Chemistry 241 Worksheet for Chapters 12 and 13
Page 1:
Index of Hydrogen deficiency for C11H12O2: 6° of “unsaturation” Probably a
benzene ring. Chemical shift of peaks near 7 ppm confirms this.
IR: carbonyl at 1711 cm-1, can also see sp2 and sp3 C-H’s.
H-NMR:
signal
chem shift
integration
12 H’s
split
what
a
7.9
1
1H
d?
not aromatic H
max of 5 benz
H’s.
CH
b
7.4
5
5H
m
benzene H’s
c
6.3
1
1H
d
CH
d
4.2
2
2H
quartet
CH2
e
1.4
3
3H
t
CH3
next
to(split)
(chem
shift)
1H
lots of H’s
1H
3 H’s, (CH3)
oxygen
2 H’s,
(CH2)
So, on one end of our molecule is a benzene ring, on the other is –OCH2CH3. In
the middle is a carbonyl (from IR). Still missing are two CH’s and a ring or double bond.
Two missing carbons cannot make a ring, so it must be a double bond. You cannot easily
tell from the spectra whether it is cis or trans or terminal. That information is in the
fingerprint, which we haven’t studied enough. Looking at p541, it looks like it is most
likely trans, but this is not definite. Putting the pieces together:
O
O
Page 2:
IHD for C7H6O2 is five bonds or rings. Probably a benzene ring. 4 H’s at ~7 ppm
confirms this.
IR shows broad –OH, maybe even a COOH, but hard to tell. It also looks like a
carbonyl at 1680 and an aldehyde H double band at ~ 2800.
NMR:
signal
chem shift
integration
6 H’s
split
what
a
11.1
1
1H
s
aldehyde H
b
9.8
1
5H
s
Ar-OH
c
7.4
4
1H
m
4 CH’s
next
to(split)
(chem
shift)
0 H’s
0 H’s
lots of H’s
So, end pieces are the aldehyde and the alcohol. The middle piece is the aromatic
ring. Whether the ring is 1,2 or 1,3 or 1,4 substituted is present in the fingerprint region,
which we have not studied much. However, we can tell from the splitting pattern that the
benzene ring is definitely not 1,4 substituted because that would mean two aromatic
doublets, which we do not have. So it is 1,3 or 1,2 substituted. I would take either
answer.
OH
OH
HO
O
or
OH
O
Page 3:
Index of H deficiency for C10H12O2: 5 pi bonds or rings. Could be benzene ring.
Aromatic H’s on NMR at 7.4 and 8.0 seem to confirm this.
IR: the usual suspects: sp2 and sp3 C-H’s and a carbonyl at 1720.
H-NMR: The signal at 0ppm is always TMS: not due to compound
signal
chem shift
integration
12 H’s
split
what
next
to(split)
(chem
shift)
a
8.0
2.4
2H
m
aromatic H
H’s. 2 CH’s
lots of H’s
b
7.4
3.5
3H
m
benzene H’s
3 CH’s
lots of H’s
c
5.3
1.2
1H
m
CH
d
1.4
7.8
6H
d
2 CH3’s
lots of H’s
1H
So we have five aromatic H’s on our benzene ring, which means we have a
benzene ring at one end of our molecule. On the other end, my two methyls (signal d)
CH
must be next to my one
, making an isopropyl end. In the middle is my
carbonyl (remaining degree of unsaturation) and oxygen. The only thing left to settle is
which is next to the CH. Since the chemical shift of the CH is 3.5 ppm, that puts it next
to the O, and not the carbonyl. Final structure:
O
O
Page 4:
O
O
Page 5:
Index of H deficiency for C9H11NO2:
(21-11)/2 = 5 degrees “unsaturation” again. probably a benzene ring, which can be
confirmed by the signals near 7 ppm.
IR:
at 1708 cm-1 is a carbonyl (C=O) again. Also visible is the two peaks of an –NH2 near
3300-3500 cm-1. sp2 and sp3 C-H’s show again
H-NMR:
signal
ch. shift
int.
split
what
next to
g
7.4
1H
d
benz. H
1H
f
7.3
1H
s
benz. H
0H
e
7.2
1H
t
benz. H
2H
d
6.8
1H
d
benz. H
1H
c
4.4
2H
q
CH2
3H’s
CH3 &O
b
3.8
2H
s (broad)
NH2
a
1.4
3H
t
CH3
2H’s
CH2
One can see that the CH2 and CH3 must be together to make an ethyl. this is one of our
end pieces. the other end piece is the NH2. In the middle go the carbonyl, benzene ring,
and the O of the ether or ester.
From the splitting pattern on the benzene ring (1 singlet), this ring must be 1,3 substituted
(meta)
from the chemical shift of the CH2, it must be next to the O. given our knowledge, we
can come up with two possibilities. with more experience, we might be able to
differentiate between the amide and the amine, but that isn’t necessary right now. The IR
carbonyl absorption is a little closer to an ester than an amide.
O
O
H2N
O
H2N
or
O