Download Definition: the term “alkaloid” (alkali

4. Alkaloids
Definition:
the term “alkaloid” (alkali‐like) is
commonly used to designate basic heterocyclic
nitrogenous compounds of plant origin that are
physiologically active.
Classification:
•
•
•
•
True (Typical) alkaloids that are derived from amino acids and have nitrogen in a heterocyclic ring. e.g Atropine
Protoalkaloids that are derived from amino acids and do not have nitrogen in a heterocyclic ring. e.g Ephedrine
Pseudo alkaloids that are not derived from amino acids but have nitrogen in a heterocyclic ring. e.g Caffeine
False alkaloids are non alkaloids give false positive reaction with alkaloidal reagents.
Nomenclature:
Trivial names should end by "ine". These names may refer to: • The genus of the plant, such as Atropine from Atropa belladona.
• The plant species, such as Cocaine from Erythroxylon coca.
• The common name of the drug, such as Ergotamine from ergot.
• The name of the discoverer, such as Pelletierine that was discovered by Pelletier.
• The physiological action, such as Emetine that acts as emetic, Morphine acts as narcotic.
• A prominent physical character, such as Hygrine that is hygroscopic.
Prefixes and suffixes:
Prefixes: •
"Nor‐" designates N‐demethylation or N‐demethoxylation, e.g.
norpseudoephedrine and nornicotine.
•
"Apo‐" designates dehydration e.g. apomorphine.
•
"Iso‐, pseudo‐, neo‐, and epi‐" indicate different types of
isomers.
Suffixes:
•
"‐dine" designates isomerism as quinidine and cinchonidine.
•
"‐ine" indicates, in case of ergot alkaloids, a lower
pharmacological activity e.g. ergotaminine is less potent than
ergotamine.
Physical Properties:
I‐ Condition:
•
•
•
Most alkaloids are crystalline solids. Few alkaloids are amorphous solids e.g. emetine. Some are liquids that are either:
Volatile e.g. nicotine and coniine, or
Non‐volatile e.g. pilocarpine and hyoscine.
II‐ Color:
The majority of alkaloids are colorless but some are
colored e.g.:
• Colchicine and berberine are yellow. • Canadine is orange.
• The salts of sanguinarine are copper‐red.
III- Isomerization:
• Optically active isomers may show different physiological activities. • l‐ephedrine is 3.5 times more active than d‐ephedrine.
• l‐ergotamine is 3‐4 times more active than d‐ergotamine.
• d‐ Tubocurarine is more active than the corresponding l‐
form. • Quinine (l‐form) is antimalarial and its d‐ isomer quinidine is antiarrythmic. • The racemic (optically inactive) dl‐atropine is physiologically active.
Chemical Properties:
I‐ Nitrogen:
• Primary amines R‐NH2 e.g. Norephedrine
• Secondary amines R2‐NH e.g. Ephedrine
• Tertiary amines R3‐N e.g. Atropine
• Quaternary ammonium salts R4‐N e.g d‐
Tubocurarine
II‐ Basicity:
• R2‐NH > R‐NH2 > R3‐N
• Saturated hexacyclic amines is more basic than aromatic amines.
According to basicity Alkaloids are classified into:
• Weak bases e.g. Caffeine
• Strong bases e.g. Atropine
• Amphoteric * Phenolic Alkaloids e.g. Morphine with Carboxylic groups e.g. Narceine
• Neutral alkaloids e.g. Colchicine
*Alkaloids III‐ Oxygen:
• Most alkaloids contain Oxygen and are solid in nature e.g. Atropine.
• Some alkaloids are free from Oxygen and are mostly liquids e.g. Nicotine, Coniine.
Extraction, Purification and Isolation of Alkaloids from Powdered plants
•
Extraction and purification
Method I:
The powder is treated with alkalis to liberates the free bases that
can then be extracted with water immiscible organic solvents.
Method II:
The powdered material is extracted with water or aqueous alcohol
containing dilute acid. Alkaloids are extracted as their salts
together with accompanying soluble impurities.
Method III:
The powder is extracted with water soluble organic solvents such as
MeOH or EtOH which are good solvents for both salts and free
bases.
• Liberation of the free bases:
Alkalis are used to liberate free bases. Alkalis must be strong enough
to liberate free bases. However, choice of strong alkalis must be
avoided in some cases:
1‐ Ester Alkaloids e.g. Solanaceous Alkaloids
2‐ Amide Alkaloids e.g. Colchicine
3‐ Phenolic Alkaloids e.g. Morphine
4‐ Lactone Alkaloids e.g. Pilocarpine
5‐ Fatty Drugs due to saponification and emulsion formation. ¾Separation of Alkaloidal Mixtures:
• Fractional Crystallization:
Ephedrine & Pseudoephedrine Oxalates
Crystallization from water
Ephedrine Oxalate
Crystals
Pseudoephedrine Oxalate
Solution
Atropine & Hyoscyamineine Oxalates
Crystallization from
Acetone/Ether
Atropine Oxalate
Crystals
Hyoscyamine Oxalate
Solution
• Preparation of Derivatives:
Separation of Primary, Secondary and Tertiary Alkaloids.
O
Mixture + p-toluenesulphonyl chloride
Cl
Add HCl and filter
O
Filtrate
tertiary alkaloids as salt
(no reaction with reagent
S
Precipitate
1ry alk derivative
2ry alk derivative
O
O
R-HN S
R-N S
R
O
keto form
O
insoluble in alkalis
acidic hydrogen
OH
R-N S
enol form
soluble in alkalis
O
NaOH, filter
Filtrate
Precipitate
1ry alk derivative
2ry alk derivative
• Fractional Liberation:
Atropine & Hyoscyamine & Hyoscine
the form of HCl salts
1- Alkalinize by NaHCO3 pH 7.5
2- Extract with Ether
Ether
Hyoscine free base
(pKa = 6.2)
Aqueous layer
Atropine & Hyoscyamine HCl
(pKa = 9.3)
• Fractional Distillation:
e.g. Separation of Nicotine and Anabasine • Chromatographic Separation.
• Phenylalkylamines:
e.g. Ephedrine
CH2 CH
CH3
NH2
• Pyridine and piperidine
e.g. lobeline, nicotine
N
N
H
• Tropane
e.g. Atropine.
NCH3
OH
• Quinoline
e.g.quinine and quinidine
N
• Isoquinoline
e.g. papaverine
N
• Phenantheren
e.g. Morphine
• Indole
e.g.ergometrine
N
H
• Imidazole
N
e.g. pilocarpine
N
• Purine
e.g. caffeine
6
1 N
5
7
N
H
8
2
N 4
3
Purine
N
9
• Steroidal
e.g. Solanum and
alkaloids
• Terpenoid
e.g. Taxol
Veratrum
Introduction:¾ As the molecular structure of alkaloids is quite complex , very little progress was achieved in the elucidation of their structures during 19th century .
¾ But now the new methods for the identification of unknown substances are known.
¾ The following pattern of procedure is adopted to establish the molecular structure of an alkaloid:
1) Molecular Formula determination: ¾ After a pure specimen has been obtained , its elemental composition , and hence the empirical formula , is found by combustion analysis. ¾ Then , its molecular weight is determined by the Rast procedure( depression of the freezing point) to establish its molecular formula.
¾ Its calculation is based upon the simple fact that introduction of a double bond or cyclisation of the chain decreases the molecular formula by two hydrogen atoms relative to the corresponding saturated aliphatic hydrocarbon.
¾ For example, the difference between hexene (C6H12) from hexane (C6H14) is two hydrogen's and this difference is called a double bond equivalent.
¾ Similarly, the difference between benzene (C6H6) and hexane (C6H14) is eight hydrogen’s which will correspond to 8/2 or 4 double bond equivalents (accommodated by the three double bonds and one ring).
¾ The above procedure is valid for simpler compounds only. However, for complex formulae, where elements other than hydrogen and carbon are present, the simpler method is that for any formula CaHbNcOd the number of double bond equivalents is given by the following expression: a – 1/2b + 1/2c + 1
¾ The above method for the calculation of double bond equivalents is useful to calculate the number of rings in a given compound. For example, hygrine has the molecular formula, C8H15NO which corresponds to
8 – 15/2 + ½ + 1 = 2
double bond equivalents. However, chemical tests reveal that hygrine contains only one carbonyl group (one double bond equivalent) and does not show other form of unsaturation. Thus hygrine must be monocyclic
to account for the other double bond equivalent.
¾ The presence of unsaturation in an alkaloid may also be ascertained by treating the alkaloid with bromine or halogen acid or alkaline potassium permanganate when a glycol is obtained
2) Functional Group Analysis:
¾ Application of classical techniques of organic analysis (especially if the alkaloid is available in appreciable amounts) and/or infra‐red examination
(especially if the alkaloid is available only in small amounts) can reveal the nature of the functional groups present.
¾ This will also reveal the aromatic or aliphatic
nature of the alkaloid and the unsaturation, if present.
3) Functional Nature Of Oxygen:
¾ If an alkaloid contains oxygen, it may be present as –OH (phenolic or alcoholic), methoxy (–OCH3 ), acetoxy (–
OCOCH3 ), benzoxyl (–OCOC6H5 ), carboxylic (–COOH), carboxylate (–COOK) or carbonyl (=C=O).
¾ Occasionally, lactone ring systems have also been encountered (e.g., narcotine, hydrastine). These functions have been detected by the usual methods of organic analysis including infra‐red examination.
¾ Various oxygen functional groups can be characterised according to the following characteristics:
a) Hydroxyl group: 9 Its presence in an alkaloid can be ascertained by the formation of acetate, on treatment with acetic anhydride or acetyl chloride or by the formation of benzoate on treatment with benzoyl chloride in the presence of sodium hydroxide. 9 However, the above test for oxygen should be applied carefully because primary amines if present in an alkaloid also yield acetyl and benzoyl derivatives.
¾ Then the number of hydroxyl groups is determined by acetylation or Zerewitnoff’s method. In the former method, the number of hydroxyl groups is determined by acetylating the alkaloid and hydrolysing the acetyl derivative with a known volume of IN NaOH.
¾ The excess of alkali is estimated by titration with a standard solution of HCl. The number of acetyl groups or hydroxyl groups can be calculated from the volume of alkali used for hydolysis.
b) Carboxylic group: 9 The solubility of an alkaloid in aqueous sodium carbonate or ammonia reveals the presence of carboxylic group. The formation of ester on treatment with an alcohol also reveals the presence of carboxylic group. 9The number of carboxylic groups may be determined by volumetrically by titration against a standard barium hydroxide solution using phenolphthalein as an indicator or gravimetrically by silver salt method.
c) Oxo group: 9 The presence of this group is ascertained by the reaction of an alkaloid with hydroxylamine, semicarbazide or phenylhydrazine when the corresponding oxime, semicarbazone or phenylhydrazone are formed. 9 Distinction between an aldehyde and a ketone can be made on the basis of reduction and oxidation reactions.
d) Methoxy group: 9 The detection of this group and its number may be determined by the Zeisel determination, analogous to the Herzeg‐Meyer method for N‐methyl groups. 9 In this method, a known weight of alkaloid is heated with hydriodic acid at its boiling point (126 °C) when the methoxyl groups are thereby converted into methyl iodide
which is then absorbed by ethanolic silver nitrate and the precipitated silver iodide is filtered, dried and weighed. From the weight of silver iodide, the number of methoxyl groups may be calculated.
9 For example, papavarine, C20H21O4N, when treated with hydrogen iodide, consumes 4 moles of hydrogen iodide, producing 4 moles of silver iodide and thus confirming the presence of four –OCH3 groups.
e)Ester and amide groups: 9These groups can be detected and estimated by observing
The products of their alkali or acid hydrolysis. 4) Nature of Nitrogen: ¾ All alkaloids contain nitrogen . But in the majority of alkaloids it is present as a part of a heterocyclic system. Therefore, it must be either a secondary (=NH) or tertiary(=N–CH3 or =N–).
¾ However, there are phenylalkyl amine type of alkaloids
(adrenaline, ephedrine, etc) which do not contain nitrogen as a part of a heterocyclic ring but in the form of a primary amino (–NH2) group.
a) The general reactions of the alkaloid with acetic anhydride, methyl iodide and nitrous acid often show the nature of the nitrogen. 9If the alkaloid reacts with one mole of methyl iodide to form an N‐methyl derivative, it means that a secondary nitrogen atom is present. For example, coniine, C8H17N reacts with one mole of methyl iodide to form an N‐ methyl derivative, indicating that coniine must contain secondary nitrogen atom.
9If an alkaloid reacts additively with one mole of methyl iodide to form crystalline quaternary salt, this indicates that nitrogen atom present in this alkaloid is tertiary. For example, nicotine reacts additively with two moles of methyl iodide, indicating that it contains both nitrogen atoms as tertiary.
9 One can detect the tertiary nitrogen atom in an alkaloid by treating it with 30 % hydrogen peroxide when tertiary nitrogen is oxidised to amine oxide.
b) Herzig‐Meyer’s method is used to detect by distillation of alkaloid with soda‐lime when methyl amine is obtained. For example, nicotine on heating with soda‐lime yields methylamine indicating that it must contain a N‐methyl group.
9 NMR spectroscopy may also be utilized for the rapid detection of N‐ methyl and N‐ethyl groups in alkaloids.
5) Estimation of C‐Methyl groups:
¾ C‐methyl groups are quantitatively estimated by the Kuhn‐Roth oxidation, the acetic acid formed being distilled off and distillate titrated against standard base.
6) Degradation Of Alkaloids:
¾ The reactions used in degradation of alkaloids are as follows:
(a) Hofmann exhaustive methylation method
(b) Emde’s degradation
(c) Reductive degradation and zinc dust distillation
(d) Alkali fusion
(e) Oxidation
(f) Dehydrogenation
a) Hofmann’s Exhaustive Methylation Method: ¾ The principle of this method is that compounds, which +
‐
contain the structural unit =CH=C–N R3OH , eliminate a trialkylamine on pyrolysis at 200 °C or above to yield an olefin.
¾ If the nitrogen atoms forms a part of a cyclic structure, two or three such cycles are essential to liberate the nitrogen and expose the carbon skeleton. ¾However , this method is applicable only to reduced ring systems such as piperidine and actually fails with analogous unsaturated compounded such as pyridine and therefore the latter should be first of all converted into the former ¾ When a molecule of water is eliminated from quaternary ammonium hydroxide, hydrogen atom is always eliminated from the β‐position, if this hydrogen is not available, the reaction fails
¾ The hofmann’s degradation method can be applied to hordenine methyl ether which yields p‐methoxy styrene.
b) Emde’s degradation:
¾ If the alkaloid does not contain a β‐hydrogen atom, the Hofmann’s exhaustive methylation method fails. In such cases, Emde’s method may be employed.
¾ In this method, the final step involves reductive cleavage of quaternary ammonium salts either with sodium amalgam or sodium in liquid ammonia or by catalytic hydrogenation:
¾ Emde’s method can be demonstrated by considering the case of isoquinoline:
c) Reductive Degradation and Zinc Dust Distillation:
¾ In some cases the ring may be opened by heating with hydiodic acid at 300 °C, e.g.,
HI
300 C
¾ Zinc dust distillation produces simple fragments from which one can draw the conclusion about the carbon framework of the alkaloid molecule.
¾ Zinc dust also brings about dehydrogenation or removal of oxygen if present. For example, ¾ As conyrine is formed by loss of six hydrogen atoms, it means that coniine must contain a piperidine ring
d) Alkali fusion:
¾ This is very drastic method which is often employed to break down the complex alkaloid molecule into simpler
fragments, the nature of which will give information on the type of nuclei present in the alkaloid molecule. For example, adrenaline when fused with solid potassium hydroxide yields protocateochic acid, indicating that adrenaline is a catechol deravative.
e) Oxidation:
¾ This method gives useful information about the structure of alkaloid. By varying the strength of the oxidising agents, it is possible to obtain a variety of oxidation products. For example,
(i) In order to carry out mild oxidation, hydrogen peroxide, iodine in ethanolic solution, or alkaline potassium ferricyanide are usually used.
(ii) In order to carry out moderate oxidation, acid or alkaline potassium permanganate or chromium trioxide in acetic acid are generally used. (iii) For carrying out vigorous oxidation, potassium dichromate‐sulphuric acid, chromium trioxide‐sulphuric acid, concentrated nitric acid or manganese dioxide‐
sulphuric acid are used. These reagents usually break up an alkaloid into smaller fragments whose structures are either already known or can be readily ascertained. For example, ¾ From this reaction, it can be concluded that nicotene
contains a pyridine ring having a side chain in β‐position.
¾ This classification of oxidising agents is not rigid because the ‘strength’ of an oxidising agent depends to some extent on the nature of the alkaloid which is being oxidised.
(f) Dehydrogenation:
¾ When an alkaloid is distilled with a catalyst such as sulphur, selenium or palladium, dehydrogenation takes place to form relatively simple and easy recognisable products which provide a clue to the gross skeleton of the alkaloid
¾ During dehydrogenation, there occurs the elimination of peripheral groups such as hydroxyl
and C‐methyl.
(7) Synthesis:
¾ The structure of the alkaloid arrived at by the exclusive analytical evidence based on the foregoing methods is only tentative. The final confirmation of the structure must be done by the unambiguous synthesis.
(8) Physical Methods:
¾ In alkaloid chemistry, the most important instrumental methods are as follows:
(a) Ultraviolet spectroscopy
(b) Infra‐red spectroscopy
(c) Mass spectroscopy
(d) Optically rotatory dispersion and circular dichroism
(e) Conformational analysis, and
(f) X‐ray diffraction
(a) Ultraviolet Spectroscopy: ¾ This is mainly used to establish the class and/or structural type to which the alkaloid being investigated belongs. Such assignments are made because ultraviolet spectrum of a compound is not a characteristic of the whole molecule but only of the chromophoric system(s) present.
¾ The usual practice is to record the ultravoilet spectra of a very large number of different types of alkaloids. Then, the data are analysed and categorised with respect to structure correlation. ¾ Each group of alkaloids having a particular chromophoric system benzene, pyridine, indole, quinoline, etc . yields characteristic absorption maxima and extinction coefficients.
¾ Therefore, the comparison of these data with those observed for a new alkaloid may allow the identification of the exact nature of the aromatic or heterocyclic system in the new compound.
(b) Infra‐red spectroscopy:
¾ In alkaloid chemistry, it is mainly used to ascertain the presence and sometimes the absence of particular functional group.
¾ The presence of aldehyde, ketone, alcohols, phenols, ester, amide, lactone, carboxylic acid, carbonyl grroups and primary and secondary amines can rapidly be identified and distinguished by comparison of the observed frequencies with those reported for structural related compounds.
¾ One can also ascertain the presence of O‐methyl, N‐
methyl and aromatic groups from the infra‐red spectrum of an alkaloid but the quantitative analysis of such groups is best accomplished by NMR spectroscopy.
(c) Mass spectroscopy:
¾ This technique is quite useful because it gives quite useful information about the alkaloid like.
(i) The molecular weight.
(ii) The empirical formula by accurate mass measurement of the molecular ion, and
(iii) Knowledge of the molecular structure by comparison of the fragmentation pattern with those of analogous system.
¾ Most of the success has been achieved in the case of polycyclic indole alkaloids because the indole nucleus of these substances gives rise to an abundant, stable molecular ion which subsquently undergoes decomposition by highly specific bond fusion involving the acyclic portion of the molecule containing the other nitrogen atom(s).
d) Optically rotatory dispersion and circular dichroism:
¾ These are only instrumental methods which are mainly used for elucidation of the stereochemistry of alkaloids but their application is restricted to those compounds which are optically active, i.e., to those in which a rotation‐
reflection symmetry axis is absent.
¾ Due to this reason, few alkaloids of the yohimbine, aprophine, morphine and benzlisoquinoline series have been examined so far by these techniques.
e) Conformational Analysis:
¾ The principles of conformational analysis have been widely used to establish the stereochemistry as well as physical properties and chemical reactivity of alkaloids.
¾ The approach is mainly experimental which involves determination, correlation and interpretation of the kinetics and product ratios obtained from simple chemical transformations such as reduction of double bonds and carbonyl groups, hydrolysis and esterfication, oxidation of alcohols and quaternization of amines, epimerization, etc
f) X‐ray Diffraction:
¾ This technique is widely used to study alkaloids because it gives the exact structure of he molecule, including bond angles and bond lengths; it also gives the information about the relative stereochemistry, including information on overcrowding twisted bonds, etc. ¾ X‐ray diffraction method is also useful to reveal the absolute configuration of the molecule.
“ Structural elucidation of RESERPINE ”
Content:‐
•
•
•
•
Introduction
Constitution of Reserpine
Structure of Yobyrine & reserpic acid confirmation
Structure of reserpine confirmation by synthesis Introduction:‐
• Reserpine is the main constituent of Rauwolfia species,perticularly R.serpentina & R.vomitoria.
• It is mainly used for the treatment of hypertension, headache, tension, asthma & dermatological disorders.
Constitution of reserpine:‐
1. Molecular formula:‐ C33H40N2O9 2. Presence of five methoxy groups:‐
By zeisel method i.e. when treated with HI yields five molecules of methyl iodide indicating the presence of five methoxy groups.
3. Nature of ‘N’ atom:‐
a) Secondary ‘N’ – Formation of monoacetyl derivative with acetic anhydride indicates secondary ‘N’ .
b) Tertiary ‘N’‐ Reserpine forms quaternary ammonium salts with CH3I that indicating one of ‘N’ is tertiary.
4. Hydrolysis:‐ Hydrolysis of reserpine yields mixture of i) Methyl alcohol
ii) 3,4,5‐trimethoxy benzoic acid
iii) Acid (A) of composition C22H28N2O5
C33H40N2O9
+
2 H2 O
NaOH
Hydrolysis
CH3OH
+ C22H28N2O5 +
COOH
MeO
OMe
OMe
•
•
As reserpine does not contain ‐COOH & ‐OH groups, hence its hydrolysis product reveal that reserpine is a di‐ester.
The ester linkage in reserpine has been further confirmed by its reduction with LiAlH4
5. Reduction:‐
CH2OH
C33H40N2O9
LiAlH4
Reserpic Alcohol +
C22H30N2O4
MeO
OMe
OMe
#Structure of Reserpic acid:‐
a) Molecular formula:‐
C 22 H 28 N 2 O 5
b) Presence of one carboxyl group:‐
By usual tests e.g. silver salt method
c) Presence one –OH group:‐
Reserpic acid on oxidation yields a secondary alcoholic group.
ketone that means it has d) Nature of two methoxy groups:‐
By zeisels method.
e) Nature of two ‘N’ atom:‐
It shown that it contains two ‘N’ atoms in heterocyclic ring in the form of‐ i) Secondary ‘N’
ii) Tertiary amino group
f)
Reduction of reserpic acid:‐
On reduction with yields reserpic alcohol.
LiAlH
g) Oxidation of reserpic acid:‐
4
On oxidation with it gives 4‐methoxy N‐oxalyl anthranilic acid.
KmnO4
C OOH
C 22 H 28 N 2 O 5
K mnO 4
MeO
NHC O C OOH
oxidation
Reserpic
acid
Thus one of the methoxyl group is present in meta position to –NH group
4-methoxy N-oxalyl
Anthranilic acid
h) Fusion with KOH :‐
When reserpic acid is fused with potash it yields 5‐hydroxy isophthalic acid.
One of the acidic groups of isophthalic acid must be COOH
C22H28N2O5
KOH
Fusion
HOOC
OH
present in m‐ position to each other this confirm by the fact that 5-hydroxy isophthalic acid
reserpic acid when heated with acetic anhydride yields a gamma lactone.
C 22H28N2O 5
A C 2O
O
OC
v-lactone
i)
Dehydrogenation:‐
When methyl reserpate is dehydrogenated with selenium it yields a hydrocarbon of
molecular formula ‐
C H N
19
16
2
This hydrocarbon is also obtained by dehydrogenation of Yohimbine with selenium & was therefore named as Yobyrine.
M ethyl r e s e r p at e
Se
D e hy d r o g e n ati o n
C 19H16N2
Yobyrine
Se
D e hy d r o g e n ati o n
Yohimbine
Structure of Yobyrine:_
1. Molecular formula:‐
2. Zinc Distillation:‐
C19H16N2
C 19H 16N 2
H
N
Zn dust
Distillation
3. Oxidation:‐
N
+
C2H5
Isoquinoline
3-ehtyl indole
When yobyrine is oxidized with permangnate it yields phthalic acid
C 19H 16N 2
K mnO4
COOH
COOH
cr2 o 3
Pthalic acid
C H3
COOH
o-Toulic acid
4. Condensation with aldehydes:‐
Yobyrine gives condensation products suggesting the presence of pyridine ring with a methylene substitution
adjacent to the nitrogen.
On the basis of above facts following structure has been postulated for yobyrine.
N
H
Yobyrine
N
C H3
Synthesis of Yobyrine:‐
CH3
NH2
N
H
2-(3-indolyl)ethyl amine
+
ClOC
CH2
O-tolyl acetyl chloride
1) -HCl
2)Reduction
N
H
N
CH3
CH2
POCl3
N
H
NH
CO
CH2
- 2H
dehydrogenation
N
H
Yobyrine
N
C H3
CH3
j)
As Yobyrine is formed from reserpic acid it means that possesses the following types of skeleton structures.
N
H
k)
reserpic acid may N
SKELETON STRUCTURE
From the fact 5(g) it follows that one of the methoxyl group is present in m‐
position to the NH group of indole i.e. on C‐11 i.e. Ome is at C‐11.
•
Reserpic acid when dehydrogenated yields 11‐hydroxy‐16‐methyl yobyrine,this may be only formed if –COOH group is present on C‐16
C22H28N2O5
Se
HO
N
H
dehydrogenation
H3C
•
N
CH2
CH3
From the step (h) it follows that –COOH & ‐OH group are in m‐position to each 11-hydroxy-16-methyl yobyrin
other but –COOH group is present at C‐16
therefore –OH group must be at C‐18. •
•
From purely biogenic reasons, the 2nd methoxyl group has been assigned position C‐17.
On the basis above mentioned facts the structure of reserpic acid may be MeO
N
H
N
HOOC
OH
OMe
Reserpic acid
Structure of Reserpine:‐
•
As reserpine is di‐ester of reserpic acid, it means that, COOH
MeO
N
H
N
+
CH3OH
+
-2 H2O
MeO
OMe
OMe
HOOC
OH
OMe
Reserpic acid
MeO
N
H
N
OMe
O
H3COOC
OMe
Reserpine
CO
OMe
OMe
Synthesis of Reserpine:‐
•
The structure of reserpine has been confirmed by its synthesis given by Woodward et.al.
O
O
H2C
C
+
CH
reaction
C
CH
CH
O
CH
COOH
O
COOH
p-Benzoquinone
Adduct
Reduction NaBH4
of less
hindered
( =CO)
OH
OH
O
Ac2O
CH2
Diels Alder
C6H5COOH
phenoxy benzoic
acid
O
COOH
O
COOH
Synthesis continue…..
O
O
O
CO
O
-H2O
Aluminium isopropoxide
O
OH
O
O
CO
O
O
CH3ONa
Br
CH3OH
NBS- H2SO4
HO
OMe
O
CO
Cro3
CO
OMe
O
CO
Synthesis continue…..
O
OH
Br
Zn / CH COOH
3
O
O
OMe
O
OMe
COOH
CO
iii)
COOCH3 COOCH3
MeO
N
H
H2N
6-methoxy
tryptamine
i) CH2N2
ii) Ac2O Pyridine
MeO
CH2
OAc
CHO
COOCH3
ii) CH2N2
MeO
i) HIO4
OAc
O
OH
OH
OSO4-H2O
Synthesis continue…..
MeO
N
N
H
H3COOC
H
C
i) NaBH4
CH2
H3COOC
OAc
OMe
ii) CH3OH
N
MeO
N
H
O
H3COOC
OAc
OMe
i) POCl3
ii) NaBH4
Synthesis continue…..
OMe
i)Resolve
N
N
H
ii)NaOH
MeO
iii)Hcl
iv)DCC
N
H3COOC
OMe
OC
N
H
O
OAc
OMe
Synthesis continue…..
Isomerisation
MeO
(CH3)3CCOOH
N
N
H
OMe
CH3OH
OCO
H3COOC
OMe
OMe
Reserpine
OMe
MeO
N
N
H
OMe
HOOC
OMe
OMe
OC
O
OMe
Reserpic acid Lactone
References :‐
1)“ Organic Organic Chemistry of Natural Products ”‐ by Gurudeep Chatwal ,Vol‐I.
2)“ Organic Chemistry ”by I.L.Finar , Vol‐II .
3)“Pharmacognosy” by C.K.Kokate ,
A.P.Purohit , S.B.Gokhale
‘O.D’