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Organic Synthesis 5th (2017) - Michael B. Smith - Solutions Manual PDF

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Complete solutions covering all 19 chapters. Master retrosynthetic analysis, stereocontrol, protecting groups, carbon-carbon bond formation, heterocyclic synthesis, and modern synthetic methods. Step-by-step mechanisms for organic chemistry students. Organic Synthesis solutions, Michael Smith manual, Organic chemistry exercises, Retrosynthetic analysis problems, Stereocontrol synthesis solved, Protecting groups answers, Carbon-carbon bond formation, Heterocyclic synthesis solutions, Modern synthetic methods, 5th edition solutions PDF, Smith organic synthesis, Advanced organic chemistry, Synthetic methodology manual, Organic mechanisms solved, Smith solutions download, Organic chemistry homework

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ALL 19 CHAPTERS COVERED




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 (CCMe) 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 CCMe) 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

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