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Introduction to Bioorganic Chemistry and Chemical Biology: ANSWERS TO CHAPTER 2
Answer 2.9 [HA] Henderson–Hasselbalch:
log [HA]
Rearranged:
= pK a – pH
[A–]
= 10(pK a – pH)
[A–]
A
10(9.2–7.2) = 100; [HCN]/[CN –] = 100:1
B
10(9.2–7.2) = 100; [NH 4+]/[NH3] = 100:1
C
10(4.2–7.2) = 0.001; [PhCO 2H]/[PhCO2–] = 1:1000
Answer 2.10 OH Ca +O H
HO
(HO) 2Ca
..
O
OH
O
HO - OH Ca
. .
R
:
B
O
.
. .
R
OH
OH -Ca + O O
R
OH
OH Ca +O O
R
OH
CaOH O
R
OH
.
..
O
R
OH OH Ca +O O
O
O
. .
OH
R
OH
R
OH
OH
Note that the actual equilibrium will involve all possible diastereomers. Answer 2.11 O
: A
O: NH
H 3C
H3C
N H2+
H3C
NC:
+ OH2 NH2
H 3C
H
3
H3C
..
H
- O:
NH2
A
A
H
: OH
H3C
NH2
CN NH2+
H 3C
NH2
Answer 2.12 A
Use the following equation. E
4 q1 × q2
ε
+
x r 12
r
–
y r 6
Only the Coulombic, first term is relevant for the charge–charge repulsion. Substitute the distances given as r into the Coulombic equation as follows. 4 q1 × q2 ε
3
4 q1 × q2 ε
= 2/3
2
In other words, the repulsive interaction potential decreases by about 33%. B
Here, 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
= (2/3)6 = 0.088
y 26
Thus, the attractive interaction potential decreases by more than 90%.
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Introduction to Bioorganic Chemistry and Chemical Biology: ANSWERS TO CHAPTER 2
