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Файл:Ординатура / Хирургия / Библиотека им академика М.И. Перельмана / Книга_5949_Библиотеки_им_академика_М_И_Перельмана
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Miconazole 291
O
HO
Cl Cl
Cl Cl
N
Cl
N
N
Cl Cl
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N
N
N
2
N
CH
OH
3
O2N
CH
N
H
3
O
or
N
CH
N
H
Cl
3
Extended Discussion
Suggest (draw the structure of) a side product which might form in the final step. Is this side product formed?
Miconazole
Dermatological Medicines (Topical)/Antifungal Medicines
An ether is often formed from an alkyl halide (X=Cl, Br, I)
N
O
and an alcohol. The C.O bond formation is most efficient
when the alkyl halide is primary, allylic, or benzylic.
Discussion. Miconazole is a 1 : 1mixture of the (R)- and (S)- enantiomers. The ether is formed in the final step by displacement
of the benzylic chloride from 2,4hol is formed by reduction of the ketone. A C-N bond is formed by displacement of bromide of the α-
bromoketone is formed by bromination of 2′,4′- dichloroacetophenone.
The α-
O
Cl
N
N
O
dichlorobenzyl chloride by the alcohol (Williamson Ether Synthesis). The secondary alco-
bromoketone by imidazole.
Cl
N
Cl Cl
Br
O
Cl Cl
N
HN
Cl Cl
HO
O
N
Cl Cl
CH
3
Cl Cl
M292
Cl
F
CH
Cl
O
3
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The α- chloroketone, 2,2′,4′- trichloroacetophenone, formed from 1,3- dichlorobenzene and chloroacetyl chloride (Friedel–
Crafts Acylation), is used in an alternative route to miconazole.
Extended Discussion
List the pros and cons for both routes. Is one route preferred?
Midazolam
Anesthetics, Preoperative Medicines, and Medical Gases/Preoperative Medication and Sedation for Short- term Procedures
Medicines for Pain and Palliative Care/Medicines for Other Common Symptoms in Palliative Care
Anticonvulsants/Antiepileptics
3
N
N
A 5- aryl- 1,3- dihydro- 2H- 1,4- benzodiazepin- 2- one is
often formed in two steps from a 2-
aminobenzophenone,
chloroacetyl chloride, and ammonia/hexamethy
lenetetramine.
N
Discussion. Midazolam is separated from the mixture of midazolam (major) and isomidazolam (minor) produced by
decarboxylation of the imidazolecarboxylic acid. Isomidazolam is also converted to midazolam. (Draw the structure of
isomidazolam and provide details on the conversion of isomidazolam to midazolam.) The imidazolecarboxylic acid is
formed by hydrolysis of the ester.
CH
3
N
N
CH
3
N
N
OH
CH
3
N
N
O
OCH2CH
N
Cl
F
N
Cl
F
N
F
Midazolam 293
O
Cl
Cl
O
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The 2- methylimidazole is formed by cyclization of the amine and the acetamide with elimination of water. The alkene with
the amine and acetamide substituents is formed by cleavage of the αbyproducts of this cleavage?) The α-
substituted acetamidomalonate is formed by displacement of the leaving group dimor-
substituted acetamidomalonate. (What are the likely
pholinophosphinate by diethyl acetamidomalonate. Reaction conditions are selected so that the three steps (leaving group
displacement, cleavage of the α-
substituted acetamidomalonate, and cyclization to form the imidazole) are accomplished
without isolation of the two intermediates.
CH
3
CH
3
N
N
N
O CH
HN
N
N
O
OCH2CH
F
3
O
O
OCH2CH
OCH2CH
3
Cl
O
HN CH
CH3CH2O OCH2CH
3
3
O O
Cl
HN
H
N
1
2
3
4
5
N
3
3
N
O
OCH2CH
F
O
N
P
O N
O
N
3
The leaving group is introduced by reaction of dimorpholinophosphinyl chloride with the 1,4- benzodiazepin- 2-
one. 7- Chloro- 5- (2- fluorophenyl)- 1,3- dihydro- 2H- 1,4- benzodiazepin- 2- one (norflurazepam) is formed after a chloride displacement from the 2- chloroacetamide by ammonia/hexamethylenetetramine. The 2- chloroacetamide is formed from
chloroacetyl chloride and 2- amino- 4- chloro- 2′- fluorobenzophenone. The 2- aminobenzophenone is formed by hydrolysis
of the acetamide.
F
F
M294
Cl
2
Cl
Cl
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O
O
Cl P N
O
N
N
P
O N
O
N
O
O
H
N
O
H
N
O
NH
N
Cl
N
Cl
O
NH
F
CH
3
C6H12N
F
4
H
N
O
NH
F
O
O
Cl
Cl
2
Cl
O
Cl
F
O
Cl
F
O
F
The benzophenone is formed from 4- chloroacetanilide and 2- fluorobenzoyl chloride (Friedel–Crafts Acylation).
2-
Fluorobenzoyl chloride is formed from the carboxylic acid. 2- Fluorobenzoic acid is formed from anthranilic acid (Balz–
Schiemann Reaction).
H
O
CH
3
N
CH
3
NH
Cl
O
O
O
F
Cl
O
F
OH
F
O
OH
NH
2
Mifepristone 295
O
3
CH
CH
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Extended Discussion
Draw the structures of a retrosynthetic analysis of an alternative route to midazolam that does not proceed via
decarboxylation of the imidazolecarboxylic acid. List the pros and cons for both routes and select one route as the
preferred route.
Mifepristone
Oxytocics and Antioxytocics/Oxytocics
3
N
3
CH
OH
3
A singlebons is often formed by modification of a natural product which has most or all of the chiral carbons already
CH
in place. Steroids missing the methyl group at position
enantiomer molecule with multiple chiral car-
19 (19- nor- steroids) are often formed from sitolactone.
Sitolactone is produced by microbial degradation of
H
H
phytosterols, including β-
sitosterol.
Discussion. Mifepristone (RU- 486) is manufactured in four steps from 3,3- ethylenedioxyestra- 5(10),9(11)- diene- 17- one
and in nine steps from β-
The 4,9-
9(11)-
diene- 3- one is formed in the final step by hydrolysis of the acetal and elimination of water. The reaction of the
ene- 5α(10)- epoxide with the arylmagnesium halide results in C-C bond formation at C11 and epoxide ring- opening
(Grignard Reaction). (The 11α-
sitosterol.
aryl side product is also formed in this reaction. What is the ratio 11β- aryl product to 11αaryl side product? How is the 11β- aryl product separated from the 11α- aryl side product?) The Grignard reagent is formed
from 4-
bromo- N,N- dimethylaniline. The 9(11)- ene- 5α(10)- epoxide is formed by epoxidation of the 5(10),9(11)- diene. (The
ene- 5β(10)- epoxide is also formed in the epoxidation. What is the ratio of the 5α(10)- epoxide to the 5β(10)- epoxide?
9(11)How is the 5α(10)acetylide from propyne to the C17ketone of 3,3-
epoxide separated from the 5β(10)- epoxide?) The C17 tertiary alcohol is formed by addition a metal
ethylenedioxyestra- 5(10),9(11)- diene- 17- one.
M296
CH
CH
CH
3
O
O
CH
O
O
O
O
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3
N
3
CH
OH
3
CH
3
H
H
CH
CH
3
N
3
Br
CH
OH
3
CH
CH
O
3
N
CH
CH
3
N
3
CH
OH
3
3
MgBr
CH
3
H
CH
H
O
O
OH
3
CH
3
O
O
OH
H
10
O
H
O
3
4
5
11
H
O
H
CH
O
3
17
H
9
H
CH
3
The acetal of 3,3- ethylenedioxyestra- 5(10),9(11)- diene- 17- one is formed by the reaction of estra- 4,9- diene- 3,17- dione with
ethylene glycol. The 4ketone and cyclohexenone of the secothe seco-
steroid is released by hydrolysis of the acetal and carried directly into the condensation.
ene- 3- one and the A ring of estra- 4,9- diene- 3,17- dione are formed by condensation of the methyl
steroid (Aldol Condensation) followed by elimination of water. The methyl ketone of
CH
3
OH
HO
H
CH
3
H
O
O
O
3
H
H
O
O
CH
3
H
CH
H
CH
CH
3
3
CH
O
3
HOO
3
H
Mifepristone 297
CH
CH
CH
ββ
OH
O
Cl
O
CH
CH
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The cyclohexenone and the B ring are formed by condensation of a ketone on a side chain and the cyclohexanone (Aldol
Condensation) with elimination of water. The cyclohexanone is formed by oxidation of the cyclohexanol and carried
directly into the condensation. The ketone on the side chain and the cyclohexanol are both formed by reaction of an alkylmagnesium chloride with the lactone carbonyl of sitolactone (Grignard Reaction). Sitolactone is formed by microbial
degradation of phytosterols, including β- sitosterol.
3
CH
CH
3
3
HOO
CH
O
3
H
CH
O
3
O
O
O
H
CH
H
H
sitolactone
MgCl
O
3
H
O
CH
3
CH
3
O
CH
3
O
CH
3
O
H
HO
H
CH
O
3
O
CH
3
3
O
3
O
CH
CH
3
3
O
O
CH
3
CH
3
3
CH
CH
3
CH
3
H
3
H
H
CH
HO
-sitosterol
The alkylmagnesium chloride is formed from 2- (3- chloropropyl)- 2,5,5,- trimethyl- 1,3- dioxane. The 1,3- dioxane/acetal is
formed by reaction of 5- chloro- 2- pentanone with 2,2- dimethyl- 1,3- propanediol (neopentyl glycol).
3
CH
3
MgCl
3
O
CH
CH
3
3
OH
CH
CH
CH
3
3
CH
3
O
O
Cl
3
M298
HO
O
estrone
O
3
3
3
O
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Extended Discussion
Draw structures of the retrosynthetic analysis of an alternative route to the intermediate 3,3- ethylenedioxyestra- 5(10),9(11)-
diene-
17- one from estrone. List the pros and cons for both routes to this intermediate and select one route as the preferred route.
CH
3
H
H
H
Miltefosine
Anti- infective Medicines/Antiprotozoal Medicines/Antileishmaniasis Medicines
O
P
O
O
CH
Discussion. Two routes will be presented for comparison. In Route A, the quaternary salt of miltefosine is formed in
thefinal step by alkylation of trimethylamine using a phosphate ester. The phosphate ester is formed by displacement
of chloride from 2- chloro- 2- oxo- 1,3,2- dioxaphospholane (ethylene chlorophosphate) by hexadecane- 1- ol (cetyl alcohol).
Chloro- 2- oxo- 1,3,2- dioxaphospholane is formed by oxidation of 2- chloro- 1,3,2- dioxaphospholane (ethylene chloro-
2phosphite). 2-
Chloro- 1,3,2- dioxaphospholane is formed from ethylene glycol.
O
CH
3
O
CH
N
A quaternary ammonium salt is often formed
by alkylation of a tertiary amine.
CH
CH
3
O
P
O
O
O
P
O
N
CH
3
(CH3)3N
CH
CH
3
3
cetyl alcohol
CH
3
OH
CH
3
O
O
Cl
O
Cl
O
P
O
HOP
OH
Misoprostol 299
O
HO
3
O
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In Route B, the quaternary salt of miltefosine (hexadecylphosphocholine) is formed in the final step by exhaustive methylation of the primary amine (hexadecylphosphoethanolamine) with dimethyl sulfate. The primary amine is formed by hydrolysis of the phosphoric acid amide. The 1,3,2-
oxazaphospholane ring is formed by reaction of cetyl dichlorophosphate with
ethanolamine. Cetyl dichlorophosphate is formed from cetyl alcohol.
CH
O
P
O
CH
3
O
CH
3
O
O
P
O
3
O
O
O
P
OH
HO
OH
N
CH
NH
CH
CH
3
2
3
3
CH3OS
NH
2
CH
3
O
OCH
3
O
O
Cl
P
O
Cl
CH
3
cetyl alcohol
Extended Discussion
List the pros and cons for both routes. Select one route as the preferred route.
Misoprostol
Oxytocics and Antioxytocics/Oxytocics
Prostaglandin analogs with the cyclopentanone core of
O
OCH
CH
3
OH
CH
3
prostaglandin E are often formed by conjugate addition
of an alkenylcuprate reagent to a cyclopentenone. An
alkoxy trialkylsiloxy group on the α- face of the cyclopentenone directs the conjugate addition at the adjacent carbon to the β- face.
M300
O
O
P
3
3
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Discussion. Misoprostol is a mixture of (16R)- and (16S)- diastereomers. Misoprostol is formed in just three steps from
(R)-
norprostol. (R)- Norprostol is formed in seven steps from suberic acid (1,8- octanedioic acid).
The C11 and C16 alcohols of misoprostol are released in the final step. (List the protecting groups P
1
C11 and C16 alcohols. Select one option for P
and one option for P2 to use in the analysis.) The C12.C13 bond is formed by
1
and P2 used for the
conjugate addition of an alkenylcuprate to a cyclopentenone. (Draw the structure of one side product which is formed inthis
reaction.) The alcohol of (R)acetate ester. (R)-
Norprostol is separated from the (S)- acetate and the (S)- acetate is racemized and recycled. The racemic
acetate ester is formed by reaction of a mixture of the 3-
norprostol is protected. (R)- Norprostol is formed by enzyme- mediated hydrolysis of the racemic
hydroxycyclopent- 4- en- 1- one and the 4- hydroxycyclopent- 2- en- 1- one
(norprostol) with acetic anhydride.
O
1
OCH
3
HO
9
11
12
13
CH
3
OH
CH
16
3
O
O
OCH
3
CH
O
1
Cu
2
OP
3
CH
3
P1O
HO
O
OCH
3
CH
2
3
OP
CH
3
O
O
OCH
3
(R)-norprostol
O
O
OCH
O
3
O
OCH
3
OH
O
O
OCH
3
O
CH
3
O
The mixture of the 3- hydroxycyclopent- 4- en- 1- one and norprostol (3 : 1) is formed by rearrangement of the furfuryl alcohol (Piancatelli Rearrangement). The secondary alcohol is formed by reduction of the ketone. The ketone is formed by
acylation of furan with an active ester (Friedel–Crafts Acylation). The active ester is formed in situ from the carboxylic
acid. (List the active esters used in this acylation of furan.) Suberic acid monomethyl ester is formed from suberic acid
(Fischer Esterification).
HO
norprostol
CH
O
O
O
CH
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