Chapter 1 solutions - University of Minnesota

Organic Chemistry
With a Biological Emphasis
Solutions Manual
Tim Soderberg
University of Minnesota, Morris
2010

In-chapter exercises
E1.1: Mass numbers are:
a) 31 P
b) 32 P
c) 37 Cl
E1.2:
E1.3:
E1.4:
a) a neutral magnesium atom: 1s 2 2s 2 2p 6 3s 2
b) a Mg 2+ atom: 1s 2 2s 2 2p 6
c) a neutral potassium atom: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1
d) a K + ion: 1s 2 2s 2 2p 6 3s 2 3p 6
Chapter 1
Chapter 1 solutions
1

e) a Cl - ion: 1s 2 2s 2 2p 6 3s 2 3p 6
E1.5:
E1.6:
E1.7:
E1.8:
2
Chapter 1

E1.9:
E1.10:
E1.11:
a) C 5H 12
b) C 4H 10O
Chapter 1
3

c) C3H9N
E1.12:
a) 2-deoxycytidine: IHD = 5 (2 rings, 3 double bonds)
b) histidine: IHD = 4 (2 rings, 2 double bonds)
c) C9H10NOI: IHD = 5
d) an ion with the molecular formula C6H11O7 - : IHD = 1
E1.13:
a) carboxylate, sulfide, aromatic, two amide groups (one of which is cyclic)
b) tertiary alcohol, thioester
c) carboxylate, ketone
d) ether, primary amine, alkene
E1.14:
The following molecules fit the descriptions in the problem:
4
Chapter 1

E1.15:
acetic acid: ethanoic acid
chloroform: trichloromethane
acetone: 2-propanone
E1.16:
fig3
E1.17:
E1.18:
The bonding in chloroform consists of three of the sp 3 hybrid orbitals on the central
carbon atom overlapping with 3p orbitals on clorine. The fourth bond is an overlap
between the fourth sp 3 orbital on carbon and the 1s orbital of hydrogen.
fig 3
E1.19:
Chapter 1
5

6
Chapter 1
Both the carbon and the nitrogen atom in CH3NH2 are sp 3 -hybridized. The C-N σ bond
is an overlap between two sp 3 orbitals.
fig 4
E1.20:
fig 4
E1.21:
a) Csp 2 -Csp 2
b) Csp 2 -Csp 3
c) Csp 2 -Nsp 2
d) Csp 3 -Nsp 3
E1.22:
The three bonds that are drawn as solid wedges should be drawn as lines – they are all
coming out of planar, sp 2 -hybridized carbons and should be shown lying in the plane of
the page.
E1.23:
fig 4
The carbon and nitrogen atoms are bond sp 2 -hybridized. The carbon-nitrogen double
bond is composed of a σ bond formed from two sp 2 orbitals, and a π bond formed from
the side-by-side overlap of two unhybridized 2p orbitals.
E1.24:

a) The σ bond indicated by this arrow is formed by the overlap of an sp 2 orbital of a
carbon atom and an sp 3 orbital of a carbon atom.
b) The σ bond indicated by this arrow is formed by the overlap of an sp2 orbital of a
carbon atom and an sp 2 orbital of a carbon atom.
c) The σ bond indicated by this arrow is formed by the overlap of an sp2 orbital of a
nitrogen atom and an s orbital of a hydrogen atom.
d) The σ bond indicated by this arrow is formed by the overlap of an sp 2 orbital of a
carbon atom and an sp 3 orbital of a carbon atom.
d) The σ bond indicated by this arrow is formed by the overlap of an sp 3 orbital of a
carbon atom and an sp 3 orbital of a carbon atom.
End-of-chapter problems
P1.1: (it will help to refer to color versions of the figures below, in the pdf file)
fig 4
Chapter 1
7

fig 5
P1.2:
a) a lithium cation (Li + ): 1s 2
b) a calcium cation (Ca 2+ ): 1s 2 2s 1 2p 6 3s 2 3p 6
c) a iron cation in the ferric (Fe +3 ) state: 1s 2 2s 1 2p 6 3s 2 3p 6 4s 2 3d 3
8
Chapter 1
P1.3: There are many possible correct answers to this problem – check your answer with
your TA or instructor.
P1.4:

Chapter 1
fig 5
Notice in doing this problem that it is much easier to determine the IHD from the
structure rather than the molecular formula – just count rings and multiple bonds! But it
would still be good practice to use the molecular formulas to calculate IHDs, then check
that they match what you determined from the structures.
P1.5: In the structures shown below:
a-i) Describe the orbitals involved in the bonds indicated by the arrows.
j) Fill in all formal charges
k) Give the molecular formula for arginine
a-i; j) see structures below
9

fig 6
k) the molecular formula of arginine is C 6H 15N 4O 2.
P1.6:
a) Csp 3 – Osp 3
b) Csp 2 – Csp 3
c) Csp 2 – Nsp 2
d) Csp 2 – Csp 2
e) Csp 3 – Csp 3
f) Csp 2 – Csp 2
g) Csp 3 - Csp 3
h)Csp 2 – H 1s
i) Csp 2 – Osp 2
j) Csp 2 – Cl 3p
k) Nsp 3 – H 1s
10
Chapter 1
l) Compound 3 contains two aldehydes, compound one contains an ether, compound 2
contains an amide, compound 3 contains a terminal alkene, and compound 4 contains a
secondary amine.
m) The molecular formula of compound 3 is C 10H 14O 2.
P1.7:
fig 6
P1.8: Draw structures for four different amides with molecular formula C 3H 7NO.
fig 6

P1.9: shortest e

12
Chapter 1
Notice that the central carbon is sp-hybridized, while the two end carbons are sp 2 -
hybridized: the left one with the three sp 2 hybrid orbitals in the plane of the page, the
right one with two if the three orbitals pointing into or out of the plane of the page.

In-chapter excercises
Chapter 2
Chapter 2 solutions
E2.1: Arrows point to the isolated double bonds. The other four double bonds are all part
of a single conjugated system.
fig 1E2.
E2.2:
13

E2.3:
14
fig 1
fig 1
The structure on the right above is the minor contributor by rules 5, 6, and 7: in this
structure the carbon does not have a complete octet (rule 5), there is a separation of
charge (rule 6), and there are overall fewer bonds (rule 7).
E2.4:
fig 2
E2.5:
Chapter 2

fig 2
E2.6
fig 2
E2.7:
a)
fig 3
b) The conjugated π system in this carbocation is composed of seven p orbitals
containing six delocalized π electrons.
E2.8:
Chapter 2
15

E2.9:
16
fig 3
c) The conjugated π system in structure a) is composed of seven p orbitals
containing eight delocalized π electrons.
Chapter 2
Horizontal trend: A fluorine atom has one more proton in its nucleus than does oxygen,
which in turn has one more proton than does nitrogen, and so on. An additional proton
means an additional positive charge, so the (negatively charged) electrons are pulled in
closer to the nucleus. This trend holds within a row of the periodic table, because the
outermost electrons of all atoms in a row are all in the same energy ‘shell’.
Vertical trend: Atoms furthur down a column have increasing numbers of electrons
,which occupy energy ‘shells’ that are further away from the nucleus (eg. 3p for sulfur vs.
2p for oxygen). In other words, the ‘electron cloud’ of sulfur is larger than that of
oxygen, meaning that the electrons are inherently further away from the nucleus in sulfur.
Sulfur is less electronegative than oxygen, phosphorus less electronegative than nitrogen,
and chlorine less electronegative than fluorine.
E2.10: Molecules a, c, and d have dipole moments.
E2.11:
fig 3

E2.12:
Vitamin C has many potential hydrogen-bonding sites, and thus is water-soluble.
Vitamin B 3 has carboxylate and amine groups that are expected to be charged at
physiological pH (we’ll learn more about this is chapter 6), thus it is water-soluble.
Chapter 2
Vitamin A has only one hydrogen-bonding group and a large hydrophobic component,
thus it is fat-soluble.
E2.13:
Analine is a base and will accept a proton from hydrochloric acid. The resulting positive
charge makes it water-soluble.
Phenol does not act as a base in this context, and remains uncharged and water-insoluble.
fig 3
E2.14:
Boiling points ( o C)
benzene: 80
benzaldehyde: 178
phenol: 182
benzoic acid: 249
Benzoic acid has the highest boiling point because it has a very polar, hydrogen-bonding
carboxylic acid group with two electronegative oxygens (benzoic acid is a crystalline
solid at room temperature, the other three compounds are liquids).
Phenol has a hydrogen-bonding hydroxyl group, but only one oxygen as opposed to two.
Benzaldehyde has a polar carbonyl group, but because a carbonyl can only serve as a
hydrogen bond accepter and not a donor, a benzaldehyde molecule cannot hydrogen bond
to another benzaldehyde. Benzene is capable only of hydrophobic intramolecular
interactions, and has the lowest boiling point.
17

End-of chapter problems
18
Chapter 2
P2.1: Both bonds in O 3 have a bond order of 1.5, and are the same length – the two
resonance contributors shown below are of equal importance in illustrating the bonding in
the molecule.
fig 4
P2.2:
P2.3:

fig 4
P2.4:
fig 5
P2.5:
fig 5
P2.6:
Chapter 2
19

fig 5
20
Chapter 2
The He 2 molecule has four electrons to account for in the MO diagram: two are placed in
the σ bonding orbital and two in the σ* antibonding orbital. Thus there is no net
attractive force to hold the He 2 molecule together (in fact, there is a net repulsive force
due to the relative energy level of the σ* molecular orbital).
P2.7:
fig 5,6
P2.8:

fig 6
P2.9:
fig 7
P2.10:
Chapter 2
21

fig 7
P2.11:
fig 7
P2.12:
P2.13:
22
Chapter 2

1= lowest bp
fig 8
Reasoning:
Chapter 2
a) 2 and 3 have two fluorines and are more polar than 1, so they have stronger
intermolecular interactions. 3 has one more carbon than 2, and therefore stronger van der
Waals interactions. 4 is capable of hydrogen bonding, so it has the strongest
intermolecular interactions and the highest boiling point.
b) 1 and 2 have only van der Waals interactions, but 2 has more carbons so these
interactions are slightly stronger. 3 has a polar carbonyl group, and 4 is capable of
hydrogen bonding.
c) 1 is not capable of hydrogen bonding. 2 and 3 both have hydrogen bonding groups,
but 3 has one more carbon and therefore stronger overal van der Waals interactions.
d) 1 has only van der Waals interactions. 2 has a polar thiol group, but 3 has a hydroxyl
group which is capable of hydrogen bonding. 4 is a salt: the charge-charge interactions
are very strong and lead to a very high boiling point.
P2.14:
a) The one on the right is more soluble (fewer hydrophobic carbons)
23

) The one on the left is more soluble (ionic phosphate group)
c) The one on the left is more soluble (fewer hydrophobic carbons)
d) The one on the left is more soluble (capable of hydrogen bonding)
e) The one on the right is more soluble (fewer hydrophobic carbons)
24
Chapter 2
P2.15: Toluene has one more carbon than benzene, and therefore has stronger overal van
der Waals interactions and a higher boiling point. The melting point trend, however, is a
different story. Remember that melting involves a transition from an ordered solid state,
where molecules are stacked together, to a disordered liquid state, where molecules move
freely relative to one another. Benzene is a flat molecule and is able to stack very
effeciently (like plates) in the solid phase. The methyl group of toluene prevents this
efficient packing, and thus the overal intermolecular van der Waals interactions in the
toluene solid phase are lower than those for benzene, and it requires less energy (heat) to
break the tolune molecules apart.
fig 8
P2.16:
fig 9
P2.17: For each structure, the atoms are indicated (with a black dot) which could bear the
corresponding negative formal charge, and one example of a resonance contributor is
provided.

fig 9
Chapter 2
P2.18: For each structure, the atoms are indicated (with a black dot) which could bear the
corresponding positive formal charge, and one example of a resonance contributor is
provided.
fig 9
P2.19: For each structure, the atoms to which the positive charge can be delocalized are
indicated with a black dot.
25

In-chapter exercises
E3.1:
fig 1
E3.2:
fig 1
E3.3:
26
Chapter 3
Chapter 3 solutions
Mannose is a diastereomer of glucose: compare the major conformer shown below to the
chair form of B-glucose (section 3.2C) and notice that they differ only in the
configuration at C2.

Chapter 3
fig 1
In the major chair form of mannose, all substituents except the C2 hydroxyl are equitorial.
E3.4.
fig 1
E3.5: Stereocenters are marked with a bold dot.
fig 2
E3.6: Fumarate, aspirin, and acetominophen are achiral. Malate and ibuprofin are chiral;
their stereocenters are marked below with a bold dot.
27

fig 2
E3.7:
fig 2
E3.8:
fig 3
28
Chapter 3
E3.9: Because the observed optical rotation is negative, we know that there is more or the
R enantiomer than the S enantiomer (if it was a 50:50 mixture, the optical rotation would
be 0 o , while if it were pure (R)-carvone the optical rotation would be -61 o ).

ee = (23 x 100)/61 = 37.8%
. . . so the mixture is about 69% R and 31% S.
E3.10:.
fig 3
E3.11:
There are many possible correct answers to this problem. Two are shown below.
Chapter 3
fig 3
D-gulose is a diastereomer of D-glucose (configuration is different at two carbons). Dallose
is an epimer of D-glucose (configuration is different at one carbon). Notice that the
relationship between D-gulose and D-allose is that of epimers.
E3.12:
a) chiral
b) meso (achiral)
c) achiral (not meso)
d) achiral (not meso)
e) meso (achiral)
f) chiral
E3.13:
29

fig 3
E3.14:
fig 4
E3.15
30
Chapter 3
First, we assign R/S designations to the three stereocenters in the Fischer projection of Dribose,
and find that all three are R. Then, we draw the zig-zag structure, and decide
whether each hydroxyl group should be drawn with a solid or dashed wedge in order to
corectly show the R stereochemistry.
fig 4
With D-ribose accomplished, it is a simple matter to draw the zig-zag structures of the
other five-carbon aldoses – just start with D-ribose, then invert the appropriate
stereocenter.

fig 4
E3.16
Chapter 3
In the figure below, homotopic hydrogens are labeled ‘h’, enantiotopic hydrogens ‘e’,
and diastereotopic hydrogens ‘d’.
fig 5
E3.17:
31

32
Chapter 3
fig 5
The two prochiral pairs on leucine are the two hydrogens on Ca and the two methyl
groups on Cb. Because there is already a true chiral center on the molecule, these pairs
are both diastereotopic. Here’s how we assigned stereochemical designations to the
diastereotopic methyl groups: if we pick one of the CH3 groups and give it priority #2
(the H is of course priority #4), then the configuration at Cb becomes S – this methyl
group is thus designated pro-S (and the other methyl group of course must be pro-R).
E3.18
fig 5
End-of-chapter problems
P3.1:

P3.2:
fig 6
P3.3:
a) Stereocenters are marked with a bold dot.
fig 6
Chapter 3
33

) The two fluorinated derivatives of Epivar are enantiomers.
P3.4:
These two structures are actually enantiomers – they are non-superimposable mirror
images of each other. Notice that, even though there is no sp 3 -hybridized carbon, the
overall geometry of the allene structure creates a multi-atom stereocenter.
P3.5:
34
Chapter 3
If we imagine looking down on the rings from above, we can redraw as shown below. We
can also double-check the R and S designations of stereocenters to ensure that we did not
draw enantiomers by mistake.
a)
b)
c)
fig 7

P3.6:
a) There is only one stereocenter, and it is in the R configuration.
b)
fig 7
P3.7:
fig 7
P3.8:
fig 8
P3.9: The two are diastereomers: two of the four stereocenters are different.
Chapter 3
35

P3.10:
P3.11:
a) enantiomers
b) same compound
c) constitutional isomers
d) daistereomers
e) same compound
f) completely different compounds
P3.12:
a) diastereomers
b) diastereomers
c) epimers
d) constitutional isomers
All of the sugars contain three stereocenters except D-ribulose, which contains two.
P3.13:
a) enantiomers
b) epimers
c) same molecule
d) completely different molecules
P3.14:
36
Chapter 3

Chapter 3
fig 8
a) There are two chiral carbons in the molecule, indicated by bold dots.
b) There are four pairs of prochiral hydrogens, the positions of which are indicated with
arrows.
P3.15:
a) The molecule contains two stereocenters and has no alkene groups that can be
classified E or Z; therefore, there are four possible stereoisomers.
b) Positions of prochiral hydrogens are indicated with arrows.
fig 8
P3.16:
a) The two structures are diastereomeric.
b) The five-membered ring is sandwiched between the aromatic, seven-membered, and
six-membered rings, with the ether oxygen as the free corner.
P3.17:
a) Both alkene groups are E.
b) In addition to the two alkene groups, there are 10 chiral carbons in the molecule.
Therefore, there are 2 12 possible stereoisomers.
c) The structure shown is a diastereomer of bistramide A – the configuration of one of the
alkene groups is changed from E to Z.
P3.18: The two structures are exactly the same - structure B is just structure A flipped
end-over-end.
37

P3.19:
a)
38
Chapter 3
b) The gauche conformation makes possible the formation of an energetically-favorable
intramolecular hydrogen bond (a model will help you to see this).
c) When the hydroxy groups are changed to methoxy groups, intramolecular hydrogen
bonding is no longer possible, and the anti conformation is expected to be lowest in
energy.
P3.20
The highest energy conformation of 2-methylbutane is the one in which the methyl
groups are eclipsed.
fig 8
P3.21:
In the lower energy chair conformation of trans-1,2-dimethylcyclohexane, the two
methyl groups are both equitorial.
P3.22:
In the lower energy chair conformation of cis-1-ethyl-2-methylcyclohexane, the larger
ethyl group is equitorial.

P3.23:
The trans diastereomer of 1,4-diethylcyclohexane, in which both ethyl groups can be
equitorial, is expected to be lower in energy.
fig 9
P3.24:
The cis diastereomer of 1,3-diethylcyclohexane, in which both ethyl groups can be
equitorial, is expected to be lower in energy.
fig 9
P3.25:
Chapter 3
cis-1,2-dimethylcyclohexane is a meso compound, but the mirror plane can only be seen
when the ring is in a higher-energy boat conformation (recall that, when thinking about
stereochemistry, we consider all possible conformations).
fig 9
P3.26:
39

a) A and B are enantiomers
b) A and C are diastereomers
c) B and C are diastereomers
d) D and D are constitutional isomers
P3.27:
40
Chapter 3

In-chapter exercises
E4.1:
Chapter 4
Chapter 4 solutions
Using λυ = c, we first rearrange to υ = c/λ to solve for frequency.
For light with a wavelength of 400 nm, the frequency is 7.50 x 10 14 Hz:
In the same way, we calculate that light with a wavelength of 700 nm has a frequency of
4.29 x 10 14 Hz.
To calculate corresponding energies:
Using hc/λ, we find for light at 400 nm:
Using the same equation, we find that light at 700 nm corresponds to 40.9 kcal/mol.
E4.2:
A wavenumber of 3000 cm -1 means that 3000 waves fit in one cm (0.01m):
41

42
Chapter 4
We want to find the length of 1 wave, so we divide numerator and denominator by 3000:
So 3000 cm -1 is equivalent to a wavelength of 3.33 µm.
E4.3:
The π → π* transition in 4-methyl-3-penten-2-one is at 236 nm, which corresponds to
121 kcal/mol:
E4.4:
Molecule B has a longer system of conjugated π bonds, and thus will absorb at a longer
wavelength. Notice that there is an sp 3 -hybridized carbon in molecule B which isolates
two of the π bonds from the other three.
E4.5:
We use the formula:
recall from your high school algebra that if y = log(x), then x = 10 y , so:
. . . so the intensity of light entering the sample (I 0) is 10 times the intensity of the light
(at a particular wavelength) that emerges from the sample and is detected (I). Because
%T is simply the reciprocal of this ratio multiplied by 100, we find that A = 1.0
corresponds to 10% transmission.
E4.6: Using ε = A/c, we plug in our values for ε and A and find that c = 3.27 x 10 -5 M, or
32.7 µM.
E4.7:

Chapter 4
Using ε = A/c and plugging in the given values for ε and A, we find that c = 16.3 ng/µL.
However, this is the concentration of the diluted solution - the original solution was 20
times higher (50 µL were removed and diluted to 1000 µL to take the UV measurement).
Thus the concentration of the original solution is (20)(16.3) = 326 ng/µL..\
E4.8:
The sample is propanal. The peak at m/z = 58 is the molecular ion (parent peak), and the
peak at m/z = 29 is the formyl acylium ion. The M+1 peak at m/z = 59 is a molecular ion
in which one of the three carbons is a 13 C.
End-of-chapter problems
P4.1:
Without doing any calculation, we can answer the first part of the calculation:
electromagnetic waves at 3400 cm -1 are shorter than those at 1690 cm -1 (more waves fit
into one centimeter) and thus correspond to a higher frequency.
3400 cm -1 = 2.94 µm = 1.02 x 10 14 Hz
1690 cm -1 = 5.92 µm = 5.07 x 10 13 Hz
P4.2:
1720 cm -1 corresponds to a wavelength of .01/1720 = 5.81 x 10 -6 m, and an energy of 4.92
kcal/mol.
P4.3:
The triple bond in compound I is symmetric, and therefore is not IR-active, so there
would be no absorbance in the carbon-carbon triple bond range (2100-2250 cm -1 ). In
compound II the presence of the fluorines makes the triple bond asymmetric and IRactive,
thus the alkyne peak will be observed. In compound III, we should see not only
43

44
Chapter 4
the carbon-carbon triple bond peak but also an absorbance at approximately 3300 cm -1
due to stretching of the terminal alkyne carbon-hydrogen bond.
P4.4:
All three spectra will have a strong carbonyl stretching peak, but the ester (compound C)
carbonyl peak will be observed at a shorter wavelength compared to the ketone
(compound B) and the carboxylic acid (compound A). In addition, Compound A will
show a broad absorbance centered at approximately 3000 cm -1 due to carboxylic acid O-
H stretching, whereas in the spectrum of compound B we should see the broad
absorbance centered at approximately 3300 cm -1 from stretching of the alcohol O-H
bond. Compound C will have no broad O-H stretching absorbance.
P4.5:
All three compounds contain alkene functional groups. However, in compound Y the
alkene is symmetric and thus we would not expect to see an absorbance from C=C
stretching in the 1620-1680 cm -1 range. We would expect to see this peak in the spectra of
compounds X and Z; in addition we would expect to see, in the compound X spectrum, a
peak in the 3020 - 3080 cm -1 range due to stretching of the terminal alkene C-H bonds.
P4.6:
P4.7:
The change in A 340 is ΔA = 0.220. Using the expression ε = A/c, we can calculate that
this represents a change in the NADH concentration of 3.50 x 10 -5 M. This is in a 1 mL
solution, so 3.50 x 10 -8 mol have been used up over the course of five minutes, or 7.00 x
10 -9 mol (7.00 nmol) per minute on average.
P4.8:
Both starting compounds contain systems of conjugated π bonds which absorb in the UV
range. The condensation reaction brings these two conjugated systems together to create
a single, longer conjugated π system, which absorbs in the blue part of the visible
spectrum.
P4.9:

P4.10:
Chapter 4
The molecular ion and the fragments from all three of the typical ketone fragmentation
patterns would all have m/z = 98.
P4.11:
Both molecules contain alkene and ketone functional groups, however the degree of π
bond conjugation is different. Therefore, UV would be the more useful technique to
distinguish the two.
P4.12:
Both molecules are straight-chain alkanes with a single ketone group, so their IR spectra
are expected to be very similar and neither will absorb strongly in the UV range.
However, the different positions of the ketone (at the C 4 vs C 5 position) will result in the
formation of fragments of different masses in an MS experiment.
45

In-chapter exercises
E5.1:
a) 8 b) 8 c) 5 d) 18
E5.2:
46
Chapter 5
Chapter 5 solutions
a) 1 ppm = 300 Hz, so 4.56 ppm corresponds to 1368 Hz on the 300 MHz instrument.
On the 200 MHz instrument, the chemical shift is 912 Hz.
b) 300 MHz instrument: 300,001,368 Hz. 200 MHz instrument: 200,000,912 Hz.
E5.3: the acetone:dichloromethane ratio is 0.76:1. If it was a 1:1 ratio, the signal
integration ratio would be 3:1 because acetone has 6 protons while dichloromethane has
2. So you need to divide the 2.3 integral value by a factor of 3 to get 0.76.
E5.4: There are three peaks: two from para-xylene and one from acetone. The acetone
peak and the para-xylene methyl peak both represent six protons, so the ratio of their
integration values is simply 64 to 36 or 1 to 0.56. The ratio of the para-xylene methyl
peak to the para-xylene aromatic peak is 6 to 4, or 0.56 to 0.37. So the final integral ratio
of acetone:methyl:aromatic signals should be 1 to 0.56 to 0.37.
E5.5: The molecule contains two groups of equivalent protons: the twelve pointing to the
outside of the ring, and the six pointing into the center of the ring. The molecule is
aromatic, as evidenced by the chemical shift of the ‘outside’ protons’ (and the fact that
there are 18 π electrons, a Hückel number). The inside protons are shielded by the
induced field of the aromatic ring current, because inside the ring this field points in the
opposite direction of B0. This is the source of the unusually low (negative relative to
TMS) chemical shift for these protons.
E5.6:

a)
H b signal
Chapter 5
b) The figure above demonstrates that the quartet subpeaks integrate to 1:3:3:1 (eg., there
are three ways for two spins to be aligned with B0 and one against B0). As an analogy, if
you flip three coins at once, you have a 1 in eight chance of getting all heads, 1 in 8
chance of all tails, a 3 in 8 chance of two heads and a tail, and a 3 in 8 chance of two tails
and a head.
E5.7:
Because of the symmetry in the molecule, there are only four proton signals. Predicted
splitting is indicated.
E5.8:
(s)
H b
(s)
OH a
O
OH a
H b
Cl Cl
Hd Hd Hc (d)
Hc (d)
d
d
H
OH
N
NH 3
(recall that splitting is generally not seen with protons on heteroatoms)
E5.9:
t
t
47

48
H b
5.64 ppm
1692 Hz
J ab = 10.5 Hz
J bc = 1.5 Hz
chemical shifts of subpeaks (Hz)
1698.0
1696.5
1687.5
1686.0
E5.10: We can think of this signal as being a triplet of triplets, but because the two
coupling constants are very close, what we would actually see is a 1,2,3,2,1 pentet.
E5.11:
H b
J ab = J bc
a) 8 signals (each carbon is different)
b) 11 signals (the two enantiotopic CH2CH3 groups are NMR-equivalent)
c) 6 signals (each carbon is different)
d) 16 signals (the fluorobenzene group only contributes 4 signals due to symmetry)
E5.12:
a) 5 signals
b) 5 signals
c) 3 signals
3
3
1 2 2 1
3 3
Chapter 5

d) 6 signals
E5.13:
E5.14:
E5.15:
29.29
O
H 3C C C
O
O
pyruvate: spectrum b
18.95
H 3N
53.25
CH 3
O
207.85
O
alanine (spectrum c)
Cl
CH 3
125.4
172.69
H 3C CH 3
compound A
178.54
O
H 2C C
136.5
C OCH3
CH 3
18.4
compound B
167.9
O
51.8
182.63 45.35 73.06
O
OH
O
O
(S)-malate: spectrum d
168.10
Cl
O
N
NH 2
N
H
159.91
cytosine (spectrum a)
Cl
compound C
183.81
95.79
144.05
Chapter 5
E5.16: This benzylic carbocation is especially stable, as the positive charge is stabilized
by two benzene rings.
49

End-of-chapter problems
50
Chapter 5
P5.1: (structures n and o are chiral, and thus we must consider diastereotopic protons)
structure # 1 H peaks # 13 C peaks
a 4 5
b 2 3
c 4 4
d 5 4
e 2 3
f 3 4
g 4 6
h 3 6
i 1 2
j 2 2
k 3 4
l 5 5
m 4 3
n 6 4
o 10 5
P5.2: (valine and leucine both have diastereotopic (and thus nonequivalent) methyl
groups.
Amino acid # 13 C signals
glycine 2
alanine 3
valine 5
leucine 6
isoleucine 6
phenylalanine 7
tyrosine 7
tryptophan 11
methionine 5
cysteine 3
serine 3
threonine 4
arginine 6
lysine 6
histidine 6
proline 5
glutamate 5
aspartate 4
glutamine 5
asparagine 4

P5.3:
Chapter 5
a) On a 300 MHz instrument, 1 ppm = 300 Hz, thus a chemical shift of 2.9634 ppm
corresponds to 889.0 Hz. The downfield subpeak is 889.0 Hz plus ½ of the coupling
constant, or 893.1 Hz. Correspondingly, the upfield subpeak is at 889.0 Hz minus ½ of
the coupling constant, or 884.9 Hz.
b) In a 100 MHz instrument, 1 ppm = 100 Hz, so the chemical shift of 2.9634 ppm
corresponds to 296.34 Hz. Adding and subtracting ½ of the coupling constant gives us
chemical shift values for the two subpeaks of 300.4 Hz and 292.2 Hz.
P5.4: The chemical shift, in Hz, of the overall signal is (1.7562 ppm)(300 Hz/ppm) =
526.9 Hz. The furthest downfield subpeak has a chemical shift of 538.3 Hz (see the figure
below). The resonance frequency is this number plus 300 million Hz, or 300,000,538.3
Hz.
P5.5:
538.3 Hz
7.6 Hz 3.8 Hz
A pair of doublets indicates para disubstitution:
H a
H b
1.7562 ppm
(526.9 Hz)
X
Y
H a
Hb
51

52
Chapter 5
Note: In problems 6-13, references are given to the structure in the Aldrich Library of
NMR Spectra (volume - page)
P5.6:
Spectrum 1: 1-941C (structure D)
Spectrum 2: 1-941B (structure F)
Spectrum 3: 1-940B (structure C)
Spectrum 4: 1-939C (structure B)
Spectrum 5: 1-938C (structure A)
Spectrum 6: 1-940A (structure E)
P5.7:
Spectrum 7: 1-735B, (structure K)
Spectrum 8: 1-735A ( structure J)
Spectrum 9: 1-733C (structure I)
Spectrum 10: 1-729B (structure G)
Spectrum 11: 1-729A (structure H)
Spectrum 12: 1-895C (structure L)
P5.8:
Spectrum 13: 1-723B (structure M)
Spectrum 14: 1-723C (structure O)
Spectrum 15: (1-724A, structure P)
Spectrum 16: (1-724C, structure N)
Spectrum 17: (1-724B, structure R)
Spectrum 18: (1-722-C, structure Q)

P5.9:
P5.10:
2.12
H3C 7.15-7.29
7.10
O
Spectrum 23: 2-794B
V
(spectrum 22: 2-789A
6.86 3.61 3.78
O
O
Cl
S T U
2.73
2.86
2.12
Spectrum 25: (2-939A, structure FF)
Spectrum 26: (2-943C, structure BB)
Spectrum 27: (2-790A, structure AA)
Spectrum 28: (2-1201A, structure CC)
Spectrum 29: (2-455B, structure EE)
Spectrum 30: (3-85C, structure DD)
P5.11:
Spectrum 31: (2-265C, structure HH)
Spectrum 32: (2-266A, structure KK)
Spectrum 33: (2-1212B, structure LL)
Spectrum 34: (2-1366A, structure GG)
Spectrum 35: (2-1366B, structure JJ)
Spectrum 36: (2-1368C, structure II)
O
Spectrum 21: 2-940B
O
7.50
8.38
7.75
W
7.81
O
8.17
Spectrum 20: 2-938B
9.98
10.14
7.23-7.30
Spectrum 24: 2-791C
7.31
2.43
H3C 7.77
X
3.53
Chapter 5
O
O
9.94
Spectrum 19: 2-938C
53

P5.12: References to the Aldrich catalog of 1 H-NMR spectra are provided for each
structure)
a ) 1-734A
54
O
b) Molecular formula: C7H14O2
1-905B
O
O
c) Molecular formula: C5H12O
1-170B
OH
d) Molecular formula: C10H12O
2-789B
O
P5.13: 13 C-NMR data is given for the molecules shown below. Complete the peak
assignment column of each NMR data table.
a) (1-895C)
1
O
O
2
3
4
Chapter 5

δ (ppm) carbon #(s)
161.12 1
65.54 2
21.98 3
10.31 4
b) (1-735A)
1
2 3
O
5
4
6
δ (ppm) carbon #(s)
194.72 4
149.10 3
146.33 2
16.93 5
14.47 1
12.93 6
c) (1-938)
1
2
O O
O 3 4 5 O
8 9
10 11
δ (ppm) carbon #(s)
171.76 3, 5
60.87 2, 6
58.36 4
24.66 8, 9
14.14 1, 7
8.35 10, 11
6
7
Chapter 5
55

d) (2-1212B)
1
H3C 56
O
O
2
δ (ppm) carbon #(s)
173.45 2
155.01 7
130.34 6, 8
125.34 4
115.56 5, 9
52.27 1
40.27 3
e) (2-455B)
3
4
2
5
1
8
6
N
3
10
δ (ppm) carbon #(s)
147.79 1
129.18 2, 6
115.36 3, 5
111.89 4
44.29 7, 8
12.57 9, 10
5
7
4
9
6
9
OH
7
8
P5.14: The structure is 3-methyl-2-butanone
Chapter 5
P5.15: First, assign the peaks. 1-bromopropane: Ha is the triplet at 3.4 ppm, Hb is the
sextet at 1.9 ppm, Hc is the triplet at 1.0 ppm. 2-bromopropane: Hd is the doublet at 1.7
ppm, He is the septet at 4.3 ppm. Now, just add up the integrations for each molecule

Chapter 5
and figure the percentage. 1-bromopropane: 0.661 + 0.665 + 1.00 = 2.326. 2bromopropane:
0.0735 + 0.441 = 0.5145. Percent 2-bromopropane is (0.5145)*100 /
(0.5145 + 2.326) = 18.1%.
Notice that even if some peaks overlapped, you could still get this number as long as you
could integrate one signal on each molecule: you could, for example, do the same
calculation (and get the same result) by comparing integrations of Hc and Hd signals, and
just scale based on the fact that the Hc signal represents three protons while the Hd signal
represents six protons.
H a
H a
H b H b
C C C
Br
H c
integration:
Challenge problems
H c
H c
H c: 1.0 ppm (t)
1.00
3 protons
percent 2-bromopropane =
= 0.33
0.0735
0.0735 + 0.33
H d
H d
H e Br
C C C
H d
integration:
H d
H d
x 100 = 18.2 %
H d
H d: 1.7 ppm (d)
0.441
6 protons
= 0.0735
C5.1: The 13 C-NMR signal for the CDCl3 solvent appears as a 1:1:1 triplet due to 1-bond
coupling with the deuterium. Protons and 13 C nuclei both have a spin quantum number I
= ½, which means that they can assume two spin energy levels (+ ½ and – ½ ) when
placed in a strong external magnetic field. Deuterium, however, has a spin number I = 1,
meaning that it can assume three spin energy level (-1, 0, +1) in the external field. So
while a proton splits the signal from a neighboring nucleus into a doublet, a deuterium
splits its neighbor’s signal into three sub-peaks of equal height.
C5.2: See J. Chem. Educ. 2000, 77, 363.
C5.3: First, convert the chemical shift data from ppm to Hz, plot the data, then simply
compute the distance between the appropriate sub-peaks. For example, the distance
between the two subpeaks in the Ha doublet (9.67 ppm) is about 7.8 Hz, which is the
value of Jab. To get Jbc, it would make sense to look at the Hc doublet – except that this
signal overlaps with the aromatic signals. Alternatively, we can look at the Hb signal,
which is centered at 6.69 ppm. This is a doublet of doublets, but as we already know Jab
we can measure Jbc (approximately 16 Hz).
57

58
H b
J bc = 16 Hz
J ab = 7.8 Hz
Chapter 5

In-chapter exercises
E6.1:
fig 1
E6.2:
E6.3
Chapter 6
Chapter 6 solutions
59

fig 1
End-of-chapter problems
P6.1:
Left side: the nucleophile is an amine, the electrophile is the methyl carbon, and the
leaving group is a sulfide.
60
Chapter 6
Right side: The nucleophile is a thiolate ion, the electrophile is the carbon atom of an
alkyl diphosphate, and diphosphate is the leaving group.
P6.2:
a)
fig 2
b) In step 1, the nucleophile is the hydroxide oxygen, and the electrophile is the carbonyl
carbon of the thioester.
P6.3:
a)

)
c)
d)
fig 2
Chapter 6
61

e)
fig 3
f) Bold dots show the two carbons that form a new bond in this reaction step.
P6.4:
The C to D step has the highest activation energy, and thus is the slowest, ratedetermining
step.fig 3
P6.5:
62
Chapter 6

In-chapter exercises
E7.1:
fig 1
E7.2:
a) pKa ~ 12 (ammonium)
b) pKa ~ 36 (amine)
c) pKa ~ 10 (thiol)
d) pKa ~ 4.5 (carboxylic acid)
e) pKa ~ 4.5 (carboxylic acid)
Conjugate bases are shown below:
Chapter 7
Chapter 7 solutions
63

fig 1
E7.3:
a)
K eq ~ 10 (10.5 - 17) = 10 -6.5 = 7.1 x 10 -5
b)
K eq = 10 (4.8 - 19) = 10 -14.2 = 6.3 x 10 -15
c)
K eq ~ 10 (18-10) = 10 8
d)
64
Chapter 7

Chapter 7
K eq ~ 10 (15.7-10) = 10 5.7 = 5.0 x 10 5
fig 1
E7.4:
Problem: If an arginine side chain in a protein is exposed to buffer with pH = 8.5,
to what extent (expressed as a percentage) is it deprotonated?
If we approximate the pK a of an arginine side chain as 12.5, and use the Henderson -
Hasselbalch equation:
. . . so the arginine side chain is essentially 100% protonated in this buffer.
E7.5:
Using the Henderson-Hasselbalch equation, we find that the ratio [HA]/[A-] = 10 (5.2-4.8) =
10 0.4 = 2.5. If we express this ratio as 5:2, then we can see that for every 7 molecules of
acetic acid in this buffer, 5 are protonated (uncharged) and 2 are deprotonated (negatively
charged). From this we can easily calculate that about 29% of acetic acid molecules are
negatively charged.
E7.6:
The thiol is the most acidic group, and thus is the group that loses the proton.
65

fig 2
E7.7:
66
Chapter 7
C is an amine, and thus is less acidic than A and B, which are (protonated) ammonium
ions. A is more acidic than B, because the electron lone pair on the conjugate base of A
is stabilized by resonance.
fig 2
E7.8:
The acidity ranking is A>C>B. In acid A, the inductive effect stabilizing the negative
charge on the conjugate base is strongest, because chlorine is more electronegative than
bromine. Acid C is stronger than B because in C, the electron-withdrawing bromine is
closer by one bond to the negative charge on the conjugate base.
E7.9:
Carbons on which the negative charge can be localized are indicated with a bold dot.
fig 2
E7.10:
The negative charge on the conjugate base of picric acid can be delocalized to three
different nitro oxygen atoms (in addition to the phenolate oxygen).

fig 2
E7.11:
Chapter 7
A is a weaker base than B: the lone pair on the amine nitrogen of A can be delocalized to
the aldehyde oxygen, whereas this is not possible with B.
Fig 2
E7.12:
a) The α-protons are circled:
fig 3
b)
67

fig 3
E7.13:
68
Chapter 7
fig 3
The non-bridging oxygens on the β (outer) phosphate are most basic, because protonation
here results in there being a greater distance between the two remaining negative charges.
E7.14:
In this situation, the lysine side chain is in a hydrophobic environment. Therefore, the
protonated (positively charged ammonium) state is highly destabilized, and the
ammonium group is more acidic (because by giving up a proton it will rid itself of its
positive charge). The overall effect of the protein microenvironment is pK a lowering in
this case.
E7.15:
The metal ion acts as a Lewis acid, accepting electron density from the negativelycharged
oxygen of the enolate that forms upon deprotonation. This, of course, is a

Chapter 7
stabilizing effect, and thus the proximity of the metal ion serves to decrease the pK a of the
α-protons.
fig 4
End-of-chapter problems
P7.1:
The interior Asp has a higher pK a (is less acidic). It is in a hydrophobic environment,
meaning that that the charged (aspartate) form of the side chain will be highly
destabilized relative to the uncharged aspartic acid form.
P7.2:
In all cases the pK a of the amino acid side chain (or of water, for part e) is expected to be
lower due to the proximity of the cationic magnesium ion. The positive charge on the
metal ion is expected to stabilize the negatively-charged conjugate base form of Glu, Tyr,
and water, and to destabilize the positively charged, conjugate acid forms of Lys and His.
P7.3:
a) The base on the right is stronger
b) The base on the left is stronger
c) The base on the left is stronger
d) The base on the right is stronger
e) The base on the left is stronger
f) The base on the right is stronger
g) The base on the right is stronger
h) The base on the right is stronger
i) The base on the left is stronger
j) The base on the left is stronger
69

70
Chapter 7
P7.4: The positive charge on the protonated form of arginine can be delocalized by
resonance to all three of the nitrogens – this stabilizes the conjugate acid form (ie. makes
it a weaker acid). On the protonated form of lysine, by contrast, the positive charge is
‘stuck’ on the single nitrogen (see the structure of lysine in chapter 6). Because the
positive charge of of lysine is not stabilized by resonance, lysine is more likely to give up
a proton and lose the charge.
fig 2
P7.5: The ester oxygen acts as an electron-donating group by resonance. This electrondonating
property destabilizes the negative charge on the enolate form, making the αproton
less acidic. This argument also holds true for thioesters.
fig 3
P7.6:
a) The most acidic proton on tetracycline is indicated below. Notice that the negative
charge on the conjugate base can be delocalized to two carbonyl oxygens.

Chapter 7
fig 4
When the second most acidic proton is abstracted by a base, the resulting negative charge
can be delocalized to one carbonyl oxygen (in addition to three aromatic carbons). The
2 nd conjugate base of tetracycline is shown below (having fully reacted with two molar
equivalents of strong base).
fig 4
P7.7:
71

fig 5
P7.8:
fig 5
P7.9:
72
Chapter 7
One nitrogen is simply a primary amine, and as such is basic. The other nitrogen is
‘pyrrole-like’, meaning that its lone pair is part of an aromatic sextet, and is not available
for bonding to another proton.

fig 5
P7.10:
The alkyl amine nitrogen is most basic, the ‘pyrrole-like’ nitrogen is least basic.
fig 5
P7.11:
Chapter 7
Proton H a is more acidic. The negative charge on the conjugate base that results from the
abstraction of H a can be delocalized to both oxygens plus a carbon, whereas the negative
charge from abstraction of H b can be delocalized just to the two oxygens.
fig 5
P7.12:
73

74
Chapter 7
Of the four protons, H a is the least acidic. The negative charge that results from
abstraction of H a can be delocalized to only one oxygen atom, whereas the charge
resulting from abstraction of either H b, H c, and H d can be delocalized to two oxygens in
each case.
fig 6
P7.13:
The total charge will be very close to –3.0. At pH = 7.3, the N-terminus proline (pK a =
10.6) will be fully protonated, and will contribute a charge of +1. This is balanced,
however, by a negative charge on the terminal glutamate (pKa = 2.2). Three of the seven
amino acids in the peptide have ionizable side chains: an aspartate (D) and two
glutamates (E). The pK a values for these side chains are 3.7 and 4.3, respectively, so at
pH 7 all three will be fully ionized, leading to a total peptide charge of –3.
P7.14:
There are three ionizable groups on this peptide: the terminal amino group on Asp (pK a ~
9.6), the side-chain carboxylate group on Asp (pK a ~ 3.7) and the terminal carboxylate on
Ile (pK a ~ 2.4). For each buffer, we can use the Henderson-Hasselbalch equation to
determine the charged / uncharged ratio for each group.
a) At pH = 4.0:
Asp (terminal amino) [HA + ] / [A] = 10 (9.6-4.0) = 10 5.6 = 4.0 x 10 5 . At this pH the terminal
amino group is essentially 100% protonated and positively charged, so this group
contributes a charge of +1.
Asp (side chain): [HA] / [A - ] = 10 (3.7 - 4.0) = 10 (-0.3) = 0.50. Approximately 2 out of every 3
side chains is deprotonated and negatively charged, so overall this group contributes a
charge of -0.67.
Ile (terminal carboxylate): ): [HA] / [A - ] = 10 (2.4 - 4.0) = 10 (-1.6) = 0.025. Most, but not all of
the terminal carboxylates are deprotonated and negatively charged. We can calculate the
percentage that are protonated:

Chapter 7
. . .thus about 97.6% are deprotonated and negatively charged. This group contributes an
overall charge of -0.98.
In a buffer of pH 4.0, the total charge on the dipeptide will be:
(+1) + (-0.67) + (-0.98) = -0.65.
b) in a buffer with pH = 7.3, the total charge on the dipeptide will be close to -2 (the
terminal amino group is 100% protonated, both carboxylate groups are 100%
deprotonated)
c) in a buffer with pH = 9.6, the total charge on the dipeptide will be close to -2.5 (in this
basic buffer, the terminal amino group is 50% deprotonated, and so only contributes a
charge of +0.5).
P7.15:
K eq = 10 -8.2 = 6.3 x 10 -9
75

76
Chapter 7
fig 6
How did we pick the most basic group on the Y species? We have four choices: a
primary amine, a 'pyrrole-like' amine, and two carboxylates. We know that pyrrole-like
amines are not basic, and we can look at our pK a table to remind ourselves that primary
amines are more basic than carboxylates.
fig 7
K eq = 10 (10-19) = 10 -9
P7.16:

fig 6
Chapter 7
P7.17: Aqueous solutions of 'Tris' and 'HEPES' are very commonly used as buffers in
biochemistry and molecular biology laboratories. You make two buffer solutions: One is
50 mM Tris at pH 7.0, the other 50 mM HEPES at pH 7.0. For each solution, calculate
the concentration of buffer molecules that are in their charged (ionic) protonation states.
Tris: Using the Henderson-Hasselbalch equation, we find that the ratio [HA + ] / [A] at this
pH is 10 (8.1-7.0) = 10 1.1 = 12.6. The percentage of HA + is thus (12.6/13.6)*100 = 93%. The
concentration of protonated (positively charged) Tris is (0.93)(50 mM) = 46.5 mM.
HEPES: the most acid proton is on the sulfate group (the negative charge on the
conjugate base can be delocalized to two oxygens).
[HA] / [A - ] = 10 0.5 = 3.2. This means 76% is protonated (uncharged) at this pH, and 24%
is deprotonated (negatively charged). Thus the concentration of negatively-charged
HEPES is (0.24)(50mM) = 12 mM.
77

In-chapter exercises
E8.1:
fig 1
E8.2:
78
Chapter 8
Chapter 8 solutions
Notice that the reaction occurs with inversion of configuration: the leaving group (I) was
pointing out of the plane of the page, while the nucleophile (CH 3S - ) attacks from behind,
and ends up pointing into the plane of the page.

fig 1
E8.3:
The intermediate species is a carbocation that forms after the bromine leaves:
E8.4:
fig 1
E8.5:
Chapter 8
Reaction A is a concerted (one-step) reaction, essentially a collision between two species.
Therefore, if the concentration of nucleophile (CH 3S - ) were doubled in the solution, the
rate of the reaction should also double. You should recall learning about rate expressions
in your general chemistry course - this is an example of a second order rate expression,
where the rate depends on some rate constant (k) and the product of two concentration:
rate = k[CH 3I] [CH 3S - ]
(If this is unfamiliar to you, now would be a good time for a quick review of rate
expressions in your general chemistry textbook!)
Reaction B is a two-step reaction. The first step – the breaking of the C-Br bond – is the
slow, rate determining step, and does not involve the CH 3SH nucleophile at all.
Therefore, changing the concentration of the nucleophile should have no effect on the
rate of the reaction. (If the concentration of CH 3SH were doubled, the second step would
occur twice as fast – but the rate of the first, slower step would not change, and so the
overall rate of the reaction would not be affected). The rate expression in this case is first
order (it depends on the concentration of only one reactant):
rate = k[C(CH 3) 3Br]
79

E8.6:
80
Chapter 8
The side chain on serine – a primary alcohol – is in general a better nucleophile then the
side chain on tyrosine, which is a phenol. Recall (section 7.4A) the argument for why
phenolate (deprotonated phenol) is less basic than ethanoate (deprotonated ethanol) –
some of the electron density of the oxygen atom in phenolate is delocalized to the
aromatic ring. The same argument holds true when comparing the nucleophilicity of
serine to tyrosine: electron density on the tyrosine oxygen is stabilized somewhat by
resonance with the benzene ring, and is therefore less reactive. Steric factors also come
into play - the large phenyl ring makes nucleophilic attack more difficult for the tyrosine.
E8.7:
A cysteine, with its thiol side chain, is very nucleophilic - much more so than a
methionine with its sulfide group. The thiol sulfur atom is less hindered than the sulfur
atom on a sulfide, and in addition the thiol proton can be removed by a base as the
nucleophilic attack takes place. The methyl group on a methionine side chain, in
contrast, cannot be removed by a base, so that the product of a nucleophilic attack by
methionine will contain a positive formal charge on the sulfur.
fig 2
E8.8:
a) phenolate is the stronger base (compare the pK as of the conjugate acids), and is also the
stronger nucleophile – the electrons on the nucleophilic oxygen atoms are more reactive.
b) water is nucleophilic, hydronium ion is not – the difference is in the protonation state
(deprotonation increases nucleophilicity)
c) trimethyl amine is less hindered, and therefore is the better nucleophile.
d) iodide ion is more polarizable, and therefore is a better nucleophile (even though it is
the weaker base).
e) CH 3NH - is anionic and deprotonated – it is the stronger nucleophile.
E8.9:

Chapter 8
In the carbocation on the left, the positive charge is located in a position relative to the
nitrogen such that the lone pair of electrons on the nitrogen can be donated to fill the
empty orbital. This is not possible for the carbocation species on the right.
fig 2
E8.10:
fig 2
E8.11:
a) 1 (tertiary vs. secondary carbocation)
b) equal
c) 1 (tertiary vs. secondary carbocation)
d) 2 (positive charge is further from electron-withdrawing fluorine)
e) 1 (lone pair on nitrogen can donate electrons by resonance)
f) 1 (allylic carbocation – positive charge can be delocalized to a second carbon)
E8.12:
81

82
Chapter 8
a) B (thiolate is a weaker base/better leaving group than alcohol)
b) A (sulfide is better leaving group than thiolate (sulfide is neutral and not at all basic)
c) A (bromide ion is a weaker base/better leaving group than acetate – compare pK a
values of HBr and acetic acid).
d) B (inductive electron-withdrawing effect of fluorines make trifluoroacetate a weaker
base/better leaving group).
E8.13:
E8.14:
a) Notice that the oxygen nucleophile attacks from behind the plane of the page, while the
chlorine is pointing out of the plane of the page – this is a backside attack by the
nucleophile.
fig 3

Chapter 8
b) Because both the nucleophilic oxygen and the chlorine leaving group are oriented on
the same side of the ring, backside attack by the nucleophile is impossible. In order for a
displacement to occur, it would have to be a stepwise (S N1) mechanism. The observation
that compound A forms an epoxide under these conditions but compound B does not
strongly suggests that the reaction proceeds by an S N2 mechanism, with a requirement for
backside attack by the nucleophile.
E8.15:
fig 3
End-of-chapter problems
P8.1:
All of the molecules in question are primary alkyl bromides, and the nucleophile is a very
powerful thiolate ion – we are talking here about SN 2 reactions. The rate of substitution
depends on the amount of steric hindrance in the electrophile – the less hindrance, the
faster the substitution reaction. The order is:
fastest D > B > A > C > slowest
P8.2:
fig 4
P8.3:
83

84
Chapter 8
Carbocation A is more stable: it has a longer system of conjugated π bonds, and as a
consequence the charge can be delocalized over five different carbons (as opposed to four
carbons for B).
P8.4:
Here we are looking at periodic trends and steric hindrance. Nucleophilicity increases
going down a column of the periodic table, so A and C, with phosphorus atoms, are
expected to be more nucleophilic than B and D. C is more nucleophilic than A, because
the three methyl groups on C are the cause of less steric hindrance than the three ethyl
groups on A. Using the same reasoning, we can see that B should be more nucleophilic
than D, because of the bulky phenyl group on D. The trend is:
most nucleophilic C > A > B > D least nucleophilic
P8.5:
fig 4
P8.6:
The reaction shown in part e) is intramolecular (one molecule reacting with itself), and
thus will likely have a first order rate expression.
P8.7:
a) water or hydroxide ion
b) CH 3S - or CH 3OH
c) CH 2S - or CH 3SH
d) acetate ion or hydroxide ion
e) diethyl sulfide or diethyl ether
f) dimethylamine or diethylether

g) trimethylamine or 2,2-dimethylpropane
P8.8:
Chapter 8
The major product will be dimethyl sulfide (CH 3SCH 3), because CH 3S - is a better
nucleophile than CH 3O - and will react faster with methyl bromide in an S N2 displacement.
P8.9:
a) the compound on the left
b) the compound on the right
c) the compound on the right
d) the compound on the left
e) the compound on the left
f) the compound on the left
g) the compound on the right
P8.10:
fig 4
P8.11:
85

fig 5
P8.12:
fig 5
P8.13:
86
Chapter 8

fig 6
P8.14:
Chapter 8
The first step in an S N1 reaction is carbocation formation. However, the rigid ‘bicyclo’
structure of the starting material prevents a hypothetical carbocation intermediate from
adopting trigonal planar geometry (make a model to see this better). Consequently, there
is a large energy barrier for carbocation formation.
fig 6
Challenge problems
C8.11:
See the following references for a solution:
March, Jerry, Advanced Organic Chemistry: Reactions and Mechanisms, and Structure,
4th ed. (1992), p. 296. (Use the 'author index' to find a discussion of this paper in more
recent editions of the book).
J. Chem Soc. 1935, 1525. (This is the original paper, but may be hard to locate.)
87

In-chapter exercises
E9.1:
88
Chapter 9
Chapter 9 solutions
The electrophilic carbon is primary, and a carbocation intermediate would not be
stabilized by resonance. Therefore it is reasonable to predict an S N2 mechanism.
fig 1
Note: in later chapters, especially in chapters 10 and 12, we will see many more reactions
involving ATP. As we will see then, the SAM-forming reaction above is somewhat
unusual – most reactions with ATP involve nucleophilic attack at one of the phosphate
groups, rather than at the methylene (CH 2) carbon.
E9.2:

Chapter 9
Due to the relatively high nucleophilicity of the amine group, it does not need to be
deprotonated prior to attack. The deprotonation step takes place after the nucleophilic
attack, when an ammonium ion is the proton donor (an ammonium ion is a relative strong
acid and does not require a strong base for deprotonation). The alcohol, in contrast, is less
nucleophilic and needs to be deprotonated by a strong base before acting as a nucleophile
in an S N2 reaction.
fig 1
E9.3:
The nucleophile in this reaction is an amine, as opposed to a thiolate in the protein
farnesyltransferase reaction. A thiolate is a very powerful nucleophile, an amine
somewhat less so. Therefore the N-alkylating reaction, with the weaker nucleophile, is
likely to be more SN1 in character, and thus would be expected to exhibit higher
sensitivity to fluorine substitution on the electrophilic substrate (in other words, more
positive charge develops on C1 of the electrophile in the AMP reaction)
E9.4:
The S N1 model is a solvolysis reaction (water is the nucleophile), which as a rule proceed
through carbocation intermediates. In the S N2 model the nucleophile is a relatively
powerful iodide ion, a factor which favors a concerted mechanism.
E9.5:
Notice that in the second step, when water enters the picture, stereochemistry is
conserved. If the step were indeed a direct nucleophilic displacement by water, the
observed stereochemistry would be different: either inversion as show below in the case
of an SN2 mechanism, or racemization in the case of an SN1 mechanism.
89

fig 1
E9.6:
90
Chapter 9
This is essentially a nucleophilic (or ‘base-catalyzed’) ring-opening, with the nucleophilic
thiol attacking the less hindered carbon of the epoxide. The right-side epoxide carbon is
‘protected’ from nucleophilic attack by the negatively charged oxygens of the
phosphonate group.
End-of-chapter problems
P9.1:
a)

fig 2
P9.2:
(See Biochem 42, 1564)
fig 3
P9.3:
Chapter 9
a) Because the reaaction involves the transfer of a methyl group to an amine, the most
likely biomolecule would be S-adenosylmethionine (SAM – see section 9.1).
b) A mechanism with an abbreviated version of the SAM structure is shown below. This
is an SN2 mechanism, similar to other SAM-dependent methyltransferase reactions.
91

fig 3
P9.4:
92
Chapter 9
a) Because the process involves two S N2 displacements, it is reasonable to predict that it
is the thiol group – the most nucleophilic group in glutathione – that is playing the key
role. For this reason, we will use the abbreviation GSH for glutathione in the mechanism
in part b) below.
b)
fig 3
P9.5:
(See Acc Chem. Res. 2001, 34, 681.) A likely mechanism could involve two successive
S N2 methyl transfer steps, the first from 5-methyltetrahydrofolate to cobalamin, and the
second from cobalamin to homocysteine, the direct precursor to methionine. This would
account for the observed retention of configuration. (McMurry p. 282)

fig 3
P9.6:
Challenge problems
Chapter 9
C9.1: See Biochem 2005, 44, 8989; Structure 2004, 12, 927.
c) (solvent isotope effect - H 2O deprotonation should not be rate-limiting if dissociative).
C9.2: See Biochem 1983, 22, 806; Chem Res Toxicol 1997, 10, 2 (scheme 1).
C9.3: See Chem Res Toxicol 1997, 10, 2 (fig 13)
C9.4: See Biochem 43, 7187 (look at compound #6 just above the Materials and Methods
section):
C9.5: Silverman p. 261
a,b) See J. Am. Chem. Soc. 1994, 116, 11594.
c) See J. Am. Chem. Soc. 1987, 109, 7530.
93

In-chapter exercises
E10.1:
94
Chapter 10
Chapter 10 solutions
We’ll use methyl phosphate as a simple example of a phosphate monoester. The three
major resonance forms show how the three non-bridging oxygens share a double bond.
Another way of putting this is that each of these bonds is a single σ bond plus 1/3 of a π
bond – a bonding order of 1.33. However, if we also consider the minor resonance form
in which the bridging oxygen is double-bonded (it's a minor form because of the
separation of charge), we can also think of the double bond being shared, in a small part,
by the bridging oxygen. We can approximate these ideas by saying that the bridging bond
order is slightly more than 1, and the non-bridging bond order is slightly less than 1.33.
O
O P
fig 1
O
O
O CH 3 O P
O CH 3
O
major resonance contributors
O
O P
E10.2: figs for problem in solnfig01a
O
O CH 3
minor resonance contributor
O
O P
O
O CH 3
A,E, and I are all representations of an alkyl- or acyl-adenosine phosphate ester (the
aspartyl adenosine phosphate species in the first figure of section 10.2E is an actual
example of an acyl-adenosine phosphate ester)

Chapter 10
F and N are both representations of an organic monophosphate ester, such as glucose-6phosphate
(section 10.2B).
C and G are both representations of an organic diphosphate ester, such as mevalonate 5diphosphate
(section 10.2C).
B, H, K, R, and S are all representations of adenosine triphosphate (ATP – section
10.2A).
D, L and O are all representations of inorganic diphosphate (also known as inorganic
pyrophosphate).
J, M, P, and Q are all representations of adenosine diphosphate (ADP).
E10.3:
We can use a number of alternative abbreviations for ATP and ADP in showing this
mechanism, but the ones used below are relatively compact and still show the chemistry
taking place at the γ-phosphate of ATP:
fig 1b
E10.4:
B:
H
protein
O
O
O P
O
O
O P
O
O
O AMP +
O
protein
O
P
O
O
O P
O
O AMP
The phosphorylation reaction does not involve any bonds to the two carbon stereocenters,
so we do not expect any changes to stereochemical configuration.
fig 1
E10.5:
OH
NH
O
a threonine residue
O O
P
O O
NH
O
a phosphorylated threonine residue
95

fig 1b
96
O O
H 3N CO 2
O
O P
O
End-of-chapter problems
O
O P
P10.1 (McMurry p. 195, step 5)
fig 1
ribose-A
ribose-A
ribose-A
O
O
P
O
O
O P
α
O
O
O
O P O O
O
O
O AMP + O P
O
β
O P
ATP
O
P
O
O
O
δ −
O O
α β
O P O P
O
ADP
P10.2: (McMurry p. 287 step 6)
O
O
O
O
γ
P
O
O
O
O
P
O
O
O
P
O
O
O
O
H 3N CO 2
O
δ −
O
O
OH
O
OH
OH
O
O P
O
O
O
O P
O
O P
Because no bonds to stereocenters are affected, the stereochemistry is unchanged.
+
O
O
O
O
Chapter 10
O
O AMP

fig 2
ribose-A
P10.3: McMurry p. 317 top
fig 2
P10.4:
B:
OPP
HO
OPP
HO
O
O
O
OH
O
P
O
O
H
OH
O
ATP
OPP
+
O
P
O
O
O
O
P
O
O P O
O
O
O P
O
O
O
P
O
AMP
H
PO
:B
O P
ATP
O
O
O
CO 2
OH
O ribose-A
CO 2
OH
OH
OH
O ribose-A
+ ADP
Chapter 10
97

98
A H
H 3N
O
O
O
P
O
O
O
fig 3
P10.5: McMurry p. 318,
a) (323 bottom)
b)
fig 3
PPO
B:
ADP
PO
O
OH
O
H
O
N
O
P
O
OH
O
OH
O
N
O
OH
H
O
H
N
H
O
NH
:B
O P
N
O
O
H
O
:B
ADP
A H
H 3N
H 3N
PO
O P
OH
O
O
O
O
O
O O
+
O
O
O
P
O
H
H
OH
Chapter 10
:B
O N
PPO + ADP
O
O
OH
O
OH
OH
O
P
O
H
N
N
OH
O
NH
O
N
O
P
O
O
+ ADP

P10.6:
Chapter 10
This is a transfer of a phosphate group to a nitrogen rather than to an oxygen, but the
mechanism is analogous.
fig 4
B:
H
N
enz
N
HO
HO
P10.7: (McMurry p. 345)
H 3N
O
O
O
O
P
O
HO
O
O
O P O
O
O P O
O
O P O
O
ribose-A
O
OH
H A
fig 4
P10.8: (From J. Biol. Chem. 280, 10774).
HO
HO
fig 4
OH
HO
O
O
O
P
O
O
O
O P O
O
O P O
O
O P O
O
ribose-C
P10.9: (From Biochemistry 2000, 39, 8603.)
H 3N
HO
HO
O
N
enz
O
OH
HO
O
O
O
P
N
O
O
P
O
O
+
AMP
HO
HO
+
+
O
O P
O
OH
HO
O
O P O
O
O P O
O
O
O
O P O
O
O ribose-C
OH
O P O
O (PPi)
99

100
Chapter 10
a) The stereochemistry of the substitution and location of the 18 O label in the product
strongly suggests that this is an SN (probably SN1-like) displacement at the anomeric
carbon atom, rather than attack by the water molecule at a phosphorus.
fig 5
HO
HO
OH
HO
O
O
O
P
O
O
O P
O
O ribose-G
HO
HO
HO
HO
OH
HO
OH
HO
O
O
+
18
OH
O
O
P
O
18
O
H
H
O
O P
O
:B
O
ribose-G
b) If water were to attack at the β-phosphorus of the GDP group, the expected product
would be:
HO
HO
OH
HO
O
OH
+
O
18 O
P
O
O
O P
O
O
ribose-G
fig 5
Notice the different stereochemistry at the anomeric carbon, and the different location of
the 18 O label.
P10.10: (From Biochemistry 44, 11476.)

Tyr
fig 5
DNA
5'
O
O H
B:
O
O
P
O
3'
O
O
5'
O
DNA
Base n
H A
O
3'
Base n+1
knicking
re-ligating
DNA
O
5'
O
O
O
P
O
3'
O
O
5'
Chapter 10
O
DNA
Base n
P10.11: (From Biochem 41, 9279). Using radiolabelled water, we could determine the
course of the reaction. In the first case, the AMP would be radioactive, in the second
case the sugar would be radioactive.
H
ribose-A
O
*
ribose-A
fig 5
H
O
O
P10.12:
O
P
O
O
P
O
:B
O
O
O
P
O
H
O
P
O
O
O
*
O
HO
H
HO
O OH
:B
OH
O OH
OH
ribose-A
ribose-A
O
O
O
P
O
O
P
O
O*
O
A H
+
Tyr
+
O
*O
O
P
O
O
P
O
O
H
O
HO
HO
:B
O
3'
O OH
OH
Base n+1
O OH
OH
101

attack at γ-phosphate:
102
Nu:
O
O P O
O
attack at β-phosphate:
Nu:
O
O P O
O
O
O P O
O
attack at α-phosphate:
O P O
O
O
O
O P O
O
Nu:
O
O
P
O
P
O
O P
O
O P
O
O
O
O
P
O
ATP
O
P
O
O
P
O
O
O P
O
O
O P
O
O
O P
O
O ribose-A
O ribose-A
eg. phosphoribosyl diphosphate (PRPP), section 10.2C
O ribose-A
:Nu
O ribose-A
:Nu
O ribose-A
Nu
O
P
O
O
+ ADP
eg. glucose-6-kinase, section 10.2B
O
O P O
Nu
O
(no examples so far - but look at UTP
reaction in section 11.4A)
(no examples in text)
O
O P O
O
O
P
O
+
+ AMP
O
P
O
O
O P
HO ribose-A
O
O
O
P
O
O
O P
+ Pi
O
Nu
Nu
Nu P O ribose-A + PPi
O
eg. generation of acyl adenoside phosphate, section 10.2E
Chapter 10
O ribose-A

attack at 5'-carbon:
O
O P O
O
O
P
O
O
O P
O
:Nu
O O
N
N
HO OH
NH 2
N
N
Nu ribose-A + PPPi
eg. excercise 9.1
fig 6
Here’s an example of this last possibility (the solution to exercise E9.1):
Challenge Problems
C10.1: See Biochem J. 1982, 201, 665.
C10.2: See J. Biol. Chem. 257, 14795.
C10.3: See J. Mol. Biol. 1999, 286, 1507.McMurry p. 176
Chapter 10
103

In-chapter exercises
104
Chapter 11
Chapter 11 solutions
E11.1: The β-phosphate of UTP is attacked by the nucleophilic hydroxyl group, and
inorganic phosphate is expelled. This is the first such phosphoryl transfer reaction that we
have seen (we have seen the β-phosphate of ATP attacked by an alcohol in section 10.2C,
but that attack was from the 'other side' - it resulted in expulsion of AMP and formation of
an organic diphosphate).
E11.2:
a) cycic ketal formation:
fig 1
I
I
I
H A
O
O O
O O
H
:B
I
I
O
H
HO OH
O
OH
B:
I
I
H
O
O
OH
OH
OH
OH 2
H A

) cyclic ketal hydrolysis:
I
I
I
fig 1
E11.3:
O O
O
HO OH
O
H
H A
:B
I
I
HO
O O H
OH
O H
A
H
I
I
O
O
OH
H 2O
OH
O
H
H
:B
Chapter 11
105

106
B:
fig 1a
R
H
PO
N
H
CO 2
H
PO
R
N
N
N
H A
OH
CO 2
N
CH 3
End-of-chapter problems
OH
CH 3
P11.1: (McMurry p. 250 bottom)
fig 2
enz
O 2C O
A H
H 2N
enamine form
O
H
O
O 2C O
O
:B
B:
PO
B:
R
H
H
PO
N
R
H
CO 2
N
N
H
N
CO 2
N
A H
OH
CH 3
H A
H 2N
OH
CH 3
NH 2
O 2C O
imine form
H
O H
O
:B
O 2C O
B:
NH 3
O
H
O
enz
Chapter 11
enz

P11.2: (McMurry p. 255 bottom, 2nd step)
fig 2
B:
H
O
H
N
H
H
A
CO 2
P11.3: (McMurry p. 320, step 10)
PO N
O
HO
fig 2
Mechanism:
N
OH
O
N
H
NH 2
O
B:
H
O
H 2O
H
N
H
CO 2
PO N
O
HO
O
O
N
H
N
H
:B
N
N N
H
O
N N
ribose-5-P
H A
ribose-5-P
B:
H
fig 3
P11.4: (McMurry p. 307)
IMP
N
N
OH
H
OH
H
A
O
O
N
N
H
NH 3
IMP
Chapter 11
CO 2
107

fig 3
108
A H
O
N
NH 2
N
R
Cytidine
P11.5: (McMurry p. 311 middle)
H
O H
R
N
N
N
:B
NH 2
H
O
fig 3
P11.6: (McMurry p. 264 step 1
Lys
O
H
N H
O
:B
O
N
O
H
O
A
H
H
A
:B
R
H
N
N
B:
O
O
H
O
N
Lys
O
H
H 2N
N
NH 2
N
R
NH
N
OH
O
H
HN
H
OH
O
A
O
N
R
Uridine
A
O
H
A
R
R
N
N
H
H
N
N
H
O
A
H3N O
N
:B
N
O
N
N
R
O
O H
NH 3
NH
NH
O
H
Lys
Chapter 11
:B
A
N
O
O

fig 3
P11.7: (McMurry p. 265 2nd step)
O
fig 4
O
O
O
N
H
H
O H
H A
:B
P11.8: (McMurry p. 277 step 4)
O 2C
B:
fig 4
O
H
OH
N
H
P11.9: (McMurry p. 277 step 7)
fig 5
O
O
NH 3
O
O2C N CO2 P11.10 (McMurry p. 287 step 2)
O
O
O
O
O
O2C HO
N
H
O
O
A H
H 2O
OH
O
O 2C
O
NH 3
O
O
+
O
H 2N
O
N
H H
A
O H
H
O
CO 2
O
:B
NH 3
Chapter 11
O
OH
N
NH 3
O
O
O
O
O
109

fig 5
110
H
HO
CO 2
O
H
OH
OH
OP
P11.11 (McMurry p. 288 step iv)
fig 5
H 2C
HO
O
OH
CO 2
OH
reverse of hemiketal
formation
H 2C
OH
P11.12: (McMurry p. 291 bottom first step)
fig 5
OP
A H
O
HO OH
H
N
P11.13: (J. Biol. Chem. 280, 13712)
CO 2
OH
OP
O
HO
O
OH
OH
CO 2
OP
CO 2
OH
anomeric carbon
enol-keto
tautomerization
HO HO
OP
OH N
H
H
OH N
HO OH
H
H 3C
:B
O
CO 2
CO 2
OH
Chapter 11
O
OH
CO 2

fig 6
A H
O
C
H H
H
N
N
H
H
:B
N R
B:
A H
P11.14: (J. Biol. Chem. 280, 12858, scheme 2 part 2)
B:
fig 6
A H
P11.15:
HO
O
H
H
H
O
N
O
H
O
N
O O
R
O
O
hemiketal
formation
HO
O
R
H
N
N
H
OH
N R
O
HO
HN
H
N
H 2C
R
N
N
H
HN
HO
HN
A H N
N
H
O
H
O
O O
OH O
O
H
H
O
hydrate
formation
HO
R
O
O
H
Chapter 11
N
H
R
O
O O
:B
OH
OH OH
O
111

Challenge problems
C11.1: (McMurry p. 296 step 5)
1)
2)
112
H 2N
N
H
O
N
O
N
R 1
N NH
O
R 2
R 2
NH
A
H
H
N
H
H
:B
H 2O
imine formation
H 2N
N
O
N
R 1
H
N NH
O
H
N H
C11.2: See Biochemistry 1994, 33, 13792, mechanism II.
C11.3: See Biochemistry 2000, 39, 8603.
C11.4: See J. Am. Chem. Soc. 2005, 127, 16412.
H
A
N
R 2
N
R 2
H
:B
:B
H 2N
N
tautomerization
O
N
R 1
HN
NH 2
R 2
N
Chapter 11
+
O
NH
R 2
NH

In-chapter exercises
E12.1:
a) carboxylic acid
b) thiol
c) water
Chapter 12
Chapter 12 solutions
E12.2: Amines are bases, and thus the amine would simply deprotonate the carboxylic
acid rather than attacking the carbonyl carbon.
E12.3:
O
OH
SOCl 2
O
R NH2 O C H
R
E12.4: Excess acetic acid is used in order to drive the equilibrium towards product
formation (by Le Chatelier’s principle).
O
E12.5: Base-catalyzed esterification (starting with a carboxylic and an alcohol) will not
work. The base will simply deprotonate the carboxylic acid, and carboxylate groups are
very unreactive towards acyl substitution reactions.
O
R O
Cl
H
+ R OH X no reaction
E12.6: This reaction would be described as the acid-catalyzed hydrolysis of an ester.
N
O
N
113

H 3C C O
114
O
E12.7:
H 3C
H 3C
H
O
O
O
OH
H
O
OH +
OH
O
+
+
OH
O
S
O
H 3C
HO
+
HO
HO
OH
O
H
O
+
CH 3CH 2OH
O
H
H 3C C O
H 2O
HO
H 2SO 4
H 2SO 4
H 2SO 4
H 2SO 4
O
H3C O
propyl acetate
(pear)
O
H 3C O
benzyl acetate
(peach)
O
H O
isobutyl formate
(raspberry)
O
O
ethyl butanoate
(pineapple)
H
O
H3C C O
O
H H
H
O
H3C C O
O
H H
O
O
S
O
Chapter 12
OH

OH
O
OH
+
CH 3OH
H 2SO 4
OH
O
O
methyl salicylate
(wintergreen)
CH 3
Chapter 12
E12.8: The leaving group in this acyl transfer reaction is a resonance-stabilized phenolate
ion - the negative charge is delocalized to the carboxylate group.
O
E12.9:
O
E12.10:
a)
H 3C
O
O CH 3
O
O
O
C SCoA
enzyme
+
HS
O
Ph O
enz
HN
O
enz
O
HN
O
H
O
O
S
O
O
CO 2
+
H 3C
O
O
enzyme
115

)
O
116
O HO
CH 3
O
H
End-of-chapter problems
P12.1: A thioester (acetyl-CoA) is converted to an amide in this reaction:
(Biochem 44, 16275)
H3C O
C
SCoA
H
:B
H3C O
C
H
SCoA
A
H3C R HN R HN
R HN
O
HO
HO
NH2 O
HO
HO
NH2 O
HO
HO
P12.2: (McMurry p. 191)
O
O 2C SCoA
O
P12.3: (McMurry p. 203?? step 3)
HO
HO
OP
HO
O
O
P
O
O
O
:B
H H
O
HO
HO
O OP
O 2C SCoA
OP
HO
O
H A
O
H
O
Chapter 12
O
O
NH 2
O 2C OP
HO
HO
OP
HO
OH
O
O

P12.4: (McMurry p. 229 top)
O
HO OP
O
HO NH 2
O
PO NH 2
A B C
Chapter 12
P12.5: This reaction can be described as the hydrolysis of an amide. (McMurry p. 244
step 2)
O 2C
NH 3
H
O
O
N
H
O
H 2O O
H O
P12.6: The second step of this epoxide-opening reaction is hydrolysis of the ester formed
in the first step. The bond to the chiral carbon is not broken, thus stereochemistry is not
inverted. Notice also that the oxygen from the active site aspartate residue is incorporated
into the product.
R
6
CO 2
O 2C
NH 3
H 2N
O O OH
HO
O O OH
O
:B
H A
H
H
Asp Asp Asp
H O
B:
P12.7: (Silverman p. 263, J. Am. Chem. Soc. 115, 10466)
a) Yes, the fact that the 18 O label from the active site aspartate is incorporated into the
product indicates that this residue is the ring-opening nucleophile as in the mechanism
above.
R
b) In a hypothetical alternate mechanism, water could open the ring directly. This would
result in an unlabeled product. If the water attacked at the more substituted side of the
epoxide, as shown below, the result would be the enantiomer of the product that was
actually observed by the investigators.
6
CO 2
R
OH
O
6
O
CO 2
117

B:
118
H
*
O
O
*
O
H H
O
Asp
P12.8: (McMurry p. 277 step 9)
O 2C
H
O
H
O
N
H
:B
CO 2
P12.9: (Stryer 6 p. 623)
H 3C
CH 3
N
CH 3
O
R C SCoA
H
O
:B
CO 2
H 3N CO 2
H 3C
P12.10: (McMurry p. 323 bottom)
O
N
O
R
CH 3
CH 3
O
B:
O 2C
O 2C
C SCoA
O
H
H A
CO 2
HO
O
O
O
N
H
O
OH
H A
CO 2
( 18 O label stays on Asp)
+
H 3C
H 3N CO 2
H 2N
CH 3
N
CH 3
CO 2
O
Chapter 12
O
O
H 3N CO 2
O
C R
CO 2
fatty acyl-carnetine

H N
R
NH
H
N
R
O
N
N
:B
H
O
O
P
O
NH 2
O
ADP
R
H N
N
N
R
NH
H
NH
N
H A
:B
OP
H
=
B:
PO
PO
R
N
Chapter 12
H A
H N NH
P12.11: (McMurry p. 308 step 3; crystal structure/mech in J. Biol. Chem. 281, 13762,
scheme 2)
a)
H
b)
H
H
O
H
O
:B
:B
O
N
H
O
N
H
N
N
H
O
H
O
B:
O
N
H
H
N
O
O
b
H A
H
O
N
H
O
H
N
H
O
H A
a a
:B
b
H A
O
CO2 NH2 N
H
N CO 2
H
O
NH 2
NH 2
N
H
CO 2
O
N
R
H
N
:B
NH
119

120
Chapter 12
c) Very often, enzymes activate a carbonyl for nucleophilic attack by coordinating a
metal cation (usually Zn 2+ or Mg 2+ ) to the carbonyl oxygen, thus withdrawing electron
density and increasing the electrophilicity of the carbon.
O
N
H
In this particular case, x-ray crystal structures show that two bound Zn 2+ ions participate
in catalysis and help to direct nucleophilic attack at the proper carbonyl (see J. Biol.
Chem. 2006, 281, 13762, scheme 2 for details).
P12.12: (McMurry p. 296, step 3)
A
H
H
H
O
HN
:B
R
N
N
N
N
R
P12.13: (McMurry p. 318)
B:
H
O
H 2N
R
N
H A
N
N
N
N
H
O
R
O
N
NH 2
N
N
HN
R
R

B:
H
O
N
O
O
O
N
R
O
P
UTP
O ADP
P12.14: (McMurry p. 322)
OP
OP
O
HO OH
O
H
N
HO OH
H
N
O
O
glycinamide ribonucleotide
O
N
H
N
H
H A
N
H
formylglycinamide ribonucleotide
O
OP
N
R
H
+
H
N
H
H
H 2N
H
:B
N
H
H 2N
O
B:
H
N
N N
H
R
PO
H
N
O
O
10-formyl-THF
N
O
N N
THF
P12.15: (McMurry p. 320 step 9; J. Biol. Chem 280, 10881 for archaea)
H
N
H
N
R
N
N
R
NH 2
N
R
CTP
O
NH 2
H
N H
Chapter 12
121

122
Chapter 12
The formate molecule is first converted to an acyl phosphate at the expense of one ATP.
The formate is now activated for an acyl transfer reaction.
N
O
H O
R
O
N
H
O
NH 2
H
:B
O
P
O
O ADP
O
H OP
O
H OP
P12.16: (McMurry p. 323 top)
H
B:
N
H
ribose-P
N
ribose-P
NH
O
O
P
O
O H
H
N H
O
O ADP
N H
O
A H
P12.17: (McMurry p. 320 step 7, p. 326)
B:
H
N
N
R
ribose-P
N
H
O
O
N
H
OP H
NH 2
H
OP
H N H
H H
PO N H
ribose-P
N H
:B
O
N H
O
N
O
R O
NH
NH 2
H

R
N
N
A
P12.18:
O
NH 2
O
H
C C N
H
H O H
C C NH 3
H O
H
H 2O
H
C
H
H
H O
H O
OH
C
R
O2C HN
P12.19: (McMurry BA p. 665)
P12.20:
O
H
H
N
N
H
B
N
N
CO 2
H
H
C C NH
H
H
O H
H O H
C C NH 2
H O
H H
H 2O
H O H H
C C NH
H
H O H
C C NH 2
H
H 2O
Chapter 12
H
H O
H
123

H 3C C O
O
S
O
124
O
Cl
+
H
Cl
H
O
H 3C C
O
O
H 3C C
S
Cl
Cl
Cl
The PCl 3 mechanism:
H 3C C OH
fig 12
O
Cl
P
Cl
Cl
Challenge problems
+ HCl
C12.1: (McMurry p. 191)
O
H
H 3C C O
HO
Cl
H 3C C O
H 3C C O
Cl
O
H
2 more times . . .
O
S
P
Cl
Cl
O
S
Cl
Cl
Cl
OH
P
Cl
H 3C C O
O
H 3C C O
Cl
O
H 3C C O
Cl
Cl
O
+
H
P
O
S
H
Cl
Cl
H
Cl
Cl
O
O
S
H 3C C Cl
Chapter 12
Cl

CoAS
O
O
O
P
O
enz
H
O
P
O
O
O
N
O P
GTP
+
O
O
N
O
O
O
GMP
H A
CoAS
O
O
O P O
A
O
H
enz
N
O
B:
O
O
N P
enz
O
H
O
O
N
O
P
O
O
N
O
O
P
O
O
O
O
O
succinate
GMP
Chapter 12
C12.2: (See J. Biol. Chem 1968, 243, 853 for experimental details). (Silverman p. 69)
a) The cysteine could attack the 14 C atom, with subsequent loss of N 2 gas (this would be a
very entropically favored step - we will see similar reaction types when we study
decarboxylation mechanisms). This process would result in the 14 C label becoming
attached to the active site cysteine.
+
O
O
O
O
125

enz
126
R
S
O
H
*
C
H
N N
:B
enz
R
O
N N
*
C
S H
enz
R
active site cysteine is
radioactively labeled
enz
O
*
C +
S H
R
O
H
*
C
S H
H A
Chapter 12
b) The cysteine could also attack the carbonyl carbon, in what is essentially an acyl transfer
reaction. In this case, the cysteine would not receive the radioactive 14 C label.
enz
S
R
O
H
*
C
H
:B
N N
R
enz
O
S
active site cysteine is not
radioactively labeled
R
*
C
H
O
H A
N N
S
enz
+
R
enz
O
S
N N H
C*
H
N N
H
C*
H
C12.3: See J. Biol. Chem 2000, 275, 40804 (Scheme 1 for the mechanism in part a, Scheme
2 for part b)
N N
C12.4: (See also Biochemistry 2001, 40, 6989, Scheme 2) (McMurry p. 313, 314)
Condensation step:

B:
H 2N O
H
H
O
O
P
O
O
CO 2
N CO 2
aspartate
Cyclization step:
Zn 2+
O
O 2C
Zn 2+
O
N
H
O
N
H
carbanoyl aspartate
H
:B
O 2C
O 2C
O 2C
H 2N
N
H
O
O
Zn 2+ Zn 2+ H A
O
N
H
O
N
H
O
O
P
O
O
O 2C
H
O
N
O
N
H
=
O
N
H
N
O 2C
O 2C
dihydroorotate
N
H
Chapter 12
O
NH 2
carbanoyl aspartate
H
O
(flipped horizontally)
CO 2
127

In-chapter exercises
E13.1:
fig 1
E13.2:
128
a)
OH
d)
OH
Chapter 13
Chapter 13 solutions
OH
b) c)
H 2C C H
There are many possibilities: if there is no α-proton present, there is no enol form
possible. Here are three such compounds:
E13.3:
fig 1
fig 1
O
H H
OP
O
OH
and
OH
OH
OH
OH
O
F 3C CF 3
OH
OH

Chapter 13
E13.4: Propose a likely mechanism for glutamate racemase, showing stereochemistry
throughout.
fig 1
E13.5:
a)
fig 2
b)
fig 2
B:
OP
H
H 3N
OH
OP
OH
O
A H
OH
OH
O
CO 2
O
OH
O OH
H
OH
:B
H 2O
OP
H 3N
OH
OP
OP
H A
OH
OH
A H
OH
O
OH
O
OH
OH
N
CO 2
O
enz
H H
:B
OH
OH OH
O H
H A
OH
:B
H 2O
H 3N
OP
OP
OH
H
O
OP
CO 2
O
OH
H
OH
OH
OH
enz
OH
N
A H
N
OH
H H H
H
enz
OH OH
O
O
OH
OH
129

E13.6:
fig 2
E13.7:
fig 2
E13.8:
fig 3
E13.9:
fig 3
E13.10:
130
PO
N
H
O
H H
B:
H
OH
2
O
A H
O
OP
PO
OH
OH
O
OP
O
OH
O
O
O
CH 2
B:
N
H
CO 2
OP
O 2C CO 2
O O
N
H
OH
O
H
Lys
H
O
:B
OH
OP
OH
OP
O
OH
OH
O
O
O
OP
O
O
O
OH
OH
OH
OP
+
OP
OH
O
O
N
H
Chapter 13
H A
OH
OH
OP

fig 3
E13.11:
a)
b)
A
A
A
H
O
H
O
H
O
A
O
O
H
O
O
O
O
O
Lys
N
O
O
O
CO 2
CO 2
O
CO 2
A
O
HO
A
H :B
Lys
NH
H
+
O
O
C
O
O
CO 2
CO 2
Chapter 13
fig 3
c) The relationship is tautomeric: A and B are keto forms of the actual product. Usually
keto forms are more stable than enols, but in the case of phenols the enol form is aromatic
and is thus much lower in energy.
E13.12:
H
O
H
O
CO 2
A
B
131

O
fig 8
E13.13:
132
O O
O
H H
H 3C C CH 3
O N H
H 2O H
fig 8
H 2O
H 2O
H A
H3C C
O
H H
H 3C C
E 13.14:
O
N
H
O O
O
OCH 2CH 3
O
H
HO N
O O
O H OH 2
H 3C C CH 3
O
H 3C C
H 2O
O O
O
N
O
Br
O O
O
N
O
H 3C C CH 2
O
H
:B
O O
O
N
H 3C C
O
OH
O O
O
N
O
O
H
H 3C C CH 2
O
CH 3
Cl
Chapter 13
H 3C
Cl
OH
C
O

PPh 3
A B
End-of-chapter problems
Ph
Ph
P
Ph
C
O
Chapter 13
P13.1: Propose a mechanism for this early reaction in the biosynthesis of the isoprenoid
building blocks: (McMurry p. 129)
O
H
C SCoA
H H
:B
A H
enz
O
O
S
O
O
O
HO CH 3
H
O
HO CH 3
:B
SCoA
P13.2: (McMurry p. 121)
enz
SCoA
S
O
H A
O
enz
B:
O
H 2C SCoA
S
O
O
HO CH 3
H O H
SCoA
133

fig 4
134
O
O
C
O
N N
S
H
R
P13.3: (McMurry p. 238)
O
O S
O
O
O
fig 4
P13.4: (McMurry p. 239)
fig 4
P13.5:
O
H 3N
CH 3
O
a) McMurry p. 250
O
H SCoA
:B
A H
O
N N
O
S
O
O
H 3N
O
O
S
+
O
H
R
H 2C
A H
CH 3
SCoA
O
O
SCoA
O
O
O
H H
N N
O
S
O
C
O
R
H 3C SCoA
H 3N
H 3N
O
O
O
H 2C SCoA
O
O
H 3C
O
Chapter 13
O
O
H A
O

O
A
H
O
B:
H 2N
O
H
O
O
fig 5
b) J. Bact. 192, 1249 (c also)
B:
H
solnfig 11
c)
CoAS
O
solnfig 11
H 3C
O
H
O
SCoA
CO 2
A H
O 2C
:B
O 2C
O
OH O
O
O
A
O
O
O
H
CoAS CH 2
O
NH 2
O
SCoA
H
CoAS CH 3
acetyl-CoA
O
O
O
O 2C
H3C CO2 pyruvate
A
O 2C
O
OH
A H
O
OH HO
O
SCoA
SCoA
Chapter 13
NH 3
H
O
H
O
H
H
A
O
:B
:B
135

P13.6: McMurry p. 258, bottom)
fig 5
136
H
:B
O
SCoA
O
P13.7: (McMurry p. 277 step 3)
fig 5
O 2C
O
H
pyruvate
:B
O 2C
C O
P13.8: (McMurry p. 286 first step)
H 2C SCoA
H
O
:B
HO
O
CO 2
O
O
H 3N
fig 5
P13.9: (McMurry p. 258 fig 5.20 step 3).
O
O
CO 2
SCoA
SCoA
CO 2
H A
O
H A
O
O
O
O 2C
B:
HO
HO
O
CO 2
CO 2
OH
H 3N
H O
O
O O
H
O
SCoA
SCoA
H
Chapter 13
H A
SCoA
O
:B
O

fig 5
O
O
P13.10: (McMurry p. 198)
fig 6
O 2C
O
O P
O
O
SCoA
O GDP
P13.11: (McMurry p. 308 step 3-4)
fig 6
O
N
H
H
NH
H
O
O
H 3N
:B
O
O
O
N
H
O
O
NH
O
O
P13.12: (McMurry p. 291 bottom)
A
H
HSCoA
H :B
B:
H 3N
H
H 2N
O
O
SCoA
OP
+
O 2C CH 2
O
O
O
NH 2
NH
O
H A
O
SCoA
+ GDP
O
Chapter 13
NH
H 2N H A
H 2N
O
H
NH 2
O
H
137

fig 6
138
CO 2
N
H
HO
H
:B
OH
OH
OP
P13.13: (McMurry p. 362 last step)
fig 7
P13.14:
H 3C
O
O
O
O
H :B
N
H A
CH 3
A
H H
H
CO 2
N
HO
H O
H 3C
O
O
O
H
OH
CO 2
N
OP
:B
HO
H O
N
Chapter 13
a) This is a Wittig reaction, followed by a transesterification (J. Org Chem 1985, 50,
3420)
H 3CO
O
O
O
O
CH 3
Ph 3P CH 2
THF, -100 o C
O
K 2CO 3
methanol
H 3CO
CH 2
O
OH
CH 3
OH
OP

Chapter 13
P13:15: Reactions a and b are acetoacetic ester syntheses; c is a malonic ester synthesis
(section 13.6A).
O
O
HO
a b c
P13.16: Reaction a is an acetoacetic ester synthesis, b is a malonic ester synthesis, and c
is a Stork enamine alkylation (section 13.6A)
fig 9
P13.17:
O
O Br
O
A B C D
OH
O
O
N
E F G
a) b) c)
OH
d)
f)
F 3C C CH 3
fig 9
P13.18:
O
A
O
N
H
F 3C
CH 3
O
g)
CH 3
O
e)
O
O
Br
O
O
H 3C N 3
O
Br
139

a)
fig 10
c)
140
O
fig 10
Br
Mg
O
N
H
TsOH
1)
2) H 3O +
MgBr
Br
CO 2
O
H 3O +
HO CH3
C CH3
O
C
H 2C Ph 3P
OH
H 3O +
SOCl 2
CH 2
Chapter 13
O
C Cl
CH 3MgBr
(2 equiv)
O CH3
C CH3

Challenge problems
C13.1: (McMurry p. 288, step iv-v)
fig 7
B:
H
O
O
OH
CO 2
OH
O
O
OH
CO 2
=
OH O
O
O
HO
H A
OH
CO 2
OH
CO 2
OH
OH
Chapter 13
141

In-chapter exercises
E14.1:
E14.2:
a)
142
O
O
N
Nu:
NO 2
F
X
O
H
:B
N
H
R
???
R
O
Chapter 14
Chapter 14 solutions
X
electrons cannot be
stabilized by resonance
with the carbonyl group
N
O
fig 1
b) The second nitro group further stabilizes the negatively-charged intermediate (the
negative charged can be delocalized to either one of the nitro groups).
O
O 2N
NO 2
F
N
H
NO 2
N
H
R
R
O
O

O
N
O
O
N
F
N
H
O
R
O
O
N
O
N
F
N
H
R
Chapter 14
c) The step in which the halogen leaves (the second step) is NOT the rate-determining
step in this reaction – the rds is the initial attack by the nucleophile (which temporarily
disrupts the aromaticity of the ring). Thus the fact that fluorine is a relatively poor
leaving group is inconsequential. A fluorine substituent actually works better than other
halogens because it is more electronegative, making the attached carbon more electronpoor
and thus a better nucleophile, increasing the rate of the rds.
fig 1
E14.3:
B:
Cys
fig 1
H
E14.4:
fig 2
H
S
O
HN
O
N
OH
H
O
N
R
O
P
OH
OH
A H
Cys
S
HN
O
N
OH
OH 2
N
R
O
O
HN
O
HN
O
N
H
O
O
N
N
R
O
B: H H A
H
H
O
O
P
OH
N
R
OH
143

The cyclohexyl phosphate could form if the phosphate attacked the carbocation
intermediate as a nucleophile rather than as a base:
E14.5:
144
H
N
H
O
O
P
OH
OH
H A
O2C CO2 B:
B:
H
H OH B:
H NH2 H
H CO2
N
PO H
PO PO
H
R
OH
OH
H
H A
O
CH 3
B: H A
Lys
H
N
H
N
H
PO
H H
R
OH
H
OP
N
H
N
H
H
N
Enz-PLP
OH
Lys
CH 3
CH 3
CO 2
H
B:
Lys
CO 2
H 2N CO 2
glutamate
H
NH 2
O
R
O
P
OH
H
OP
R
OH
N
H
N
H
H H
N
CH 3
R
OH
CH 3
CO 2
H
H
H
OP
N
H
N
H
Chapter 14
H
H
N
N
R
OH
CH 3
R
OH
CH 3
CO 2
H A
CO 2

fig 2
E14.6: McMurry p. 280 step 3
The elimination phase of the γ-substitution is:
H
O
H 3N CO 2
O
OP
CO 2
Schiff base
formation
N
H
N
OH
R 1
CH 3
CO 2
OP
N
H
O R 1
The addition phase completes the substitution:
OP
N
H
N
O
H
B:
CH 3
CO 2
H
S
CO 2
NH 3
O
R 2
OP
B:
N
H
N
H
N
OH
CH 3
O
CH 3
O R 1
CO 2
CO 2
O
OP
N
H
N
OP
H
O
CH 3
N
H
CO 2
B:
N
H
H
O
O R 1
S R2 A H
A H
H
OP
O
N
H
H
Chapter 14
CH 3
N
O R 1
CO 2
OH
CH 3
O
S R 2
CO 2
145

fig 3
fig 3
E14.7:
H 3C
146
R
H 3C
R
H 3C
R
fig 4
E14.8:
OP
R
N
S
R
N
S
R
N
S
H
N
H
enz
N
A H
H 3C
O
CH 3
OH
CH 3
H
O
O
+
O
:B
H CH 3
H
S
H 3N CO 2
O
cystathionine
H 3C
R
R
N
S
CO 2
H 3C
R
NH 3
R
N
S
H A
OH
CH 3
C O
The mechanism for the hydrolysis reaction can be depicted:
O
OH
CH3 O
OP
O
H 3C
R
N
H
N
R
N
S
H
Chapter 14
OH
CH 3
S R 2
CO 2
OH
CH3 O
O

S
R
C
S
R
H H
O
H
HS SH +
S
R
C
S
R
O
H H
O
H
R R
H
H
O
H
HS
HS
H
H
S O
H
C
R R
S
R
C
H
O
R
Chapter 14
O H
H
O H
The methyl iodide helps to prevent the reverse reaction (thioacetal formation) by serving
as an electrophilic 'trap' for the thiol groups:
I
H
C
H
H
End-of-chapter problems
HS SH
H
H
P14.1: (See Biochemistry 1997, 36, 12526) (McMurry p. 177)
Lys
O
NH 2
O
H
OP
OH
=
Mg 2+
Lys
O
NH 3
O
H
C
OP
I
OH
H
O
Glu
O
S S
CH 3
CH 3
H
147

148
Lys
NH 3
O
O
O
H H
OP
O
Glu
O
Chapter 14
P14.2: This reaction is an elimination followed by enamine to imine tautomerization
followed by hydrolysis of the imine. (McMurry p. 239, fig 5.8 bottom path)
H 3N H
OH
O
:B
O OH
A H
O
O
O
fig 5
P14.3: (McMurry p. 252 top)
fig 5
O 2C
H 3N CO 2
NH 3
O
H 3N
H
O
A H
O
:B
O
NH 4
O
O 2C
H
A H
A H
H
O
H H
CO 2
aspartate fumarate
P14.4: A likely mechanism for cysteine 'tagging' is by a Michael addition.Michael addition)
N
H
O
NH 2
:B
O
O
:B
O

fig 6
Cys
S
H
:B
O
N
O
Cys
A H
P14.5: This is a dehydration followed by tautomerization. McMurry p. 285 bottom
H 3C OH
H 3C O
HO
H
O
:B
fig 6
P14.6: (McMurry p. 260 3rd step)
O
H H
H 2C
:B
O
O OH
A
P14.7: (McMurry p. 286)
fig 6
CO 2
H 3C OH
S
H 3C O
HO
A
CO 2
H A
OH
H
O
O
O
O OH
A
N
O
CO 2
intermediate product
OH
A
CO 2
Cys
CH 3
CH 3
S
O
H 3C O
O
O
Chapter 14
O
H 3C O
H
HO
O H
O
N
O
O OH
:B
A
149

150
Chapter 14
P14.8: This is likely an E1 mechanism, with a resonance-stabilized carbocation
intermediate. Notice that the more common E1cb mechanism is not possible, as there is
no adjacent carbonyl group. (McMurry p. 291)
O
CO 2
CO 2 CO2
NH 3
H
CH 2
H A
O
CO 2
NH 3
H
fig 6
P14.9: This is an E1cb elimination - AMP is the leaving group. McMurry p. 326)
R
N
N
O 2C
HN CO 2
N
N
R
N
N
CH 3
N
AMP
:B
NH 2
N
+
O 2C
CO 2
NH 3
CO 2
fumarate
fig 7
P14.10: All of the following reactions are PLP-dependent. Draw mechanisms, showing
the ‘electron sink’ action of PLP. In each case, begin the substrate-PLP adduct, and end
with the product-PLP adduct (in other words, you do not need to show the Schiff base
being formed and later hydrolyzed)
a) This is a retro-aldol reaction. (McMurry p. 239 fig 5.8 middle path)

N
H
enz
N
PLP-enzyme
OH
CH 3
enz
H3N CO2 N glycine
OP OP
N
H
OP
PLP-enzyme
OH
CH 3
+
H 3C
OH
H 3N CO 2
threonine
N
H
N
OH
CH 3
CO 2
OP
OP
A H
N
H
H 3C O
N
H
N
OH
CH 3
PLP-threonine
N
OH
CH 3
CO 2
H
CO 2
Chapter 14
:B
H 3C O
H
acetaldehyde
fig 8
b) This is something we haven't seen before - a PLP-dependent retro-Claisen. Notice that
a carbon-carbon bond is broken. (McMurry p. 248)
HO
NH 2
O
R
NH 3
O
O
OP
:B
O2C H
R
N
H
N
OH
PLP-substrate
O
CH 3
OP
B:
O 2C R
N
H
N
H
OH
O
O
CH 3
H
151

152
OP
fig 8-9
OP
O 2C
N
H
O 2C CH 3
N
N
H
PLP-alanine
N
CH 3
OH
CH 3
c) (McMurry p. 283)
PO
N
H A
OH
CH 3
OP
O 2C
N
H
OP
N
CH 2
H
O
H A
CH 3
N
H
enz
N
PLP-enzyme
O
O R
OH
CH 3
+
OP
O 2C
H
O
O2C R
N
H
NH 3
alanine
OP
N CO2 OP
N CO2 H
OP
OH
O
H
PLP-substrate
H
CH 3
:B
N
H
H
PO H
CH 3
:B
N
CH 3
PO
N
H
O
H
CH 3
N
H
O
Chapter 14
:B
CH 3
O
CO 2

fig 9
P14.11:
OP
OP
O O
H
fig 10
H
A H
O
N
H
N
H
N
N
OCH 3
OH
CH 3
OH
CH 3
OH
CO 2
OH
CO 2
OP
O O
B:
N
H
O
N
H O
H H
O
OH
CH 3
CO 2
A H
OP
N
H
N
OH
CH 3
Chapter 14
CO 2
O O
O O
O O
O O
H OH 2
153

P14.12: McMurry p. 284)
H 3C
154
H 3C
R
H 3C
R
R
(
fig 10
P14.13:
R
N
S
R
N
S
R
N
A H
S H3C O
+
H 3C
H
H 3C
O
CO2 OH
:B
O
O
O
HO CO 2
H 3C
R
A
H
R
N
S
H 3C
R
O
O
R
N
S
OH
CH 3
O
OH
CH3 O
CO 2
Br Br OH
O
H 3C
R
R
N
S
Chapter 14
OH
CH3 O
O

fig 12
Chapter 14
P14.14: Notice that in order for the (E)-alkene to form, the C-H and C-Br bonds must be
anti and co-planar.
B:
H
H
H3C CH 3
H
Br
H
H
H3C :B δ-
CH 3
H
δ
Br
-
B H
H 3C H
fig 12
P14.15: Recall that for an E2 reaction to occur on a cyclohexane ring, the leaving group
and the abstracted hydrogen must both be in axial positions.
Ph
Ph
OTs
three large groups axial - very unfavorable!
P14.16:
a)
b)
I
+
O
O
O CF 3
dimethyl sulfoxide
H 3C
CH 3
C O
CH 3
methanol
H
Br
If bromine leaving group is axial,
there are no axial hydrogens on
neighboring carbons
E2
O
S N2
CH 3
Br
155

c)
d)
e)
P14.17:
H
156
O
OH
fig 14
P14.18:
H 3C
Br
OH
OTs
I
OH
O
SH
acetone
O
H
H
O
OH
F
OH
methanol
methanol
65 o C
S N2
O
O
SH
HO H
O
OCH 3
S N1
O
CH 3
E1
HO H
O
F
+ enantiomer
O
O
H
O
OH
Chapter 14

a) b)
fig 14
P14.19:
a)
b)
O
O
fig 14
P14.20:
O
+
O O
O
+
O
O
O
CH 3
a) b) c)
Challenge problems
C14.1: (McMurry p. 317)
O
O
O
O
O
O
O
N(CH 3) 2
+ C 2H 4
Chapter 14
a) It is a decarboxylation reaction, but there is no obvious way for the electron pair to be
stabilized.
fig 11
O
HN
O
N
R
???
O
O
157

) A carbene intermediate would imply a very hydrophobic active site pocket – see
Chemical and Engineering News, May 12, 1997, p. 12; Science 1997, 276, 942.
c) See Chemical and Engineering News, March 13, 2000, p. 42.
C14.2: See Biochemistry 2003, 42, 248 (scheme 1) McMurry p. 330)
158
Chapter 14
C14.3: The first part is cyclic imine formation, then we see an E1cb dehydration adjacent
to the imine. McMurry p. 277, steps 4,5)
O 2C
B:
(fig 11
C14.4:
H A
O OH
H
N
H
CO 2
A
H
HO
O2C N CO2 OH
O2C N CO2 H
O 2C
H
H
H
:B
OH
N
H A
OH
O
O2C N CO2 B:
H
O
H A
a) Lacking any other information, it is reasonable to expect a mechanism that parallels
that of asparagine synthetase (section12.2B): conversion to an acyl phosphate, then acyl
transfer with ammonia nucleophile. (McMurry p. 320 step 7-8)
b) The nitrogen comes from the amino group on aspartate, and is freed by the final
elimination step.

ibose-P
fig 11
O
O
NH 2
ATP ADP
O
ribose-P
OP
NH 2
aspartate
P i
E1cb
ribose-P
O
ribose-P
O
O 2C
NH
NH 2
N
H
NH 2
Chapter 14
CO 2
CO 2
CO 2
159

In-chapter exercises
E15.1:
fig 1
E15.2:
fig 1
E15.3:
a)
b)
fig 1
160
δ +
δ
E
+
H Br :Br
Chapter 15
Chapter 15 solutions
Br
H 3C OH
δ
Nu
-
δ +
E
Br

E15.4:
Chapter 15
a) The regiochemistry of both electrophilic addition steps is governed by the relative
stabilities of the carbocation intermediates. Intermediate 1 is stabilized by
hyperconjugation with the neighboring methyl group, while intermediate 2 is stabilized
by resonance with the bromine.
b)
H3C C C H
H Br
H3C C C H
Br
H OH 2
H 3C
Br
H 2O
:Br
H
H3C C C
slowest
H
step intermediate 1
C CH 3
CH 3
H3C C C H
C CH 3
O
H
H 2O
:Br
H Br
Br H
C C
H3C H
Br
H
Br
H
C C H C C
H3C H
H3C H
intermediate 2
H 2O
H3C C C H
O H
H H
H3C C C H
H
O
H
H
H OH 2
c) Alkynes are less reactive towards electrophilic addition relative to alkenes. This is
because the vinylic carbocation intermediate involved is very high energy. In the case of
HBr addition, for example, once the monobrominated alkene forms, the second HBr will
add very rapidly.
E15.5:
H
161

H OH 2
fig 1
E15.6:
Br
fig 1
E15.7:
162
O
OH
O H OH 2
R R
O
O
H 2SO 4
Br
O
O
HO
O
R R
O
OH
Mg O
MgBr
O
H 2O
O
H
H 3O +
OH
O
Chapter 15
OH
R R
This is an intramolecular reaction – thus there is inherently a much lower entropy
component to the energy barrier. In other words, the electrophilic atom is held close to
the aromatic π bonds, which makes initiation of the electrophilic attack easier.
End-of-chapter problems
P15.1: (McMurry p. 388)
OH
OH

fig 2
R 1
S
CH 3
R 2
R
R
H 3C
N
N
H R
H
P15.2: (McMurry p. 383 step 4)
fig 2
P15.3:
O 2C
B:
R1 N
N R2 HO
O 2C
H
CO 2
N
H
H
A
CO 2
H
CO 2
R
H
CO 2
R
O 2C
R
R
R
O 2C
O 2C
H 3C
H
N
R
R
Chapter 15
R
R1 N
N R2 O 2C
N
H
B:
H
+
CO 2
O
H H
CO 2
N
H
H
CO 2
R
CO 2
163

O
164
mechanism A:
O
P
18 O
O 2C
O
mechanism B:
O
O
P
18 O
O 2C
O
O 2C
O
O
A H
O
A H
OH
OH
OH
OH
OH
OH
OH
OH
OH
OH
OP
OP
OP
B:
H O
B:
H
O
18 O
O
H O
O
18 O
P O
O
18 O
CO 2
CO 2
P O
H
CO 2
18 O
OH
OH
OH
OH
P O
OH
OH
OH
CO 2
fig 3
c) Mechanism B: See Biochem. Biophys. Res. Commun. 1988, 157, 816.
P15.4: (JBC 264, 2075) this exp not described
a)
B:
O
H
O
O
OH
OH
OH
OH
OH
OH
OH
OP
OH
OP
OP
OH
Chapter 15
OP

PPO
OPP
OPP
Chapter 15
b) If the labeled substrate shown below were to undergo a concerted reaction, the label
would necessarily be found on the β (outside) phosphate group of the product. If, on the
other hand, the label were actually to be observed on the α (inside) phosphate of the
product, this would rule out a concerted mechanism.
fig 4
O
O
O P O
O
P
O
18
O R
P15.5: (Arch Biochem Biophys 2003, 417, 203)
OPP
fig 4
P15.6: Fig 1, Microbiol. 151, 2199
a)
O
O
O H
H
O
P
O
O
P
18
O O no label here
:B
R
O
H A
OH
O H
:B
165

)
166
PPO
N
H
R
fig 5
P15.7: JBC 264, 2075
R
N
H
a) (+)-bornyl diphosphate
HN
O
O
N
H
N
H
R
H
:B
N
H
N
H
HN
R
H
O
O
N
H
Chapter 15
N
H
:B
R

) (+)-sabinene.
OPP
OPP
H
OPP
(+)-bornyl diphosphate
(+)-sabinene
OPP
Chapter 15
fig 6
c) This is an anti-Markovnikov addition, because the secondary carbocation, rather than
the tertiary carbocation, forms during the addition.
P15.8: Essentially, the cysteine is 'tricked' into acting as a nucleophile rather than as a
base. (See J. Am. Chem. Soc. 2005, 127, 17433 for a complete description of the
experiments referred to here).
H
:B
167

168
PPO
P15.9:
fig 7
P15.10:
A
H 3C
H
C C H
H OH 2
PPO
PPO
H 2O
H 3C
C C H
H 3C
H
O
S
H
enz
C CH 3
:B
B:
H 3C
H 3C
PPO
H H
O
C C H
H
O H
H 2O
C CH 3
Chapter 15
S
H OH 2
The NMR data shows that the main product is from anti-Markovnikov addition of HBr.
This regiochemistry is due to the electron-withdrawing effect of the carbonyl group,
which makes the primary carbocation intermediate more stable than the secondary
carbocation.
O O O H C C C
3.5 ppm (t)
Br
H
H
H
O
CH 3
3.0 ppm (t)
2.2 ppm (s)
If the reaction were to proceed with Markovnikov regiochemistry, the NMR spectrum of
the product would look very different:
O O O
Br
d, 3H
q, 1H
H 3C
H
O
C C
Br
CH 3
s, 3H

(see J. Chem. Educ. 1990, 67, 518 for more details on this experiment).
P15.11: (Silverman p. 277 bottom, similar to McMurry p. 289)
fig 7
O2C C
PO
H 2C
HO
C
CH 2
O
CO 2
O
H A
OH
HN
P15.12: (Silverman p. 282)
O
O
CH 3
O2C C
PO
ribose-U
CH 3
B:
H
H 2C
PO
HO
C
B:
O
CO 2
HO
H
O
O
OH
HN
O
OH
HN
O
O
CH 3
O
O
CH 3
ribose-U
ribose-U
Chapter 15
We would expect (and in fact we observe) alkylation to occur ortho to the phenol group,
because the associated carbocation intermediate can be stabilized by resonance with the
phenol oxygen.
fig 8
O
OH
O
O
R
OH
intermediate
O
product
O
R
OH
O
R
169

170
Chapter 15
P15.13: This is a methylation reaction, so it is reasonable to expect (and in fact it is the
case) that SAM is the required cofactor. (McMurry p. 388).
R 1
SAM
O 2C
S R 2
CH 3
R 2
CO 2
NH
R 1
B:
fig 8
P15.14: (J. Biol. Chem. 279, 39389)
fig 9
OP
HO
O
OPP
OH
OP
HO
O 2C
H 3C
H
R 2
CO 2
NH
resonance-stabilized
cationic intermediate
O
OH
NH 2
CO 2
R 1
O
O 2C
H 3C
NH 2
O
OH
P15.15: See Science 1997, 277, 1815 for a detailed discussion of these reactions.
McMurry p. 144
a)
OP
HO
OP
HO
O
OH
R 2
O
NH 2
CO 2
NH
R 1

OPP
common intermediate
methyl shift
H
hydride shift
:B
H
epi-arisolochene
A H
B:
H
Chapter 15
fig 9
b) The mechanism is same as in part a) up to the point after the hydride shift and before
the methyl shift.
common intermediate
H
:B
vetispiradiene
fig 10
P15.16: This is an acyloin rearrangement (section 15.7C). (McMurry p. 137)
171

172
A H
H 3C
B:
O
2
H
3 4
O
OH
OP
HO
H 3C
OH
OP
4 =
2 3
O
H
O
OH
2
3 4
H 3C OH
Chapter 15
fig 10
P15.17: A tyrosine in that position is able to play an additional role in stabilizing the
carbocation intermediate, through a cation-π interaction.
P15.18:
a)
fig 11
b)
H Cl
Br
OTs
OH
c) H
H
H
Cl
H 2O
H
H
O
O H
H
TsO
Cl
H
OP

d)
fig 12
15.19:
a)
H Cl
O
HO
OH (+ enantiomer)
(4 diastereomers)
c) d)
e)
g)
i)
Cl
j)
O
(major)
O
Br
O
NO 2
CH 3
Br
+
HNO 3
H 2SO 4
+ enantiomer
O
O
NO 2
O
Cl
O
CH 3
NO 2
h)
Cl
b)
NaOH(aq)
f)
NO 2
OCH 3
NO 2
H 3O +
Cl
O
Br Br
OH
Cl
OtBDMS
NO 2
Chapter 15
173

k)
P15.20:
a)
b)
c)
P15.21:
a)
174
O
O CH 3
O
Br 2
FeBr 3
Br 2
FeBr 3
SO 3
H 2SO 4
B 2H 6
O
O
O CH 3
SO 3H
Mg CO 2
O
H Cl
AlCl 3
(dry ice)
H 2O 2, NaOH
H 2O
+ enantiomer
c) d)
COOH
COOH
NaOH(aq)
NaH
SOCl 2
O
HOCH 2CH 2OH, H + Mg
b)
CH 3
Ph
H 3O +
O
O
N H
OH
SO 3H
H 3O +
O
H 3O +
O
O
N
Chapter 15
O
OH
OH

P15.22:
a)
b)
c)
d)
P15.23:
a)
b)
c)
O
O
+
O 2N
+
+
+
CH 3
O
O
N
CH 3
OCH 3
CH 3
Br
O
O
CH 3
NO 2
CH 3
OCH 3
O
Br
N
O
CH 3
Chapter 15
175

d) Note that the initial product of Claisen rearrangement has lost aromaticity - but
aromaticity is restored by a subsequent keto-enol tautomerization step.
176
O O
Challenge problems
H
C15.1: get NMR data from Ted, include J if available. provide ref
OH
Chapter 15
Addition can take place with either Markovnikov or anti-Markovnikov regiochemistry:
H 2C
CH 3
H
O
Br
OCH 3
H 3C
H 2C
O
CH 3
H 3C
H
OCH 3
O
OCH 3
H 3C
H 3C
O
Br
H 3C
OCH 3
Markovnikov product
Br
H 2C
H
O
OCH 3
anti-Markovnikov product
NMR data shows that it is the anti-Markovnikov product that forms, due to the electronwithdrawing
effect of the ester carbonyl. (The 1 H spectrum of the Markovnikov product
would be expected to contain just two singlet signals.)
Peak assignments are given below. Notice that the the H R and H S protons are
diastereotopic (there is a stereocenter in the molecule) and have different chemical shifts.
These signals show dd splitting because H R and H S are coupled to each other, and also to
the proton at 2.3 ppm.

fig 12
3.5 ppm
3.6 ppm
H S
H R
1.3 ppm
Br
C
H 3C
H
O
OCH 3
2.3 ppm
C15.2: See Biochem. Biophys. Acta 1998, 1384, 387.
C15.3:
a) See Nature 2003, 422, 185.
b) See J. Am. Chem. Soc., 127 3577.
C15.4: McMurry p. 294
HO
O
CO 2
chorismate
CO 2
CH 2
fig 10
C15.5: (Silverman p. 285)
A H
O
prephenate
O
HO CO 2
O
3.7 ppm
Chapter 15
O
phenylpyruvate
CO 2
177

A H
178
NH 3
R 1 R 2 R 1 R 2
N
H
NH 3
R 1
N
H
R 2
fig 10
C15.6: (McMurry p. 285)
N
H
NH 3
NH 3
R 1 R 2 R 1
N
H
N
H
B:
H
N
H
N
H
R 2
R 1 R 2 R 1 R 2
The condensation of two pyruvate molecules requires the participation of thiamine
diphosphate coenzyme:
H 3C
R 2
H 3C
R 2
N
S
R 1
N
S
R 1
A H
H 3C
+
O
H 3C
O
O
O
CO 2
H 3C OH
H 3C
R 2
The second step is an acyloin rearangement:
A H
H 3C
O
CO 2
H3C O H
fig 11
C15.7: See J. Org. Chem. 2003, 68, 5433.
B:
N
S
H 3C
R 2
HO
H 3C
R 1
CH 3
CH 3
OH
O
R 1
O
H 3C
R 2
:B
N CH3 O
H
S CO2 H3C OH
O
CO 2
=
OH
O
A H
N
S
CO 2
R 1
Chapter 15
CH 3
OH
H 3C CO 2
O

In-chapter exercises
Chapter 16
Chapter 16 solutions
E16.1: An aldol condensation is not a redox reaction. There is no transfer of electrons
from one species to another, either in the form of a hydrogenation/dehydrogenation step
or in terms of an insertion of an oxygen atom (or other electronegative atom).
E16.2: Epoxidation reactions are redox transformations: both carbon atoms lose a bond
to carbon and gain a bond to oxygen. The cyclase reaction, however, is not redox.
E16.3: This is a transformation that is thermodynamically favorable, but the kinetics are
very slow. The activation energy, in other words, is too high for the transformation to
occur uncatalyzed at room temperature at a rate that is noticeable.
E16.4: The anomeric carbon (in other words, the carbon that is the center for the
hemiacetal) has two bonds to oxygen – thus it is in the carbonyl oxidation state. This
becomes clear when we look at the straight-chain form of glucose, in which this carbon is
an aldehyde.
E16.5: Saturated fat is in a more highly reduced state, and therefore contains more
electrons to transfer to oxygen and more energy to release when oxidized to CO 2 during
respiration.
E16.6: In grape fermentation, glucose is converted to two equivalents of ethanol and one
of carbon dioxide (this is the bubbling that is observed). When wine turns to vinegar, the
ethanol is being oxidized to acetic acid. Grape juice (full of sugar) contains the most
calories, and vinegar the least.
E16.7:
a) For every molecule of substrate that is oxidized, one molecule of NADP + is reduced to
NADPH. We can determine the concentration of NADPH formed by using its extinction
coefficient, ε = 6290 M -1 cm -1 .
179

180
6290 M -1 cm -1 =
.096
(1 cm)(c)
c = 1.53 x 10 -5 M
Chapter 16
So the amount of substrate oxidized is (1.53 x 10 -5 M)(1 x 10 -3 L) = 1.53 x 10 -8 moles.
b) (100 x 10 -6 M) – (1.53 x 10 -5 M) = 8.47 x 10-5 M = 84.7 µM.
c)
E16.8:
fig 5
E16.9:
a)
(4.0 x 10-13 1.53 x 10
moles enz) (300 sec)
-8 moles substrate reacted
= 128 reactions per second
per enzyme molecule
Br Br
Br Br
S S
Cl
H 2O
Br
Br
HO
HO
Br
OH
H 2O
Br
O
H
+ enantiomer
S
S
Br
Cl
H
+ enantiomer

)
S
S
protein
protein
fig 6
H
S
H A
:B
OH
OH
End-of-chapter problems
P16.1: (McMurry p. 179)
A H
SH
HS
H 3C CO 2
H
N
R
H
O
O
protein
H A
protein
NH 2
S
S
OH
S
H
OH
:B
HO
H 3C
H
N
R
O
O
O
NH 2
HS
HS
protein
protein
Chapter 16
S S
fig 1
b) The hydride adds to the front of pyruvate as drawn in the figure. The is the re face.
fig 1
P16.2:
a) (McMurry p. 250)
O
H 3C CO 2
H
OH
OH
181

B:
fig 1
b)
182
H
H
O NH 3
O H
B:
H
H
O
S CoA
fig 1
P16.3: (McMurry p. 277 steps 1-2)
O
O
H 3N
O
O
O
O
R
O O H
H R
:B
H
N
R
O
NH 2
CO2 H
S
S
CoA
CoA
ATP ADP
fig 1
P16.4: (McMurry p. 277 steps 5-6)
PO
O
H 3N
O
A
CO 2
NAD +
O
O
NADPH
O
Pi
NADP +
Chapter 16
O NH 3
H
O
H
N
R
H
+
O
CO 2
H NAD
O
H 3N
B
O
NH 2
O
O
O

fig 2
P16.5:
O
HN
O
N
H
B:
H
OH
O2C N CO2 H
H
CO 2
H A
:B
P16.6: (McMurry p. 327)
A
H
O2C N CO2 OH
O2C N CO2 H
O
HN
O
N
H
N
N
R
H
FMN
CO 2
O
N
H A
N
A
H
O
H NADP
O2C N CO2 H
H
O2C N CO2 H :B
O
HN
H
N
N
R
O
N
H
+
FMNH 2
Chapter 16
O
N
H
CO 2
N
H
O
183

R
184
N
N
B:
fig 2
P16.7:
O
N
R
N
N
NH
H
H
S
O
N
H
NH
O
enz
R
N
N
R
O
N
N
N
NH
S
A
H
O
H
N
NAD +
NH
enz
O
H :B
R
R
N
N
N
N
O
N
O
N
NH
H
S
NH
S
Chapter 16
O H
a) First an imine (Schiff base) linkage forms, then the imine is reduced to an amine by
NADPH.(McMurry p. 264 step 1)
O 2C CO 2
B:
H A
O
H N CO2
H
lysine
NH 3
A H
Lys
HO N
H
O 2C CO 2
:B
CO 2
O 2C N H
saccharopine
O
OH
NH 3
O
O 2C N
CO 2
enz
H NADP
fig 3
b) The amine formed in the previous reaction (part a) is oxidized to an imine (not the
reverse of the reduction step in part a – it occurs on the other side of the molecule!).
Hydrolysis of the imine results in the products. (McMurry p. 265)
H
:B
enz
H A
H A
Lys

CO 2
NH 3
O 2C N CO 2
B:
CO 2
H
H
NAD +
+
O
NH3
O2C NH2 CO2
glutamate
CO 2
O H
Chapter 16
NH 3
O 2C N CO 2
A
H
CO 2
O 2C N H
fig 3
P16.8: This reaction is analogous to that catalyzed by the pyruvate dehydrogenase
complex (section 16.12B). The additional cofactors are thiamine diphosphate, lipoamide,
and FAD. (McMurry p. 186 step 4)
P16.9: (McMurry p. 257)
CoAS
O
O
SCoA
CO 2
H 2O
OH O H2O HSCoA
CoAS
SCoA
NADH
A
H
NAD +
CoASH
OH
H
O H
C
B
fig 4
P16.10: See Biochemistry 39, 6732 for the original report on this experiment.
a)
O
O
O
O
O
A
O
O
:B
:B
NAD +
NADH
O
NH 3
CO 2
185

186
H H
S S
DsbA
S S
DsbA
:B
S S
DsbB
H H
S S
DsbB
H A
B:
A
H
H S
S S
DsbA
DsbB
SH
Chapter 16
fig 4
b) The bromoalanine side chain prevents formation of the key disulfide bond in DsbB,
and provides an alternative carbon electrophile for the cysteine in DsbA to attack. The
result of this SN2 displacement is a stable sulfide linkage beetween the two proteins, and
isolation of this species provides evidence for the existance of the unstable disulfidelinked
DsbA-DsbB intermediate in the normal reaction.
H H
S S
DsbA
:B
Br
DsbB
SH
proteins linked by
sulfide bond
H
S S
DsbA
DsbB
fig 4
P16.11 This reaction is an electrophilic aromatic substitution, with the flavin peroxide
oxygen as the electrophile. The substitution takes place ortho to the electron-donating,
ortho/para-directing exocyclic amine, and meta to the electron-withdrawing, metadirecting
carbonyl. McMurry p. 247
P16.12: (See also J. Am. Chem. Soc. 2004, 126, 15060; J. Nat. Prod. 2001, 64, 597).
fig 5
P16.13:
H 3C
CH 2
CH3 OH
Br
CH 3
H 3C
CH 2
CH3 OH
Cl:
Br
CH 3
H 3C
SH
CH 2
CH3 OH
Br
CH 3
Cl

fig 5
P16.14:
HO
H 3C COOCH3
O
O
O OH
OH OH
A H
H SG
a) b)
c)
d)
BrMg
P16.15:
a)
O
COOCH 3
O
A
O
O
Br
:B
O
A O
B
OH O
C
O
O
HO
O
O
O OH
stable hemiacetal
HO
O
HO
A H
B
O
O
O
HO OH
OH
O
O
SG
OH OH
OH
+ GSSG
OH
H SG
:B
Chapter 16
187

fig 6
b)
fig 7
c)
188
H 3C
P16.16:
OH
O
O
O
FeCl 3
+
Cl
Na
H 2
Pd/CaCO 3
NaNH 2
O
PCC
(CH 3CH 2) 3N
Zn(Hg)
H 3C
H 3O +
Cl
+_
O
HCl
O
DMSO
Cl
O
Cl
FeCl 3
OH
Pd/CaCO 3
O
OOH
HO OCH 3
O
H 3C Cl
H 2
OH
H 3O +
O
NaOCH 3
Chapter 16

a)
R OH
b)
H 3C S CH 3
O
R OH
Cl
O
Cr
O
Cl
Cl
H
R O Cr
H
O R
H 3C S CH 3
O
O
O
H
R O Cr
H
R O
H
H
H
O R
H 3C S CH 3
Cl
O
O
OH
H
O
R
:B
Chapter 16
189

In-chapter exercises
E17.1:
fig 1a
E17.2:
Cl
Cl
fig 1a
E17.3:
190
H
Cl
Cl
Cl
Chapter 17
Chapter 17 solutions
O O O O
Cl
C C
Cl
C
Cl Cl
H
H Cl
X
Cl
Cl
Cl
Cl
H
Cl
Cl
H
C C
X
H
Cl
H
C C
H
H Cl
Cl
Cl
H
Cl
Cl
Cl
Cl
H
C C
X
H
Cl
Cl
C
Cl Cl
H
C C
H
H
Cl
Cl
Cl
Cl
Cl
etc
Cl
Cl

End-of-chapter problems
P17.1:
To simplify matters, we'll use 'R' to abbreviate the methyl ester group:
H 2C
CH 3
O
O
CH 3
R
=
CH 3
H 2C C R
The first two propagation steps, forming a acrylamide dimer, are shown below:
X
CH 3
The polymer is represented by:
P17.2:
a)
O
O S O O S
O
O
O
R
2
CH 3
=
X R
R
H
etc.
C C
H
CH 3
C
O
OCH 3
O
O O S O
O
n
X
CH 3
R
CH 3
R
CH 3
R
CH 3
Chapter 17
191

192
O
O S O
O
O
2- O4S NH 2
O
O
NH 2
NH 2
2- O4S
O
2- O4S NH 2
O
O
NH 2
n
O
NH 2
polyacrylamide
NH 2
Chapter 17
b) The bis-acylamide molecule has two alkene groups, at either end, that can participate
in radical chain elongation reactions. This allows it to tie two growing polyacrylamide
strands together (ie to form a cross-link):
strand 1
strand 1
O
N
N
H H
O
strand 2
strand 2
Here is a more detailed mechanism. We start with a growing polyacrylamide strand
(strand 1) reacting in a chain propagation reaction with a bis-acrylamide molecule. In the
next, step, the remaining alkene group on bis-acrylamide reacts adds to a second growing
polyacrylamide strand (strand 2).

strand 1
strand 1
strand 1
O
strand 1 contiues to
grow from this point
strand 1
O
strand 1
O
O
NH 2
O
O
NH 2
NH 2
O
NH 2
N
N
H H
O
NH 2
O
NH 2
N
O
N
NH 2
O
O
N
H H
N
H H
O
N
N
H H
O
O
H 2N O
H 2N O
O
H 2N O
O
NH 2
NH 2
strand 2
strand 2
strand 2
strand 2 continues to
grow from this point
Chapter 17
193

P17.3:
194
HO
OH
O
Chapter 17
Notice that the radical form of resveratrol shown above is more stable, compared to a
radical species in which the unpaired electron is located on one of the 'lower' phenoxy
groups. The extra stability is due to resonance - the unpaired electron can be delocalized
over both of the aromatic rings. (Consider the two other alternative radical species
below - in these cases, the unpaired electron cannot be delocalized over both rings! It all
comes back to para vs. meta positioning on the aromatic ring.)
P17.4:
P17.5:
H 3C
CH 3
C
CN
O
N N
OH
CH 3
C
CN
O
OH
OH
HO
a) b)
c) d)
CH 3
HO
CH3 H3C C N N
CN
HO
O
N 2 gas
OH
CH 3
C
CN
S
CH 3
OH
O

Chapter 17
The driving force for homolytic cleavage is the formation of nitrogen gas, which is very
entropically favorable.
P17.6:
a) Ege p. 770
Cl Cl 2 Cl
H
H 3C C
H
H
H Cl H 3C C
H
H 3C C
+ H Cl
b) Butane could form from an alternate chain termination step:
c)
H
H 3C C
H
H
C
H
H
H
Cl
CH 3
Cl
H
H 3C C C
H
H
H
CH 3
195

196
H 3C
Cl
H 3C
H 2C
H
C C
CH 2
H
H
H
CH 3
H 3C
H 3C
H 3C
H
C C
H
H 3C
H
C C
CH 3
H
H
CH 3
H 3C
+ Cl
CH 3
C C
H
H
CH 3
H 3C
H 3C
H
C C
A B C D
C C
H
H
CH 3
H 3C
H 3C
C C
Cl
H
H
CH 3
H 3C
H 3C
C C
H
H
CH 3
H 3C
H 3C
H
C C
Chapter 17
d,e) Radical intermediate B is the most stable (it is tertiary), but the resulting alkyl
chloride is not necessarily the most abundant product. Even though radical A is less
stable than radical B, there are six possible pathways for its formation (the chlorine
radical could abstract six different hydrogens to form radical A), as opposed to only one
pathway for the formation of B. There are two pathways to radical C, and three to D.
P17.7:
a) The peroxide contaminant acts as a radical chain initiator:Vollhardt 5th p. 531
RO OR 2 RO
The regioselectivity arises from the higher stability of the secondary alkyl radical
intermediate (compared to the alternate possibility, which would be a primary radical).
RO
H Br ROH + Br
Cl
Br
H Br
H
H
H
H
+ Br
CH 2
CH 2
Cl
Br

)
RO
H SCH 2CH 3 SCH 3CH 3
P17.8: (McMurry BA p. 146)
O
O
H
Fe
O
H
R 1
R 2
CO 2
O
O
H
S
H SCH 2CH 3
R 1
R 2
R 1
R 2
O
O
S
+
SHCH2CH3 Chapter 17
P17.9: Silverman p. 204, J. of Pharmacol. Exp. Therapeutics Fast Forward 2007, 321,
590
a)
O
O
H
H
R 1
R 2
197
R 1
R 2

)
198
N
N
O
Fe
S
enz
HO H
N
N
N
O
HO H
N
O
H
C
H
H
N
N
P17.10: (xxSilverman p. 222)
O
Fe
S
enz
N
N
N
N
O
Fe
S
enz
N
N
R 1
R 2
Chapter 17
N
N
HO H
N
O
CH 2
N
N
HO H
N
N
N
O
Fe
S
enz
O
Fe
S
enz
OH
Fe
S
enz
OH
N
N
R 1
R 2
N
N
N
N

HO
HO
P17.11:
fig 7
enz
Cu
O
OH
H
N
N
enz
Cu
OH
O
N
R H
enz
enz
Cu
O
O
H H
N
NH 2
H
O
OH
enz
O O
R
A H
H
N
N
R
R
(semiquinone)
enz
Cu
O
O
N
O
NH 2
enz
O OH
N
NH 2
H
O
H 2O
OH
enz
enz
Cu
O
R
H
N
N
R
H
enz
Cu
O
R
OH
O
O
N
Chapter 17
O
NH 2
N
enz
H
O
OH
NH 2
enz
199