Alkaloids

General methods of structural determination

 Introduction
 Alkaloids are a group of naturally occurring chemical compounds that mostly contain
basic nitrogen atoms.
 The term alkaloid was coined by Meissner, a German pharmacist, in 1819.
 Alkaloids are cyclic organic compounds containing nitrogen in a negative state of oxidation
with limited distribution among living organisms.

Properties
 Most alkaloids contain oxygen in their molecular structure; those compounds are usually
colorless crystals at ambient conditions.
 Some alkaloids are colored, like berberine (yellow) and sanguinarine (orange).
 Most alkaloids are weak bases, but some, such as theobromine and theophylline, are
amphoteric.
 Many alkaloids dissolve poorly in water but readily dissolve in organic solvents.
 Most alkaloids have a bitter taste or are poisonous when ingested.
 Classification
 The alkaloids, as an important and enormously large conglomerate of naturally occurring
nitrogen containing plant substances having very specific as well as most diversified
pharmacological properties may be classified in a number of modes and means.
Based on the precursor from which the alkaloids are
produced in the plant biosynthetically.
1) Indole alkaloids derived from tryptophan.
2) Piperidine alkaloids derived from lysine.
3) Pyrrolidine alkaloids derived from ornithine.

based on the presence of the basic heterocyclic nucleus

i. Biosynthetic Classification (i.e., the chemical entity).
1) Pyrrolidine alkaloids e.g., Hygrine
ii. Chemical Classification 2) Piperidine alkaloids e.g., Lobeline
iii. Pharmacological Classification
iv. Taxonomic Classification pharmacological characteristics
1) Morphine as Narcotic analgesic
2) Quinine as Antimalarial
3) Strychnine as Reflex excitability

Based on the naming

1) Cannabinaceous Alkaloids
Ex: Hemp, Marijuana
2) Rubiaceous Alkaloids
Ex: Quinine, Katum, Kratum, Yohimbe
 Chemical tests
Dragendorff’s test: To 2–3 mL of the alkaloid
solution add few drops of Dragendorff’s reagent
(potassium bismuth iodide solution). An orange
brown precipitate is formed.
Mayer’s test: To 2–3 mL of the alkaloid solution add
few drops of Mayer’s reagent (potassium mercuric
iodide solution). White brown precipitate is formed.
Wagner’s test: To 2–3 mL of the alkaloid solution
add few drops of Wagner’s reagent (iodine-
potassium iodide solution). Reddish brown
precipitate is formed.
Biological activity
 In Plants
 They may act as protective against insects and herbivores due to their bitterness and
toxicity.
 They are, in certain cases, the final products of detoxification (waste products).
 Source of nitrogen in case of nitrogen deficiency.
 They sometimes act as growth regulators in certain metabolic systems.
 They may be utilized as a source of energy in case of deficiency in carbon dioxide
assimilation.

 In Humans
 High biological activity.
 Produce vary degrees of physiological and psychological responses – largely by
interfering with neurotransmitter.
 In large doses – highly toxic – fatal.
 In small doses – many have therapeutic value.
 Muscle relaxant, pain killers, tranquilizers, mind altering drugs, chemotherapy.
 General methods of structural
determination of alkaloids
 Molecular Formula :
 determination of molecular formula -Elemental composition -empirical formula
is found by combustion analysis.
 optical rotatory power. and hence the
 Determination of Unsaturation:
 adding bromine water
 by hydroxylation with KMnO4
 by reduction (using either LiAlH4 or NaBH4).
 Number of Double bond: - Number of Rings present in an alkaloids can be
determine by following formula- Ca Hb Nc Od

Number of No. of hydrogen No.of hydrogen

double bond present = in alkane in formula

2
 Functional Group Determination
 Functional groups containing oxygen
By using the usual standard chemical tests or by infrared (IR) spectroscopy
 Hydroxyl group: - Formation of Acetate on treatment with Acetic anhydride
/Acetyl chloride or benzoate on treatment with Benzyl chloride.

 By determining the amount of Acetic anhydride /Acetyl chloride or benzoate

that reacted with alcohol to form an ester, the number of hydroxyl groups can
be determined.
 Distinguish between primary, secondary and tertiary alcohol
 Lucas Test
 Oxidation with Cu/570K
 Excess of Alkali is estimated by titration with standard HCl. Number of -
OH group can be calculated from the volume of Alkali used for Hydrolysis.

PHENOLOC GROUP

 Neutral FeCl3 Test

 Soluble in NaOH = phenolic and regenerated with HCl
 Acetylation or Benzoylation
 Estimation of acetylation
 Carboxylic group: -
 soluble in aqueous solution sodium carbonate Na2CO3 or
ammonia NH3 and regenerated with HCl
 on treat with alcohol form ester.
 determined by volumetrically by titration against a standard
Ba(OH)2 or NaOH solution using phenolphthalein as an
indicator.
 Silver Salt Method

 Specific IR and NMR signals

 Carbonyl group:
 The presence of aldehydes and ketones can be detected by their
reaction with hydroxylamine to form the corresponding oxime.

 Tollens and Fehlings test can be used

 The aldehydes and ketones are distinguished by their oxidation
or reduction products
 IR, NMR and UV Spectra
 METHOXYL GROUP:
DETERMINATION BY ZEISEL METHOD
 When methoxy group present in a alkaloids treated with HI at
1260C perform methyl iodide which can treated further with
silver nitrite to perform silver iodide precipitate. Which
estimated gravimetrically : e.g.. Papavarine.
 Nature of Nitrogen:
 General reactions of alkaloids with acetic acid, methyl
iodide and nitrous acid indicates the nature of nitrogen. If all
reactions are negative – N2 probably tertiary

 Majority of nitrogen presence in alkaloids are secondary

and tertiary. (Nitrogen is in the ring)
 If tertiary when treated with H2O2 (30%) form amine
oxide.
 Nature and No. alkyl group attached to Nitrogen: Distillation
with Aq. KOH, formation of methylamine, dimethylamine and
trimethylamine (Vol. products)
 Herzig- Meyer method: presence and number of N- methyl group.
Estimation of N-alkyl groups
 Degradation of alkaloids
 Study of degradation of alkaloids gives rise to some identifiable
products of known structure.

 Knowing structure of the degraded products and the changes

occurred during the degradation, it is convenient to know the
structure of the original molecule.
 Different degradation reactions
 Hoffman exhaustive methylation method
 Emde’s method
 Von Braun’s (VB) method for 3° cyclic amines
 Reductive degradation
 Oxidation
 Zinc distillation
 Alkali fusion
1. Hoffman exhaustive methylation method
 The method was applied by Willstater in 1870 and was further developed by
Hoffmann.
 Heterocyclic rings are opened with elimination of nitrogen and the nature of
the carbon skeleton can be obtained.
 Hydrogenation of heterocyclic ring (if it is unsaturated)
 Convert the saturated compound to the quaternary methylammonium
hydroxide which is then heated.
 In this stage, a molecule of water is eliminated, a hydrogen atom in the β
position with respect the N atom combining with –OH group.
 The ring is opened at the N atom on the same side as the β-H atom
eliminated.
 This process is repeated on the same product . This resulted in the complete
removal of N atom from the molecule, leaving an unsaturated hydrocarbon,
which is isomerizes to a conjugated diene.
 HEM fails if there is no β-H atom available for elimination as water.
In such cases the Emde modification may be used
 As β-H atom is needed to cleave C–N bond and eliminate water molecule, the HEM
fails on the ring system that does not have β-H atom.
 For example, in the degradation of isoquinoline, the cleavage of N atom does not
occur at the final step as there is no β-hydrogen with respect to ‘N’.
 2. Emde’s method
 Emde’s modification may be used where HEM failed. In this method,
 4° ammonium halide is reduced with sodium amalgam in aqueous ethanol or
catalytically hydrogenated.
 3. Von Braun’s method for
tertiary cyclic amines
 Tertiary amine, which contains at least one-alkyl substituent, is treated with cyanogen
bromide.
 The results in cleavage of an alkyls nitrogen bond to give an alkyl halide and a substituted
Cyanamide.
 This method is often applicable to such compounds that do not respond to Hofmann's
method.

CNBr HBr
Boil + CH2Br
N Br
N

N N
H3C CN H3C CN
CH3

CH2Br
NH

CH3
 3. Von Braun’s method - II
 Secondary cyclic amine is treated with Benzoyl chloride in presence of NaOH to yield
the Benzoyl derivative which on treatment with phosphorus bromide followed by
distillation under reduced pressure yield α, ω - dihalo compound.

PBr5
Distil
Under red Br-(CH2)5 Br
C6H 5COCl PBr2-Br2
pressure

NaOH C6H5CN
N N
N Br
H
C6H5 C6H5
Benzoylation O Br
Other methods
• Oxidative degradation
• Zinc dust distillation
• Alkali distillation
• Catalytic dehydrogenation
 5. Oxidation
 Oxidation gives valuable information about the fundamental structure of
alkaloids and the position and nature of functional groups, side chains, etc.
 For example, picolinic acid obtained upon oxidation of coniine indicates that the
coniine is an α-substituted pyridine.

 By varying the strength of oxidizing agents, a variety of products may be

obtained. Different types of oxidizing agents used are as follows:
 For mild oxidation: H2O2, O3, I2.
 For moderate oxidation: acid or alkali KMnO4, CrO3 in CH3COOH.
 For vigorous oxidation: K2Cr2O7–H2SO4, concentrated HNO3 or MnO2–H2SO4.
5. Zinc distillation
 Distillation of alkaloid over zinc dust degrades it into a stable aromatic
derivative.

phenanthrene
morphine

2 propyl pyridine
 6. Alkali fusion
 Fusion of alkaloids with solid KOH gives simple fragments
from which the nature of alkaloid can be derived.

 Means papaverine is containing isoquinoline nucleus.

So adrenaline is a monosubstituted catechol derivative.

 7. Dehydrogenation
 Distillation of alkaloid with catalysts such as S, Se and Pd yields
simple and recognizable products from which the gross skeleton
of the alkaloid may be derived.
 Thus with the help of degradation, nature of various fragments
obtained, nature of nucleus and type of linkages are established.
 The fragments obtained are arranged in the possible ways with the
possible linkages and the one that will explain all the properties is
selected and confirmed by synthesis.
 Optical activity of an alkaloid helps greatly in establishing the
structure of alkaloid
 8. Physical Methods
The important physical methods used in structural elucidation
of alkaloids are as follows:
 IR spectroscopy: Identify functional groups
 UV spectroscopy: Characteristic of chromophoric system
 NMR spectroscopy: Detect protons
 Mass spectroscopy: Know molecular weight & fragments
 X-ray analysis: Distinguish the various possible structures
 Optical rotatory dispersion (ORD) and circular dichroism :
Optically active stereoisomers.
 Conformational analysis: Stereochemistry
 9. Synthesis
The above-mentioned chemical and analytical work
helps to propose a tentative structure (or structures)
of the alkaloid under investigation.
Synthesis always gives additional evidence for the
assigned structure even though the physical methods
(mentioned above) provide final proof of the proposed
structure.
STRUCTURAL ELUCIDATION

Reserpine
Papaverine
 Reserpine ச ர ் ப க ந ் த ா (rauvolfia
serpentina; ப ா ம் ப ு க ் க ள
Synonym: Sarpagandha. ா / ப ா ம் ப ு க லா , ச ி வன
் அமல் ப ொ டி )
Biological source: It is obtained from dried
roots of Rauwolfia serpentine & Rauwolfia
vomitoria.
Family: Apocynaceae.
Parts used: Roots.
Chemical constituent: Alkaloids- reserpine,
yohimbmine, ajmaciline & ajmaline.
• Medicinal Importance
• Controls
• Hypertension
• Epilepsy
• insomnia
 Constitution

1.Molecular formula:
C H N0
33 40 2 9

2. Presence of five methoxyl group:

When treated with Hl yields five molecules of methyl iodide
indicating the presence of five methoxy groups.
 Nature of Oxygen – Test shows No –COOH and -OH

3. Nature of ‘N’ atom:

It is weak base.
So both N atoms are inside the ring system.
forms monoacetyl derivative It does not have any hydroxyl group
So one N atom is present as NH group.

IR spectra reveals indole nucleus

4. Hydrolysis: Hydrolysis of reserpine with alkali yields mixture of
i) Methyl alcohol
ii) 3,4,5-trimethoxy benzoic acid
iii) Reserpic acid.

3,4,5-trimethoxy benzoic acid

Mono hydroxy carboxylic acid

 As reserpine does not contain -COOH & -OH groups, hence its
hydrolysis product proves that reserpine is a di-ester.
5. Reduction:

3,4,5-trimethoxy
benzoic acid
LiAlH4 LiAlH4

3,4,5-trimethoxy benzyl alcohol

7.Structure of Reserpic acid:
a)Molecular formula- C H N O
22 28 2 5

b)Presence of one carboxyl group

Silver salt method - Reserpic acid gives only one
mole of Ag/ mole of Reserpic acid
c) Presence one -OH group
Reserpic acid on oxidation yields a ketone
that means it has secondary alcoholic
group.

d) Nature of methoxy groups

By zeisels method. 2 moles of AgI formed per mole of
Reserpic acid
It shows that reserpic acid contains 2 methoxy groups.
e) Nature of two ‘N' atom
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 LiAlH4 yields reserpic alcohol.

Contains
Two –OCH3,
One –OH
One CH2OH
 g) Oxidation of reserpic acid
On oxidation with KMnO4 it gives 4-methoxy N-oxalyl
anthranilic acid.

KMnO4

Thus one of the methoxyl group is present in meta position to -

NH group
h) Fusion with KOH
When reserpic acid is fused with KOH it yields 5-hydroxy
isophthalic acid. One of the acidic groups of isophthalic acid
must be One –COOH group is that of acid
so –OH and –COOH are meta
 Present in m-position to each other this confirm by the fact that
reserpic acid when heated with acetic anhydride yields a gamma
lactone.

α γ
β

i) Dehydrogenation
When methyl reserpate is dehydrogenated with selenium it
yields a yobyrine of Molecular formula C19H16N2

This yobyrine is also obtained by dehydrogenation of Yohimbine

with selenium & was therefore named as Yobyrine.
8. Structure of Yobyrine
a) Molecular formula: C19H16N2

a) Zinc Distillation
When distilled with zinc it yields 3 ethyl indole &
isoquinoline.
c) Oxidation
When yobyrine is oxidized with pot. permangnate it yields
phthalic acid.

KMnO4

Chromic acid /
Jones reagent
d) Condensation with aldehydes
Yobyrine gives condensation products suggesting the presence of pyridine
ring with a -CH2 substitution adjacent to the nitrogen.
On the basis of above facts following structure has been postulated for
yobyrine.
e)Synthesis of yobyrine

dehydration

Pd/C

Yobyrine
f) As yobyrine is formed from reserpic acid it means that
reserpic acid may possesses the following types of skeleton
structures.

g) We have found
one methoxyl group is meta to the NH
group of indole
Reserpic acid - dehydrogenated yields 11-hydroxy-
16-methyl yobyrine, So -COOH group is present on C-16
 Also -COOH & -OH group are in m-position to each 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
9. Structure of reserpine
As reserpine is di-ester of reserpic acid, the structure of
reserpic acid is written as follows
Synthesis of Reserpine

KNOWN We will synthesise

10. Synthesis of Reserpine
The structure of reserpine has been confirmed by its
synthesis given by Woodward et. Al., (1956, 1958)

AcONa
 Allylic
Bromination
In polar solvent
gives addition
product

CrO3
Check this you tube video to understand the thought process behind the
synthesis. Run it in 0.5 speed

https://www.youtube.com/watch?v=oCKCkuh_ISQ
PAPAVARINE

ISOQUINOLINE ALKALOID
OPIATE ALKALOID
PLANT PAPAVER SOMNIFERUM
PRODUCED SYNTHETICALLY
MEDICINAL BENEFITS
• direct-acting smooth muscle
relaxant
• antiviral activity against
respiratory syncytial virus,
cytomegalovirus, and HIV.
• vasodilator agent
• antispasmodic drug.
• MOLECULAR FORMULA C20H21NO4
• Optically inactive compound
• Molecular Weight 339.4 g/mol
Quaternary ammonium
So nitrogen is tertiary
salt

Contains four
OCH3 groups

[O]

C9H12O2 C11H11NO2
 C9H12O2 C11H11NO2

MeO
C11H11NO2
MeO
Nitrogen is found to be tertiary
Zeisel method- two methoxy group

demethylation Zn Dust
C11H11NO2 C9H7NO2
HI
-CH3I

Compound may be a dimethoxy isoquinoline

C11H11NO2 [O]
Synthesis of 6,7 - dimethoxy isoquinoline
C9H12O2

1. From Zeisel method- two OCH3 groups are present

2. On oxidation – Veratric acid- 3,4 dimethoxy benzoic acid

C9H12O2

3. On demethylation + oxidation it gives Protocatechuic acid –

3,4 dimethoxy benzoic acid

HI
C9H12O2
[O]
 Lets connect C9H12O2 & C11H11NO2

Papaverine contains 4 methoxy groups so we cannot disturb the OCH3

So connection through benzene ring or –CH3
Oxidation products - 6,7- dimethoxyisoquinoline-1-carboxylic acid
2,3,4 pyridine carboxylic acid

HOOC
MeO

HOOC
MeO
COOH COOH
6,7- dimethoxyisoquinoline-1-carboxylic acid
2,3,4 pyridine carboxylic acid
 CO

Oxidation

with cold with hot

dilute KMnO4 dilute KMnO4 ketone
Secondary alcohol

CHOH PAPAVARINIC ACID

HOOC

HOOC
CO

So a -CH2 – group must be present

Synthesis of PAPAVARINE

Step 1 Synthesis of homoveratryl amine

Step 2 Synthesis of homoveratroyl Chloride
Step 3 Condensation of homoveratryl amine
and homoveratroyl Chloride
Step 1 Synthesis of homoveratryl amine
 Hydrolysis
Reduction
PCl5
H2/ Ni
Condensation of homoveratryl amine and homoveratroyl Chloride