40 Suggested solutions for Chapter 40 PROBLEM 1 Suggest mechanisms for these reactions, explaining the role of palladium in the first step. O
O
O
AcO
O
OEt
OEt
(Ph3P)4Pd
O
NBS
H2O
O
O
Br
O
1. Ph3P
O
2. K2CO3
Purpose of the problem Revision of enol ethers and bromination, the Wittig reaction, and, of course, first steps in palladium chemistry. Suggested solution The first step is a reaction of an enol with an allylic acetate catalysed by palladium(0) via an η3 allyl cation. There is no regiochemistry to worry about as the diketone and allylic acetate are both symmetrical. (Ph3P)4Pd
AcO
O
AcO
OEt
HO
H
PdLn
allyl cation complex in various satisfactory ways—some are mentioned on p. 1089 of the textbook. LnPd
OEt
OEt
O
LnPd
OEt
OEt
O
O
O
■ You might have drawn the η3 NBS in aqueous solution is a polar brominating agent, ideal for reaction with an enol ether. The intermediate is hydrolysed to the ketone by the usual acetal style mechanism. 2 Solutions Manual to accompany Organic Chemistry 2e O
OEt
O
Br
O
O
O
OEt
N
O
O
H2O
O
Br
Br
O
Finally, an intramolecular Wittig reaction. This is a slightly unusual way to do what amounts to an aldol reaction but the 5,5 fused enone system is strained and the Wittig went under very mild conditions (K2CO3 in aqueous solution). The stereochemistry of the new double bond is the only one possible and Wittig reactions with stabilised ylids generally give the most stable of the possible alkene. ■ This process is a general way to make 5,5 fused systems devised by B. M. Trost and D. P. Curran, J. Am. Chem. Soc., 1980, 102, 5699. O
O
O
Ph3P
O Wittig
K2CO3
O
O
H 2O
O
PPh3
O
PPh3
PROBLEM 2 This Heck-­‐style reaction does not lead to regeneration of the alkene. Why not? What is the purpose of the formic acid (HCO2H) in the reaction mixture? PhI, cat. (Ph3P)2Pd(OAc)2
HCO2H, R2NH, DMF
Ph
H
Purpose of the problem Making sure you understand the steps in the mechanism of the Heck reaction. Suggested solution The reaction must start with the oxidative addition of Pd(0) into the Ph–I bond. The reagent added is Pd(II) so one of the reduction methods on page 1081 of the textbook must provide enough Pd(0) to start the reaction going. The oxidative addition gives PhPdI and this does the Heck reaction on the alkene. Addition occurs on the less hindered top (exo-­‐) face and the phenyl group is transferred to the same face. Solutions for Chapter 40 – Organometallic chemistry Ph
I
Pd(0)
Ph
I
Ph
Ph
Pd
PdI
PdI
oxidative
insertion
Normally now the alkyl palladium(II) species would lose palladium by β-­‐elimination. This is impossible in this example as there is no hydrogen atom syn to the PdI group. Instead, an external reducing agent is needed and that is the role of the formate anion: it provides a hydride equivalent by ‘transfer hydrogenation’ when it loses CO2. O
R2NH
Ph H
Pd
3 HCO2H
Ph
O
Ph
Pd
I
H
H
PROBLEM 3 Cyclization of this unsaturated amine with catalytic Pd(II) under an atmosphere of oxygen gives a cyclic unsaturated amine in 95% yield. How does the reaction work? Why is the atmosphere of oxygen necessary? Explain the stereochemistry and regiochemistry of the reaction. How would you remove the CO2Bn group from the product? H
NHCO2Bn
cat. Pd(II)
O2
N
H
CO2Bn Purpose of the problem Introducing you to ‘aminopalladation’: like oxypalladation, nucleophilic attack on a palladium π-­‐complex. Suggested solution The π-­‐complex between the alkene and Pd(II) permits nucleophilic attack by the amide on its nearer end and in a cis fashion because the nucleophile is tethered by a short chain of only two carbon atoms. Nucleophilic attack and elimination of Pd(0) occur in the usual way. The removal of the CO2Bn group would normally be done by hydrogenolysis but in this case ester hydrolysis by, say, HBr would be preferred to ■ A heterocyclic version of this reaction was part of a synthesis of the natural analgesic epibatdine by S. C. Clayton and A. C. Regan, Tetrahedron Lett., 1993, 34, 7493. 4 Solutions Manual to accompany Organic Chemistry 2e ■ This general synthesis of avoid reduction of the alkene. The free acid decarboxylates spontaneously. heterocycles was introduced by J.-­‐
E. Bëckvall and group, Tetrahedron Lett., 1995, 36, 7749. H
NHCO2Bn
Pd
X
H
X
Pd(0)
H
+ PdX
N
H
PdX
N
H
CO2Bn
CO2Bn
PROBLEM 4 Suggest a mechanism for this lactone synthesis. O
Br
CO, Bu3N
OH
O
cat. Pd(OAc)2 Ph3P
Purpose of the problem Introducing you to carbonyl insertion into a palladium (II) σ-­‐complex. Suggested solution Oxidative insertion into the aryl bromide, carbonylation, and nucleophilic attack on the carbonyl group with elimination of Pd(0) form the catalytic cycle. No doubt the palladium has a number (1 or 2?) of phosphine ligands complexed to it during the reaction and these keep the Pd(0) in solution between cycles. ■ M. Moru et al., Heterocycles, 1979, 12, 921. Br
Pd(0)
OH
Br
Br
PdLn CO
Pd
OH
O
O
Br
OH
O
Pd
OH
O
5 Solutions for Chapter 40 – Organometallic chemistry PROBLEM 5 Explain why enantiomerically pure lactone gives syn but racemic product in this palladium-­‐catalysed reaction. CO2H
O
MeO2C
CO2Me
O
CO2Me
(Ph3P)4Pd
(–)-lactone
syn but
racemic
CO2Me
Purpose of the problem Helping you to understand the details of palladium-­‐catalyzed allylation. Suggested solution Following the usual mechanism, the palladium complexes to the face of the alkene opposite the bridge. The ester leaves to give an allyl cation complex. This is attacked by the malonate anion from the opposite face to the palladium. So the overal result is retention of configuration, the syn starting material giving the syn product. O
O
CO2Me
CO2Me
LnPd
LnPd
CO2H
CO2
CO2
LnPd
CO2Me
CO2Me
CO2Me
The racemization comes from the structure of the allyl cation complex. It is symmetrical with a plane of symmetry running vertically through the complex as drawn. Attack by the malonate anion occurs equally at either side of the plane giving the two enantiomers of the syn diasterereoisomer in equal amounts. MeO2C
MeO2C
CO2Me
CO2H
CO2
CO2H
MeO2C LnPd
CO2Me
CO2Me
CO2Me
CO2Me
CO2Me
■ This investigation helped to establish the mechanism of these reactions: B. M. Trost and N. R. Schmuff, Tetrahedron Lett., 1981, 22, 2999. 6 Solutions Manual to accompany Organic Chemistry 2e PROBLEM 6 Explain the reactions in this sequence, commenting on the regioselectivity of the organometallic steps. O
O
R
OSiMe3
R
MgBr
R
CHO
CHO
Pd(0)
Cu(I), O2
H, H2O work-up
R
N
NH3
Purpose of the problem Revision of allylic Grignard reagents, the synthesis of pyridines, and the mechanism of the Wacker oxidation. Suggested solution The allylic Grignard reagent does direct addition from the end remote to the magnesium atom, as often happens. Hydrolysis of the silyl enol ether reveals an aldehyde. O
R
MgBr
OSiMe3
O
MgBr
OSiMe3 H, H2O
R
R
OH2 O
H
R
SiMe3
OH OSiMe3
R
OH2
O
SiMe3
R
CHO
Now the Wacker oxidation, by whatever detailed mechanism you prefer, must involve the addition of water to a Pd(II) π-­‐complex of the alkene and β-­‐elimination of palladium to give Pd(0) which is recycled by oxidation with oxygen mediated by copper. Solutions for Chapter 40 – Organometallic chemistry CHO
O
OH
Cl
H 2O
R
Pd(II)
R
Pd
R
Cl
R
PdCl
CHO
CHO
CHO
Finally, the pyridine synthesis is simply a double enamine/imine formation between ammonia and the two carbonyl groups. Probably the aldehyde reacts first. R
R
O
NH3
CHO
R
O
H
OH
N
R
–H2O
N
■ M. A. Tius, Tetrahedron Lett., 1982, 23, 2819 NH
PROBLEM 7 Give a mechanism for this carbonylation reaction. Comment on the stereochemistry and explain why the yield is higher if the reaction is carried out under a carbon monoxide atmosphere. Ph3P
Ph3P
Bu3Sn
Pd
Cl
O
CH2Ph
Ph
PhCOCl
Hence explain this synthesis of part of the antifungal compound pyrenophorin. O
Bu3SnH
Bn
Bu3Sn
Ph3P
Pd
Bn
O
O
Ph3P
O
AIBN
Cl
O
CH2Ph
O
COCl
OSiPh2t-Bu
OSiPh2t-Bu
Purpose of the problem More carbonylation with a Stille coupling. Bn
O
7 8 Solutions Manual to accompany Organic Chemistry 2e Suggested solution The tin-­‐palladium exchange (transmetallation) occurs with retention of configuration at the alkene. The exchange of the benzyl group for the benzoyl group is necessary to get the reaction started. Ph
Bu3Sn
+
Ph3P
Ph3P
Pd
Cl
Ph3P
CH2Ph
Ph3P
Ph
Ph3P
Ph3P
Pd
Ph
Pd
+
Ph3P
PhCOCl
+ Bu3Sn
O
+ PhCH2Cl
Pd
Ph3P
Cl
Now the coupling can take place on the palladium atom producing the product and Pd(0) which can insert oxidatively into the C–Cl bond. Transmetallation sets up a sustainable cycle of reactions. It is better to have an atmosphere of carbon monoxide because the acyl palladium complex can give off CO and leave a PdPh σ-­‐complex. The atmosphere of CO reverses this reaction. Ph
Ph3P
Ph3P
O
O
+
Pd
Ph3P
Ph3P
Ph
O
Pd
+
Ph3P
Ph3P
Cl
Ph3P
Pd
Ph3P
Bu3Sn
Ph
PhCOCl
Ph
Ph
Ph3P
Ph3P
Pd
O
Pd
Cl
O
+ PhCH2Cl
The second sequence starts with a radical hydrostannylation (chapter 37) giving the E-­‐vinyl stannane preferentially if a slight excess of Bu3SnH is used. SnBu3
Bu3Sn
OBn
O
H
Bu3Sn
H
SnBu3
OBn
O
H
H
CO2Bn
Bu3Sn
Bu3Sn
CO2Bn
Now the coupling with the acid chloride takes place as before though this time we have an aliphatic carbonyl complex. There is no problem with β-­‐elimination as that would give a ketene. Again, the stereochemistry of the vinyl stannane is retained in the product. 9 Solutions for Chapter 40 – Organometallic chemistry OSiPh2t-Bu
(Ph3)2PdClBn
COCl
+
OSiPh2t-Bu
CO2Bn
Bu3Sn
O
Ph3P
Ph3P
OBn
Pd
CO2Bn
OSiPh2t-Bu
PROBLEM 8 The synthesis of an antifungal drug was completed by this palladium-­‐catalysed reaction. Give a mechanism, explaining the regio-­‐ and stereochemistry. Me
NHMe
t-Bu
N
OAc
(Ph3P)4Pd
Purpose of the problem A simple example of amine synthesis using palladium. Suggested solution The palladium forms the usual allyl cation complex and the nitrogen nucleophile attacks the less hindered end thus also retaining the conjugation. Attack at the triple bond would give an allene. The E stereochemistry of the palladium complex is retained in the product. (Ph3P)4Pd
product
OAc
RNHMe
PdL2
O
O
10 Solutions Manual to accompany Organic Chemistry 2e PROBLEM 9 Work out the structures of the compounds in this sequence and suggest mechanisms for the reactions, explaining any selectivity. OH
CHO
Pd(II)
CHO
heat with
catalyitc acid
A
CuCl, O2
KOH
B
H2O, THF
C
B has IR: 1730, 1710 cm–1, δH 9.4 (1H, s), 2.6 (2H, s), 2.0 (3H, s), and 1.0 (6H, s). C has IR: 1710 cm–1, δH 7.3 (1H, d, J 5.5 Hz), 6.8 (1H, d, J 5.5 Hz), 2.1 (2H, s), and 1.15 (6H, s). Purpose of the problem An intramolecular aldol reaction (p. 636) and a Wacker oxidation (p. 1096). Suggested solution B clearly has aldehyde and ketone functional groups with nothing but singlets in the NMR. On the other hand C has a cis disubstituted alkene with a small (and therefore cis) J value and is a cyclpentenone. O
Pd(II)
CHO
A
CuCl, O2
KOH
CHO
B
H2O, THF
O
C
Solutions for Chapter 40 – Organometallic chemistry PROBLEM 10 A synthesis of the Bristol-­‐Meyers Squibb anti-­‐migraine drug Avitriptan (a 5-­‐HT receptor antagonist) involves this palladium-­‐catalysed indole synthesis. Suggest a mechanism and comment on the regioselectivity of the alkyne attachment. N
O
O S
NHMe
N
OMe
I
+
NH2
cat. Pd(OAc)2, Ph3P
LiCl, Na2CO3
H2O, MeCN
SiEt3
O
O S
NHMe
N
N
H
SiEt3
N
MeO
Purpose of the problem A new reaction for you to try – a palladium-­‐catalysed indole synthesis. Suggested solution Although palladium(II) is added to the solution, the aryl iodide tells you that this is an oxidative insertion of Pd(0) produced by one of the methods described on p. 1081 of the textbook. The resulting Pd(II) species complexes to the alkyne and the amine can now attack the triple bond. This gives a heterocycle with the Pd(II) in the ring. Coupling of the two organic fragments extrudes Pd(0) to start a new cycle. The nitrogen attacks the more hindered end of the alkyne so that the palladium can occupy the less hindered end. 11 12 Solutions Manual to accompany Organic Chemistry 2e O
O S
NHMe
O
O S
NHMe
I
Pd(0)
NH2
Pd
N
H
I
O
R
SiEt3
Pd
O S
NHMe
Pd(0)
R
NH2
SiEt3
O
O S
NHMe
N
H
■ This is the Larock indole synthesis (R. C.Larock and E. K. Yum, J. Am. Chem. Soc., 1991, 113, 6689) and its usedin the synthesis of Avitriptan is described in P. D. Brod fuehrer et al, J. Org. Chem., 1997, 62, 9192). SiEt3