Solutions for Organic Synthesis, 5th Edition by Michael Smith
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Course
General Chemistry
Institution
General Chemistry
Solutions for Organic Synthesis, 5e 5th Edition by Michael Smith.
Retrosynthesis, Stereochemistry, And Conformations
Acids And Bases and Addition Reactions
Functional Group Exchange Reactions. Aliphatic and Aromatic Substitution, and Elimination Reactions
Acids, Bases, And Functional Group E...
CHAPTER 1
1. The calculations are shown for each molecule using values from Table 1.3 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 = 1+K , via
eq
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 only one, an initial glance suggests that B will be lower
in energy 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 only 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, GC and GCHR2.
,2 Organic Synthesis Solutions Manual
Cl CH3
O
OMe
H3C O Cl
MeO
Cl H3C
MeO O
A B
3
(b) H°A = 4 (ACH2R + ACl) = 0.1.8 + 0.4 = 1.65 kcal mol-1 If A + B = 1
3 3
H°B = (AOR + GCH2R + GCl) =
4 4 (1.8 + 0.4 + 0.5) = 2.03 kcal mol-1 then, A = 1-B
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 Keq
log Keq = -1.364 = -0.279 and B = 1+K
eq
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 actually lower in energy because the axial chlorine has a
lower interaction that the combined G value interactions in B. It accounts for only 35% of the
population of chair conformers. Conformation B has two adjacent and diequatorial groups, so GCH2R
and GCl are used from Table 1.4. Since B has one axial methoxy 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 = -1.364 = –0.66 and B = 1+K
eq
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 sterically
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 = Aaryl + 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 = -1.364 = -6.598 and B = 1+K
eq
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 essentially 100%.
2. The absolute configuration for each chiral center in the following molecules is shown beside the
appropriate chiral center.
(S) OCH3
(S)
O O CH3
(S) OCH3
OH OH OH O O O
(R) (R) NH
(S) (R)
H OH NH
OH (S)
(S) (R) (S) (R)
(R)(R) (S)
(S)
(a) S OSO3- OH (S) (c)
(b) O H O O (Z)
HO (R) H3C
H
(S) (S) N
HO OH Kotalanol (E)
CH3 (Z)
CH3 Pericoannosin A (+)-Neopeltolide
3. Determine the absolute configuration for every stereogenic center in the following molecules.
, 4 Organic Synthesis Solutions Manual
H3C (S) (R)
O N
(R) (R) (S)
(R) O
H (E) (c)
(S) (S) (b) (S) N (R) (R)
O (S) (R) OH OH
(a) CH3 (R)
MeO2C
OHC (E)
CH3 O O
(+)-Lapidilectine B
see J. Org. Chem. 2004, 69, 9109
(S)
OH (S)
Cochlearol B (E) (E)
O (R) (S)
see Angew Chem. Int. Ed. 2022, 62, e202201213
(E) (E) (E)
O Mycolactone C OH OH
see Org. Lett. 2004, 6, 4901
4. Draw both chair conformations and one twist-boat conformation for trans-1,2-
di(triphenylmethyl)cyclohexane.
CPh3 Ph3C
H
H
H H
CPh3 H
Ph3C
H CPh3 CPh3
5. Using the reaction wheel (Fig. 1.3) give reasonable syntheses, including reagents and all
intermediates (no mechanisms).
There is more than one "correct" answer. One possible solution is shown for each
transformation. The letters (a), (f), etc., beside each reagent refer to the lettered transformations from
Figure 1.2 in Chapter 1.
The first reaction is a poor one since elimination will be a major process. Limiting
the choices to those in Table1.1 and Figure 1.3 is a problem since there are other ways to do this.
(d) PBr3 KOH , EtOH 1. O3 2. H2O2 CO2Me
OH Br
(d) (f) 2. SOCl2 ; MeOH CO2Me
(x), but there are no reactions shown for conversion to acid derivatives
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