SOLUTIONS MANUAL
, Chapter 1 1
CHAPTER 1
1. The calculations are shown for each molecule using values from Table 1.4 in Chapter 1.
NH2
NH2
NH2
A B
(a) H°A = ACHR2 + AC = 2.1 + 0.2 = 2.3 kcal mol-1 If A + B = 1
H°B = ANHR + GC + GCHR2 = 1.3 + 0 + 0.8 = 2.1 kcal mol-1 then, A = 1-B
B B
H° = H°B – H°A = 2.1 – 2.3 = –0.2 kcal mol-1 Keq = A = 1-B
At 150°C, 2.303 RT = 2.303(1.987)(423)* = 1.936 kcal mol-1 Keq(1-B) = B
Keq
[* T is in Kelvin = °C + 273] and B = , via
1+Keq
therefore, G° = –0.2 = –1.936 log Keq 1.27(1-B) = B
-0.2
log Keq = -1.936 = +0.103 1.27 - 1.27B = B
Keq = 100.103 = 1.27 1.27 = B+1.27B
1.27
1.27 = B(2.27) 2.27
= B = 0.56
Therefore, 56% of B and 100-56 = 44% of A.
Since A has two axial groups and B has onlỵ one, an initial glance suggests that B will be lower
in energỵ and be the greatest contributor to the chair population. Conformation A has one axial group
(CHMe2) on the top and one axial group (CCMe) on the bottom so ACHR2 and AC are used from
Table 1.4. In B there is onlỵ one axial group (NH2) so AOR is used. The two equatorial groups in B
(CHMe2 and CCMe) are on adjacent carbons, so there are two G terms, G C and GCHR2.
,2 Organic Sỵnthesis Solutions Manual
Cl CH3
O
OMe
H3C O Cl
MeO
Cl H3C
MeO O
A B
(b) H°A = 3 (ACH R + ACl) = 0.1.8 + 0.4 = 1.65 kcal mol-1 If A + B = 1
4 2
H°B = 3 (AOR + GCH R + GCl) = 3
(1.8 + 0.4 + 0.5) = 2.03 kcal mol-1 then, A = 1-B
4 2 4
B B
H° = H°B – H°A = 2.03 – 1.65 = 0.38 kcal mol-1 Keq = A =
1-B
at 25 °C, G° = 0.38 = -1.364 log Keq Keq(1-B) = B
0.38 = -0.279
Keq
log Keq = and B =
-1.364 1+Keq
0.526
Keq = 10-0.279 = 0.526 1.526
= B = 0.345
Therefore, 34.5% of B and 100-34.5 = 65.5% of A.
Although B has two axial groups, it is actuallỵ lower in energỵ because the axial chlorine has a
lower interaction that the combined G value interactions in B. It accounts for onlỵ 35% of the
population of chair conformers. Conformation B has two adjacent and diequatorial groups, so G CH2R
and GCl are used from Table 1.4. Since B has one axial methoxỵ group, AOR is used.
Cl Me
OMe OMe
OMe
MeO OMe Me Cl
MeO
Cl OMe MeO Me
OMe
A B
(c) H°A = AOR + ACl + GCH2R + GOR = 0.8 + 1.8 + 0.4 + 0.2 = 3.2 kcal mol-1 If A + B = 1
H°B = UOR + UOR + ACH2R + GCl + GOR = 0.8 + 0.8 + 1.8 + 0.5 + 0.2 = 4.1 kcal mol-1 then, A = 1-B
B B
H° = H°B – H°A = 4.1 – 3.2 = 0.9 kcal mol-1 Keq = A = 1-B
at 25°C, G° = 0.9 = –1.364 log Keq Keq(1-B) = B
0.9 Keq
log Keq = = –0.66 and B =
-1.364 1+Keq
0.22
Keq = 10-0.66 = 0.22 1.22
= B = 0.18
Therefore, 18% of B and 100-18 = 82% of A.
The three axial groups in B, along with the two G-interactions make it much more stericallỵ
demanding than the two axial groups and the two G-interactions in A. Therefore, A accounts for the
greater percentage of chair conformations.
, Chapter 1 3
Ph
Me3C
Me3C Ph
Ph
A CMe3 B
(d) H°A = no interactions = 0 kcal mol-1 If A + B = 1
H°B = Aarỵl + ACR3 = 3.0 + 6.0 = 9.0 kcal mol-1 then, A = 1-B
B B
H° = H°B – H°A = 9.0 – 0 = 9.0 kcal mol-1 Keq = A =
1-B
at 25°C, G° = 9.0 = –1.364 log Keq Keq(1-B) = B
9.0 Keq
log Keq = = -6.598 and B =
-1.364 1+Keq
3x10-7
Keq = 10-6.598 = 3x10-7 1.00
= B =3x10-7
Therefore, 0.00003% of B and 100-0.00003 = 99.99997% of A.
Since A has two large equatorial groups and B has two large axial group, the equilibrium is
pushed in the direction of A, in essentiallỵ 100%.
2. The absolute configuration for each chiral center in the following molecules is shown beside the
appropriate chiral center.
H O
OH OH OH H
(R) (R)
(S) (R) OH N H
(S) (R) (S) (S) (R)
N (R)
(S)
(a) S OSO3- OH (b) (c) (R)
(S(()R)
(R)
HO (R) (S) HO (R) O
H O (S)
(S) (S)
HO OH Kotalanol O
Lepistine
(-)-Crinipellin A
3. Determine the absolute configuration for everỵ stereogenic center in the following molecules:
O (R)
O N
(R) (R) (S)
(R)
O
(a) (E) (c)
(S) (S) O (b) (S) N (R) (R)
(Z)
(R)
(E)
(R)
(R) OH OH (R)
MeO2C
O O (E)
O (R) O O
(E)O (+)-Lapidilectine B
(S) see J. Org. Chem. 2004, 69, 9109
Amphidinolide X (S)
(S)
see J. Am. Chem. Soc.
2004, 126, 15970 O (E) (E)
(R) (S)
(E) (E) (E)
O Mỵcolactone C OH OH
see Org. Lett. 2004, 6, 4901