url stringclasses 71
values | filename stringclasses 71
values | success bool 1
class | page_count int64 1 917 | pdf_text stringlengths 0 4.81k |
|---|---|---|---|---|
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 33 – Diastereoselectivity
367
Me3SiO
O
LDA
Me3SiO
OLi
RCHO Me3SiO
Me
O
Li O
R
Li
O
O
Me3SiO
R
H
H
Me3SiO
O
OH
R
PROBLEM 6
Explain how the stereochemistry of this epoxide is controlled.
CO2H
1. I2, NaHCO3
2. MeO
CO2H
O
Purpose of the problem
An example of the important iodolactonization ... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | 368
Solutions Manual to accompany Organic Chemistry
PROBLEM 7
Explain how these reactions give different isomers of the same product.
Ph
Ph
O
Ph
Ph
OH
Ph
H
O
PhMgBr
LiAlH4
Purpose of the problem
Practice at analysing stereochemical control using the Felkin-Anh model.
Suggested solution
In each case we have ... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 33 – Diastereoselectivity
369
PROBLEM 8
Explain the stereoselectivity of this reaction. What isomer of the epoxide would be
produced by treatment of the product with base?
O
Cl
NaBH4
OH
Cl
Purpose of the problem
A stereoelectronically controlled Felkin-Anh analysis.
Suggested solution
... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | 370
Solutions Manual to accompany Organic Chemistry
PROBLEM 9
How could this cyclic compound be used to produce the open-chain compound
with correct relative stereochemistry?
O
O
O
O
?
OHC
CO2Me
OH
Purpose of the problem
Practice at relating the stereochemistry of cyclic and open-chain
compounds.
Suggested ... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 33 – Diastereoselectivity
371
PROBLEM 10
How would you transform this alkene stereoselectively into either of the
diastereoisomers of the amino alcohol?
?
OH
NH2
Purpose of the problem
A more difficult extension of problem 1.
Suggested solution
Opening the epoxide with a nitrogen nucleop... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | 372
Solutions Manual to accompany Organic Chemistry
transition state contains a chair in which the methyl group has no choice but
to be axial while the aldehyde’s R substituent chooses to be equatorial.
OSiMe3
O
Li
R
O
O
Li
O
R
OSiMe3
R
OSiMe3
O
OH
We have inevitably drawn the syn aldol product as one enantiomer ... |
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373
PROBLEM 12
The following sequences show parts of the syntheses of two different HIV
protease inhibitors. What reagents are required for steps 1–4? (For steps 1 and 3,
consider carefully how the stereochemistry of the product might be controlled.)
NHBoc
Ph
OH
NH... |
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Solutions Manual to accompany Organic Chemistry
diastereoisomer is formed. Drawn below is the expected reactive
conformation for a reaction involving chelation control: note that the acidic
NH proton must be removed by any basic nucleophile. The outcome is
correct: we need to achieve chelation control, so a ... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 |
Suggested solutions for Chapter 34
PROBLEM 1
Predict the structure of the product of this Diels-Alder reaction.
CO2Me
OMe
Me3SiO
+
?
Purpose of the problem
Can you deal with a moderately complicated Diels-Alder?
Suggested solution
The diene is electron-rich and will use its HOMO in the cycloaddition. It
w... |
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Solutions Manual to accompany Organic Chemistry
MeO2C
H
OMe
Me3SiO
H
[4 + 2]
endo
MeO
Me3SiO
CO2Me
H
O
CO2Me
H
work-up
PROBLEM 2
Comment on the difference in rate between these two reactions.
O
155 °C
half-life
15 hours
O
O
65 °C
half-life
1 hour
O
Purpose of the problem
More details of the intramolecula... |
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377
the ends of their chains close together and probably lowers the LUMO
energy by conjugation with the ketone.
PROBLEM 3
Justify the stereoselectivity in this intramolecular Diels-Alder reaction.
O
O
CO2H
O
H
O
HO2C
Purpose of the problem
... |
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Solutions Manual to accompany Organic Chemistry
PROBLEM 4
Explain the formation of single adducts in these reactions.
HO
O
O
OMe
O
O
OMe
H
BnO
HO
BnO
+
OR
H
O
O
Cl
180 °C
OR
H
H
O
O
H
H
Cl
Purpose of the problem
Investigating the regio- and stereoselectivity of one inter- and one intra-
molecular Diels-Alder... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 34 – Pericyclic reactions 1: cycloadditions
379
HO
largest
HOMO
coeff.
O
O
OMe
OBn
largest
LOMO
coeff.
O
O
OMe
OBn
O
O
OMe
OBn
deactivating
conjugation
The second example is intramolecular so the regiochemistry is
determined by that alone: the ester linkage between the diene and the
dienophi... |
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Solutions Manual to accompany Organic Chemistry
O enone
aldol
1,4-diCO
O
Cl
CHO
O
2
3
4
1
CHO
+
Diels-
Alder
CHO
The Diels-Alder reaction has the right (‘para’) regioselectivity, especially if
we use a Lewis acid catalyst such as SnCl4, and we shall need a non-basic
specific enol equivalent for the alkylation... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 34 – Pericyclic reactions 1: cycloadditions
381
PROBLEM 6
Draw mechanisms for these reactions and explain the stereochemistry.
1.
N
Ph
O
2. LiAlH4
Ph
OH
NH2
Purpose of the problem
Exploration of stereochemical control by 1,3-dipolar cycoaddition reactions.
Revision of the importance of c... |
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Solutions Manual to accompany Organic Chemistry
PROBLEM 7
Give mechanisms for these reactions and explain the regio- and stereochemical
control (or lack of it!). Note that MnO2 oxidizes allylic alcohols to enones.
N
O
Ph
N
O
Ph
H
H
mixture of diastereoisomers
1. Zn,
HOAc
2. MnO2
N
O
H
Ph
+
Purpose of the pro... |
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383
more stable), but that means the phenyl group has to be axial, which is
presumably not the more stable arrangement. It seems likely that this is the
kinetic product. It looks as though the ring closes with the best overlap
between the nitrogen l... |
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Solutions Manual to accompany Organic Chemistry
PROBLEM 9
Photochemical cycloaddition of these two compounds is claimed to give the
diastereoisomer shown. The chemists who did this work claimed that the
stereochemistry of the adduct is simply proved by its conversion into a lactone on
reduction. Comment on th... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 34 – Pericyclic reactions 1: cycloadditions
385
O
H
Me
CO2Me
CO2Me
H
BH3
O
Me
CO2Me
H
OMe
O
Me
CO2Me
H
O
O
H
MeO2C
O
=
PROBLEM 10
Thioketones, with a C=S bond, are not usually stable. However, this thioketone is
quite stable and undergoes reaction with maleic anhydride to give an addition
p... |
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Solutions Manual to accompany Organic Chemistry
S
O
O
O
[8 + 2]
cycloaddition
S
O
H H
H
O
O
PROBLEM 11
This unsaturated alcohol is perfectly stable until it is oxidized with Cr(VI): it then
cyclizes to the product shown. Explain.
OH
Cr(VI)
O
H
H
Purpose of the problem
Discovery of a common effect in int... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 34 – Pericyclic reactions 1: cycloadditions
387
PROBLEM 12
Give mechanisms for these reactions, explaining the stereochemistry.
O
O
H
H
H
Et3N
O
COCl
Et3N
H
O
Purpose of the problem
Looking at [2 + 2] cycloadditions of ketenes.
Suggested solution
Treatment of acid chlorides with tertia... |
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Solutions Manual to accompany Organic Chemistry
The stereochemistry at the remaining centre comes from the way the two
molecules approach one another. The two components are orthogonal and
the dotted lines in the middle diagram below show how the new bonds are
formed. The carbonyl group of the ketene will pref... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 |
Suggested solutions for Chapter 35
PROBLEM 1
Give mechanisms for these steps, commenting on the regioselectivity of the
pericyclic step and the different regioselectivity of the two metals.
CHO
1.
2. MnO2
MgBr
O
P2O5
MeOH
O
Me2CuLi
aqueous
work-up
O
H
Purpose of the problem
A gentle introduction to an elect... |
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Solutions Manual to accompany Organic Chemistry
The final cuprate addition goes in a conjugate fashion as we should expect
as this is what Cu(I) cuprates do. The cis 5,5 ring junction is much preferred
to trans and can equilibrate on work-up by enolization.
PROBLEM 2
Predict the product of this reaction. ... |
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391
PROBLEM 3
Give mechanisms for this alternative synthesis of two fused five-membered rings.
ClOC
+
AlCl3
O
SnCl4
O
Purpose of the problem
Exploring another aspect of the Nazarov cyclization.
Suggested solution
The first stage... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | 392
Solutions Manual to accompany Organic Chemistry
PROBLEM 4
Explain what is going on here.
OTs
base
200 °C
Purpose of the problem
Two pericyclic reactions: a sigmatropic shift and a cycloaddition in one
reaction scheme.
Suggested solution
The aromatic anion of cyclopentenone displaces tosylate from the alky... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 35 – Pericyclics 2: sigmatropic and electrocyclic reactions
393
PROBLEM 5
A tricyclic hydroxyketone was made by hydrolysis of a bis silyl ether. Further
reaction gave a new compound. Explain these reactions including the
stereochemistry. The diene has the proton NMR spectum: δH 6.06 (1H, dd, ... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | 394
Solutions Manual to accompany Organic Chemistry
Heating
this
product
leads
to
a
retro
Diels-Alder
reaction:
cyclopentadiene is released and a cyclobutene is formed stereospecifically
trans. This now decomposes by a four-electron conrotatory electrocyclic
reaction that could give either the E,E- or the ... |
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395
electrocyclic reaction closes the five-membered ring and this must be
disrotatory, moving both phenyl groups up (or down).
Ph
N
N
Ph
Ph
H
base
N
N
Ph
Ph
Ph
N
N
Ph
Ph
Ph
NH
N
Ph
Ph
Ph
PROBLEM 7
This problem concerns the struct... |
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Solutions Manual to accompany Organic Chemistry
PROBLEM 8
Propose a mechanism for this reaction that accounts for the stereochemistry of
the product.
N
Ph
Ph
Ph
H
H
EtO2C
CO2Et
heat
N
EtO2C
CO2Et
Ph
Ph
Ph
Purpose of the problem
Another electrocyclic/cycloaddition combination for you to work out.
Suggest... |
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397
Purpose of the problem
An unusual electrocyclic reaction on an anion with stereochemistry and
NMR revision.
Suggested solution
The first anion A is formed by removal of the only possible proton: one
from the NCH2 group. This an... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | 398
Solutions Manual to accompany Organic Chemistry
PROBLEM 10
How would you make the starting material for these reactions? Treatment of the
anhydride with butanol gives an ester that in turn gives two inseparable
compounds on heating. On treatment with an amine, an easily separable mixture
of an acidic and a ne... |
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399
Treatment with the tertiary amine forms the anions of the carboxylic
acids. The one from B can do a conjugate addition to the unsaturated ester
and form a lactone but that of A is too far away and cannot react.
CO2H
CO2Bu
R3N
O
... |
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Solutions Manual to accompany Organic Chemistry
PROBLEM 12
Explain the following observations. Heating this phenol brings it into rapid
equilibrium with a bicyclic compound that does not spontaneously give the final
product unless treated with acid.
H
H
OH
heat
OH
acid
OH
Purpose of the problem
A sigmat... |
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401
PROBLEM 13
Treatment of cyclohexa-1,3-dione with this acetylenic amine gives a stable
enamine in good yield. Refluxing the enamine in nitrobenzene gives a pyridine
after a remarkable series of reactions. Fill in the details, give... |
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Solutions Manual to accompany Organic Chemistry
O
N
H
O
NH
C
H
O
NH
C
H
H
O
NH2
C
Now we can transfer a proton from nitrogen to the middle of the allene.
This is formally a [1,5]-H shift and is, of course, allowed, but it may be an
ionic reaction as nitrogen is involved. This gives a diene that can twist roun... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 |
Suggested solutions for Chapter 36
PROBLEM 1
Rearrangements by numbers: just draw a mechanism for each reaction.
R
O
O
O
BF3
O
O O
R
O
O
OH
O
O
OH
H
CO2Me
Br
Zn
O
H
H
OHC
BF3
Purpose of the problem
This problem is just to help you acquire the skill of tracking down
rearrangements by numbering (arbitrarily) ... |
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Solutions Manual to accompany Organic Chemistry
There is more than one reasonable mechanism: here are two possibilities,
the second being perhaps the better.
R
O
O
O
BF3
R
O
OBF3
O
O
O
R
O
F3B
O
O O
R
BF3
O
O
BF3
R
O
O
O
O
BF3
R
The second reaction is even easier to work out. Atoms 2 and 3 are easy to
find ... |
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405
CO2Me
Br
Zn
O
H
H
2
3
4
5
1
OHC
7
6
8
9
8
9
7
6
5
4
3
2
1
Zn
H
H
OMe
O
O
CO2Me
ZnBr
O
H
PROBLEM 2
Explain this series of reactions.
O
1. NH2OH
2. TsCl, pyridine
H
N
O
Purpose of the problem
Working out the stereochemistry and mecha... |
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Solutions Manual to accompany Organic Chemistry
PROBLEM 3
Draw mechanisms for the reactions and structures for the intermediates. Explain
the stereochemistry, especially of the reactions involving boron. Why was 9-BBN
chosen as the hydroborating agent?
1. 9-BBN
2. H2O2, NaOH
3. MsCl, Et3N
OMs
Na2CO3
H2O
H
H... |
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407
OMs
H
H
=
H
H
HO
H
H
H2O
Elimination of the alcohol (E1 of course as it is tertiary) gives the alkene
and a repeat of the hydroboration from the outside (convex face) of the
folded molecule gives the final alcohol with five new stereoge... |
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Solutions Manual to accompany Organic Chemistry
Suggested solution
The expected three-membered lactone would have a very high carbonyl
stretching frequency because of ring strain. Three-membered cyclic ketones
have carbonyl stretches at about 1815 cm–1 and lactones have higher
frequencies than ketones. So it ... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 36 – Participation, rearrangement and fragmentation
409
N
Me
Br
Br
N
Me
Br
HO
N
Me
OH
Br
N
O
Me
N
Me
O
H
PROBLEM 6
A single enantiomer of the epoxide below rearranges with Lewis acid catalysis to
give a single enantiomer of the product. Suggest a mechanism and comment on
the stereochemistry... |
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Solutions Manual to accompany Organic Chemistry
•
Surely the carbocation intermediate is planar and the product
would be racemic? Answer. This was the purpose of the
investigation. One chiral centre is lost in the reaction so only
absolute stereochemistry is relevant. One explanation is that the
cation is sh... |
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411
Reduction to the alcohol is trivial and then acid treatment allows the loss
of water and ring expansion of the remaining four-membered ring. You may
well have drawn this as a stepwise process. Elimination gives the most
substituted alken... |
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Solutions Manual to accompany Organic Chemistry
PROBLEM 9
Suggest mechanisms for these rearrangements, explaining the stereochemistry in
the second reaction.
H2N
OH
HCl
H2N
Br2
NH
Br
NaHCO3
NH
Purpose of the problem
Unravelling one rearrangement after another with some stereochemistry.
Suggested solution ... |
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413
Finally, the weak base bicarbonate (HCO3–) is enough to remove a proton
from the nitrogen atom and allow participation in nitrogen migration by
displacement of bromide. This alkene is formed because the C–N+ bond to
tertiary carbon is br... |
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OH
HO
TsOH
OH
H2O
OH
–H
O
?
OH
O
Breaking open one of the three-membered rings gets us off to a better
start. This gives a hydroxy-ketone that can rearrange in a pinacol fashion
with ring expansion of the remaining cyclopropane.
O
HO
H
H
HO
O
H
O
HO
H
OH
O
P... |
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415
PROBLEM 12
Treatment of this hydroxy-ketone with base followed by acid gives the enone
shown. What is the structure of intermediate A, how is it formed, and what is the
mechanism of its conversion to the final product?
OH
O
base
A
acid
... |
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Solutions Manual to accompany Organic Chemistry
PROBLEM 13
Just to check your skill at finding fragmentations by numbers, draw the
mechanism for each of these one-step fragmentations in basic solution with
acidic work-up.
MsO
HO
O
O
H
CO2H
O
HO
CO2H
O
CO2H
1. base
2. acid
1. base
2. acid
Purpose of the prob... |
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417
O
H
CO2H
O
HO
CO2H
O
CO2H
2
3
4
5
6
7
8
9
1
7
8
9
4
5
6
3
2
1
Clearly the OH proton is removed and one of the carboxyls is a leaving
group. The stereochemistry disappears in the fragmentation but it is
important, as the conformational d... |
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O
O
O
OH
O
O
O
O
O
O
The second compound follows the same sequence and a different enolate
emerges, but it is simply another enolate of the same ketone. Both
compounds give the same basic structure.
O
OAc
KOH
EtOH
O
O
OH
O
O
O
But what about stereochemistry?... |
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419
PROBLEM 15
Suggest a mechanism for this fragmentation and explain the stereochemistry of
the alkenes in the product. This is a tricky problem, but find the mechanism and
the stereochemistry will follow.
O
O
TsO
CO2
H
165 °C
O
O
Pur... |
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O
O
O
O
O
H
CO2
H
trans
O
O
PROBLEM 16
Suggest a mechanism for this reaction and explain why the molecule is prepared
to abandon a stable six-membered ring for a larger ring.
O
O
OH
NaH
O
O
O
Purpose of the problem
A simple example of fragmentation used to c... |
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421
O
O
O
O
O
O
O
O
O
H
O
O
O
O
O
OH
NaH
= product
PROBLEM 17
Give mechanisms for these reactions, commenting on the fragmentation.
O
1.
2.
CHO
H
H
H
H
N
O
1. MeI
2. NaOH
H
H
CO2H
N
H
Purpose of the problem
Revision of conjugate ad... |
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Solutions Manual to accompany Organic Chemistry
there are various ways you could draw details. Cyclization of this compound
now gives the intermediate we are looking for.
OH
N
N
OH
N
O
O
N
=
The last two diagrams show where the stereochemistry comes from. The
final product has a chair six-membered ring. The... |
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423
Purpose of the problem
Simple fragmentations involving the opening of three-membered rings.
Suggested solution
The first reaction is a fragmentation without any ‘push’ but that is all right
because the bond that is being broken is in a... |
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Solutions Manual to accompany Organic Chemistry
PROBLEM 19
What steps would be necessary to carry out an Eschenmoser fragmentation on
this ketone, and what products would be formed?
O
CO2Me
Purpose of the problem
Revision of an important and complex reaction involving fragmentation.
Suggested solution
The... |
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425
PROBLEM 20
Revision content. Suggest mechanisms for these reactions to explain the
stereochemistry.
O
HBr
Et2O
Br
OH
AgOAc
AcOH
CHO
Purpose of the problem
Rearrangements and a fragmentation.
Suggested solution
The ring opening and t... |
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Suggested solutions for Chapter 37
PROBLEM 1
Give a mechanism for the formation of this silylated ene-diol and explain why the
Me3SiCl is necessary.
CO2Me
CO2Me
Me3SiCl
Na
OSiMe3
OSiMe3
Purpose of the problem
Reminder of an important radical reaction.
Suggested solution
This is an acyloin condensation li... |
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Solutions Manual to accompany Organic Chemistry
PROBLEM 2
Heating the diazonium salt below in the presence of methyl acrylate gives a
reasonable yield of a chloroacid. Why is this unlikely to be nucleophilic aromatic
substitution by the SN1 mechanism (p. 520 of the textbook)? Suggest an
alternative mechanism ... |
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429
Cl
N
N
Cl
Cl
CO2Me
Cl
CO2Me
Cl
MeO2C
Cl
N
N
Cl
Cl
MeO2C
Cl
PROBLEM 3
Suggest a mechanism for this reaction and comment on the ring size formed.
What is the minor product likely to be?
CO2Et
CN
(PhCO2)2
CN
CO2Et
+ minor
product
Purpose of the problem
Activat... |
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Solutions Manual to accompany Organic Chemistry
preferred because the orbital alignment is better. The minor product has a
five-membered ring.
CO2Et
CN
CO2Et
CN
CH2
CO2Et
CN
PROBLEM 4
Treatment of this aromatic heterocycle with NBS (N-bromosuccinimide) and AIBN
gives mainly one product but this is difficul... |
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431
N
N
H
CN
N
N
Br
Br
N
N
Br
+ Br
All subsequent hydrogen abstractions are carried out by bromine atoms,
either of the kind we have just seen or to remove a hydrogen atom from the
other methyl group. This reaction provides the HBr that generates more
bromine from NBS... |
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Solutions Manual to accompany Organic Chemistry
Suggested solution
This time AIBN abstracts the hydrogen from Bu3SnH and the tin radicals
carry the chain along. First they remove the bromine atom from the starting
material to make a vinyl radical that cyclizes onto the unsaturated ketone to
give a radical sta... |
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433
PROBLEM 6
An ICI process for the manufacture of the diene used to make pyrethroid
insecticides involved heating these compounds to 500 °C in a flow system.
Propose a radical chain mechanism for the reaction.
Cl
+
500 °C
Purpose of the problem
Learning how to a... |
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Solutions Manual to accompany Organic Chemistry
PROBLEM 7
Heating this compound to 560 °C gives two products with the spectroscopic data
shown below. What are they and how are they formed?
O
Cl
560 °C
A + B
A has IR 1640 cm–1; m/z 138 (100%) and 140 (33%), δH (ppm) 7.1 (4H, s), 6.5 (1H, dd,
J 17, 11 Hz)... |
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435
the four-membered ring next to the benzene ring. Breaking these bonds
releases strain and allows one of the radical products to be secondary and
delocalized.
O
Cl
b
a
a
Cl
O
A
b
Cl
O
B
PROBLEM 8
Treatment of methylcyclopropane with peroxides at very low temperat... |
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Solutions Manual to accompany Organic Chemistry
Suggested solution
The peroxide is a source of t-BuO• radicals and these abstract a hydrogen
from the methyl group of the hydrocarbon. The first spectrum is that of the
cyclopropylmethyl radical. The odd electron is in a p orbital represented by
a circle and the... |
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437
PhS
H
CN
PhS
Ph
Ph
PhS
PhS
Ph
SPh
H
PhS
Ph
PROBLEM 9
The last few stages of Corey’s epibatidine synthesis are shown here. Give
mechanisms for the first two reactions and suggest a reagent for the last step.
N
NHCOCF3
Br
Br
Cl
t-BuOK,
THF, –78 °C
N
COCF3
Br
N
Cl... |
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Solutions Manual to accompany Organic Chemistry
N
COCF3
Br
N
Cl
Bu3Sn
N
COCF3
N
Cl
Bu3Sn
H
product
Finally we need to hydrolyse the amide. This normally requires strong
acid or alkali but the CF3 group makes this amide significantly more
electrophilic than most and milder conditions can be used. Corey actuall... |
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439
The problem asks for a conformational drawing of the product and
indeed that is necessary. The 1,3-lactone bridge must be diaxial as that is the
only way for the carboxylate to reach across and therefore it must attack
from an axial direction too.
I
O
O
=
O
O
I
O
... |
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Solutions Manual to accompany Organic Chemistry
PROBLEM 11
Suggest a mechanism for this reaction explaining why a mixture of
diastereoisomers of the starting material gives a single diastereoisomer of the
product. Is there any other form of selectivity?
O
OEt
Br
1. Bu3SnH, AIBN
2. Cr(VI), H2SO4
O
O
Purpo... |
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441
PROBLEM 12
Reaction of this carboxylic acid (C5H8O2) with bromine in the presence of
dibenzoyl peroxide gives an unstable compound A (C5H6Br2O2) that gives a stable
compound B (C5H5BrO2) on treatment with base. Compound B has IR 1735 and
1645 cm–1 and NMR δH 6.18 (... |
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Solutions Manual to accompany Organic Chemistry
CO2H
H
PhCO2
Br
Br
CO2H
CO2H
Br
initially
Br
thereafter
CO2H
Br
Br
H
Br
Br
CO2H
Br
CO2H
Br
Br
CO2H
Br
Br
unstable compound
A
base
Br
Br
O
O
O
O
Br
stable compound
B
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Suggested solutions for Chapter 38
PROBLEM 1
Suggest mechanisms for these reactions.
H
OMe
MeOCH2Cl
BuLi
Br
Br
CHBr3
t-BuOK
Purpose of the problem
Two simple carbene reactions initiated by base.
Suggested solution
Going to the right we must remove the rather acidic proton from CHBr3 to
give the carbanion... |
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Solutions Manual to accompany Organic Chemistry
PROBLEM 2
Suggest a mechanism for this reaction and explain the stereochemistry.
CO2Me
O
N2
CuSO4
CO2Me
O
Purpose of the problem
Another important carbene method used in the synthesis of a natural
antibiotic.
Suggested solution
The diazo compound decompose... |
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445
Suggested solution
The first reaction is a variation on Simmons-Smith cyclopropanation.
Though strictly a carbenoid rather than a carbene, it delivers a CH2 group
from an organozinc compound bound to an oxygen atom, in this case the
OMe group. Onl... |
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Solutions Manual to accompany Organic Chemistry
Suggested solution
The carbene formed by loss of nitrogen from the diazoketone rearranges
with the migration of either C–C bond to give a ketene picked up by
methanol.
O
N2
O
hν
–N2
O
O
C
O
O
MeOH
H
OMe
O
MeO
O
CO2Me
O
PROBLEM 5
Suggest a mechanism for the ... |
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447
OH
C
OH
C
PROBLEM 6
Decomposition of this diazo compound in methanol gives an alkene A (C8H14O)
whose NMR spectrum contains two signals in the alkene region: δH 3.50 (3H, s),
5.50 (1H, dd, J 17.9, 7.9), 5.80 (1H, ddd, J 17.9, 9.2, and 4.3), 4.20 ... |
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Solutions Manual to accompany Organic Chemistry
?
?
H
H
H
?
?
?
MeO
H
?
?
A
C8H14O
C4H8
C4H5
C2H4O
= C10H12O
But these add up to C2H3 too much! Clearly the CH attached to OMe and
the CH attached to the alkene are the same atom and the CH2 at the other
end of the alkene must be one end of the chain of four CH2... |
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449
H
OMe
OMe
H
PROBLEM 7
Give a mechanism for the formation of the three-membered ring in the first of
these reactions and suggest how the ester might be converted into the amine
with retention of configuration
Ph
N2
CO2Et
+
Cu(I)
Ph
CO2Et
Ph
NH... |
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Solutions Manual to accompany Organic Chemistry
PROBLEM 8
Explain how this highly strained ketone is formed, albeit in very low yield, by
these reactions. How would you attempt to make the starting material?
CO2H
Me3Si
Me3Si
1. (COCl)2, 2. CH2N2
3. Cu(I)Br
O
Me3Si
Me3Si
Purpose of the problem
To show that... |
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451
SiMe3
SiMe3
+
CO2Me
N
N
Cu(I)Br
CO2Me
Me3Si
Me3Si
ester
hydrolysis
acid
PROBLEM 9
Attempts to prepare compound A by phase-transfer catalysed cyclization
required a solvent immiscible with water. When chloroform (CHCl3) was used,
compound B was fo... |
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Solutions Manual to accompany Organic Chemistry
PROBLEM 10
Revision content. How would you carry out the first step in this sequence?
Propose mechanisms for the remaining steps explaining any selectivity.
O
?
OSiMe3
Cl3CCO2Na
OSiMe3
Cl
Cl
HCl
H2O
Cl
O
MeNH2
H2O
N
Me
O
Cl
H
Purpose of the problem
Revisio... |
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453
You can draw the ring expansion in a number of ways. All start with the
removal of the Me3Si group with water. You might then simply use a one-
step mechanism (a) but an electrocyclic process via the cyclopropyl cation
(b) might be better. This is a... |
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Solutions Manual to accompany Organic Chemistry
Purpose of the problem
Applications of this important and powerful method.
Suggested solution
Metathesis is usually E-selective and these are both E-alkenes so prospects
are good. We must disconnect each compound at the alkene and add
something to the end of ea... |
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455
PROBLEM 12
Heating this acyl azide in dry toluene under reflux for three hours gives a 90%
yield of a heterocycle. Suggest a mechanism, emphasizing the role of any reactive
intermediates.
N3
O
NH2
heat
toluene
N
H
H
N
O
Purpose of the problem... |
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Solutions Manual to accompany Organic Chemistry
Suggested solution
Decomposition of trichloroacetate ion releases the Cl3C– carbanion. Loss of
chloride gives dichlorocarbene and addition to one of the double bonds in
the pyrrole gives a bicyclic intermediate.
Cl
O
Cl
Cl
O
CO2 +
Cl
Cl
Cl
Cl
Cl
N
H
Cl
Cl
N
H
C... |
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Suggested solutions for Chapter 39
PROBLEM 1
Propose three fundamentally different mechanisms (other than variations of the
same mechanism with different kinds of catalysis) for this reaction. How would (a)
D labelling and (b) 18O labelling help to distinguish the mechanisms? What other
experiments would you c... |
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Solutions Manual to accompany Organic Chemistry
Finally, the ester can be transformed into an enolate, using hydroxide as a
base. Elimination gives a ketene that can be attacked by hydroxide as a
nucleophile to give the product.
O
O
O2N
H
H
OH
O
O
O2N
O2N
C
O
O
OH
O2N
O
CO2H
O2N
OH
CO2
O2N
O
OH
O
HO
H
Mechani... |
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459
PROBLEM 2
Explain the stereochemistry and labelling pattern in this reaction.
t-Bu
O
O
enantiomerically
pure
* = 18O
H2SO4
HOAc
*
t-Bu
O
O*
*
racemic
Purpose of the problem
A combination of labelling and stereochemistry reveals the details of a
su... |
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Solutions Manual to accompany Organic Chemistry
PROBLEM 3
The Hammett ρ value for migrating aryl groups in the acid-catalysed Beckmann
rearrangement is –2.0. What does that tell us about the rate-determining step?
Me
N
OH
X
H
N
H
X
Me
O
Purpose of the problem
The Hammett relationship gives an intimate pic... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 39 – Determining reaction mechanisms
461
Me
N
OH2
X
N
Me
X
N
H
X
Me
OH2etc.
amide
The Hammett ρ value of –2.0 gives very definite evidence that
participation does not occur. If it did the closure of the unstable three-
membered ring would be the slow step and a positive charge would form on
... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | 462
Solutions Manual to accompany Organic Chemistry
nucleophilic attack on the carbonyl group will be unusually fast. This is why
there is a region of pH-independent hydrolysis not found with EtOAc. You
might have suggested that acetate is a nucleophile or a general base catalyst
but the solvent deuterium isotope ... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 39 – Determining reaction mechanisms
463
Ar
N
C
N
Ar
H
fast
Ar
N
C
N
Ar
H
H2O
rate-
determining
step
Ar
N
N
Ar
H
OH2
fast
Ar
N
N
Ar
H
O
H
H
fast
Ar
N
N
Ar
H
O
H
H2O
Ar
N
OH
H
O
–CO2
ArNH2
The expected equilibrium Hammett ρ value for the protonation would be
about –2.5 to –3 so the kinetic Ham... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | 464
Solutions Manual to accompany Organic Chemistry
X
Cl
X
Cl
Ph
OH
SN2
rate-
determining
step
X
OH
SN1
X
Ph
OH
OH2
or
fast
X
OH
Ph
rate-
determining
step
Since SN1 produces a cation delocalized round the benzene ring in the
slow step, a large negative Hammett ρ value is reasonable. It is not obvious
what sign th... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | Solutions for Chapter 39 – Determining reaction mechanisms
465
starting material. It cannot be a specific base as it is too weak, even in
acetonitrile, to remove a proton from methanol. But it can act as a general
base. As methanol attacks the carbonyl group its proton becomes more
acidic and, in the transition st... |
http://lib.ysu.am/disciplines_bk/ea4b336028cd91ba7265865d8fde153c.pdf | ea4b336028cd91ba7265865d8fde153c.pdf | true | 519 | 466
Solutions Manual to accompany Organic Chemistry
oxygen atom and ends up with the hydrolysis of an imine. Again, the second
or third step could be rate-determining.
Ph
N
t-Bu
O
H
fast
Ph
N
t-Bu
O
H
Ph
N
t-Bu
OH
H2O
Ph
N
t-Bu
OH
H2O
Ph
N
t-Bu
OH
OH
H
Ph
O
+
HO
N
H
t-Bu
PROBLEM 9
Explain how both methyl groups... |
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