22 Suggested solutions for Chapter 22 PROBLEM 1 What is the structure of the product of this reaction and how is it formed? It has δC 191, 164, 132, 130, 115, 64, 41, 29 and δH 2.32 (6H, s), 3.05 (2H, t, J 6Hz), 4.20 (2H, t, J 6Hz), 6.97 (2H, d, J 7Hz), 7.82 (2H, d, J 7Hz), 9.97 (1H, s). You should obviously interpret the spectra to get the structure. CHO
OH
Me2N
F
C11H15NO2
NaH
νmax (cm–1) 1730
Purpose of the problem Revision of NMR with an exercise in nucleophilic aromatic substitution. Suggested solution Summing the formulae of the two starting materials shows that this is a substitution of fluoride (the product is the sum of the starting materials less HF). The aldehyde is still there (from the IR and the proton at 10 ppm) so the spectra are best interpreted by this structure: δC
O
132/130/115
191
δH
7.82
6.97
H
29
164
Me2N
41
64 O
Me2N
2.32
4.20
3.05
O
H
9.97
O
That suggests a simple nucleophilic aromatic substitution by the addition-­‐elimination mechanism with both F and CHO assisting the first step. 2 Solutions Manual to accompany Organic Chemistry 2e O
H
Me2N
OH
NaH
O
Me2N
F
O
O
H
Me2N
H
Me2N
O
O
F
PROBLEM 2 Draw a mechanism for this reaction. Why is base unnecessary? CN
PhPH2
CN
+
Ph
excess
P
CN Purpose of the problem Simple example of conjugate addition with a nucleophile from the second row of the periodic table. Suggested solution The phosphine is a good soft nucleophile with a high energy lone pair, well able to add in a conjugate fashion without help. In particular, the neutral phosphine does not need to be converted into its anion. The intermediate is a good base and removes a proton from itself, not necessarily intramolecularly. N
C
N
C
PhPH2
Ph
P
H
H
CN
Ph
P
and
H repeat
Solutions for Chapter 22 – Conjugate Addition and Nucleophilic Aromatic Substitution 3 PROBLEM 3 Which of the two routes suggested here would actually lead to the product? 1. EtMgBr, 2.HCl
O
OH
?
Cl
1. HCl 2. EtMgBr
Purpose of the problem Do you understand the essentials of conjugate addition? Can you say when it won’t happen? Suggested solution To get the product, the chloride must add in a conjugate fashion and ethyl Grignard in a direct fashion that removes the carbonyl group. Conjugate addition can happen only if the carbonyl group is intact so HCl must be added first. O
HCl
OH
O
OH
EtMgBr
Cl
Cl
Cl
In the other sequence, EtMgBr is likely to add to the carbonyl group direct and further addition of HCl may either substitute on the allylic alcohol or add the ‘wrong way round’ to the alkene. O
EtMgBr
?
OH
HCl
?
or
OH
Cl
Cl
OH
Cl 4 Solutions Manual to accompany Organic Chemistry 2e PROBLEM 4 Suggest reasons for the different outcome of each of these reactions. Your answer must of course be mechanistically based. Me3SiO
Et3N, Me3SiCl
O
OH
LiAlH4
O
R2NH, RCO2H
NR2 Purpose of the problem A reminder of the reactions possible with enones. Suggested solution The three reactions are: enolization and trapping with silicon, direct addition with a hard irreversible nucleophile, and conjugate addition with a softer reversible nucleophile. Cl
O
H3Al
Et3N
H
SiMe3 O
Me3SiO
O
O
work-up
OH
H, H2O
O
OH
O
NHR2
±H
NR2
±H
NR2
Solutions for Chapter 22 – Conjugate Addition and Nucleophilic Aromatic Substitution 5 PROBLEM 5 Suggest a mechanism for this reaction. O
O
AlCl3
+
aqueous
work-up
Purpose of the problem Combination of conjugate addition and electrophilic aromatic substitution. Suggested solution The weakly nucleophilic benzene has evidently added in conjugate fashion to the enone in a kind of Friedel-­‐Crafts reaction and we can use the Lewis acid to make the enone into the necessary cation. O
AlCl3
O
AlCl3
AlCl3
O
O
aqueous
work-up
H
PROBLEM 6 Suggest a mechanism for this reaction, explaining the selectivity. F
F
O
F
CO2Et
Me
CO2Et
NH
+
F
F
O
F
F
F
N
N
Me
N
F
Purpose of the problem Introduction to the mechanism and selectivity of nucleophilic aromatic substitution. 6 Solutions Manual to accompany Organic Chemistry 2e Suggested solution Both ortho and para positions are activated by the ketone towards nucleophilic attack by the amine, but the para position is preferred because of steric hindrance between the large heterocyclic ring and the ketone. The substitution works because those five fluorine atoms make the ring very electron-­‐deficient. F
F
O
F
CO2Et
NH
Me
N
N
F
F
Me
F
O
F
N
CO2Et
F
product
F
F
PROBLEM 7 Pyridine is a six-­‐electron aromatic system like benzene. You have not yet been taught anything systematic about pyridine (that will come in chapter 29) but see if you can work out why 2-­‐ and 4-­‐chloropyridines react with nucleophiles but 3-­‐chloropyridine does not. Cl
RNH2
RNH2
N
Cl
NHR
N
N
NHR
N
4-chloropyridine
2-chloropyridine
Cl
RNH2
no reaction
N
3-chloropyridine
Purpose of the problem Extension of the ideas on nucleophilic aromatic substitution into new compounds. Suggested solution The problem is to find somewhere to park the negative charge in the intermediate and the only possible place is on the pyridine nitrogen atom. This is easy with 2-­‐ and 4-­‐choropyridine but impossible with 3-­‐
chloropyridine. Using a general nucleophile: Solutions for Chapter 22 – Conjugate Addition and Nucleophilic Aromatic Substitution 7 Cl
X
Cl
X
X
X
N
X
N
Cl
N
Cl
X
N
N
N
X
?
Cl
N
can't get
electrons
to N
Amine formation by this reaction is particularly important as you will see in chapters 29 and 30. The mechanism is the same with a few proton transfers. Cl
H2NR
N
N
Cl
N
R
N
N
H
H
H
R
N
H
NHR
PROBLEM 8 How would you carry out these two conversions? OMe
OMe
?
OMe
CN
?
H2N
Me
Me
Me
Purpose of the problem Application of nucleophilic aromatic substitution in synthesis. Suggested solution Usually you would think of introducing NH2 by nitration and reduction (chapter 21), but the regioselectivity is wrong for the first reaction: the methoxy group will direct nitration ortho to itself. An alternative is to introduce both NH2 and CN as nucleophiles, but the ring is unactivated so we can’t use the addition-­‐elimination mechanism (there is nowhere for the negative charge to go). The successful alternatives are electrophilic aromatic substitution followed by diazonium salt formation and the benzyne method. Here are two possible routes. 8 Solutions Manual to accompany Organic Chemistry 2e Nitration will insert the nitro group ortho to the more strongly electron-­‐
donating MeO group. Reduction, diazotization and substitution with copper cyanide by the SN1 mechanism gives one product. OMe
OMe
OMe
NO2
HNO3
OMe
NH2
H2/Pd
OMe
N2
HONO
CN
CuCN
H2SO4
Me
Me
Me
Me
Me
The other product could come from chlorination, elimination to give a benzyne, addition of amide anion to put the anion ortho to MeO (p. 524 in the textbook) and protonation. OMe
OMe
Cl2
Me
Cl
OMe
OMe
OMe
NaNH2
Me
NH2
H2N
H2N
Me
Me
Me
PROBLEM 9 Suggest mechanisms for these reactions, pointing out why you chose the pathways. CN
CN
Cl
NaNH2
OMe
OMe
Br
O2N
NaCN
NH2
NC
O2N
NH2
Purpose of the problem Studies in selectivity and choosing the right mechanism. Suggested solution In the first reaction, the nucleophile adds in the ‘wrong’ position (i.e. where the leaving group isn’t) so a benzyne mechanism is likely. Notice Solutions for Chapter 22 – Conjugate Addition and Nucleophilic Aromatic Substitution 9 that the nucleophile and the benzyne are formed with the same strong base, that the anion is recycled and that the nucleophile adds to the benzyne to put the negative charge next to OMe (p. 524 in the textbook). CN
H
NH2
anion is recycled
CN
Et
product
H
Cl
OMe
OMe
OMe
Et
CN
The second reaction is a straightforward substitution by the addition-­‐
elimination mechanism activated by the nitro group. The amino group is a spectator. CN
Br
Br
O
O
N
O
NH2
N
O
CN
NC
NH2
O2N
NH2