Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
Chapter 16
Nuclear Magnetic Resonance Spectroscopy
Review of Concepts
Fill in the blanks below. To verify that your answers are correct, look in your textbook at
the end of Chapter 16. Each of the sentences below appears verbatim in the section
• When two protons are interchangeable by rotational symmetry, the protons are
said to be ___________.
• When two protons are interchangeable by rotational symmetry, the protons are
said to be ___________.
• The left side of an NMR spectrum is described as _____field, and the right side is
one peak, a __________has two, a __________has three, a __________has four,
and a __________has five.
• Multiplicity is the result of spin-spin splitting, also called __________, which
follows the n+1 rule.
• When signal splitting occurs, the distance between the individual peaks of a signal
362
CHAPTER 16
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 16. The answers appear in the section entitled
SkillBuilder Review.
16.1 Determining the Relationship between Two Protons in a Compound
16.2 Identifying the Number of Expected Signals in a
1
H NMR Spectrum
H
H H H O
HHOH
H H H H H
FOR EACH OF THE FOLLOWING COMPOUNDS, DETERMINE WHETHER THE TWO INDICATED PROTONS ARE CHEMICALLY
EQUIVALENT.
CHEMICALLY EQUIVALENT?
16.3 Predicting Chemical Shifts
16.4 Determining the Number of Protons Giving Rise to a Signal
STEP 1 - COMPARE THE
RELATIVE __________________
VALUES, AND CHOOSE THE
LOWEST NUMBER
STEP 2 - DIVIDE ALL INTEGRATION
VALUES BY THE NUMBER FROM
STEP #1, WHICH GIVES THE RATIO
OF _____________
STEP 3 - IDENTIFY THE NUMBER OF PROTONS IN THE COMPOUND
(FROM THE MOLECULAR FORMULA) AND THEN ADJUST THE
RELATIVE INTEGRATION VALUES SO THAT THE SUM TOTAL
EQUALS THE NUMBER OF __________________________________.
16.5 Predicting the Multiplicity of a Signal
CHAPTER 16
363
16.6 Drawing the Expected
1
H NMR Spectrum of a Compound
STEP 3 - DETERMINE THE ___________
OF EACH SIGNAL BY COUNTING THE
NUMBER OF ________ GIVING RISE TO
EACH SIGNAL
STEP 4- PREDICT THE
_______________ OF
EACH SIGNAL
STEP 5 - DRAW
EACH SIGNAL
STEP 1 -
IDENTIFY THE
NUMBER OF
___________
STEP 2- PREDICT
THE __________
__________ OF
EACH SIGNAL
1
16.8 Analyzing a
1
H NMR Spectrum and Proposing the Structure of a Compound
STEP 3 - ANALYZE EACH SIGNAL (____, ____,
AND ____), AND THEN DRAW FRAGMENTS
CONSISTENT WITH EACH SIGNAL. THESE
FRAGMENTS BECOME OUR PUZZLE PIECES
THAT MUST BE ASSEMBLED TO PRODUCE A
MOLECULAR STRUCTURE
STEP 4 -
ASSEMBLE THE
FRAGMENTS
STEP 1 - USE THE _________
___________ TO DETERMINE
THE HDI. AN HDI OF _____
INDICATES THE POSSIBILITY
OF AN AROMATIC RING
STEP 2 - CONSIDER THE
NUMBER OF SIGNALS AND
INTEGRATION OF EACH SIGNAL
(GIVES CLUES ABOUT THE
______________ OF THE
COMPOUND)
16.9 Predicting the Number of Signals and Approximate Location of Each Signal in a
13
C NMR
Spectrum
16.10 Determining Molecular Structure using DEPT
13
C NMR Spectroscopy
BROADBAND
DECOUPLED
CH
2
CH
3
CH C
COMPLETE THE FOLLOWING CHART BY DRAWING THE EXPECTED SHAPE OF EACH SIGNAL:
364
CHAPTER 16
Solutions
16.1.
a) homotopic
16.2.
a) All four protons can be interchanged either via rotation or reflection.
16.3.
16.4.
16.5. The presence of the bromine atom does not render C3 a chirality center because
16.6. This compound will exhibit two signals in its
1
H NMR spectrum:
CHAPTER 16
365
16.7.
a)
O
CH
3
H
3
C
methylene protons (CH
2
) = 1.2 ppm
alpha to the oxygen = + 3.0 ppm
4.2 ppm
methyl protons (CH
3
) = 0.9 ppm
methyl protons (CH
3
) = 0.9 ppm
beta to the carbonyl = + 0.2 ppm
1.1 ppm
H H
b)
methylene protons (CH
2
) = 1.2 ppm
alpha to the oxygen = + 2.5 ppm
alpha to the carbonyl = + 1.0 ppm
4.7 ppm methyl protons (CH
3
) = 0.9 ppm
beta to the oxygen = + 0.5 ppm
1.4 ppm
366
CHAPTER 16
c)
O
CH3
methyl protons (CH
3
) = 0.9 ppm
alpha to the carbonyl = + 1.0 ppm
1.9 ppm methyl protons (CH
3
) = 0.9 ppm
d)
methyl protons (CH3) = 0.9 ppm
alpha to the oxygen = + 2.5 ppm
3.4 ppm
methyl protons (CH3) = 0.9 ppm
beta to the carbonyl = + 0.2 ppm
1.1 ppm
e) All four methylene groups are equivalent, so the compound will have only one signal
in its
1
H NMR spectrum. That signal is expected to appear at approximately (1.2 + 2.5 +
0.5) = 4.2 ppm.
16.8.
O
O
O
methylene protons (CH
2
) = 1.2 ppm
alpha to the oxygen = + 3.0 ppm
alpha to the oxygen = + 2.5 ppm
6.7 ppm
H H
16.10.
a)
H
O
H H
H
H
H H
H H
H
~ 10 ppm
~ 3 ppm ~ 1.2 ppm
~ 4.5 - 6.5 ppm
~ 4.5 - 6.5 ppm
~ 2 ppm
~ 2.5 ppm
368
CHAPTER 16
d)
H
3
CO
H
3
C CH
3
~ 3.4 ppm
~ 1.4 ppm
~ 2.5 ppm
H
16.11.
The signal at 4.0 ppm represents two protons.
16.12.
The signal at 9.6 ppm represents one proton.
16.13. Each signal represents two protons.
16.14.
Cl
CHAPTER 16
369
16.15.
a)
H
3
CO
O
CH
3
singlet
singlet quartet
triplet
HH H H
c)
H
3
C
OCH
3
CH
3
CH
3
H
3
C CH
3
singlet
H H
singlet
singlet
singlet
16.16.
H
16.17.
a) The spectrum exhibits the characteristic pattern of an isopropyl group.
16.18.
a)
J
ac
J
ab
H
a
J
ab
(17 Hz)
CHAPTER 16
371
c)
16.19. Draw the expected
1
H NMR spectrum for each of the following compounds
a)
O
O
372
CHAPTER 16
16.20.
O
O
O
16.21.
a) The first compound will have only three signals in its
1
H NMR spectrum, while the
second compound will have six signals.
16.22. The presence of peroxides caused an anti-Markovnikov addition of HBr:
HBr
B
r
CHAPTER 16
373
16.23.
OH
O
HO
O
O
16.24.
O
OO
O
16.25.
a) Four signals. Three appear in the region 0 – 50 ppm, and the fourth signal (the C=O)
appears in the region 150 – 220 ppm.
b) Five signals. All five appear in the region 0 – 50 ppm.
c) Six signals. Two appear in the region 0 – 50 ppm, and four signals appear in the
region 100-150 ppm.
d) Nine signals. Two appear in the region 0 – 50 ppm, one appears in the region 50 – 100
ppm and six signals appear in the region 100 – 150 ppm.
374
CHAPTER 16
16.26. The first compound lacks a chirality center. The two methyl groups are
enantiotopic and are therefore chemically equivalent. The second compound has a
16.27.
16.28.
O
16.29.
OH
16.31.
16.32.
CHAPTER 16
375
16.33. This compound will exhibit three signals in its
13
C NMR spectrum:
16.36. The first compound will have five signals in its
13
C NMR spectrum, while the
second compound will have seven signals.
16.37.
O
CH
3
CH
3
H
3
C
H H
triplet
doublet
16.38.
a) The first compound will have four signals in its
13
C NMR spectrum, while the second
compound will have twelve signals.
The first compound will have two signals in its
1
H NMR spectrum, while the second
compound will have eight signals.
d) The first compound will have three signals in its
13
C NMR spectrum, while the second
compound will have five signals.
The first compound will have two signals in its
1
H NMR spectrum, while the second
compound will have four signals.
376
CHAPTER 16
16.39. This compound will exhibit two signals in its
13
C NMR spectrum:
16.40.
a) homotopic b) enantiotopic c) enantiotopic d) homotopic
16.41.
16.42.
a) Four signals are expected in the
1
H NMR spectrum of this compound.
CHAPTER 16
377
16.43.
16.44.
a) Nine b) Eight c) Six
16.45.
HH H H H
6.5 - 8 ppm
~ 3.7 ppm
~ 5 ppm
16.46.
a) Six signals, all of which appear in the region 100 – 150 ppm.
b) Seven signals. One appears in the region 150 – 220 ppm, and the remaining six
16.47. The
1
H NMR spectrum of the Markovnikov product should have only four
signals, while the anti-Markovnikov product should have many more signals in its
1
H
NMR spectrum.
16.48.
378
CHAPTER 16
16.49.
H
H
H
H
HH
Increasing chemical shift in
1
H NMR spectroscopy
16.51. 16.52. 16.53.
OH
16.56.
HO
16.57. 16.58.
OH
O
CHAPTER 16
379
16.60.
OH
16.61.
O
H
16.62.
O
OH
16.65. N,N-dimethylformamide (DMF) has several resonance structures:
H N
O
CH
3
CH
3
H N
O
CH
3
CH
3
H N
O
CH
3
CH
3
16.66. In a concentrated solution of phenol, the OH groups are engaged in extensive,
intermolecular hydrogen-bonding interactions. These interactions cause the average
16.67. The methyl group on the right side is located in the shielding region of the π bond,
so the signal for this proton is moved upfield to 0.8 ppm.
16.68. Bromine is significantly larger than chlorine, and the electron density of a
bromine atom partially surrounds any carbon atom attached directly to the bromine,
