integral next. The peak at 0.6 is a methyl
group so it has an integral of 3 so the integral for the peak at 1.6 is 2. The
singlet at 1.3 is probably a methyl; group, but its integral is much bigger than the one for the methyl at
0.6ppm. In fact it is
twice as big, so we have 2 equivalent methyl groups to give an integral of 6.So
these three upfield peaks integrate to 2+6+3=11
hydrogen. So the signal at 7.2 (the phenyl group) must be 5 hydrogens (a total of 16 – 11 = 5!).
So the phenyl group has only one group attached to it:
So we now we have a phenyl ring (6 carbons) an ethyl group (2
carbons) and two equivalent methyl groups (2 carbons) which totals to 10
carbons but the molecule has 11 carbons. There must be one more carbon and it
can’t have any hydrogens attached to it ore there
would be another peak in the 1H NMR!
So the answer has to be:
1)Provide the IUPAC name for the following compounds.
|
a)
|
b)
|
5) Predict the product(s) of the following reactions
a)
b)
c)
d)
3) The IR and 13C NMR spectra below are for a molecule with molecular formula C7H14O. Determine the structure of this compound, match each carbon with its peak, and identify all important IR signals.
From the IR: signal at 3000cm-1 is the C-H bond stretch, 1710 cm-1 is the C=O stretch, and the signals around 1400 cm-1 are the C-C stretches. So most important for solving the structure is the compound has a C=O group.
The degree of unsaturation is (16-14)/2 = 1. Since we know there has to be a C=O grouop, that accounts for the one degree of unsaturation>. So the rest of the molecule must be alkyl only!
For the 13C NMR, we see only 4 peaks, indicating only four unique carbons. The signal at 210ppm is the C=O carbon. The three other peaks are in the typical alkyl region, with the lowest one typical for a methyl group.
Since we have 7 carbons, one of which is the C=O carbon, and only 3 more unique carbons, and there cannot be a ring (because of the DOS) the molecule must have the C=O in the middle of a symmetric system. This can be done in only 2 ways:
The compound on the left (A) has 4 unique carbons, while the compound on the right has only 3 unique carbons (the four methyl groups are identical!). So the correct answer is compound A.
4) How would you distinguish the following two isomer using 1H and 13C NMR?
13C NMR: compound A has 5 peaks. Compound B has 5 peaks. The chemical shifts in A and B are likely to be quite similar. So 13C NMR is not by itself really solve this problem for us.
1H NMR: compound A has 3 unique Hydrogens and compound B also has three unique hydrogens.
For compound A, we would expect a singlet at ~5 (2H), a singlet at ~1.5 (6H), and a singlet at about 2.0ppm (2H).
For compound B, we would expect a triplet at ~5 (2H), a triple at ~2 (2H) and a singlet around 1.8 (6H)
So the most obvious thing is the difference in the splitting. A would have three singlets, while B would have one singlet and two triplets.
2)Provide the appropriate reagent to affect the following reactions.
a)
<
b)
6) The 1H NMR below is for a compound with molecular formula C11H16. Determine the structure of the molecule and justify it by matching each proton with its corresponding peak.
Degree of unsaturation = (24-16)/2 = 4
The peaks at 7.2 indicate a phenyl ring, which accounts for all 4 degrees of unsaturation.
Let’s look at the signals at 0.6 and 1.6 – a triplet and a quartet. This looks to be an ethyl group with the neighboring carbon having NO hydrogens.
Let’s look at the integral next. The peak at 0.6 is a methyl group so it has an integral of 3 so the integral for the peak at 1.6 is 2. The singlet at 1.3 is probably a methyl; group, but its integral is much bigger than the one for the methyl at 0.6ppm. In fact it is twice as big, so we have 2 equivalent methyl groups to give an integral of 6.So these three upfield peaks integrate to 2+6+3=11 hydrogen. So the signal at 7.2 (the phenyl group) must be 5 hydrogens (a total of 16 – 11 = 5!).
So the phenyl group has only one group attached to it:
So we now we have a phenyl ring (6 carbons) an ethyl group (2 carbons) and two equivalent methyl groups (2 carbons) which totals to 10 carbons but the molecule has 11 carbons. There must be one more carbon and it can’t have any hydrogens attached to it ore there would be another peak in the 1H NMR!
So the answer has to be: