organic chemistry with a biological emphasis: solutions to
TRANSCRIPT
University of Minnesota Morris Digital Well University of Minnesota Morris Digital Well
University of Minnesota Morris Digital Well University of Minnesota Morris Digital Well
Chemistry Publications Faculty and Staff Scholarship
7-2019
Organic Chemistry with a Biological Emphasis: Solutions to Organic Chemistry with a Biological Emphasis: Solutions to
Selected End-of-Chapter Problems Selected End-of-Chapter Problems
Timothy Soderberg University of Minnesota, Morris
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Organic Chemistry With a Biological Emphasis
Solutions to selected end-of-chapter problems
Tim Soderberg University of Minnesota, Morris
This work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike 4.0
International License.
https://creativecommons.org/licenses/by-nc-sa/4.0/
P1.1:
a) Formal charges are located as shown.
b) There are 16 hydrogen atoms:
c) The structure contains a nucleotide segment and an amino acid segment:
P1.4:
a)
Reaction A: aldehyde to primary alcohol
Reaction B: Secondary alcohol to ketone; aldehyde to primary alcohol
b) The second structure from the right is an appropriate abbreviation. The part of the
molecule in the box does not change in the reaction, and this can be abbreviated with 'R'.
OOP
O
O
O
O
O
NH3
O
O
N
N
NN
NH2
OH
OOP
O
O
O
O
O
NH3
O
O
N
N
NN
NH2
OH
H H
H
H
H
H
HH
H
H
OOP
O
O
O
O
O
NH3
O
O
N
N
NN
NH2
OH
nucleotide
amino acida-carbon of amino acid
c) The part of the molecule in the box does not change in the reaction, and this can be
abbreviated with 'R'.
P1.5:
a) Threonine contains a secondary alcohol.
b) Glutamine and asparagine contain amides.
c) Cysteine contains a thiol.
d) Methionine contains a sulfide.
e) Tyrosine contains a phenol.
f) The lysine side chain contains a primary ammonium.
g) The glutamate and aspartate side chains contain carboxylates.
h) Proline contains a secondary amine.
H
O
O
O
NH3
HOO
O
NH3
H R
O
OH R
H
O
OH
OH
OH
OH
O
OH
OH
OH OH
H
O
OH
R
OH
O
R
P1.6:
Note that according to VSEPR theory, ozone has bent geometry, azide ion is linear, and
the geometry around the oxygen and carbon atoms of bicarbonate is bent.
P1.8:
P1.10:
N N N
azide ionO
OO
C
O
O OH
ozone bicarbonate ion
H3NNH2
O
OO
N
O
OH
O
F
NH2
F
N
N
H3C
H
CH3
H
NN
NHO
F
F
H3C
N
N
F
NH3
N
O
NN
N
CF3
F
F
F
16 hydrogen atoms
10 hydrogen atoms 23 hydrogen atoms
14 hydrogen atoms
carboxylate
carboxylic acid
cyclopropyl
amide
amide
ketone
secondary
ammonium
tertiary alcohol
H N
O
CH3
CH3
H3C N
O
H
CH3
N
O
H
H
H N
O
H
Chapter 2
P2.1:
a) this is a bond formed by the overlap of an sp3 orbital on one carbon and an sp2
orbital on another carbon.
b) this is a bond formed by the overlap of an sp2 orbital on one carbon and an sp2
orbital on another carbon.
c) this is a bond formed by the overlap of an sp2 orbital on a carbon and an sp2 orbital
on a nitrogen, combined with a bond formed by the overlap of a 2p orbital on a carbon
and a 2p orbital on a nitrogen.
d) This is a bond formed by the overlap of an sp2 orbital on a nitrogen and a 1s orbital
on a hydrogen.
e) this is a bond formed by the overlap of an sp2 orbital on one carbon and an sp3 orbital
on another carbon.
f) this is a bond formed by the overlap of an sp3 orbital on one carbon and an sp3 orbital
on another carbon.
N
OH
CH3
HO
H
pyridoxine
(vitamin B6)
OH
b
a
c
O N NOH
O
H
O
H
H3C CH3
e f
pantothenate
(vitamin B5)
d
P2.2:
a)
b)
Top: the contributor on the right is minor due to separation of charge.
Middle: the contributor on the left is minor due to one carbon not having a complete
octet.
Bottom: The contributors shown are roughly equivalent.
P2.5:
a) This is a bond formed by the overlap of an sp2 orbital on one carbon and an sp3
orbital on another carbon.
b) This is a bond formed by the overlap of an sp2 orbital on a carbon and an sp2 orbital
on an oxygen, combined with a bond formed by the overlap of a 2p orbital on a carbon
and a 2p orbital on an oxygen.
c) This is a bond formed by the overlap of an sp3 orbital on a carbon and an sp3 orbital
on another carbon.
NCH3H
OCH3O
NCH3H
OCH3O
N
HN
H
OOOO
OO
d) This is a bond formed by the overlap of an sp3 orbital on a carbon and an sp3 orbital
on an oxygen.
e) This is a bond formed by the overlap of an sp3 orbital on a carbon and an sp3 orbital
on an oxygen.
f) This is a bond formed by the overlap of an sp2 orbital on a carbon and an sp3 orbital
on a nitrogen.
g) This is a bond formed by the overlap of an sp3 orbital on a carbon and an sp2 orbital
on a nitrogen.
h) This is a bond formed by the overlap of an sp3 orbital on a carbon and an sp3 orbital
on a nitrogen.
i) This is a bond formed by the overlap of an sp2 orbital on one carbon and an sp3
orbital on another carbon.
P2.6:
a) Csp3 – Osp3 b) Csp2 – Csp3 c) Csp2 – Nsp2
d) Csp2 – Csp2 e) Csp3 – Csp3 f) Csp2 – Csp2
g) Csp3- Csp3 h)Csp2 – H1s i) Csp2 – Osp2
j) Csp2 – Cl3p k) Nsp3 – H1s
l) The walking stick compound 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 the walking stick compound is C10H14O2.
P2.7:
shortest
bond e (triple bond)
bond c (double bond)
bond d (single bond between sp2 and sp hybridized carbons)
bond f (single bond between sp and sp3 hybridized carbons)
bond b (single bond between sp2 and sp3 hybridized carbons)
bond a (single bond between two sp3 hybridized carbons)
longest
ab
cd e f
P2.11:
shortest
bond c (double bond)
bond d (single bond between two sp2 hybridized carbons)
bond b (single bond between sp2 and sp3 hybridized carbons)
bond a (single bond between two sp3 hybridized carbons)
longest
P2.12: The amide below is not capable of acting as a hydrogen bond donor (it does not
have any N-H bonds), and thus is expected to be less soluble in water. The other three
amides of the same formula have one or more N-H bonds, and can thus participate in
hydrogen bonding with water as both donor and acceptor.
P2.13:
H N
O
CH3
CH3
N N
Cl
Cl
Cl
N
O
N
Ha)N N
Cl
Cl
Cl
N
O
N
H
b)N N
Cl
Cl
Cl
N
O
N
H N N
Cl
Cl
Cl
N
O
N
H
P2.14:
P2.15:
N
N
N
N
Br Br
NH2
H
H
N
N
N
N
Br Br
NH2
H
H
NN
HO
HN
N
HO
H
R
O
R
O
Ph(OC)5W
SS
Ph(OC)5W
SS
N
N
N
NH2N
H
O
H
H
RN
N
N
NH2N
H
O
H
H
R
O
HO
OOCH3
OH
P2.16:
P2.17:
P2.18:
O
HO
OOCH3
OH
genipin
isolated double
bond
N
NN
O
N
HO
H
isolated double
bond
PAC-1
N
H
N
H
N
H
N
H
N
HN
H
NH2O
O N
O
H
O N
O
H
H2N
O
a) b)
P2.19:
a)
2 and 3 have two fluorines and are more polar than 1, so they have stronger
intermolecular dipole-dipole 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 overall van der Waals interactions.
d)
CH3FCH2F2 HFCH3CHF2
23 14
NH2O
2 3 14
NH2NH2
N
CH3
12 3
OH SH O Na
42 13
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.
P2.20:
a) The compound on the right is more soluble (fewer hydrophobic carbons)
b) The compound on the left is more soluble (ionic phosphate group)
c) The compound on the left is more soluble (fewer hydrophobic carbons)
d) The compound on the left is more soluble (capable of hydrogen bonding)
e) The compound on the right is more soluble (fewer hydrophobic carbons)
P2.22: The lone pair electrons on the peptide nitrogen are conjugated to the carbonyl
bond, and thus are not available to act as hydrogen bond acceptors.
P2.23: Both bonds are the same length, and have a bond order of 1.5 (one part single
bond, one part double bond). The central oxygen is sp2 hybridized.
P2.26: The five-membered ring is not part of the aromatic system, due to the presence of
an sp2 hybridized carbon in the ring.
P2.27:
A is not aromatic (sp3 hybridized carbon in the ring)
B is aromatic (count the lone pair and you get 10 electrons, which is a Huckel number)
C is not aromatic (the 2p orbital on the carbocation is empty, thus there are only four
electrons in the system, which is not a Huckel number)
D is not aromatic (four electrons, not a Huckel number)
E is not aromatic (sp3 hybridized carbon in the ring)
F is not aromatic (sp3 hybridized carbon in the ring)
N
O
HR
R
N
O
HR
R
N
O
HR
R
OO
O OO
O
G is not aromatic (lone pair electrons count as part of system, thus there are four
electrons which is not a Huckel number.
H is aromatic (carbocation is sp2 hybridized, the 2p orbital is empty, so there are two
electrons in the system, and 2 is a Huckel number)
I is not aromatic (there are three conjugated p bonds with six p electrons in the system,
but the compound is not cyclic).
P2.28:
P2.29:
P2.30:
N
N
N
N
N
OH
H
H
N
N
N
N
H
NH2
O OOO
OH
OH
HO
OH
O
N
NO
O
H
H
CH3
N
N
N
N
H
O
H
NH2
NH2
Chapter 3
P3.7:
a) Stereocenters are marked with a bold dot.
fig 6
b) The two fluorinated derivatives of Epivar are enantiomers.
P3.9:
a) There is only one stereocenter, and it is in the R configuration.
b)
NN
OH
H
O
O
H
N
N NH2N
H
O
H
N
O
O2C
O2C
H
Alimta
(chiral - 2 stereoisomers)SAHA
(achiral)
O
O
O N
O
H OH
NS
NH2
O O
Darunavir
chiral - 32 stereoisomers)
S
O
HON
N
NH2
O
Epivir
(chiral - 4 stereoisomers)
N
O
H
N
CH3
R
not a stereocenter!
N
O
H
N
CH3
S
enantiomer of the drug
fig 7
P3.10:
fig 7
P3.11:
fig 8
P3.12: The two are diastereomers: two of the four stereocenters are different.
N
OHHO
HO
OH
S
R
R
SHN
NH
N
S
S
O
O
O
O
Br
S S
HO CH3
a)
Cl
b)
O
O
HOOH
OH
OH
c)
OH
O
Cl
d)
Cl
e)
P3.16:
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
P3.17:
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.
P3.18:
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 212 possible stereoisomers.
P 3.19:
a)
N
NHO
H
O
N
O
H
OH
H
OH
H
H
H
OH
H
H
OH
H
gauche anti
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.
P3.21:
These two structures are actually enantiomers – they are non-superimposable mirror
images of each other. Notice that, even though there is no sp3-hybridized carbon, the
overall geometry of the allene structure creates a multi-atom stereocenter.
H
O
H
O
H
H
H H
C C CF
H
F
HCCC
F
H
F
H
enantiomers of 1,3-difluoroallene
Chapter 4
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 m = 1.02 x 1014 Hz
1690 cm-1 = 5.92 m = 5.07 x 1013 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 IR-
active, thus the alkyne peak will be observed. In compound III, we should see not only
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-1range. 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 A340 is A = 0.220. Using the expression = 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 bonds which absorb in the UV
range. The condensation reaction brings these two conjugated systems together to create
a single, longer conjugated system, which absorbs in the blue part of the visible
spectrum.
P4.9:
Both molecules contain alkene and ketone functional groups, however the degree of
bond conjugation is different. Therefore, UV would be the more useful technique to
distinguish the two.
P4.10:
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 C4 vs C5 position) will result in the
formation of fragments of different masses in an MS experiment.
# moles = MV = (4.09 x 10-5M)(0.725 x 10-3L)= 2.96 x 10-8 = 29.6 nmol
Chapter 5
P5.2:
Amino acid # 13C 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
* Valine and leucine both have diastereotopic (and thus, technically nonequivalent) methyl groups.
P5.3:
Spectrum 1: structure D
Spectrum 2: structure F
Spectrum 3: structure C
Spectrum 4: structure B
Spectrum 5: structure A
Spectrum 6: structure E
P5.4:
Spectrum 31: structure HH
Spectrum 32: structure KK
Spectrum 33: structure LL
Spectrum 34: structure GG
Spectrum 35: structure JJ
Spectrum 36: structure II
P5.5:
Spectrum 13: structure M
Spectrum 14: structure O
Spectrum 15: structure P
Spectrum 16: structure N
Spectrum 17: structure R
Spectrum 18: structure Q)
P5.6:
Spectrum 19: structure X
Spectrum 20: structure W
Spectrum 21: structure T
Spectrum 22: structure V
Spectrum 23: structure S
Spectrum 24: structure U
P5.7:
Spectrum 25: structure FF
Spectrum 26: structure BB)
Spectrum 27: structure AA)
Spectrum 28: structure CC)
Spectrum 29: structure EE)
Spectrum 30: structure DD)
P5.8: 13C-NMR data is given for the molecules shown below. Complete the peak
assignment column of each NMR data table.
a)
(ppm) carbon #(s)
161.12 1
65.54 2
21.98 3
10.31 4
b)
(ppm) carbon #(s)
194.72 4
149.10 3
146.33 2
16.93 5
14.47 1
12.93 6
O
O
1
2
3
4
O
1
2 34
5
6
c)
(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
d)
(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
O O
O O
1
2
3 4 5
6
78 9
10 11
O
O
H3C
OH
1
23
4
5
6
7
8
9
e)
(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
P5.9:
a ) 1-734A
b) 1-905B
c) 1-170B
d) 2-789B
N1
2
3
4
5
6
7
89
10
O
O
O
OH
O
P5.10: The structure is 3-methyl-2-butanone.
P5.11: 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 and figure the percentage.
1-bromopropane: 0.661 + 0.665 + 1.00 = 2.326.
2-bromopropane: 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.
CC
C
Br
Ha
Ha
Hb Hb
Hc
HcHc
CC
C
Hd
Hd
Hd
He Br
Hd
HdHd
Hc: 1.0 ppm (t)
integration:1.00
3 protons= 0.33
Hd: 1.7 ppm (d)
integration:0.441
6 protons= 0.0735
percent 2-bromopropane =0.0735
0.0735 + 0.33x 100 = 18.2 %
Chapter 6
P6.1:
a)
b)
P6.2:
a) The overall reaction (A to D) is 'downhill' (exergonic), thus we know that Keq > 1.
b) C to D is the rate-determining step (highest energy barrier).
c) B to C is faster (lower energy barrier)
d) C to D is thermodynamically 'downhill' (exergonic), while A to B is 'uphill'
(endergonic), so C to D is more thermodynamically favorable.
Cl
CH3O
E
LG
O
O
Br
Nu
E
LG
P6.3:
a)
b)
P6.4:
a)
NH3O
HO OH
OP
O
O OO
HO OH
OP
O
O O
NH3
CH3
NH3C
CH3
CO2
O
RC
SCoA
O
H O
O
enz
CH3
NH3C
CH3
CO2
O
C
RO
SCoA
HO
O
enz
N
NO
NH2
R
O
H
H N
NO
H2N
R
H
OHH
O
O
enz
HO
O
enz
Chapter 7
P7.2:
a) The base on the right is stronger: amine vs aniline.
b) The base on the right is stronger: consider conjugate acids: pKa of alcohol is ~16, pKa
of ammonium is ~10.
c) The base on the left is stronger: ketone in para position is electron withdrawing by
resonance.
e) The base on the left is stronger: positively-charged quaternary amine destabilizes
positively charged conjugate acid.
h) The base on the right is stronger: amine vs 'pyrrole-like' nitrogen
ig 3
P7.3:
fig 5
NN
O
H
F
OH
OOH OH
Lipitor
H3C
OH
CH3
O
HO O
O
Zocor
a)
(pKa ~ 4.5)
b)
c)
(pKa ~ 16)
HH
fig 5
P7.4: 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
P7.5: The alkyl amine nitrogen is most basic, the ‘pyrrole-like’ nitrogen is least basic.
fig 5
P7.6: Of the four protons, Ha is the least acidic. The negative charge that results from
abstraction of Ha can be delocalized to only one oxygen atom, whereas the charge
resulting from abstraction of either Hb, Hc, and Hd can be delocalized to two oxygens in
each case.
N
Cl
OCH3O
Plavix
d) (pKa ~ 20)
e)
H
N
H
H2N
CO2
CO2
porphobilinogen
basic
(primary amine)
not basic
('pyrrole-like')
N
NN
N
NH2
O
OHOH
S
CH3NH2
O
O
S-adenosylmethionine
alkyl amine
aryl amine
pyrrole
pyrimidine
pyrimidine
pyrimidine
fig 6
P7.7: The total charge will be very close to –3.0. At pH = 7.3, the N-terminus proline
(pKa = 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 pKa 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.
P7.8: There are three ionizable groups on this peptide: the terminal amino group on Asp
(pKa ~ 9.6), the side-chain carboxylate group on Asp (pKa ~ 3.7) and the terminal
carboxylate on Ile (pKa ~ 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) = 105.6 = 4.0 x 105. 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:
. . .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.
N
NN
N
O
Hb
HaO
Hc
O
Hd
uric acid
0.025
(1 +0.025)x 100 = 2.4 %
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).
P7.9:
Keq = 10-8.2 = 6.3 x 10-9
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 pKa table to remind ourselves that primary
amines are more basic than carboxylates.
O
OOH P
O
O
OH
+
X Y
H3O++a) H2O
O
OOH P
O
O
O
(pKa ~ 6.5)
(pKa = -1.7)
N
H
H2N
CO2
CO2
+
X
Y
H3CC
O
O
b)
H3CC
OH
O
N
H
H3N
CO2
CO2
+
(pKa = 4.8)
(pKa ~ 10)
Keq = 10(10-4.8) = 105.2 = 1.6 x 105
fig 7
Keq = 10(10-19) = 10-9
P7.10:
fig 6
P7.11:
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.
O
+
X Y
+ N
CH3
CH3
CH3
Hc) N
CH3
CH3
CH3
O
(pKa ~ 19) (pKa ~ 10)
N
N
N O
CH3H
H
lysergic acid diethylamide
(LSD)
most basic(tertiary alkyl amine)
amide - not basic
pyrrole-like - not basic
fig 4
P7.13: In all cases the pKa 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.
P7.14: 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.
OH O
H3C OH
OHOH
O
NH2
O
O
N
CH3H3C
OH O
H3C OH
OHOH
O
NH2
O
O
N
CH3H3C
OH O
H3C OH
OHOH
O
NH2
O
O
N
CH3H3C
H
:B
OH O
H3C OH
OHOH
O
NH2
O
O
N
CH3H3C
fig 2
P7.15: The ester oxygen acts as an electron-donating group by resonance. This electron-
donating property destabilizes the negative charge on the enolate form, making the -
proton less acidic. This argument also holds true for thioesters.
P7.16:
Tris: Using the Henderson-Hasselbalch equation, we find that the ratio [HA+] / [A] at this
pH is 10(8.1-7.0) = 101.1 = 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.
Imidazolium: The Henderson-Hasselbalch equation tells us that the buffer is 50%
protonated at pH 7 (this is always true when the pH of the solution equals the pKa of the
buffer compound), so the concentration of the protonated form (imidazolium) is 25 mM.
N C
NH2
NH2
H
N C
NH2
NH2
H
Arg
Arg
N C
NH2
NH2
H
Arg
H3CC
CH3
O
H3CC
O
O
CH3
H2CC
CH3
O
H2CC
O
O
CH3H2C
CO
O
CH3
Chapter 8
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 SN2 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
P8.4: 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
P8.7:
a) water or hydroxide ion
b) CH3S- or CH3OH
c) CH2S- or CH3SH
d) acetate ion or hydroxide ion
e) diethyl sulfide or diethyl ether
f) dimethylamine or diethylether
g) trimethylamine or 2,2-dimethylpropane
P8.8:
a) The major product will be dimethyl sulfide (CH3SCH3), because CH3S- is a better
nucleophile than CH3O- and will react faster with methyl bromide in an SN2
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
P8.11:
P8.12: The first step in an SN1 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
P8.14:
a) Because the reaction 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.
CN O S
O
O
O CH3+a)OTs
NaCN
O+ Br
b) BrNaOCH3
Cl
X
carbocation cannot adopt
trigonal planar geometry!
fig 3
P8.16: Phosphate is the nucleophile in the reaction. We cannot predict the
stereochemistry of the starting bond, as we were not told whether the SN1 reactions
proceeds with inversion or retention of configuration.
H3CN
CH3
OH
H3CN
CH3
OH
H3C
choline
R1 S R2
CH3
O
OH
HO OH
N N
O
O
H OP
O
O
O+
Chapter 9
P9.1: Because no bonds to stereocenters are affected, the stereochemistry is unchanged.
fig 2
P9.2:
CO2
O
OH
OHOPO
O
O
P
O
O
OP
O
O
Oribose-A
ATP
H
:B
CO2
PO
OH
OH
+ ADP
O P
O
O
O
OH
O
O
OPO
O
O
P
O
O
OP
O
O
Oa gb
ribose-A
ATP
O P
O
O
O
OH
O
OO
PO
O
O
P
O
O
OP
O
O
Oribose-Ad-
d-
P9.3:
a) (EC 6.3.4.2)
b) (EC 6.3.3.1)
fig 3
O P
O
O
O
OH
O
O
P
O
O
OOP
O
O
OP
O
O
Oa b
ribose-A
ADP
+
O
N
OHOH
N
O
O
H
PPO
O P
O
O
O
ADP
B:
O
N
OHOH
N
O
O
PPO + ADP
P
O
O
O
O
OHOH
POHN
N
NH
O
H :BOP
O
O
O
ADP
+ ADPO
OHOH
POHN
N
NH
OP
O
O
O
P9.4:
fig
fig 4
P9.5: See J. Biol. Chem. 2005, 280, 10774.
P9.6:
a) Electrophile is P
b) Electrophile is P
P9.7: See Biochemistry 2000, 39, 8603.
a) The stereochemistry of the substitution and location of the 18O 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.
H3NO
O
O
P
O
O O
PO O
O
PO O
O
ribose-A
H3NO
O
AMP
O
P
O
O O
PO O
O
+
R OH
ATP P i
R OADP
R OH
ATP AMP
R OPP
fig 5
b) If water were to attack at the -phosphorus of the GDP group, the expected product
would be:
fig 5
Notice the different stereochemistry at the anomeric carbon, and the different location of
the 18O label.
P9.8: (From Biochemistry 2002, 41, 9279). Using 18O-labelled water, we could
determine the course of the reaction. In the first case, the AMP would contain the label,
in the second case the sugar would contain the label. The authors of the study above,
working with an enzyme from E. coli, found that the reaction proceeded by the top
mechanism (attack by water at the adenosyl phosphate).
O
OHO
HOHO
OH
PO
O
O P
O
O
O ribose-G
O
PO
O
O P
O
O
O ribose-G
HO
H
:B
18
O
HOHO
HO
OH
+
O
HOHO
HO
OH
OH18
O
HOHO
HO
OH
OH
18O
PO
O
O P
O
O
O ribose-G+
fig 5
P9.9:
O OH
OHHO
OP
O
O
OP
O
O
O
O*
HH
rib
ose
-A
:B
O OH
OHHO
OP
O
O
OP
O
O
O
O*
HH
:B
O*P
O
O
Oribo
se-A
O OH
OHHO
OP
O
O
O+
rib
ose
-A
OP
O
O
Orib
ose
-A
O OH
OHHO
OP
O
O
*O+
O
OH
OH
HO
HO
O
N
NH enzP
OO
O
H A
HO
N
NH enz
P
OO
O
O
OH
OH
HO
HO
step 1 +
N
NH enz
P
OO
O
HO
H
:B
step 2
N
NH enz
PO
O
O
HO +
P9.10: (From Biochemistry 2005, 44, 11476.)
P9.11: See J. Biol. Chem 2007, 282, 21573, Figs 1 and 2.
P9.12: See Molecules and Cells 2010, 29, 397; E.C. 2.7.7.9
P9.13: See J. Mol. Biol. 1999, 286, 1507.
P9.14: See Biochem J. 1982, 201, 665.
O
O Basen
O
3'
5'
DNA
P
O
O
O
O Basen+15'
3'
O
DNA
O
O Basen
O
3'
5'
DNA
P
O
O
O
O Basen+15'
3'
O
DNA
O
H A
OH
B:
TyrTyr
H
:B
A H
knicking
re-ligating
Chapter 10
P10.1:
fig 3
P10.3: (
fig 4
O
O
O
O
O
:B
AH
O
O
O
O
OH
N H
H
Lys
N
Lys
AH
O
O
O
O
N
Lys
HB:
ON
O
O
OO
H
NH3
O
OH
H:B
H A
ON
O
O
OONH3
OH H
OH
:B
HA
ONH3
O
O
OONH3
O
O+
P10.5:
P10.6:
fig 2
P10.7:J. Biol. Chem. 280, 12858, scheme 2 part 2)
fig 6
OHO
O O
O
O
H
O O
O
HO
OH OO
HH
O O
O
HO
OH OHOHhemiketal
formation hydrate
formation
N
CO2
H
O
CO2
NH3O
H
H
B:
AH
N
CO2
HO
HH
B:
N
O
HO
RO
N
N
HN
HO
RO
HO
H
HB:
A H
N
HO
HN
HO
R
O
A H N
HO
HN
R
O
O
H
H
:B
P10.9:
fig 2
Mechanism:
fig 3
P10.10:
a)
fig 3
O
OHHO
PO N
N
N
N
O
H
H2O
IMP
O
OHHO
PO N
N
NH
NH2
O
O
N
N
NH
N
O
O
H
H
:B
AH
N
N
N
N
O
OH
H
ribose-5-P ribose-5-P H
B:AH
IMP
N
NO
NH2
HN
NO
O
R
R
Cytidine
Uridine
O
H
H
:B
A H
N
NO
H2N
R
H
OH
AH
N
NO
H3N
R
H
O H
:B
AH
b)
fig 3
P10.11:
a)
fig 5
b)
N
N
N
N
R
NH2
N
N
N
NH
R
O
AH
N
N
N
NH
R
NH2
:B
O H
H
OH
AH
N
N
N
NH
R
NH3OH
:B
AH
O
CO2HO
H2C
OH
OH
OHCO2
O
H2C
OH
OH
OCO2
O
H3C
OH
OH
reverse of hemiketal
formation
enol-keto
tautomerization
OHN
OP
HO OH
CO2
A H
OH N
OP
HO HO
CO2H
H
:B
OH N
OP
HO OH
CO2H
fig 5
P10.12: (See J. Biol. Chem. 2005, 280, 13712)
The last step is not shown in detail, but is simply a Schiff base hydrolysis.
fig 6
P10.14:
NH
N
HNR
O2C CO2
NHHN
H
:B
H A
+
NH
N
HNRH2N NHR
NH
N
HNRHN
H A
H2NR =O2C CO2
NH2
H2ONH3
NH
N
HNRO
THF
5-formyl THF
glutamate
N NH
R2
O
NN
H2NO
R1
NH2
NN
H2NO
R1
NN
R2
N
H
H
H
:B
H
A
N NH
R2
O
NN
H2NO
R1
N H
H
H
:B
NH
R2
O
NH
+
NH
R2
O
NH
H
:BAH
HNN
R2
HAN
N
R2
HHO
HA
H
B:
P10.15: See Arch. Biochem. Biophys. 2008, 474, 302, scheme 4.
P10.16: See Biochemistry 1994, 33, 13792, mechanism II.
P10.17: See Biochemistry 2000, 39, 8603.
P10.18: See J. Am. Chem. Soc. 2005, 127, 16412.
Chapter 11
P11.2:
a)
P11.4:
P11.6:
HO OP
O
HO NH2
O
PO NH2
O
A B C
H3N CO2
N
CO2O
H
O2C
O
O
O2C
H3N CO2
H2N
CO2
+
OH
H:B
H3N CO2
N
CO2O
H
O2C
OHB:
H A
CH3
NH3C
CH3
CO2
O
RC
SCoA
O
H:B
CH3
NH3C
CH3
CO2
O
C SCoAR
OH A
CH3
NH3C
CH3
CO2
OC R
O
fatty acyl-carnetine
P11.8:
a)
P11.11: (A: EC 6.3.2.6; B: EC 6.3.4.4)
P11.13: Note this is hydrolysis of an amidine, and is essentially the reverse of the reaction
type discussed in section 11.8, although in this direction the leaving group (an amine) is
stable enough that activation by phosphorylation is not required.
P11.14
N
N
O
H
O
H
O
H
H
:B
N
N
O
H
O
H
OH
B:
H A
N
NH2
CO2
O
H
N
N
RNH2
O
O
A
N
N
N
N
R
HN
CO2O2C
H
B
N
N
NNR
HN
R
O
H
H
:B
HA
N
N
NNR
R
H2N
OH
B:
H A
HN
N
NNR
R
NH2
O
P11.15 (See also Biochemistry 2001, 40, 6989, Scheme 2)
Condensation step:
Cyclization step:
H3NO
O
NH
H2N NH2
arginine
O
H
H:B
H3NO
O
NH
H2N NH2
OH
B:
H A
H3NO
O
NH3
H2N NH2
O
H A
H2N O
O
P
O
O
O
N CO2
CO2
H
H
B:
H2N
O
O
P
O
O
ONO2C
H
O2C
NH2
O
NO2C
H
O2C
aspartate
carbanoyl aspartate
N
N
O
H
O
H
CO2
dihydroorotate
N
O
NO2C
H
O
O
carbanoyl aspartate
H
H
Zn2+Zn2+
N
N
O
H
O
H
O2C
O
Zn2+ Zn2+H A
N
N
H
O
H
O2C
O
=
(flipped horizontally)
:B
P11.17:
P11.18: (See J. Biol. Chem 1968, 243, 853 for experimental details). (Silverman p. 69)
a) The cysteine could attack the 14C atom, with subsequent loss of N2 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 14C label becoming
attached to the active site cysteine.
CoAS
O
O
O
NN P
O
O
O
enz
P
O
O
O
O
CoAS
O
O
OO
H A
O
O
O
O
PO O
O
P
O
O
O
NN
enz
HB:
O
O
O
O
succinateO
PO
O
O GMP
HA
O P
O
O
O GMPP
O
O
O
GTP
NN
enz
H
+
+
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 14C label.
P11.19: See J. Biol. Chem 2000, 275, 40804 (Scheme 1 for the mechanism in part a, Scheme
2 for part b)
R
O
NN
C
H
*
S
enz
H:B
R
O
NN
C
H
*
S
enz
R
O
C
H
*
S
enz
NN
+
H A
R
O
C
H
*
S
enz
H
active site cysteine is
radioactively labeled
R
O
NN
C
H
*
S
enz
H:B
R
O
NN
C
H
*
S
enz
H A
R
O
NN
C
H
*S
enz
H
R
ON
N
C
H
*S
H
enz+
active site cysteine is not
radioactively labeled
Chapter 12
P12.2:
CO2
N
H OH
OH
HO
OP
CO2
N
H O
OH
HO
OP
H
:B
CO2
N
H O
OH
HO
OP
H :B
HA
P12.3:
d)
P12.8:
a)
C SCoA
O
H
H H:B
H2C SCoA
O
S
OO
enz
H A
SCoA
OHO CH3
S
O
enz
HO
HB:
SCoA
OHO CH3
S
O
enz
O
H:B
A H
SCoA
OHO CH3
O
O
O2C
O
SCoA
O
H
B:
SCoA
O
A H
O2C
OH
SCoA
O
O
H
H
:B
O2C
OH
SCoA
HO O
A H
H
:B
AH
O2C
OH O
O
P12.10:
fig P12.14:
fig 7
CO2
CO2HO
H2C SCoA
O
CO2
O
H:B
SCoA
O
H A
CO2HO
SCoA
O
H O
HB:
CO2HO
SCoA
OO
H :B
H A
O
CO2O
OH
OH
CO2HO
O
OH
OH
HB:
O
CO2O
OH
OH O
CO2O
OH
OH
=
H A
Chapter 13
P13.4:
fig 5
P13.9:
fig 6
P13.14:
fig 4
O
OO
O
O
H2N
H
B:HA
O
OO
O
O
NH2O
O
O
NH3
HA
O2C
O
O2C CH2
OP
O
O
O P
O
O
O GDP
+ GDP
OS
O
O
O
O
H2C
O
O
O
S
O
O
+
A H
H3C
O
O
O
P13.15:
fig 5
P13.19:
c)
fig 14
P13.20:
a) It is a decarboxylation reaction, but there is no obvious way for the electron pair to be
stabilized.
SCoA
O
SCoA
O
O
O
H:B
SCoA
O
C OO
O
OH
H
OO
O
O
HO H
O
O
HOH
O
O
OH
O
HO H
H
OH
OH
O
OHH
O
fig 11
b) 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.
P13.21: See European J. Biochem. 2002, 269, 1790, fig 7.
P13.21: The product is chalcone. Aldol addition is followed by dehydration (E1cb
elimination) which is spontaneous due to the stability afforded by the extensively
conjugated product (note that all carbons are sp2-hybridized, meaning the conjugated
system extends over the entire molecule. If the reaction stopped after the aldol stage, we
would see NMR signals in the 2-3 ppm range.
HN
N
O
O
R
O
O
???
H
O
H3C
O
+
OH O
aldol
H2O
OH
H
7.8 ppm
7.5 ppm
chalcone
dehydration
Chapter 14
P14.3: (Microbiol. 2005, 151, 2199 Fig 1)
a)
b)
fig 5
NH
HN
N
O
O
H
NH
HN
N
O
O
H
PPO
R
NH
R
H:B
NH
R
NH
R
H:B
NH
R
P14.5:
P14.6: 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).
P14.7:
N
R
R
R
N
R
R
R
H3C
H H
CH3
SR1 R2
H
B:
N
R
R
R
H3C
H
PPO
HA
PPO
PPO
S
H
enz
:B PPO
S
C C HH3C
H OH2
C C HH3C
H
H2O
C C HH3C
H
OH H
B:
H OH2
C CH3H3C
OH
H2O
C CH3H3C
O
P14.9:
fig 7
P14.10:. (Biol. Chem. 2004, 279, 39389)
fig 9
OHO
OH
OHN
O
ribose-UCH3O
CH2C
O2C
PO
OHO
OH
OHN
O
ribose-UCH3O
CH2C
CO2
H A
CH3C
O2C
POH
B:
PO
H
B:
OHO
OH
OHN
O
ribose-UCH3O
CH2C
CO2
CO2
NH2
O
OPP
OHHO
OP
O
OHHO
OP
NH2
O
OHHO
OP
O
O
NH2
O
OHHO
OP
P14.11: (Biochem. Biophys. Res. Commun. 1988, 157, 816.)
fig 3
18O
O2C
OP
O
OH
OH
OHP
O
OO
A H
OP
OH
OH
OH
OH
CO2
18O
P OO
O
H O
HB:
OP
OH
OH
OH
OH
CO2
18O
mechanism A:
18O
O2C
OP
O
OH
OH
OHP
O
OO
OP
OH
OH
OH
OH
O2C
O
A H
OP
OH
OH
OH
OH
CO2
18O
P OO
O
H O
HB:
OP
OH
OH
OH
OH
CO2
18O
P OO
O
OHB:
mechanism B:
P14.12:
a)
b) If the labeled substrate shown below were to undergo a concerted reaction, the label
would necessarily be found on the (outside) phosphate group of the product. If, on the
other hand, the label were actually to be observed on the (inside) phosphate of the
product, this would rule out a concerted mechanism.
fig 4
c)
fig 4
PPO
OPP
OPP
O RP
O
O
O18
R
O
PO
O
OP
O
O OP
O
O
O18 no label here
OPP
OH
O
H
OH
:B
H A
O H
:B
d) (+)-bornyl diphosphate
e) (+)-sabinene.
fig 6
f) This is an anti-Markovnikov addition, because the secondary carbocation, rather than
the tertiary carbocation, forms during the addition.
P14.13:
Methyl vinyl ketone:
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,
OPP
(+)-bornyl diphosphate
OPP
OPP
OPP
(+)-sabinene
H
H
:B
which makes the primary carbocation intermediate more stable than the secondary
carbocation.
If the reaction were to proceed with Markovnikov regiochemistry, the NMR spectrum of
the product would look very different:
P14.13:
Methyl methacrylate:
Addition can take place with either Markovnikov or anti-Markovnikov regiochemistry:
NMR data shows that it is the anti-Markovnikov product that forms, due to the electron-
withdrawing effect of the ester carbonyl. (The 1H spectrum of the Markovnikov product
would be expected to contain just two singlet signals.)
Peak assignments are given below. Notice that the the HR and HS protons are
diastereotopic (there is a stereocenter in the molecule) and have different chemical shifts.
These signals show dd splitting because HR and HS are coupled to each other, and also to
the proton at 2.3 ppm.
O O O C C
O
CH3C
H
H
2.2 ppm (s)
Br
H
H3.5 ppm (t)
3.0 ppm (t)
O O O
Br
C C
O
CH3H3C
H
Br
s, 3H
q, 1H
d, 3H
H2C
CH3
OCH3
O
BrH
CH3
OCH3
O
H3C
H3C
OCH3
O
H3C
Br
H3C
OCH3
O
H2C
HH3C
OCH3
O
H2C
H
Br
Markovnikov product
anti-Markovnikov product
fig 12
(see J. Chem. Educ. 1990, 67, 518 for more details on this experiment).
P14. 15:
a)
H3C
OCH3
O
C
H
Br
HR
HS
1.3 ppm2.3 ppm
3.5 ppm
3.6 ppm3.7 ppm
N
O2C
O2C
H
N
H
CO2
CO2
R1 R2
H
A
N
O2C
O2C
H
N
H
CO2
CO2
R1 R2
N
O2C
O2C
H
R
N
H
CO2
CO2
RHO
+
N
H
CO2
CO2
R
OH H
B:
P14.18:
a)
fig 9
b) The mechanism is same as in part a) up to the point after the hydride shift and before
the methyl shift.
fig 10
OPP
H
B:A H
H
methyl shift
hydride shift
common intermediate
H :B
epi-arisolochene
H :B
vetispiradiene
common intermediate
P14.21: See J. Am. Chem. Soc. 2009, 131, 14648 (Scheme 2 pathway A)
P14.22:
P14.23: See J. Org. Chem. 2003, 68, 5433
HO
O
CO2
CH2
CO2
CO2
O
chorismate
phenylpyruvateprephenate
HO CO2O
O
OA H
Chapter 15
P15.1
b)
fig 1
f)
fig 1
P15.2:
fig 1
H
O NH3
O
O
O
O NH3
O
O
N
R
NH2
OHH
OH
B:
R
H R
O O H
:B
N
R
NH2
OHH
H
O
CO2 S
O
CO2
H
SCoA
B:
H
O
CO2
SCoA
NAD+
CoA
H NAD
+
H3NO
O
O
O
ATP ADP
A
NADPH
Pi
NADP+
B
H3NO
O
PO
O
H3NO
O
H
O
P15.3:
a)
fig 2
b)
fig 2
P15.4:
a) First an imine (Schiff base) linkage forms, then the imine is reduced to an amine by
NADPH.
N
OH
O2C CO2
NO2C CO2
H
B:
H A
NH
OH
O2C CO2
HA
NH
O2C CO2
H NADP
NO2C CO2
H :B
HA
N
N
N
NH
R
O
S
enz
N
N
N
NH
R
O
O
H
B:
HN
N
N
NH
R
O
H
S
enz
NAD+
N
N
N
NH
R
O
S
enz
H
OH
:B
N
N
N
NH
R
O
OH
S
enz
H A
H :BHA
N
N
NH
NH
R
O
O
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.
fig 3
O2C CO2
O
O2C NH
CO2
CO2 NH3
saccharopine
HN CO2
NH3
H
B:
H A
O2C CO2
HO N
Lys
A H
lysine
O2C NLys
OOH
:B
H NADP
H A
H
O2C N CO2
CO2 NH3
HB:
H
NAD+
O2C N CO2
CO2 NH3
H
OH
:B
HA
O2C NH
CO2
CO2 NH3OH
:B
HA
O2C NH2 CO2
CO2 NH3O+
glutamate
P15.5: (These steps are catalyzed by enzymes EC 4.2.1.17, EC 3.1.2.4, EC 1.1.1.31,
1.2.1.27)
fig 4
P15.6:
P15.9: See Biochemistry 2000, 39, 6732 for the original report on this experiment.
a)
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
SCoA
OH2O
SCoA
OOH H2O HSCoA
O
OOH
A
NAD+
NADH
O
OO
B
NAD+
CoASHNADH
O
OO
CoAS
CO2O
CoAS
C
NH
N
R R R R
R R
H
NADP H H A
NH
NH
R R R R
R R
HH
bilirubin
S S
H H
DsbA
S S
DsbB
S S
H H
DsbB
S S
DsbA
:B
H A
S S
H S
DsbA
DsbB
SH
B:HA
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 disulfide-
linked DsbA-DsbB intermediate in the normal reaction.
fig 4
P15.10:
a)
P15.12:
a)
b)
S S
H H
DsbA
Br SH
DsbB
:B
S S
H
DsbA
SH
DsbBproteins linked by
sulfide bond
O
HO
HO
HO OH
O
O
OH
O
O
HO
OH OH
HSG
:B
A H
O
OH
O
O
HO
OH
SGH
SG
:B
A H
+ GSSG
R1
OR1
N
H
R2
+
H2N R2
NADH, H+ NAD+
R1
N
H2O
section 10.5
R2section 15.3
R1 SR2
O
R1 SR2
OH O
FADH2
section 15.4B
R1 SR2
O
FAD H2O
section 13.4
c)
d)
e)
f)
R1 SR2
OH O
R1 SR3
O
+
H3C SR2
O
NADH, H+NAD+
section 15.3
R1 SR2
O O
R3SH
section 13.3C
R1 R2
OH
R1 R2
ONADH, H+NAD+
section 15.3R1 R2
OFADH2
section 15.4B
FAD
CH3R1
O
+
R2H
O
R1
O
R2
OH
section 12.3
NADH, H+NAD+
section 15.3 R1
O
R2
O
R2
O
R1
R2
O
R1
R2
O
R1section 14.3
NADH, H+
NAD+
section 15.4A
P15.13:
a)
b)
c)
R1
R2
O
+
R3S R4
O
R1
R2
O
R4H3N
R1
R2
O
R4O
R3SHNH3
R1
R2
O
R4HN
H2O
NADH, H+
NAD+
section 13.3 section 10.5
section 15.3
R1R2
OH
OH
R1R2
OH
OH
NADH, H+NAD+
section 15.3
R1R2
O
OH
R1R2
O
OHsection 12.2B
NADH, H+
NAD+
section 15.3
R1 R2
O O
OH
R1
O
OH
O
R2
+
NADH, H+ NAD+
section 15.3
R1 R2
O OH
OH
NADH, H+NAD+
section 15.3
OH
R2
section 12.3C
d)
P15.14:
a) These are the final steps in the biosynthesis of vanillin.
b) These are the final steps in the biosynthesis of menthol.
OH
R2
O
R1
R1
ATP ADP
OP
R1
R2
O
R2
OR1
Pi
section 9.4 section 14.1F
NADH, H+
NAD+
section 15.4A
HO
H3CO
HO
H3CO
O
H2O HO
H3CO
OH
section 13.4
+H3C O
O
O
O
O
O
section 12.3C
vanillin acetate
c) These are the early steps of the 'mevalonate pathway' of isoprenoid biosynthesis in
eukaryotes.
NADH, H+
OH
OH
NAD+
Osection 15.3
NADH, H+ NAD+
section 15.4A O
O
section 12.2C
NADPH, H+NADP+
section 15.4AO
NADPH, H+NADP+
section 15.3
Chapter 16
P16.1:
To simplify matters, we'll use 'R' to abbreviate the methyl ester group:
The first two propagation steps, forming a acrylamide dimer, are shown below:
The polymer is represented by:
H2C
CH3
O
O
CH3
R
=
H2CC
R
CH3=
R
CH3
R
CH3
X
R
CH3
X R
CH3
X
CH3R
CH3
Retc.
C C
H
H
CH3
C
n
O
OCH3
P17.2: 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):
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).
N
O
N
O
H H
strand 1
stra
nd 1
strand 2
strand 2
N
O
N
O
H HNH2O
strand 1
N
O
N
O
H HNH2O
strand 1
H2N O
strand 2
NH2
O
P16.3:
a)
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.)
N
O
N
O
H HNH2O
strand 1
H2N O
strand 2
NH2O
strand 1 contiues to
grow from this pointNH2
O
N
O
N
O
H HNH2O
strand 1
H2N O
strand 2
NH2O
NH2
O
strand 2 continues to
grow from this point
strand 1
O
OH
HO
O
OH
HO
P16.4:
P16.5:
The driving force for homolytic cleavage is the formation of nitrogen gas, which is very
entropically favorable.
OH
OH
O
OH
O
HO
O
OHSHO
a) b)
c) d)
N NCH3C
CN
CH3
C
CH3
CN
CH3 N NCH3C
CN
CH3
C
CH3
CN
CH3
N2 gas
P16.8:
P16.9:
fig 7
CO2
Fe
O
R1
R2O
O
HHR1
R2
HR1
R2
H
O
O
R1
R2
H
O
OR1
R2O
O
Cu
enz
O
OH O
enz
RNH2
Cu
enz
OO
enz
H2O
RNH2
Cu
enz
OO
enz
RNH2
Cu
enz
OOH
enz
RNH2
Cu
enz
O
OH
enz
NH2
OH
OH
enz
Cu
enz
H H
H
A HHO
HO
Chapter 17
P17.1:
a) Here, the bond adjacent to a ketone carbonyl is cleaved, telling us that thiamine
diphosphate (in green below) is likely involved. The carbon-carbon bond-breaking
(decarboxylation) step is:
b) Here, a decarboxylation occurs on an amino acid, a step that requires the participation
of PLP. The decarboxylation step is:
P17.6:
a) See Biomed. Res. Int. 2013, 194371, Fig 1.
P17.7: Curr. Opin. Chem. Biol. 2005, 9, 475.
R
OH
N
S
RR
H3C
O
O
N
RO
O
N
OH
CH3
H
OP
O2C
O
H3C CO2
OH3C
O
CO2HO+
acetohydroxybutyratepyruvate
CO2
[ThDP]
2-ketobutyrate
P17.8:
c) See J. Am. Chem. Soc. 2007, 129, 15750, Scheme 2
P17.9:
a) This can be described as a PLP-dependent retro-Claisen reaction.
O
O
NH3ONH2
HO
R
N
N
OH
CH3
H
PLP-substrate
O2C R
O
H:B
N
N
OH
CH3
H
O2C R
O
O
HHB:
OPOP
N
N
O
CH3
H
O2C R
O
OH
:B
H
N
N
O
CH3
H
CH2O2C
H
H A
O R
O
N
N
OH
CH3
H
O2C CH3
H A
OPOPOP
fig 8-9
b)
N
N
OH
CH3
H
CH3O2C
PLP-alanine
N
N
OH
CH3
H
CH3O2C
enz
NH3
alanine
PLP-enzyme
+
OP
OP
N
N
OH
CH3
H
CO2
PO
PLP-substrate
H
:B
N
N
O
CH3
H
CO2
PO H
H:B
H
N
N
O
CH3
H
CO2
PO
HOP OP OP
N
N
OH
CH3
H
CO2
A H
N
N
OH
CH3
H
CO2
OH H
B:
N
N
OH
CH3
H
CO2
OHA H
OPOPOP
fig 9
c) See J. Mol. Biol. 2009, 388, 98, Scheme 1.
P17.10: See J. Biol. Chem. 2013, 288, 22985, fig 5.
P17.11: See J. Biol. Chem. 2010, 285, 18684, fig 5A
P17.12: See J. Mol. Biol. 2004, 342, 183, fig 7. Note that the electron accepting-
donating mechanism for NAD+ is the same as what we have seen in redox reactions
throughout this chapter, except that the electrons being accepted (and then given back)
come from a carbon-carbon pi bond rather than a hydride ion.
P17.13:
a) See Biochemistry 2014, 53, 796, Fig 1.
b) See J. Appl. Microbiol. 2009, 106, 534.
N
N
OH
CH3
H
CO2
OH
OP