CHEMISTRYOF
NATURAL PRODUCTS
Prepared ByMr. Dilip Tayade
(Associate Professor)Department of Chemistry
Dhanaji Nana Mahavidyalaya, FaizpurDist.-Jalgaon , Maharashtra, India- 425503
NATURAL PRODUCTS
TerpenoidsTerpenoids are naturally occurring compounds obtained from various parts of
plants, microbes and animals.Terpene-Mixture of isomeric hydrocarbons of the molecular formula C10H16
occurring in terpentine and many other essential oils. All these compoundsare composed of isoprene units C5H8
Terpene- It is defined as natural products which consist of one or more
isoprene units. They are also called as terpenoids.Types:- Terpenoids are two types 1) Simple Terpenoids- obtained from sap and tissues of certain
plants ,these are steam volatile ,occurred in essential oils 2) Complex Terpenoids – obtained from gum & resins of
plants,these are not steam volatile.
1)2)3)4)
Isolation: Four MethodsExtractionSteam DistillationSolvent ExtractionAdsorption in purified fats ( enfleurage)
Extraction Method- Barks, leaves, seeds and flowers are crushed &juice is screened to remove particles, centrifuged e.g.- Citrus ,lemon and grass oil
Steam Distillation- Volatile oil distill out along with steam when plantmaterial is kept under steam distillation. An oil then collected byextracting with organic solvents such as petroleum ether . Oilunder goes changes such as,
Acid gets decarboxylated , ester gets hydrolyzed , some ringcompounds may breaks and also affect on odor of essential oil.
Solvent Extraction- crushed plant material is extracted with petroleumether, diethyl ether and then distilled out
Adsorption in purified fats ( enfleurage)-
Classification of Terpenoids
No. ofIsoprene
units
MolecularFormula
Class of Terpenes Examples
1 C5H8 Hemi-terpenoid Isoprene itself
2 C10H16 Mono-terpenoid Citral,Limolene,α-terpenoids,menthol,dipentoneα-pineol,camphor
3 C15H24 Sequi-terpenoid Humulene,bisabolenes
4 C20H32 Di-terpenoid Sex hormons,testoteron,estradiol,vitamin-D
5 C30H48 Tri-terpenoid( Steroid)
Steroids such ascholeterol,Ergosterol
6 C40H64 Tetra-terpenoid(Caretenoid)
β-Carotene
7 (C6H8)n Poly-terpenoid Natural Rubber
At the second level, each group of terpenoids is further classified intofurther groups on the basis of number of rings present in the molecule.
Mono-terpenoids
AcyclicMonoterpenoid
Monocyclic Monoterpenoids
BicyclicMonoterpenoids
CnH2n-2Citral
C10H16O
CnH2n α- Terpineol, C10H18O Menthol C10H20O
Dipentene C10H16 Limolene C10H16
CnH2n-2 Camphor
C10H16Oα- Pinene C10H16
At the third level, each class is further subdivided according tofunctional groups present in the terpenes.
Monocyclic Monoterpenoids Bicyclic monoterpenoids Hydrocarbons Alcohols Hydrocarbons KetonesDipentene α- Terpineol α- pinene Camphor Menthol
Isoprene Rule:The skeleton structure of all naturally occurring terpenoids is built up
of two or more isoprene units called as isoprene rule
Ingold pointed out that the isoprene unit in naturallyoccuring terpenoids were joined head to tail .This isknown as special isoprene rule.
Applications of rule: i) This rule helps in determining the
framework of unknown terpenoids.ii)The framework in which the isoprene rule & special
isoprene rules are noy followed are ruled out. Thefollowing frameworks are accepted .
All these structure follows rules
C)
General Methods for Structure Determination of Terpenoids
1. Molecular Formula:-Presence of elements are determined fromwhich empirical formula is determined. From empirical formula &mol.wt. the mol.formula is calculated.
2. Detection of unsaturation:- 3.Nature of Oxygen Function:- Oxygen atom may be (A)-OH, (B) –COOH,
(C ) -C=O, (D) –CO-CH2- (E) –COCH3 can be detected asA) -OH ( Alcoholic )group:- It is detected by the action of acetic
anhydride or phenyl isocyanate R-OH + Ac2O R-O-Ac + AcOH
R-O-H + O=C=N-C6H5 R-O-CO-NH-C6H5
B ) -COOH ( Carboxylic ) group :- R-COOH +NaHCO3 R-COONa +CO2 + H2O
>C=O ( Carbonyl ) group :- >C=O + H2N-OH >C=N-OH + H2O Oxime derivative
R-CHO + 2Ag(NH3)2OH R-COOH +2Ag +2NH3 +H2O
4)
-CO-CH2- (Oxo) group :- C6H5CHO +HNO2 Oxime +Benzylidene deri. E) –CO-CH3 ( Acetyl ) group:-It forms haloform reaction or on strong
oxidation with CrO3 gives acetic acid.
R-CH2-CO-CH3 3 Br2 /NaOH R-CH2-COOH + CHBr3 Bromoform
R-CH2-CO-CH3 CrO3 ( O ) R-COOH + CH3COOH Acetic acid
Degradation Products: Oxidative oxidation is more imp in structure determination.
Oxidizing agents such as alkaline KMnO4, H2O2,O3,CrO3,NaOBr etc.used for degradation. We can find out position of olefinic bond ,primary, secondary & tertiary alcohol present in a terpenoid .
a) Alkaline KMnO4 :- Olefinic bond gives dihydroxy compound ( glycol )
ii) Trisubstituted compound gives ketones & acid iii)Primary & secondary alcohols are stepwise oxidised to acids
while tertiary alcohols are not oxidized by alkaline KMnO4 .
R1-CHOH-R2 ( O ) R1-CO-R2
b)CrO3 – It also oxidizes alcohol stepwise.
c) O3 ( Ozone )- Ozonolysis find out the position of olefinic double bond .
i) Terminal alkenes after ozonolysis gives formaldehyde as one of theproduct.
ii)Tetra-substituted alkene gives diketones:-
d) OsO4 :- Alkenes react with OsO4 gives glycol
e) Performic acid( H-COOOH):-Alkenes react with
performic acid and forms epoxide which on hydrolysisgives glycols.
5) Dehydrogenation:- Dehydrogenation of terpenoids iscarried out by heating them in S , Se , Pd & Zn.
6) Dehydration:- conc.H2SO4 ,KHSO4,P2O5, ZnCl2 are commonly
used for dehydration of terpenoids , most of theterpenoids on dehydration gives p-cymene.
7) Physical Methods:- U.V., I.R., NMR,Mass spectroscopy are
modern methods used for determining structure ofterpenoids.
8) Synthetic Methods:- The final confirmation of structure is
achieved by its synthesis in the laboratory.
1)Monoterpenoids
CitralClass:- Citral belongs to acyclic ( open chain
monoterpenoids)Occurrence:- Lemon grass oil ( 70-80 %)Isolation :- Isolated as sodium bisulphite addition product
which on hydrolysis gives citral.Properties :- 1). It is a colourless oil,B.P .2280C 2). It is optically inactive.Uses:- 1). It is mostly used as perfume 2). It is the starting material for synthesis of vitamin
A Structure determination of citral1). Molecular Formula:- C10H16O
2). Detection of unsaturation:-
3)The formula for parent saturated hydrocarbon of citralbecomes C10H22,which corresponds to a general formulaCnH2n-2 suggesting that citral is a acyclic terpenoid.
Nature of oxygen:- a) Citral forms an oxime withhydroxyl amineand bisulbhiteaddition productwith saturated NaHCO3.Therefore citral contains either aldehyde or ketonic group.
b) Citral on reduction with Na/Hg in dil.acid forms a primaryalcohol,geraniol (C10H19O ). Citral 0n mild oxidation withmoist Ag2O gives geranic acid( C10H16O2) without loss ofcarbon atom.
Geranic acid Citral Geraniol R-COOH R-CHO R-CH2-OH These reactions confirm the presence of aldehyde group.
5) Position of aldehyde group:-a)It shows λmax 238nm which is characteristic of
conjugated aldehyde group.b) Citral on hydrolysis with aq.K2CO3 soln under high
pressure gives acetaldehyde & a ketone i.e. methylheptenone( 6-methyl-hep5-ene-2one)
Both reaction proves that citral is a conjugated aldehyde.6) Carbon skeleton in citral:- Citral when heated with
KHSO4 undergoes cyclisation to form p-cymene.Formation of p-cymene proves that citral contains a C-CH3 and isopropylidenen gr. In 1:4position wrt eachother
By considering all above reactions,the possible carbon skeleton is
Does not follow isoprene rule follow isoprene rule
7) Oxidative degradation:-a) b) Formation of acetone in both reaction shows the presence of gem
dimethyl group in citral. From all above reactions following structureis assigned
Reactions of Citral:
Synthesis of Citrala)2,4 Dibromo2 methylbutane reacts with sodio acetyl
acetone forms methyl heptenone
b) Conversion of methyl heptenone into citral
ALKALOIDSAlkaloids are naturally extracted which contain at least one nitrogen atom.
These are similar to alkali hence called as alkaloids and having markedphysiological,toxic and curative action on living organism. alkaloids arevery poisonous but if taken into small quantity can act as medicine. E.g.-Nicotine stimulates the CNS, Morphine relieves pain
Occurrence: - Alkaloids are naturally extracted from leaves roots , barks,seeds and fruits
Isolation:-1). The plant material containing alkaloids are dried , powdered &
extracted with petroleum ether to remove soluble fats.2). The residue is then extracted with methyl alcohol so that alkaloids get
dissolved in methyl alcohol and residue containing cellulosic and otherinsoluble material is rejected.
3). The filtrate is distilled off & the crude plant extract is then treated withdil.HCl or H2SO4 (PH=2) when alkaloids converted into their soluble salts,which are then treated with ether or chloroform.
4). Ether layer contains non basic plant material &it is rejected .
5). The water soluble parts containig salt ofalkaloids is treated with Na2CO3 or NaOH whenalkaloids precipitate out which are thenextracted with ether or chloroform.
6). The ether layer is distilled off & the residueobtained contains a mixture of alkaloids . Theindividual alkaloids are separated by usingtechniques like fractional distillation, steamdistillation and chromatography.
Powered Plant Material Extracted with PetroleumEther
Residue Ether PartExtracted with Methyl alcohol (Soluble fats rejected)
Methyl Alcohol Part contain Alkaloids Residue Distilled off to remove methyl alcohol ( Cellulosic material rejected) Crude Plant Extract dil.HCl PH=2 Extracted with Ether/ Chloroform Water Soluble Part contains Alkaloids Ether Soluble Part Na2CO3 or NaOH Non basic plant material rejected Extracted with ether/Chloroform Ether Soluble Part Water Soluble Part Evaporated Residue contains mixture of alkaloids
1.
2.
3.
4.5.
General Properties of Alkaloids:The alkaloids are colourless, crystalline, non-volatile solids.They are insoluble in water but soluble inorganic solvents like ether,chloroform , alcohols,except coniine & nicotine are soluble in water.Most of them are optically active (laevorotatory).They have bitter taste.Being basic in nature they dissolve in acidsforming their salts that may be readilycrystallized.
Alkaloid Formula Source Properties & Uses
1.Ephedrine C10H15NO GenusEphadra Heart Stimulant used in dry fever &asthamas
2.Ricinine C9H9N2O2 Castor been seeds Less toxic than other alkaloids
3.Coniine C8H17N Hemlock ( conicummaculatum )
Paralyses the CNS causes death.
4.Piperine C7H19NO3 Black pepper (piper nigrum )
Less toxic ,optically inactive
5.Nicotine C10H14N2 Tobacco ( nicotinaTabacum )
It is deadly poison,increases B.P.
6.Atropine C17H23NO3Atropa,belladonnaDatura Hyocyamusniger
Strong poisonDilation of the pupils of theeyes of cats.
7.Cocaine C17H21NO4 Erythroxylon cocaperuvian leaves
Its hydrochloride salts is used as a localanaesthetics in eye surgery & dentistry
8.Quinine C20H24N2O2 Cnchona Bark Medicine for Maleria & swamp fever
9.Papaverine
C20H21NO4 Opium oppy,Paparersomniferum
It produces slight sleep
10.Morphine
C17H19NO3 Opium serfumerplant
Analgesic to control pain
a)
General Methods for determining Structure of Alkaloids:Molecular Formula: :-Presence of elements are determined from which
empirical formula is determined. From empirical formula & mol.wt. themol.formula is calculated.
Detection of Unsaturation: Bromination OR KMnO4
Nature of Oxygen: Oxygen may be is in the form of -OH, X=O,-COOH,esteramide ,lactone , lactum , methoxy , methylene dioxy .-OH ( Hydroxyl):- It reacts with acetyl chloride or acid anhydride. Itmay be phenolic or alcoholic.phenol is soluble in NaOH & gives blue/green/ violet colouration with FeCl3 . The alcoholic action isdetermined by the action of dehydrating agent such as H2SO4 or P2O5
b) >C=O ( Carbonyl ) group : >C=O + H2N-OH >C=N-OH + H2O Oxime derivative R-CHO + 2Ag(NH3)2OH R-COOH +2Ag +2NH3 +H2Oc)-COOH ( Carboxylic group:- R-COOH +NaHCO3 R-COONa +CO2 + H2O
d) Ester , Amide, Lactone , Lactum : These groups can be detected &estimated by observing the products after alkali hydrolysis.
e) –OCH3 ( Methoxy): The presence of methoxy gr. & their nos. are
determined by Zeisel’s method. The ppt. of AgI dried & weighed, From the weight of AgI no. of methoxy
groups are calculated
1 mole of AgI = 1 methoxy group(-OCH3)
235 gm = 31 gm.
Problem:- How many methoxy groups per molecule are present inan alkaloid , when 4.24 mgs of alkaloid is treated with hot conc. HIto yield CH3I , which is passed into an alcoholic solution of AgNO3 ,when 11.62 mg of AgI is obtained .( Mol. Wt.= 340)
0 . 3 6 1 5 x 3 4 0
3 13 . 9 6 4 4
b)
c)
d)
f) Methylene dioxy( ): Alkaloids when treated with H2SO4 orHCl gives formaldehyde. The quantitative estimation of formaldehydegives the number of methylene dioxy groups.
Nature of Nitrogen : a) All alkaloids are basic in nature & contains nitrogen .
It may be primary ,secondary or tertiary & tertiary amines aredetermined by reacting methyl iodide & H2O2.
Distillation of alkaloids in presence of KOH gives an idea of no. of alkylgroups attached to the nitrogen atom.The formation of methyl amine& dimethyl amine indicates the no. of alkyl group attached to thenitrogen atom. >N-CH3 ( N-methyl group):Herzig Meyer method- Alkaloid is heatedwith HI at 3000c gives CH3I which then passed into the alcoholicsolution of AgNO3 forms ppt. of AgI.
Nature of heterocyclic rings: By breaking C-N bond & converting it insaturated hydrocarbon which gives information about the size ofheterocyclic ring.
H 2 C
O
O
-
-
5) Oxidative degradation: Oxidation of alkaloids is carriedout in presence of oxidizing agents such as KMnO4, H2O2, CrO3 ,conc.HNO3,K2Cr2O7 etc gives benzoic acid which proves thatephedrine might contain a benzene ring with side chain.
6) Physical Method: Physical methods like U.V., I.R., NMR,X-
ray analysis & mass spectroscopy are used for determining thestructure of an alkaloids.
7) Synhetic Methods: The final confirmation of structure isachieved by its synthesis in the laboratory starting from knowncompound.
Atropine:Isolation:Properties: 1)It is optically inactive.2)It is strong poison with sharp bitter taste.3) It acts on pupils of eyes .4) It is used in ophthaquinology.5) If it is taken orally first it stimulates & then
depress the CNS.6) Its M.P.is115-1160c.
Structure Determination of Atropine.1). Its moleclar formula is C17H23NO3
2). It is on hydrolysis gives an alcohol, tropine C8H15NO, & ( )tropic acid C9H10O3indicating that atropine is an ester i.e.tropine tropinate .
It means that it is made up from tropic acid & tropine.Structure of Tropic Acid:1) M.F. is C9H10O3
2). It possess –OH & -COOH group found out by usual tests.3). The formation of benzoic acid suggest that atropic acid &
hence tropic acid both contain a benzene ring with oneside chain.
Therefore the atropic acid C9H8O2 having a benzenenucleus, one carboxylic group & adouble bond may berepresented in the following two manners. Butcompound I may be atropic acid because compd II iswell known cinnamic acid
4) Atropic acid is formed by dehydration of tropic acid.
Hence addition of water molecule to I & II compoundswould give III & IV
Out of these two tropic acid has been shown to be IV by
various synthesis & Structure III is atropic acid.
1)2)
3)
Structure of Tropin ( Tropanol) :Its molecular formula isC8H15NOIt is saturated secondary alcohol & having tertiary nitrogen in the formof NCH3 group ( Herzig-Meyer Method)Tropine on treatment with HI gives tropine iodide which on reductionwith zinc & HCl yields tropane hydrochloride which on distillation inpresence of zinc dust forms 2-ethylpyridine through nor tropane.
On the basis of these reactions ,Landenburg proposed thatTropine is a reduced pyridine derivative. 4) This reaction suggest that ketonic gr. must be present in a ring & acoholic
gr . Is attached to the ring.Thus Ladenburg structure was discarded.
Further more as the tropine forms a dibenzylidene derivative withC6H5CHO,it must have –CH2CO-CH2-
5) Tropinic acid was degraded by exhaustive methylation to piperylenecarboxylic acid (A) which on reduction gives pimelic acid.
The formation of pimelic acid suggest that tropinic acid must contain a
seven carbon chain & tropinone a seven carbon ring. Further moretropinic acid contain –COOH & -CH2COOH gr. Three & fourmembered ring structures are not possible for tropinic acid.
Its structure is confirmed by oxidation with chromic oxide in H2SO4under drastic conditions to give N-methyl succinimide.Thus it has 5membered ring.
6). Thus on the basis of the structure of tropinicacid, tropinone and tropine must be as follows
Synthesis of Tropine: a) Robinson obtained tropinone bykeeping a mixture of succindialdehyde,metyl amine andacetone in water for 30 minute
The yield of tropinone from Robinson method has beenimproved upto 40% by using calcium dicarboxylate orethyl acetone dicarboxylate instead of acetone
b)Elming synthesized tropine in81% yield by condensing
methyl amine,acetone dicarboxylic acid &succiudialdehyde,generated in situ by the action of anacid on 2,5 dimethoxy tetrahydrofuran
Synthesis of Atropine: ( Fischer-SpeierEsterification): Tropine & tropic acid heated inpresence of HCl gives atropine . It is called as Fischer-Speier Esterification.
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