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Chapter 15
Infrared Spectroscopy and Mass Spectrometry
Review of Concepts
Fill in the blanks below. To verify that your answers are correct, look in your textbook at
the end of Chapter 15. Each of the sentences below appears verbatim in the section
entitled Review of Concepts and Vocabulary.
• Spectroscopy is the study of the interaction between _______ and ________.
• The difference in energy (∆E) between vibrational energy levels is determined
by the nature of the bond. If a photon of light possesses exactly this amount
of energy, the bond can absorb the photon to promote a __________________
excitation.
• IR spectroscopy can be used to identify which _____________________ are
present in a compound.
• The location of each signal in an IR spectrum is reported in terms of a
high energy _______________, generating a radical cation that is symbolized
by (M)
+•
and is called the molecular ion, or the __________ ion.
• Only the molecular ion and the cationic fragments are deflected, and they are
then separated by their ____________________ (m/z).
• The tallest peak in a mass spectrum is assigned a relative value of 100% and is
called the __________ peak.
• The relative heights of the (M)
+•
peak and the (M+1)
+•
peak indicates the
number of ___________________.
344
CHAPTER 15
Review of Skills
Fill in the blanks and empty boxes below. To verify that your answers are correct, look
in your textbook at the end of Chapter 15. The answers appear in the section entitled
SkillBuilder Review.
15.2 Distinguishing Two Compounds Using IR Spectroscopy
STEP 1 - WORK METHODICALLY
THROUGH THE EXPECTED
________________________ OF EACH
COMPOUND
STEP 2 - DETERMINE IF ANY
__________ WILL BE PRESENT FOR
ONE COMPOUND BUT ABSENT FOR
THE OTHER
STEP 3 - FOR EACH EXPECTED SIGNAL,
COMPARE FOR ANY POSSIBLE DIFFERENCES IN
_____________, ________________, OR
________________.
Solutions
15.1.
a)
Increasing wavenumber
C
C
C
C
C
H
b)
Increasing wavenumber
C
C
C
H
CHAPTER 15
345
15.2.
a)
O H > 3000 cm
-1
b) No c) No
15.3.
a)
O
O
lower wavenumber
(carbonyl group is conjugated)
346
CHAPTER 15
15.4. The C=C bond in the conjugated compound produces a signal at lower
wavenumber because it has some single bond character, as seen in the third resonance
15.5.
a)
C
l
The presence of the Cl will cause the C=C bond in this compound to have
15.6. The C=C bond in 2-cyclohexenone has a large dipole moment, as can be
rationalized with the third resonance structure below:
15.7. The vinylic C-H bond should produce a signal above 3000 cm
-1
.
15.8. The narrow signal is produced by the O-H stretching in the absence of hydrogen
bonding effect. The broad signal is produced by O-H stretching when hydrogen bonding
15.9.
a) ROH b) neither c) RCO
2
H d) neither e) ROH f) RCO
2
H
CHAPTER 15
347
15.10.
a) ketone b) RCO
2
H c) R
2
NH d) RNH
2
e) ROH f) ketone
15.11. The C
sp
3
—H bonds can stretch symmetrically, asymmetrically, or in a variety of
15.12.
O
15.13.
1) The O-H bond of the carboxylic acid moiety (expected to be 2200 - 3600 cm
-1
)
2) The vinylic C-H bond (expected to be ~ 3100 cm
-1
)
15.14.
a) The starting material is an alcohol and is expected to produce a typical signal for an
O-H stretch –a broad signal between 3200 - 3600 cm
-1.
In contrast, the product is a
348
CHAPTER 15
c) The starting material is an unsymmetrical alkyne and is expected to produce a signal at
d) The C≡C bond in the starting material and the C=C bond in the product are both
e) The starting material will have two signals in the double-bond region: one for the
15.15. The starting material has a cyano group (C≡N) and is expected to produce a signal
-1
15.16. The C≡C bond in the starting material (1-butyne) is unsymmetrical and produces
a signal at 2200 cm
-1
, corresponding with the C≡C stretch. In contrast, the C≡C bond in
the product (3-hexyne) is symmetrical and does not produce a signal at 2200 cm
-1
.
15.17. The starting material should have a C=O signal at 1720 cm
-1
,
while the product
should have an O-H signal at 3200 – 3600 cm
-1
.
15.18. 1-chlorobutane is primary substrate. When treated with sodium hydroxide,
substitution is expected to dominate over elimination (see Chapter 8), but both products
CHAPTER 15
349
15.19.
a)
O
m
/
z
=
6
8
m
/
z
=
6
6
b)
N
m
/
z
=
7
8
m
/
z
=
7
9
15.20.
a) This compound does not have any nitrogen atoms. According to the nitrogen rule, this
compound should have an even molecular weight (m/z = 86).
15.21.
a) There must be four carbon atoms, and the molecular weight must be 72. The
molecular formula could be C
4
H
8
O.
15.22. Each nitrogen atom in the molecular formula of a compound should contribute
0.37% to the (M+1)
+•
peak. Three nitrogen atoms therefore contribute the same amount
15.23.
a) This fragment is M – 79, which is formed by loss of a Br. So the fragment does not
350
CHAPTER 15
15.24.
a) There is not a significant (M+2)
+•
peak, so neither bromine nor chlorine are present.
b) There is not a significant (M+2)
+•
peak, so neither bromine nor chlorine are present.
15.25.
a)
(M-57)
+
+
15.26. A fragment at M-29 should result from α cleavage:
HO
α
αα
α
HO HO H
2
C CH
3
+
α
-cleavage
CHAPTER 15
351
15.27. The fragment at M-43 is expected to be the base peak because it corresponds with
formation of a tertiary carbocation:
(M-43)
+
+
15.29.
a)
HO
OH
OH
m
/
z
=
1
2
6
.
0
3
1
5
m
/
z
=
1
2
6
.
1
4
0
4
b)
O
O
m
/
z
=
1
1
2
.
0
5
2
2
m
/
z
=
1
1
2
.
1
2
4
8
15.30.
a) The first compound should have a very broad signal between 3200 and 3600 cm
-1
,
15.31.
15.32. Both C
3
H
5
ClO
2
and C
3
H
6
have one degree of unsaturation.
15.33. The compound has one degree and unsaturation, which is a C=O bond:. The
H
OO
H
O
352
CHAPTER 15
15.34. The compound must have one degree of unsaturation. The broad signal between
3200-3600 cm
-1
indicates an OH group, and the absence of signals between 1600 and
15.35. A signal at 2200cm
-1
signifies the presence of a C≡C bond. There are only two
possible constitutional isomers: 1-butyne or 2-butyne. The latter is symmetrical and
would not produce a signal at 2200cm
-1
. The compound must be 1-butyne.
15.37.
a)
H
D
I
=
1
H
D
I
=
1
CHAPTER 15
353
15.38.
a)
~ 1680 cm
-1
O
~ 1600 cm
-1
15.39.
O ON H OHHH C NR
Increasing Wavenumber
354
CHAPTER 15
c) The C=C bond and the C=O bond should each produce a signal in the double bond
region, 1600 - 1850 cm
-1
, In addition, the two C≡C bonds should produce two signals
15.41.
a) The reactant should have signals at 1650 cm
-1
and 3100 cm
-1
, while the product
should not have either signals.
b) The reactant should have a broad signal from 3200 - 3600 cm
-1
, while the product
should lack this signal and instead should have a signal at ~ 1720 cm
-1
.
15.42.
a) a C=O signal at ~ 1720 cm
-1
b) a C=O signal at ~ 1680 cm
-1
(conjugated) and a C=C signal at ~ 1600 cm
-1
15.43.
a) C
7
H
8
(m/z = 92) b) C
6
H
6
O (m/z = 94) c) C
6
H
10
O (m/z = 98)
d) C
6
H
8
O (m/z = 96) e) C
6
H
15
N (m/z = 101)
CHAPTER 15
355
15.45. There are nine carbon atoms in the compound (10 / 1.1).
15.46.
15.47.
a) an OH group and double bonds.
b)
%
2
.
27
%3
.9 × 100% = 14.3%
15.48.
a) Both compounds are C
6
H
12
b) Both compounds have an HDI of 1.
15.49.
The signal at
m/z
= 111 is (M-15) which corresponds with loss of a methyl group.
The signal at
m/z
= 97 is (M-29) which corresponds with loss of an ethyl group. Both
fragmentations lead to a tertiary carbocation:
356
CHAPTER 15
15.50.
15.51.
The more substituted alkene will not produce a signal at 1650 cm
-1
nor will it
produce a signal at 3100 cm
-1
. The less substituted alkene will display both signals in its
IR spectrum:
15.52.
a) C
5
H
6
b) C
4
H
6
O
15.53.
a) the molecular ion peak appears at
m/z
= 114
15.54.
15.55
H H
CHAPTER 15
357
15.56.
Limonene has three degrees of unsaturation and is comprised of only carbon and
hydrogen atoms. The molecular weight is 136, so the molecular formula of limonene
must be C
10
H
16
.
15.58.
The compound exhibits intramolecular hydrogen bonding even in dilute solutions.
15.59.
The IR spetrum indicates that the compound is a ketone. The mass spectrum
O O
15.60.
The IR spectrum indicates the presence of a triple bond as well as a C-H bond
H H
15.61.
Cl NaOEt 1) BH
3
THF
OH
2) H
2
O
2
, NaOH
358
CHAPTER 15
a) Compound F is an alcohol and its IR spectrum will exhibit a broad signal
between 3200 and 3600 cm
-1
. Compound G is an ether and its IR spectrum
will not exhibit the same signal.
15.62.
1-butene can lose a methyl group to form a resonance stabilized carbocation
(M-15)
+
+
r
e
s
o
n
a
n
c
e
-
s
t
a
b
i
l
i
z
e
d
CH
3
15.63.
Yes, compound D is an unsymmetrical alkene:
B
C
D
15.64.
M
M+2
CHAPTER 15
359
15.66.
M
M+2
M+4
15.67.
O
O
1
3
0
0
c
m
-1
1
0
0
0
c
m
-1
360
CHAPTER 15
15.68.
HO OH H O S O
O
O
H
HOH
O
HO O
H
H
OH
HO H
HO
- H
2
O
hydride
shift
