Introduction to Bioorganic Chemistry and Chemical Biology
Answers to Chapter 2
(in-text & asterisked problems)
Answer 2.1
sp3 ..
: OH
>
Basicity:
.. sp2
N
H
N
..
sp3
Basicity:
>
sp2.. R N
S
+
Basicity:
sp2.. ..
O
- .. sp
+ C
N
R
>
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Answer
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A
B
σ C–F
*
Nu:
C
*C– O
O
F
Nu:
Cl
Nu:
D
N
E
Nu:
N
CH3
F B
CH3
pB
*N–N
N
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Answer
2.3
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A
B
C
nO
nN
..
O
..
..
N
D
nN
.. sp3
.. NH2
sp2 N
H σ B-H
B H
H
H
-
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Answer
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A
..
n-BuNH2
H
n-Bu
H
+ N - O:
O
Cl
OEt
Cl
also acceptable...
..
n-BuNH2
B
OEt
Ph
Ph
..
+
OEt
H2O:
H
+
OEt
..
H A
OEt
HO
Cl
: OEt
A H
OEt
OEt
n-Bu +
N
B:
H H
H
n-Bu
H
O
+N
O
Cl
O
n-Bu +
N
B:
H H
HO
..
Ph
O
+
HO
H
Ph
-A:
OEt
Ph
-A:
H
OEt
O
+
Ph
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OEt
O H
σ*H–O
:Nu
(:B)
1
2
Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 2
Answer 2.5
OH
H-bond acceptors = boxed
H-bond donors = circled
HO
O
+H
NH
O
H
N
3N
NH3+
O
CH3
HO
HO
HO
O-
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Note: amide bond resonance precludes the lone pair
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OH
O
OH
of the amide NH from acting as a
hydrogen-bond acceptor.
Answer 2.6
O
O
H
N
O
N
H
O
O
H
N
O
N
H
O
N
H
H
N
N
H
H
N
O
Kevlar
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Answer 2.7
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A Keq = 10[pKa (PhOH) – pKa (H20)]
Keq = 10[10 – –1.7)
k1 ≈ 1011 M–1 s–1 (diffusion limit for H3O+ in water) = 1011.7 ≈ 1012
Keq =
k–1 ≈ 1011 M–1 s–1 / 1012 = 0.1 M–1 s–1
B (H2N– anions are unlikely intermediates in water.)
Keq = 10[pKa (H3N) – pKa (H20)]
Keq = 10[38 ––1.7)
k1 ≈ 1011 M–1 s–1 (diffusion limit for H3O+ in water) = 1039.7 ≈ 1040
Keq =
k–1 ≈ 1011 M–1 s–1 / 1040 = 10–29 M–1 s–1
The deprotonation of NH3 by H2O would be inconceivably slow.
Answer 2.8
A
H
O:
A
NH2
B
B:
+
OH
N
H
OH
H
+B
H
O
NH2
-O
NH
: A-
H
..
NH2
O
NH3+
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Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 2
Answer 2.9
Henderson–Hasselbalch:
[HA]
log
[HA]
Rearranged:
= pKa – pH
[A–]
= 10(pKa – pH)
[A–]
A 10(9.2–7.2) = 100; [HCN]/[CN–] = 100:1
B 10(9.2–7.2) = 100; [NH4+]/[NH3] = 100:1
C 10(4.2–7.2) = 0.001; [PhCO2H]/[PhCO2–] = 1:1000
Answer 2.10
HO - OH
Ca
O +O H
..
(HO) 2Ca
..
OH
O
R
:B
HO - OH
Ca
O
O
R
OH
..
O
R
OH
R
OH
OH
OH
OH
OHChemical Biology | A2117
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-Ca
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Ca .
+
.
+ O&Ca
.O
+O
O.
O
O.
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R
R
OH
CaOH
O
O
OH
R
OH
R
OH
OH
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Note
that&the
actual
equilibrium will involve all possible diastereomers.
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Answer 2.11
O
H3C
: NH3
H3C
H
N +
H2
NH
.. 2
H3C
H
- O:
NH2
H3C
NC:
+
OH2
H3C
:A
O-
A
A
H
: OH
H3C
NH2
CN
NH2+
H3C
NH2
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Answer
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A Use the following equation.
E∝
4 q1 × q2
εr
+
x
r12
–
y
r6
Only the Coulombic, first term is relevant for the charge–charge repulsion. Substitute the distances given as r into the Coulombic equation as follows.
–9800 cal mol–1
4 q1 × q2
1.987 cal mol–1 K–1 × 298.15 K
ε3
= 2/3
4 q1 × q2
ε2
In other words, the repulsive interaction potential decreases by about 33%.
BHere, only the third term for the attractive interaction is relevant, because the first
two terms can be approximated as zero. We can compare the attractive energies at
those two differences by taking the ratio.
y
36
y
= (2/3)6 = 0.088
26
Thus, the attractive interaction potential decreases by more than 90%.
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3
4
Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 2
Answer 2.13
H-bond donors = circled
H-bond acceptors = boxed
net dipole
net dipole
H2N
O
O
OH
O
net dipole
arene
interaction
arene
interaction
N
arene
interaction
O net dipole
Every part of these molecules is capable of engaging in dispersion interactions (not shown)
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*Answer 2.14
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A
:O:
H3C
C
:O:
B
.. H
O
..
H3C
:O:
H
H
H
H
E
F
+
:C O:
G
:O:
.. CH3
O
..
H
D
..
H .. O
.. H
O
..
H
: O:
H
H
H
H
H
+ H3C N C :
H
H
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*Answer
2.16
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A
B
nO
..
O:
C
nO
O:
D
O
H3C
O
N
n C-
..
-CH2
C
H
NH
.. 2
nN
or
- O:
actually, these "lone pairs" are
in orbitals with symmetry
H3C
N
C
nO
CH2
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2.17
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©*Answer
A
N
HN
+
N
H
+ H
N
H
O
H2N
..
N
H
O
..
- N:
C
C
C
H
H
B N
C
- :C
C
H
H
C
- ..
N
- ..
N
N+
N+
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2.18
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© *Answer
: O:
H A
: OC:
H
OCH3 A
Ph
C
Ph
OCH3
H2O
..
H2O
..
: OH
Ph: OH
C +
3
HOC OCH
Ph
+
OCH
H 3
HO
H
OH
Ph OH
C
HOC OCH
Ph
..
OCH
..
A HO
H
A H
O
O
O
H
O
H
+ H
+O H
OC
Ph
OCH3
C
Ph
OCH3
:B :B -
O
O
..
O
..O
-
OH
Ph OH
C
HOC OCH3
Ph
+
-A:
HOH OCH3
+
-A:
H
+ H
: AO
+O H
: AOC
OC
Ph
OH
Ph
OH
C
C
Ph
OH
Ph
OH
..
-O
..
-O
O
O
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Introduction
Bioorganic
Chemistry and Chemical Biology | A2098
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O
O
..
O
..O
-
H B
H B
O
O
OH
OH
Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 2
*Answer 2.20
pH
phenol
amine
R
2
HO
R'
Me
H
N
R
R'
R'
Me
+
R N H
R'
R'
Me
R N
R'
carboxylic acid
O
+
OH
R
R"
R
7
HO
R"
R
12
-O
R"
O
OR
O
OR
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2.21
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©*Answer
O
O
HO
-O
HN
pKa 5
Cl
OH
H2N
+
pH 7
pKa' 10
Cl
OH
OH
pKa 5
N
N
O
N
OH
pKa' 10
O
OMe
N+
H
N
OH
N
pKa' 10
OH
pH 7
N
pKa 16
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*Answer 2.24
B
HO
+H
H
N
3N
OH
O
HO
O-
O
N
H
HO
NH2
OH
C
N+
H
OO
OMe
A
O
pH 7
O
-O
O
HO
O
O
P
O
NH3+
CO2-
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OH
5
Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 2
*Answer 2.28
H
-A:
HO
HO
HO
α-ribofuranose
..
HO +
O
O
O
+
H
H
-A:
β-ribofuranose
HO
HO
A H
+
HO
O
O
OH
..
HO
OH
H
OH
O
HO
+
O
HO
OH
..
HO
HO
H
HO
A H
ribose
HO
6
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*Answer 2.30
A 23 = 8 stereoisomers
B
D-ribose
H
OH
L-ribose
OH
OH
OH
H
O
O
OH
OH
OH
OH
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*Answer 2.33
N C NH2
aminonitrile
N:
: NH3
H2N
H
+N
H H
:B
- N:
H2N
N
H
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H
H NH3
+
NH
H2N
NH2
guanidine