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http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 17 – Elimination reactions
167
PROBLEM 9
Only one of these bromides eliminates to give alkene A. Why? Neither alkene
eliminates to give alkene B. Why not?
O
O
Br
base
O
O
alkene A
O
O
Br
O
O
alkene B
×
base
Purpose of the problem
Helping you to understand that cage molecules often have res... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | 168
Solutions Manual to accompany Organic Chemistry
PROBLEM 10
Account for the constrasting results of these two reactions.
Br
CO2H
base
CO2H
Br
CO2H
base
Purpose of the problem
How configuration controls mechanism, and an alternative type of
elimination.
Suggested solution
The two compounds differ only in th... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 |
Suggested solutions for Chapter 18
PROBLEM 1
A compound C6H5FO has a broad peak in the infrared at about 3100-3400 cm–1
and the following signals in its (proton decoupled) 13C NMR spectrum. Suggest
a structure for the compound and interpret the spectra.
δC (ppm) 157.4 (d, J 229 Hz), 151.2 (s), 116.3 (d J 7.5... |
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Solutions Manual to accompany Organic Chemistry
PROBLEM 2
The natural product bullatenone was isolated in the 1950s from a New Zealand
myrtle and assigned the structure A. Then authentic compound A was
synthesized and found not to be identical to natural bullatenone. Predict the
expected 1H NMR spectrum of ... |
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171
The last four are esters (cyclic esters or lactones) and they would have a
C=O frequency at 1745–1780 cm–1 so D–G are all wrong. The hydrogen on
the alkene cannot be next to oxygen as it would have a very large chemical
shift indeed whereas it is close... |
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Solutions Manual to accompany Organic Chemistry
Suggested solution
Compound A has a carbonyl group (IR) that is an acid derivative (179 ppm
in the 13C NMR). The 9H singlet in the proton NMR must be a t-Bu group
and the 3H singlet at 3.67 ppm must be an OMe group. Putting these four
fragments together we get... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 18 – Review of spectroscopic methods
173
OMe
HO
SMe
1. SOCl2, Et3N
2. H2O
×
MeO
O
CHO
Me
Me
Me
O
O
CHO
Me
Me
Only one compound is possible if we have an S atom—this fits the data
very much better and indeed is the correct structure. It has a genuine
aldehyde (not a formate ester) and SMe fit... |
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Solutions Manual to accompany Organic Chemistry
O
O
3.9
2H, s
1.02
6H, s
1.66
2H, t
2.51
2H, t
The mass spectrum of compound B shows that it has chlorine in it, the IR
shows a CN group and the proton NMR shows eight Hs. If we assume that
no carbons have been lost, the most reasonable formula is C5H8ClNS. Th... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 18 – Review of spectroscopic methods
175
PROBLEM 5
Two alternative structures are shown for the products of these reactions.
Explain in each case how you would decide which product is actually formed.
Several pieces of evidence will be required and estimated values are better
than general st... |
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Solutions Manual to accompany Organic Chemistry
1680 cm–1). The aldehyde proton and the number of protons next to oxygen
make a clear distinction. There will also be differences in the 1H and 13C
NMR signals of the benzene rings as one is conjugated to a C=O group and
the other is not. This reaction actually... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 18 – Review of spectroscopic methods
177
F
O
O
O
D2O
F
O
O
H
H
DO
OD
δC 86.1 (d)
J 171 Hz
δH 4.3 (d)
δC 176.1 (d)
J 2Hz
δC 83.5 (d)
J 22 Hz
PROBLEM 7
An antibiotic isolated from a microorganism was crystallized from water and
formed different crystalline salts in either acid or base. The sp... |
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Solutions Manual to accompany Organic Chemistry
N
HO
CO2H
NH2
N
HO
CO2H
NH3
H
H
H
H
H
H
H
8.37, 1H, m
8.02, 2H, m
4.57, 1H, t
3.67, 2H,d
azatyrosine
PROBLEM 8
Suggest structures for the products of these two reactions.
CO2H
HO2C
O
O
+
A
PhNH2
B
Compound A:
m/z 170 (M+, 1%), 84 (77%), and 66 (100%);
IR 1... |
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179
be the C of CMe2. C8H10O4 would have four double bond equivalents, so the
last two degrees of unsaturation must be rings. The cyclopropane provides
one and the other must link the two oxygen atoms in the second part-
structure. So we have:
CO2H
HO2C
O
... |
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Solutions Manual to accompany Organic Chemistry
of 3.2 ppm) or to the carbonyl group (estimated shift 2.2 ppm) of the amide.
The observed value (3.92) fits the first better. A similar estimate for the CH
gives the same answer and the first structure is indeed correct.
O
O
O
O
PhNH2
N
N
O
CO2H
O
HO2C
or
P... |
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181
O
O
TsNHNH2
H
H
H
O
H
O
2.02, 1H, t
2.28,
2H, d
J = 3 Hz
1.12, 6H, s
2.50,
2H, s
2.15
3H, s
3290 cm–1
2115
cm–1
71
82
1710 cm–1
PROBLEM 10
Reaction of the epoxy-alcohol below with LiBr in toluene gave a 92% yield of
compound A. Suggest a structure f... |
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Solutions Manual to accompany Organic Chemistry
PROBLEM 11
Female boll weevils (a cotton pest) produce two isomeric compounds that
aggregate the males for food and sex. A few mg of two isomeric active
compounds, grandisol and Z-ochtodenol, were isolated from 4.5 million
insects. Suggest structures for these... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 18 – Review of spectroscopic methods
183
PROBLEM 12
Suggest structures for the products of these reactions.
O
1. Me2PCl
2. base
A
OTs
O
MeOH
MeO
B
Compound A:
C10H13OP, IR (cm–1) 1610, 1235;
δH (ppm) 6.5-7.5 (5H, m), 6.42 (1H, t, J 17 Hz), 7.47 (1H, dd, J 17, 23 Hz), and 2.43
(6H, d, J 25... |
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Solutions Manual to accompany Organic Chemistry
O
1. Me2PCl
2. base
A
OTs
O
MeOH
MeO
B
O
MeO
H
H
Me
Me
H
H
Ph2P
O
6.5-7.5 (5H, m)
7.25 (5H, s)
2.96 (3H, s)
0.76 (3H, s)
1.26 (3H, s)
4.28 (1H, d)
3.91 (1H, d)
6.42 (1H, t)
7.47 (1H, dd, J 23, 17)
J 17
17
J 23
PROBLEM 13
Identify the compounds produced in th... |
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185
PROBLEM 14
The yellow crystalline antibiotic frustulosin was isolated from a fungus in 1978
and it was suggested the structure was an equilibrium mixture of A and B.
Apart from the difficulty that the NMR spectrum clearly shows one compound
and not an... |
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Solutions Manual to accompany Organic Chemistry
cannot be assigned with confidence. At least the revised structure is one
compound and not two.
OH
OH
CHO
H
H
H
Me
H
OH
CHO
H
Me
H
10.31
6.88
7.16
2.06
5.44, 5.52
22.8
80.8
100.6
195.6
OH
151.8 or
154.5
151.8 or
154.5
The true structure was later
described... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 |
Suggested solutions for Chapter 19
PROBLEM 1
Predict the orientation of HCl addition to these alkenes.
Purpose of the problem
Simple examples of addition with regioselectivity.
Suggested solution
The first and last alkenes have different numbers of substituents at each end
of the alkene and will give the m... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | 188
Solutions Manual to accompany Organic Chemistry
PROBLEM 2
Suggest mechanism and products for these reactions.
Br2
?
Br2
?
Purpose of the problem
Checking that you understand the bromination mechanism.
Suggested solution
The question of what product is formed is easily answered as we know
bromine adds tra... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 19 – Electrophilic addition to alkenes
189
PROBLEM 3
What will be the products of the addition of bromine water to these alkenes?
H2O
Br2
?
Br2
?
H2O
Purpose of the problem
Checking that you understand the bromonium ion mechanism with an
external nucleophile.
Suggested solution
The bromo... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | 190
Solutions Manual to accompany Organic Chemistry
PROBLEM 4
By working at low temperature with one equivalent of buffered solution of a
peroxy-acid, it is possible to prepare the monoepoxide of cyclopentadiene.
Why are these precautions necessary and why does a second epoxidation not
occur under these conditi... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 19 – Electrophilic addition to alkenes
191
PROBLEM 5
The synthesis of a tranquilizer uses this step. Give mechanisms for the
reactions.
HN
CF3
Cl
1. HBr
2. H2O
HN
CF3
Cl
OH
Purpose of the problem
An electrophilic addition followed by a substitution: revision of chapter 17.
Suggested solut... |
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Solutions Manual to accompany Organic Chemistry
PROBLEM 6
Explain this result:
1. Br2
2. KCN, base
CN
Purpose of the problem
An electrophilic addition followed by an elimination (revision of chapter 17)
and a substitution (revision of chapter 15).
Suggested solution
Addition of bromine occurs first to gi... |
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193
Suggested solution
The mechanism is formation of the bromonium ion and nucleophilic attack
by the OH group at the more substituted carbon. The product has one
carbon with two methyl groups, of which one must be axial and one
equatorial. The remainin... |
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Solutions Manual to accompany Organic Chemistry
Suggested solution
Hydration of the top alkene with aqueous acid goes via a carbocation
intermediate, and would therefore give the product on the left. To get the
product on the right we need a more roundabout method, involving
hydroboration to place boron on ... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 |
Suggested solutions for Chapter 20
PROBLEM 1
Draw all possible enol forms of these carbonyl compounds and comment on
their stability.
O
O
O
O
H
H
Purpose of the problem
Simple exercise in drawing enols with an extra twist.
Suggested solution
There is only one enol for the first compound and it might be sta... |
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Solutions Manual to accompany Organic Chemistry
PROBLEM 2
The proportions of enol in a neat sample of the two ketones below are rather
different. Why is this?
O
O
CO2Et
0.0004% enol
62% enol
Purpose of the problem
Thinking about the stability of enols bearing stabilizing substituents.
Suggested solution
... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 20 – Formation and reactions of enols and enolates
197
PROBLEM 3
The NMR spectrum of this dimethyl ether is complicated: the two MeO groups
are different as are all the hydrogen atoms on the rings. However the diphenol
has a very simple NMR spectrum—there are only two types of proton on the
... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | 198
Solutions Manual to accompany Organic Chemistry
PROBLEM 4
Suggest mechanisms for these reactions:
H
H2O
O
Br
1. Mg
2. PhCN
O
3. H+, H2O
Br2
HOAc
O
Br
O
O
Purpose of the problem
Revision of mechanisms plus exploration of reactions involving enols.
Suggested solution
The first reaction starts with acetal hyd... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 20 – Formation and reactions of enols and enolates
199
PROBLEM 5
Suggest mechanisms for these reactions and explain why these products are
formed.
O
O
O
O
HClO4
O
O
O
Br
Br2
AcOH
Purpose of the problem
Making sure that you understand why enols usually react through carbon
but sometimes ... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | 200
Solutions Manual to accompany Organic Chemistry
PROBLEM 6
1,3-Dicarbonyl compounds such as A are usually mostly enolized. Why is this?
Draw the enols available to compounds B–E and explain why B is 100% enol but
C, D, and E are 100% ketone.
OEt
O
O
A
1
3
O
O
O
O
O
O
O
O
O
O
B
C
D
E
Purpose of the problem
E... |
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201
O
O
O
O
O
O
HO
O
O
O
O
OH
O
O
O
OH
The remaining compounds have problems with enolization. Compound
C can form an enol on the side away from the other carbonyl group, but
cannot form an enol between the two ketones as it would be a ‘brid... |
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Solutions Manual to accompany Organic Chemistry
follow the mechanism through with this acid. Enolization gives the
nucleophile, which picks up NO+.
OH
O
H
OH
OH
N
O
OH
N
O
O
+H+
–H+
This nitroso compound can’t tautomerize because there are no adjacent
protons, so instead the compound forms an oxime by losin... |
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203
OH
H
O
OH
H
OH
OH
H
O
enol
HO
CHO
HO
CHO
both equatorial: more stable
CHO group axial
|
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Suggested solutions for Chapter 21
PROBLEM 1
All you have to do is to spot the aromatic rings in these compounds. It may not
be as easy as you think and you should give some reasons for questionable
decisions.
O
I
HO
I
I
I
CO2H
NH2
H
OH
O
O
OH
OH
CO2H
OH
thyroxine: human hormone
regulating metabolic rate
akla... |
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Solutions Manual to accompany Organic Chemistry
O
I
HO
I
I
I
CO2H
NH2
H
OH
O
O
OH
OH
CO2H
OH
1
2
3
Colchicine has one benzene ring and a seven-membered conjugated ring
with six electrons in double bonds (don’t count the carbonyl electrons as
they are out of the ring). It perhaps looks more aromatic if you del... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 21 – Electrophilic aromatic substitution
207
PROBLEM 2
First, as some revision, write out the detailed mechanism for these steps.
HNO3 + H2SO4
NO2
NO2
NO2
In a standard nitration reaction with, say, HNO3 and H2SO4, each of these
compounds forms a single nitration product. What is its struct... |
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Solutions Manual to accompany Organic Chemistry
NHAc
Me
O
NO2
O2N
NHAc
Me
O
NO2
NO2
PROBLEM 3
How reactive are the different sites in toluene? Nitration of toluene produces
the three possible products in the ratios shown. What would be the ratios if all
the sites were equally reactive? What is the actual r... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 21 – Electrophilic aromatic substitution
209
Me
NO2
Me
NO2
H
Me
NO2
Me
NO2
H
Me
NO2
H
etc
reaction in the ortho position
reaction in the meta position
positive charge is
never adjacent to Me
PROBLEM 4
Draw mechanisms for these reactions and explain the positions of substitution.
OH
HNO3
OH
... |
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OH
NO2
OH
H
NO2
OH
NO2
Br
Br
OH
NO2
Br
OH
NO2
Br
H
The second example has two Friedel-Crafts alkylations with tertiary alkyl
halides. The first occurs para to bromine, a deactivating but ortho,para-
directing group (see p. 489 in the textbook), preferring para be... |
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211
Suggested solution
The first product has only eight hydrogens so two nitro groups must have
been added. The molecule is clearly symmetrical and the coupling constant
is right for neighbouring hydrogens so a substitution on each ring must have
occu... |
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Solutions Manual to accompany Organic Chemistry
F
NO2
F
H
NO2
F
NO2
H
H
7.15 (2H, dd, J 7,8)
8.19 (2H, dd, J 6,8)
PROBLEM 6
Attempted Friedel-Crafts acylation of benzene with t-BuCOCl gives some of the
expected ketone A as a minor product, as well as some t-butylbenzene B, but the
major product is the subst... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 21 – Electrophilic aromatic substitution
213
The main product C comes from the addition of both these electrophiles,
but which adds first? The ketone in A is deactivating and meta directing but
the t-butyl group in B is activating and para-directing so it must be added
first.
+
Cl
O
AlCl3
O
... |
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N
H
HNO3
H2SO4
N
H
O2N
N
H
O2N
or
Ha
Hb
You could argue that NH is ortho, para-directing and so the second
structure is more likely. But this is a risky argument as the reaction is carried
out in strong acid solution where the nitrogen will mostly be protonated.... |
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215
OH2
MeO
H
MeO
MeO
H
MeO
MeO
MeO
A
B
Bromination will go either ortho or para to the methoxy group: A has
two different positions ortho to the OMe, but the para position is blocked.
The least sterically hindered position gives a 1,2,4,5-tetrasubs... |
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PROBLEM 9
On p. 479 of the textbook we explain the formation of 2,4,6-tribromophenol by
bromination of phenol in water. It looks as though we can go no further as all
the ortho and para positions are brominated. But we can if we treat the
tribromo-compound with ... |
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217
In the second reaction, one of the reactive bromines in the para position
is transferred to the amine. It could have added ortho or para to the NMe2
group but CF3 is small and NMe2 is large, because the two methyl groups lie
in the plane of the rin... |
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O
MeCOCl
AlCl3
O
NO2
HNO3
H2SO4
Sn, HCl
O
NH2
MeCOCl
AlCl3
HNO3
H2SO4
NO2
The second compound has a bromo substituent, which is ortho,para-
directing, and a meta-directing nitro group. We need the para relationship,
so we must put the bromine in first, then nitr... |
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Suggested solutions for Chapter 22
PROBLEM 1
Draw a mechanism for this reaction. Why is base unnecessary?
PhPH2
+
Ph
P
CN
CN
CN
excess
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 ... |
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PROBLEM 2
Which of the two routes suggested here would actually lead to the product?
O
1. EtMgBr, 2.HCl
1. HCl 2. EtMgBr
?
Cl
OH
Purpose of the problem
Do you understand the essentials of conjugate addition? Can you say when
it won’t happen?
Suggested soluti... |
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221
PROBLEM 3
Suggest reasons for the different outcome of each of these reactions. Your
answer must of course be mechanistically based.
O
Me3SiO
Et3N, Me3SiCl
LiAlH4
R2NH, RCO2H
OH
O
NR2
Purpose of the problem
A reminder of th... |
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PROBLEM 4
Suggest a mechanism for this reaction.
+
AlCl3
aqueous
work-up
O
O
Purpose of the problem
Combination of conjugate addition and electrophilic aromatic substitution.
Suggested solution
The weakly nucleophilic benzene has evidently added in conjugate ... |
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223
HF). The aldehyde is still there (from the IR and the proton at 10 ppm) so
the spectra are best interpreted by this structure:
O
H
O
Me2N
O
H
O
Me2N
132/130/115
191
29
41 64
164
2.32
3.05
4.20
9.97
7.82
6.97
δC
δH
That sugge... |
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substitution works because those five fluorine atoms make the ring very
electron-deficient.
F
F
F
F
F
CO2Et
O
N
NH
Me
F
F
N
F
F
CO2Et
O
N
Me
F
product
PROBLEM 7
Pyridine is a six-electron aromatic system like benzene. You have not yet been
taught anything sy... |
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225
N
Cl
X
N
Cl
X
N
X
N
Cl
X
N
Cl
X
N
X
N
Cl
X
?
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.
... |
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the more strongly electron-donating MeO group. Reduction, diazotization
and substitution with copper cyanide by the SN1 mechanism gives one
product.
OMe
Me
HNO3
H2SO4
OMe
Me
NO2
H2/Pd
OMe
Me
NH2 HONO
OMe
Me
N2 CuCN
OMe
Me
CN
The other product could come from ... |
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227
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).
Cl
OMe
H
NH2
OMe
CN
Et... |
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electron-withdrawing group. The conjugate addition must always occur first
so as to keep the carbonyl group intact for the second step.
O
NaBH4
OH
NaBH4×
OH
cyclo-
pentenone
cyclo-
pentanol
PROBLEM 11
Stirring thioacetic acid with acrolein (propenaldehyde) in ... |
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Suggested solutions for Chapter 23
Several problems in this chapter ask you to suggest ways to carry out
conversions of one molecule into another. We always give one possible
answer and sometimes comment on alternatives but you should realize that
there are usually many possible ‘right’ answers to questions of ... |
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Solutions Manual to accompany Organic Chemistry
the Grignard reagent from the aryl bromide. Then we can add to RCHO,
and deprotect with acid, and we have our product.
CHO
Br
MeOH
H
Br
OMe
MeO
1. Mg
Et2O
MgBr
OMe
MeO
2. RCHO
OMe
MeO
HO
R
H
H2O
CHO
HO
R
PROBLEM 2
How would you convert this lactone selectivel... |
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231
PROBLEM 3
Predict the products of Birch reduction of these aromatic compounds.
O
MeO
O
OMe
Purpose of the problem
Exploring the principles of Birch reduction.
Suggested solution
In each case a unique product results if you draw a dianion i... |
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Solutions Manual to accompany Organic Chemistry
Suggested solution
Every step is straightforward except the final reduction where a less reactive
ester must be reduced in the presence of a more reactive ketone. Protection
is the answer and an acetal is suitable.
H
OH
CO2Et
OH
CO2Et
O
CO2Et
H2, Ni
or PtO2
Cr... |
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233
O
NO2
CO2H
H
HO
OH
H2, Pd
HOAc
LiAlH4
NO2
CO2H
O
O
NH2
CO2H
O
O
H
N
O
O
O
H
N
O
O
H
N
O
O
H , H2O
PROBLEM 6
Why is this particular amine formed by reductive amination here?.
MeO
N
H
OMe
OMe
OMe
Na(CN)BH3
pH 5.5
N
OHC
Purpose of the problem... |
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Solutions Manual to accompany Organic Chemistry
PROBLEM 7
Account for the chemoselectivity of the first reaction and the stereoselectivity
of the second. A conformational drawing of the intermediate is essential.
O
H
HO
OH
H2, Pd/C
O
O
H
O
O
O
O
O
Purpose of the problem
Extending
chemoselectivity
into
mo... |
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235
PROBLEM 8
How would you convert this diamine to either of these two protected
derivatives?
H2N
NH2
H2N
NHBoc
BocHN
NH2
or
Purpose of the problem
Using protecting groups to reveal reactive groups selectively.
Suggested solution
The Boc group... |
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Suggested solutions for Chapter 24
PROBLEM 1
Two routes are proposed for the preparation of this amino-alcohol. Which do
you think is more likely to succeed and why?
NH
CHO
2. NaBH4
1.
N
OH
NH
CO2Me
2. LiAlH4
1.
Purpose of the problem
Practical application of the choice of reagent to ensure the correct
re... |
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Solutions Manual to accompany Organic Chemistry
Suggested solution
Both reactions involve addition of organometallic compounds to
unsaturated carbonyl compounds. The key difference is the metal. With
Cu(I) as catalyst, the Grignard reagent will give conjugate addition in the
first case. MeLi will give direct ... |
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239
Br
Br
Br2, AlCl3
Br
Br
Br
H
Br
In the presence of light, bromine’s weak Br–Br bond undergoes
homolysis, and Br• radicals are formed. One of these can abstract a
hydrogen atom, breaking the weakest C–H bond. The methyl groups’ C–H
bonds are weaker than those of th... |
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Solutions Manual to accompany Organic Chemistry
NO2
H
NO2
H
NO2
What about the alternatives? A similar cation is formed if the electrophile
attacks the position labelled ‘1’, but the nitro group is in a more hindered
position here, so we don’t expect this to contribute much to the product
mixture. Position... |
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241
ortho to the tether. The one it chooses is the less hindered. It is also para to
an electron-donating methoxy group, so the reaction works well.
MeO
MeO
NH
R
MeO
MeO
NH2
RCHO MeO
MeO
N
R
MeO
MeO
NH
R
In the second case, there is only one methoxy group, and both the... |
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Br
O
CO2H
A
O
O
O
AlCl3
Br
Br
O
CO2
H
The next step is the reduction we introduced on p. 540 of the textbook,
the ‘Wolff-Kishner’ reduction. The mechanism is there so we need not
repeat it here; the product is the acid B (or, rather, its potassium salt). Now
... |
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243
Suggested solution
Both reactions involve ortholithiation—deprotonation of the aromatic ring
to form an intermediate aryllithium. As we explain on pp. 563–4 of the
textbook, the deprotonation occurs where the BuLi can be ‘guided in’ by
coordinating oxygen atoms. Th... |
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Solutions Manual to accompany Organic Chemistry
PROBLEM 8
Comment on the regioselectivity and chemoselectivity of the reactions in the
sequence below.
N
OH
N
OH
O
MeO
BnBr
N
OH
Ph
Br
NaBH4,
EtOH
N
OH
Ph
NaHCO3
MeO
Cl
O
Purpose of the problem
Practice using simple principles of reactivity to explain why nucl... |
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245
N
OH
Ph
H
BH3
N
OH
Ph
H
N
OH
Ph
H
H3B
N
OH
Ph
In the final step, another electrophile is added: it’s an acid chloride,
though a slightly unusual one, usually called ‘methyl chloroformate’. As in
the first step, the most nucleophilic atom is the pyridine N, so we us... |
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Solutions Manual to accompany Organic Chemistry
Suggested solution
The first reaction is a Friedel-Crafts acylation. There are two rings and two
carbonyl groups, so we must explain first of all the choice between each of
these pairs. One ring is chlorinated: chlorine has a deactivating effect on
electrophil... |
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247
Suggested solution
There are lots of ortho relationships in this compound! And somehow we
have to join the two aromatic rings together to make an ether. This can only
really be done by nucleophilic aromatic substitution, so we need to look for
an electron-withdraw... |
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Suggested solutions for Chapter 25
PROBLEM 1
Suggest how these compounds might be made by alkylation of an enol or
enolate.
EtO2C
CO2Et
O
Purpose of the problem
An exercise in choosing good routes to simple compounds.
Suggested solution
As you can see from the carbonyl groups in these compounds, it is... |
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Solutions Manual to accompany Organic Chemistry
PROBLEM 2
How might these compounds be made using alkylation of an enol or enolate as
one step in the synthesis?
O
O
O
O
Purpose of the problem
An exercise in using enolate chemistry to make carbonyl compounds
disguised as acetals.
Suggested solution
The ... |
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251
EtO2C
CO2Et
H
H
1. EtO
2. PhCH2Br
3. EtO
4. PhCH2Br
Ph
Ph
EtO2C
CO2Et
LiAlH4
Ph
Ph
OH
OH
O
EtO2C
O
EtO2C
HO
HO
1. EtO
2. EtBr
LiAlH4
Me2CO
H
acetal
PROBLEM 3
How might these amines be prepared using enolate-style alkylation as part of the
synthesis?
NH2
R
NH2... |
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Solutions Manual to accompany Organic Chemistry
PROBLEM 4
This attempted enolate alkylation does not give the required product. What has
gone wrong? What products would actually be formed? How would you make the
required product?
Me
CHO
1. BuLi
2. i-PrCl
CHO
×
Purpose of the problem
An exercise in trouble-... |
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253
PROBLEM 5
Draw mechanisms for the formation of this enamine, its reaction with the alkyl
halide, and the hydrolysis of the product.
O
O
H
H2O
N
O
O
Br
O
N
O
N
H
O
cat H
O
Purpose of the problem
Exploration of the details of enamine formation and reaction. Th... |
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Solutions Manual to accompany Organic Chemistry
The reaction of the enamine with the alkyl halide goes as expected—these
very good SN2 electrophiles work particularly well with enamines and the
first product under the reaction conditions is another enamine.
O
Br
N
O
O
N
O
O
H
N
O
Finally the enamine is hydr... |
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255
Suggested solution
The last two ketones have two different α-positions so there is a good
chance of controlling enol formation from the parent ketone. But the first
ketone has two primary α-positions and the difference appears only in the
two β-positions. The ... |
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Solutions Manual to accompany Organic Chemistry
In the reaction with the primary alkyl halide, the boot is on the other foot
as there will be a good reaction with the lithium enolate but no reaction with
the more stable silyl enol ethers. Lewis acid won’t help here either as
primary cations are unstable. Preli... |
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257
PROBLEM 9
How would the imine from problem 8 react with the reagents below? Draw
mechanisms for each step: the reaction with LDA, the addition of BuBr, and the
work-up.
N
R
1. LDA
2. BuBr
3. workup
?
Purpose of the problem
Checking you know how to make and ... |
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Solutions Manual to accompany Organic Chemistry
Suggested solution
Some aldehydes can be converted directly into lithium enolates but this is
not usually very successful because the rate of reaction of the lithium enolate
with the very electrophilic aldehyde is too great and at least some aldol
reaction will ... |
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259
PROBLEM 12
How does this synthesis of a cyclopropyl ketone work?
CO2Et
O
EtO , EtOH
O
O
O
O
base
O
HBr
Br
O
Purpose of the problem
Enols and enolates are involved in an unlikely looking sequence that you can
work out if you persist.
Suggested solution
Alk... |
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Solutions Manual to accompany Organic Chemistry
PROBLEM 13
Give the structures of the intermediates in the following reaction sequence and
mechanisms for the reactions.
NH
1. NaNH2
2. MeOTs
3. LDA
4. EtBr
O
N
O
Me
Purpose of the problem
A reminder that enolate-like intermediates can be formed at nitrogen a... |
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Suggested solutions for Chapter 26
PROBLEM 1
The aldehyde and the ketone below are self-condensed with aqueous NaOH so
that an unsaturated carbonyl compound is the product in both cases. Give a
structure for each product and explain why you think this product is formed.
NaOH
H2O
?
CHO
O
NaOH
H2O
?
Purpose o... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | 262
Solutions Manual to accompany Organic Chemistry
cannot dehydrate
can dehydrate
O
O
O
O
H
H
OH
O
OH
PROBLEM 2
Propose mechanisms for the ‘aldol’ and dehydration steps in the termite defence
compound presented on p. 623 in the textbook.
R
H
O
CH2NO2
NaOMe
R
OH
NO2
Ac2O
pyridine
R
NO2
Purpose of the problem... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 26 – The aldol and Claisen reactions
263
PROBLEM 3
How would you synthesize the following compounds?
Ph
Ph
O
CO2H
Purpose of the problem
Application of the aldol reaction to make unsaturated carbonyl compounds.
Suggested solution
Just find the conjugated alkene and so find the hidden carb... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | 264
Solutions Manual to accompany Organic Chemistry
PROBLEM 4
How would you use a silyl enol ether to make this aldol product? Why is it
necessary to use this particular intermediate? What would be the products be if
the two carbonyl compounds were mixed and treated with base?
CHO
CHO
+
?
CHO
Purpose of the pro... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 26 – The aldol and Claisen reactions
265
PROBLEM 5
In what way does this reaction resemble an aldol reaction? Comment on the
choice of base. How can the same product be made without using phosphorus
chemistry?
(EtO)2P
O
O
EtCHO
K2CO3 water
O
Purpose of the problem
Showing that there are ... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | 266
Solutions Manual to accompany Organic Chemistry
O
1. LDA
2. Me3SiCl
OSiMe3
EtCHO
TiCl4
O
OH
PROBLEM 6
Suggest a mechanism for this attempted aldol reaction. How could the aldol
product be made?
expected aldol product
O
OH
OH
O
OH
CH2O
NaOH
Purpose of the problem
A demonstration of one way that aldol r... |
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