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Lecture1
Alkaloids
Originally the name alkaloids (which means alkali-like) were given to all organic
bases isolated from plants. Ladenberg proposed that alkaloid are natural plant
compounds having a basic character and containing at least one nitrogen atom in a
hetrocyclic ring.
This definition excludes any synthetic compounds and any compound obtained from
animal source. Even today it is difficult to define an alkaloid .We specify these as animal
alkaloid and plant alkaloids
The alkaloids are usually very poisonous, but are used medicinally in very small
quantity
They have basic properties, complex structure, physiological active and plant
origin is the main characteristic, which define plant alkaloids.
Seldom does a single alkaloid occur alone in given plant; rather another
accompanies it, which is closely related in structure and biosynthetic origin. The function
of alkaloids in plants is not yet understood. It has been suggested that they are simply
waste products of plants' metabolic processes, but evidence suggests that they may
serve specific biological functions. In some plants, the concentration of alkaloids
increases just prior to seed formation and then drops off when the seed is ripe,
suggesting that alkaloids may play a role in this process. Alkaloids may also protect some
plants from destruction by certain insect species.
Although alkaloids can be named by IUPAC method, the more complex ones
are known by their trivial names, which often reflect their physiological activities and end
with “ine” indicating a base.
Extraction of Alkaloids
Alkaloids are usually found in seeds, roots, leaves or the bark of the plants and
occur as salts of various plant acids e.g. acetic acid, oxalic acid, citric acid, malic
acid.tartaric acid, etc.
Method
The plant is dried and finely divided into powder and extracted with boiling
methanol or Chloroform.
The solvent is distilled off and the residue [mixed gum or oil].There may, however,
hundreds of compounds in this gum. A number these may be the and common chemicals
of general metabolism for example citric acid but more will compounds which are
characteristics of the plant, spices, extracted,[ which must be separated into its
component compounds) is treated with inorganic acid and bases are collected as watersoluble salts .
The free bases are librated with Na2CO3 and are extracted using various solvents
e.g. ether and chloroform etc. The mixtures of bases are separated by various methods
into the individual compounds including the use of chromatography.
Once a pure natural product is isolated, attention is focused on the elucidation of
the molecular structure.
General Properties
1. The alkaloids are usually colorless, crystalline, nonvolatile solids, which
are insoluble in water. Some alkaloids are liquids, which are water
soluble e.g. nicotine and a few are colored e.g. berberine, is yellow.
2. Most alkaloids have bitter taste and are optically active (levororatory).
They are usually tertiary nitrogen compounds and contain one or two
nitrogen atoms usually in ring system; most of the alkaloids contain
oxygen.
3. Alkaloids form insoluble precipitates with solutions of phosphotungstic
acid, phosphomolybdic acid, picric acid, potassium mercuric acid etc.
Many of these precipitates have definite crystalline shapes and so may be
used to help us the identification of an alkaloid.
General methods for determining of structure
Qualitative combustion analysis
After a pure specimen has been obtained it is subjected to qualitative combustion
analysis of carbon, nitrogen, and hydrogen.
Sodium fusion test
Na +C+NNaCN+FeSO4 Na4 [Fe (CN) 6] +FeCl3  Fe4 [Fe (CN) 6] 3 [For Nitrogen]
Ferri ferrocynide Blue Prussian
Na +XNa X
Na+SNa2 S +(CH 3COO) 2Pb  PbS
Quantitative analysis
In the quantitative analysis the empirical formula is obtained. The determination of
molecular
weight
finally
leads
to
the
molecular
formula
2
2
Wt of H 2O
%Age of Hydrogen = -------------------------100
18
Wt of sample
12
Wt of CO2
%Age of Carbon = -------------------------100
44
Wt of sample
Empirical Formula
%age of H
%age of C
%age of O
--------------------------1.008
12
16
=X
Y
Z
= XX
YX
ZX
= m =1
n
p
Molecular Formula Weight = {C n H m O p}N
If the alkaloid is optically active its specific rotation is measured. When an alkaloid
contains oxygen, the functional nature of oxygen is determined.
Determination of functional Groups
1.Hydroxyl Group
The presence of this group may be ascertained by action of acetic anhydride, acetyl
chloride, or benzoyl chloride on alkaloid. e.g
C6H5OH +CH3COCl  C6H5COOCH3 +HCl
CH3OH +CH3COCl  CH3COOCH3 +HCl
When is has been ascertained that hydroxyl group is present then their number can be
estimated
Hydroxyl group is Alcoholic or Phenolic?
.
The next problem is to decide where the group is alcoholic or phenolic
It is phenolic if the alkaloid is soluble in NaOH and reprecipated by CO 2
C6H 5OH +NaOH  C 6H5ONa +H 2CO3  C 6H5OH + NaHCO3
(Since
phenols
are
weaker
acids
than
carbonic
acid)
also a colouration with FeCl3 will indicate the presence of a phenolic group.
If it does not behave as a phenol the –OH group may be assumed to be
alcoholic. This may be verified by the action of dehydrating by H2SO4 or P2O 5.
2. Carboxyl Group (-COOH)
The solubility of the alkaloid in aqueous NaOH or NH3 indicates the presence of a
carboxyl group. The formation of esters also shows the presence of a carboxyl group.
CH3 COOH+ NaOH  CH3COONa +H2O
CH3 COOH + NH3 CH3COONH4
CH3 COOH+ CH3OH CH3 COO CH3
3. Oxo or Carbonyl > CO Group
The formation an oxime and phenyl hydrazone readily ascertain the presence of an
oxo group. Hydrolysis of the alkaloids and examination of the products led to information
that the compound is an ester, lactone
4. MethoxyGroup - OCH3
The presence of methoxy group and their number may be determined by the Zeisel
Method .The alkaloid is heated with Conc.HI at its boiling point 126 C.Methoxy group is
converted into methyl iodide which is then absorbed by ethanolic silver nitrate and
the AgI is weighed.
CH3 -O-CH3 +2 HI H2O + 2CH3 (Zeisel method for methoxy group )
C6H 5O CH3 + HI
 C6H 5OH +CH3 I
at 100 C
The results of hydrolysis will show the presence of an amide, lactam or betaine
5.Methylenedioxyl Group ( --0CH 2O--)
The presence of this group is indicated by the formation of formaldehyde when the
alkaloid is heated with HCl or H 2SO4
- 0CH 2O- +HCl  HCHO
6. The Functional Nature of Nitrogen
The general reaction of the alkaloid with acetic anhydride,CH3 I and HNO2
RNH2 [primary] +(CH3CO)2O CH3CONHR+ CH3COOH
RNH2 +2HNO2  ROH+N2+H 2O
RNH2+ CH3 I  RNH-CH3I
RNR 2 + HNO2  (RNR 2H) + (NO2) Salt
If all the reactions are negative, then the nitrogen is most probably tertiary. The
difficulty here is that some alkaloids undergo ring fission, the product being N-acetylated
derivatives.
If the alkaloid forms an amine oxide with 30% H2O2, then the nitrogen atom is tertiary.
CH3
CH3
\
\
CH3--N
+ H2O 2 
CH3-N=O
+
H2O
/
/
CH3
CH3
7. Determination of N-alkyl Groups
[A]Distillation of an alkaloid with aqueous KOH
Distillation of an alkaloid with aqueous KOH usually leads to information regarding
the nature and number of alkyl groups attached to nitrogen. The formation (in the
volatile products) of methylamine,dimethylamine or trimethylamine indicates
respectively the attachment of one, two or three methyl groups to a nitrogen atom;
the formation of ammonia shows the presence of an amino group.
CH3CONH-CH3 +KOH CH 3COOK + NH2 -CH3
CH3CON-CH3 +KOH CH 3COOK + NH -CH3
I
I
CH3
CH3
The presence of N-methyl groups and their number may also be determined by means of
[B]The Herzig-Meyer method.
When the alkaloid is heated with hydroiodic acid at 150-300 C under pressure, Nmethyl groups are converted into methyl iodide
CH3CONH-CH3 + HI  CH 3CO NH2 + CH3 I
CH3CON-CH3 +2 HI  CH 3CONH2 +2CH3I
I
CH3
Lecture2
8.Hofmann's exhaustive methylation method
Hofmann's exhaustive methylation method (1883) is a very important process in
alkaloid chemistry, since by its means heterocyclic rings are opened with the
elimination of nitrogen, and the nature of the carbon skeleton is thereby obtained.
The general procedure is to hydrogenate the heterocyclic ring (if this is
unsaturated), convert this compound to the quaternary methylammonium hydroxide
which is the heated. In this last stage a molecule of water is eliminated, a hydrogen atom
in the -position with respect to the nitrogen atom combining with the hydroxyl group,
and the ring is opened at the nitrogen atom on the same side as the -hydrogen atom
eliminated. The process is repeated on the product; this results in the complete
removal of the nitrogen atom from the molecule, leaving an unsaturated hydrocarbon,
In general, isomerises to a conjugated diene. e.g., pyridine gives piperylene.
CH2=CH-CH=CH-CH 3 1,3 penta diene
Although the general procedure for exhaustive methylation is to heat the
quaternary hydroxide at about 200 C, in a number of cases the reaction may be carried
out by refluxing an aquerous or ethanolic solution of potassium hydroxide containing the
methyiodide or methosulphate of the base. This procedure is usually satisfactory for bases
which contain benzene ring in the -position to the nitrogen atom. This may be explained
on the basis that benzylic hydrogen has an increased acidity (and so is more readily
removed) because of stabilisation of the transition state by conjugation (with the benzene
ring).
An interesting example of this is the case of laudanosine.
WHEN EITHER (-HYDROGEN ATOM IS ELIMINATED, A STYRENE DERIVATIVE IS FORMED, BUT IN ONE
THERE IS MORE EXTENDED CONJUGATION THAN IN THE OTHER, AND SO THE FORMER IS THE PRODUCT.
When either (-hydrogen atom is eliminated, a styrene derivative is formed, but
in one there is more extended conjugation than in the other, and so the former is the
product.
Even though the compound contains a β-hydrogen in e.g. tetrahydroquinoline ,The
exhaustive methylation may fail.
Alcohols are often obtained as by-products in this elimination product, and in some
cases no alkene is obtained (as in above example)
Emde Degradation
When bases which do not contain -hydrogen atom and Hofmann’s exhaustive
methylation method fails.
In this method, the quaternary ammonium halide is reduced with sodium amalgam
in aqueous ethanol or with liquid ammonia, or
is catalytically hydrogenated.
Examination of [I] shows that the β- hydrogen atom is absent and hence Hofmann’s
method cannot be used
It has been mentioned that exhaustive methylation fails with tetrahydroquinoline .The
ring is ,however, opened by EMde Degradation
9.Other Methods for opening Heterocyclic rings containing Nitrogen
Von Braun's Method for tertiary cyclic amines
A point of interest about the cyanogens bromide method is that it is often
successful with compounds that fail with the Hofmann’s method. Furthermore, where
both methods are applicable, ring opening occurs at different points of the ring
In general, fission of unsymmetrical amines occurs to give the bromide of
the'shorter'bromide
In the above example, the ring is opened, but in the other case dealkylation occurs with
formation of the cyclic N-derivatives, e.g.,
cocaine
Hydrolysis of the cyano compounds with HCl brings about the following changes
>NCN → {>NCOOH} →.NH
Thus, the final result is the removal of the N-methyl group without opening of the
ring.
(i)
Von Braun's Method for secondary cyclic amines
(III) In a number of cases the ring may be opened by heating With HI at 300C
10 Reduction
The presence of unsaturation in alkaloids may be ascertained by the addition of bromine
or halogen acids, or by the ability to be hydroxylated with dilute KMnO4.
OH OH
I
I
>C=C< + KMnO4 +H 2O 
>C—- C <
Reduction by means of sodium and ethanol, Zn with HCl, HI may also be used to
show presence of unsaturation .In some cases, reduction may decompose the molecule.
This often happen when catalytic reduction is used, (ring cleavage occurs). and hence
milder methods of reduction are desirable.
LiAlH4,NaBH4
11. Oxidation
This is one of the most valuable means of determining the structure of alkaloids .By
varying the strength of the oxidizing agent; it is possible to obtain a variety of products.
 Mild oxidation is usually effected with H 2O 2,I2 in ethalonic solution,O3,or alkaline
potassium ferricyanide.
 Moderate oxidation may be carried out by means of acid or alkaline KMnO4 or Cr 2O3
in Acetic acid.
 Vigorous oxidation is usually affected by K 2Cr 2O 7-H 2SO4, Cr 2O3H 2SO4, Cons.HNO3, MnO2- H2SO4
This classification is by no means rigid. The strength of an oxidizing agent depends to
some extent on the nature of the compound being oxide
12.Fusion of Alkaloid with KOH
Fusion of Alkaloid with KOH often relatively simple fragments, the nature of which will
give information’s on the type of nuclei present in the molecule.
13.Zinc Dust Distillation
This usually gives the same products as [12] except that when alkaloid contains
oxygen this is removed.