Alkaloids PDF

Alkaloids Introduction: 1- Definition of an alkaloid: Originally the name alkaloid (alkali- like) was given to all organic bases isolated from plants. On the whole, alkaloids are very poisonous, but are used medicinally in very small quantities. Complex structure, physiological action and plan...

Alkaloids Introduction: 1- Definition of an alkaloid: Originally the name alkaloid (alkali- like) was given to all organic bases isolated from plants. On the whole, alkaloids are very poisonous, but are used medicinally in very small quantities. Complex structure, physiological action and plant origin are the main characters which define plant alkaloids. 2- Extraction of alkaloids: Alkaloids are usually found in the seeds, root, leaves, or bark of plant, and generally occur as salts of plant acids, e.g., acetic, citric, tartaric, etc., the plant is dried, then finely powdered and extracted with suitable solvent. The solvent is distilled off, and the residue treated with inorganic acids, where upon the bases are extracted as their soluble salts. The free bases are liberated by the addition of sodium carbonate and extracted with various solvents, e.g., ether, chloroform, etc. The mixtures of bases thus obtained are separated by various methods ( chromatography, ion-exchange, etc.) into individual compounds. Most alkaloids are obtained from natural sources, but a few are synthesized commercially, e.g., ephedrine. 3- General properties: The alkaloids are usually colourless, crystalline, non-volatile solids which are soluble in ether, ethanol, chloroform, etc., but are insoluble in water. Some alkaloids are liquids which are soluble in water, e.g. nicotine and coniine and a few are colourled, e.g. berberine is yellow. Most alkaloids have a bitter taste and are optically active. They are generally tertiary nitrogen compounds and contain one or two nitrogen atoms usually in the tertiary state in a ring system; most of alkaloids also contain oxygen. The optical active alkaloids are very useful for resolving 1 racemic acids. The alkaloids form insoluble precipitates with solutions of some acids, e.g., picric acid. Many of these precipitates have definite crystalline shapes and so may be used in the identification of an alkaloid. 4- General methods for determining alkaloid structure; 1- Determination of molecular formula using qualitative and quantitative analysis. 2- The function nature of oxygen; * -OH group; it appear with acetic anhydride, acetyl chloride, or benzoyl chloride, then the number of hydroxyl groups is also estimated (using acetylation), phenolic or alcoholic (using sodium hydroxide or ferric chloride). The alcoholic groups are readily dehydrated with sulphuric acid or phosphorus penta chloride. * -COOH group; by treatment with aqueous sodium carbonate or formation of ester. * -CO- group; formation of oxime, semicarbazone and phenylhydrazone. * Hydrolysis of the alkaloid and examination of the products led to information that the compound is an ester, lactone, amide, etc. * -OMe group; the presence of methoxy groups and their number may be determined by the Zeisel method; AgNO3 -OMe + HI MeI AgI * Methylenedioxyl group (-OCH2O-); by heating with hydrochloric acid heat -OCH2O- CH2O HCl 2 3- The function nature of nitrogen; * The reaction of the alkaloid with acetic anhydride, methyl iodide and nitrous acid often show the nature of the nitrogen, if all these reactions are negative, then the nitrogen is tertiary. * Distillation of an alkaloid with aqueous potassium hydroxide leads to information regarding the nature and number of alkyl groups attached to nitrogen. * Hofmann's exhaustive methylation method is a very important process in alkaloid chemistry, since by its means heterocyclic rings are opened with elimination of nitrogen, and the nature of the carbon skeleton structure is thereby obtained. H2 MeI heat Pt AgOH + OH- - H2O N N N N H Me2 Me2 MeI heat AgOH + OH- - H2O + NMe3 N Me3 * Hydramine fission also indicate the nature and number of alkyl groups attached to nitrogen throught heating with hydrochloric acid. 4- The presence of unsaturation in an alkaloid may be ascertained by the addition of bromine, or halogen acids. Reduction by means of Na/Hg, Na/EtOH, also may be used to show the presence of unsaturation. In some cases, reduction may decompose the molecule, and hence milder methods of reduction are desirable, e.g. LiAlH4 and sodium borohydride. 5- Oxidation is one of the most valuable means of determining the structure of alkaloids using different oxidizing agents; - Mild oxidation: H2O2, O3, I2/EtOH - Moderate oxidation: acidic or alkaline KMnO4, CrO3/AcOH 3 - Vigorous oxidation: K2Cr2O7/H2SO4, CrO3/H2SO4, conc. HNO3, or MnO2/H2SO4 6- Fusion of an alkaloid with solid KOH produces relatively simple fragments, the nature of which will give information on the type of nuclei present in the molecule. 7- Zinc dust distillation give also the nature of nuclei, but it remove oxygen. 8- Physical methods, e.g. IR, UV, X-ray, NMR, GC-MS. 9- Synthesis of alkaloid as a means of final proof of determining structure. 5- Classification of alkaloids: 1- Phenylethylamine group. 2- Pyrrolidine group. 3- Pyridine-piperidine group. 4- Pyrrolidine-pyridine group. 5- Quinoline group. 6- Isoquinoline group. 7- Phenanthrene group. 8- Indole group. Phenyl ethylamine group. Ephedrine (C10H15NO) Ephedrine is colorless solid melting at 38 0C and occurs in the genus ephedra, its physiological action is similar to that of adrenaline, and it can be taken orally. Ephedrine has also been used in the treatment of hay fever, bronchial asthma, etc. 4 *Determination of ephedrine structure: 1- Oxidation; ephedrine gives on oxidation benzoic acid, the structure therefore contains a benzene ring with only one side- chain oxid. C10H15NO COOH 2- Action of nitrous acid; it forms a nitroso compound, thus it contains imino group ( secondary amine ) 3- With benzoyl chloride it forms dibenzoyl derivative, thus one hydroxyl group must be present (one benzoyl group is accounted for by the imino group) 4- When heated with hydrochloric acid (hydramine fission), ephedrine forms methyl amine and propiophenone heat C10H15NO HCl CH3NH2 + COCH2CH3 5- The formation of the above products can explained if the structure of ephedrine is either (A) or (B) OH OH NHCH3 CHCH2CH2NHCH3 CH-CHCH3 (A) (B) 6- It has been observed, however that compound (B) undergo hydramine fission, thus structure (B) is more likely than (A) 5  7- Hofmann's exhaustive methylation method; it forms 1,2- methyl phenyl-ethylene oxide, which cannot be formed from structure (A) + OH NHCH3 OH (CH3)3 N MeI CH-CHCH3 CH-CHCH3 OH- AgOH O heat -H2O CH-CHCH3 + (CH3)3N 8- Structure (A) contains one chiral center and so replacement of the hydroxyl group by hydrogen will result in the formation of an optically inactive compound. Experimentally it has been found that when this replacement is effected in ephedrine, the product is optically active, thus structure (B) agrees with all the know facts *Synthesis of ephedrine: OO O NCH3 C-C-CH3 + CH3NH2 C-C-CH3 1-phenylpropane-1,2-dione OH NHCH3 H2/Pt CH-CHCH3 Adrenaline (C9H13NO3) 6 Adrenaline is a non-steroid hormone and the adrenal medulla is the source of adrenaline. It is active only when given by injection, it raises the blood-pressure and it is used locally to stop haemorrhage. Adrenaline is a colorless crystalline solid melting at 2110C and dissolves in acids and alkali (insoluble in water) and it is also optically active. *Determination of adrenaline structure: 1- The phenolic character of adrenaline is indicated by its solubility in sodium hydroxide and its reprecipitation by carbon dioxide. NaOH CO2 Adrenaline Soluble Reprecipitate 2- It gives a green color with ferric chloride, thus it is a catechol derivative. 3- When boiled with aqueous potassium hydroxide, adrenaline evolves methyl amine, thus a methyl amino group is probably present. Boiling C9H13NO3 CH3NH2 aq. KOH 4- Fusion with solid potassium hydroxide gives protocatechuic acid. OH Fusion OH C9H13NO3 KOH COOH 7  5- Methylation, followed by fusion with solid potassium hydroxide, gives veratric acid and trimethyl amine which indicates that the nitrogen atom must occur at the end of the side chain. OMe OMe MeI/KOH C9H13NO3 + NMe3 Fusion COOH 6- Since adrenaline is optically active, it must contain at least one chiral centre 7- Treatment of adrenaline with benzenesulphonyl chloride gives a tribenzensulphonyl derivative, which on oxidation gives a ketone, thus the structure of adrenaline contains a secondary alcoholic group. 8- Oxidation of adrenaline gave benzoic acid derivative, thus the - CHOH- group must be attached to the benzene nucleus, had it been -CH2CHOH-, then a phenyl acetic acid would have been obtained *Synthesis of adrenaline: OH OH OH OH POCl3 + ClCH2COOH COCH2Cl OH OH CH3NH2 OH H2/Pd OH 8 COCH2NHCH3 CH(OH)CH2NHCH3  Pyrrolidine group Hygrine (C8H15NO) One of the coca alkaloids. It is a liquid boiling at 1930C. *Determination of hygrine structure: 1- Reactions of hygrine indicate the presence of a ketone group and a tertiary nitrogen atom. 2- When hygrine is oxidized with chromic acid, it gave hygrinic acid. Oxid C8H15NO C6H11NO2 Hygrinic acid was first believed to be a piperidine carboxylic acid, but comparison with the three piperidine acids showed that this was incorrect. 3- Dry distillation of hygrinic acid gives N-methyl pyrrolidine, hence hygrinic acid is an N-methyl pyrrolidine carboxylic acid. Furthermore, since the decarboxylation occurs very readily, the carboxyl group was assumed to be in the 2-position (by analogy with amino acids). Dry C6H11NO2 Dist. N CH3 9  4- The structure of hygrinic acid (1-methyl pyrrolidine-2- carboxylic acid) was confirmed by synthesis; Br(CH2)3Br + [CH(COOC2H5)2]-Na+ Br(CH2)3CH(COOC2H5)2 Br2/H2O CH3NH2 Br(CH2)3CBr(COOC2H5)2 COOC2H5 N COOC2H5 H2O/1600C CH3 N COOH CH3 Thus, possible structures for hygrine are; N CH2COCH3 N COCH2CH3 CH3 CH3 (A) (B) *Synthesis of hygrine: Condensation of γ-methylaminobutyraldehyde with ethyl acetoacetate in pH= 7 confirmed that the possible structure of hygrine is (A) COOC2H5 pH 7 NH CHO + CH2COCH3 N CH2COCH3 CH3 CH3 Pyridine-Piperidine group Coniine (C8H17N) 10  Coniine is a liquid alkaloid boiling at 1660C, it occurs in oil of hemlock. *Determination of coniine structure: 1- Distillation of coniine with zinc dust gives conyrine Zn C8H17N C8H11N Dist. 2- Oxidation of conyrine with potassium permanganate gives pyridine-2-carboxylic acid (α-picolinic acid), it follows that a pyridine nucleus is present with a side chain in 2-position. Thus coniine is probably a piperidine derivative with a side chain in 2-position. KMnO4 C8H11N N COOH 3- The side chain of coniine must contain three carbon atoms, since tow are lost when conyrine is oxidized, it is therefore either n-propyl or isopropyl, and it was actually shown to be n- propyl by the fact that when heated with hydroiodic acid at 3000C under pressure, coniine forms n-octane. Had the side chain been isopropyl, then the expected product would be isooctane. HI/3000C Zn CH3 CH2(CH2)2CH3 U.P. N CH2CH2CH3 N CH2CH2CH3 H n-Octane Coniine Conyrine 11 *Synthesis of coniine: Piperine (C17H19NO3) Piperine is a solid alkaloid melting at 128 0C, occurring in pepper, especially black pepper. *Determination of piperine structure: 1- Hydrolysis of piperine with alkali gives piperic acid and piperidine, thus piperine is the piperidine amide of piperic acid. KOH C17H19NO3 + H2O C12H10O4 + N H 2- The routine tests show that piperic acid contains one carboxylic group and two olefinic bonds 3- Oxidation of piperic acid with permanganate, gives piperonal and then piperonylic acid. Oxid. Oxid. C12H10O4 C8H6O3 C8H6O4 12 4- The structure of piperonylic acid is deduced from the fact that when heated with hydrochloric acid at 200 0C under pressure, it gives protocatechuic acid and formaldehyde 5- Since one carbon atom is eliminated in the last reaction and there are no free hydroxyl groups in piperonylic acid, the HO COOH HCl/2000C C8H6O4 U.P. + HCHO HO structure of this acid is probably the methylene ether of protocatechuic acid, i.e., piperonylic acid is 3,4-methylenedioxy benzoic acid, this has been confirmed by synthesis HO COOH COOH Heat O + CH2I2 HO NaOH O 6- Furthermore, since piperonal gives piperonylic acid on oxidation, piperonal is 3,4-methylenedioxy benzaldehyde O CHO O COOH KMnO4 O O 7- Careful oxidation of piperic acid gives tartaric acid in addition to piperonal and piperonylic acid. This indicates that piperic acid is a benzene derivative containing only one side chain that contains the two double bonds. 13  O CH=CH-CH=CH-COOH Oxid. O (I) O CHO O COOH HO -CH-COOH + + HO -CH-COOH O O Considering all this facts, one can decide that the structure of piperic acid is as shown in formula (I) *Synthesis of piperine: Piperonal (prepared via the Reimer- Tiemann reaction) is condensed with acetaldehyde in the presence of sodium hydroxide (Claisen- Schmidet reaction) and the product (a cinnamaldehyde derivative) is then heated with acetic anhydride in the presence of sodium acetate (Perkin reaction) HO HO CHO CHO NaOH CH2I2 O + CHCl3 NaOH HO HO O CH3CHO NaOH O CH=CH-CH=CH-COOH CH=CH-CHO (CH3CO)2O O O CH3COONa O When the acid chloride of piperic acid is heated with piperidine in benzene solution, piperine is formed PCl5 O CH=CH-CH=CH-COCl Piperic acid O N O CH=CH-CH=CH-CO N H 14 O  Pyrrolidine-Pyridine group Nicotine (C10H14N2) Nicotine is the best known and most widely distributed of the tobacco alkaloids, it have been isolated from tobacco leaf, liquid compound boiling at 247 0C. *Determination of nicotine structure: 1- Oxidation of nicotine with potassium permanganate gives nicotinic acid COOH KMnO4 C10H14N2 N 2- Since nicotinic acid is a product of oxidation, nicotine therefore contains a pyridine nucleus with a complex side chain in 3- position. Thus we may write the formula of nicotine as follows; C5H10N N 3- The side chain (-C5H10N) was originally believed to be piperidine, but further work showed that this was incorrect. When nicotine is distilled with zinc chloride, the products are pyridine, pyrrole and methyl amine. This suggest that the side chain is a pyrrole derivative. ZnCl2 C10H14N2 + + CH3NH2 Dist. 15 N N H 4- When nicotine is heated with concentrated hydroiodic acid at 1500C, methyl iodide is formed. HI C10H14N2 MeI 1500C Thus the side chain contains an N-methyl group. It therefore appear that the side chain could be N-methyl pyrrolidine, but its point of attachment to the pyridine nucleus could be either 2- or 3- C5H10N OR N N CH3 CH3 5- The most direct analytical evidence for the presence of the pyrrolidine nucleus has been given by Karrer, nicotine hydroiodide forms nicotine isomethyl iodide when warmed with methyl iodide and this, on oxidation with potassium ferricyanide, is converted into nicotone which, on oxidation with chromium trioxide, gives hygrinic acid. K3Fe(CN)6 CrO3 N N N COOH CH3 N O CH3 N CH3 CH3 }+ I- CH3 6- All the foregoing facts are satisfied by the following structure for nicotine 16 N CH3 N *Synthesis of nicotine: The complete synthesis of nicotine was carried out in 1928 by Elec. 1.(CH3)2SO4 (i) CO OC CO Red. CO 2. NaOH N N N H H CH3 Succinimide 2-Pyrrolidone N-methyl-2-pyrrolidone COOH COOC2H5 (ii) H2SO4 + C2H5OH N N Nicotinic acid Ethyl nicotinate Spath as follows COOC2H5 CO EtONa (iii) OC + N CO N N N CH3 CH3 COCH(COOH)CH2CH2NHCH3 HCl -CO2 CO HN 1300C N CH3 N + Zn dust 3HI H2N HN I 2I- OH 1000C + CH3 EtOH/NaOH CH3 N N H NaOH N 17 CH3 N  Quinoline group Cusparine (C19H17NO3) It is a solid isolating from angostura bark, melting at 900C. *Determination of cusparine structure: 1- It has been shown to contain one methoxy group from Zeisel method 2- When cusparine is fused with potassium hydroxide, protocatechuic acid is obtained. OH OH i) fusion with KOH C19H17NO3 + ii) H COOH 3- Controlled oxidation of cusparine gives piperonylic acid and 4- methoxyquinoline-2-carboxylic acid. OMe O CH2 CrO3 O C19H17NO3 HOOC + N COOH 4- Consideration of this information led to the suggestion of the following structure for cusparine 18  OMe O CH2 N CH2-CH2 O *Synthesis of cusparine: This has been confirmed by Spath synthesis OMe O CH2 O + OHC N CH3 OMe O ZnCl2 CH2 O N CH=CH OMe H2 O CH2 Pd-C O N CH2-CH2 Opium alkaloids Many alkaloids have been isolated from opium and they are divided into two groups according to the nature of their structure; i) Isoquinoline group: e.g. papaverine ii) Phenanthrene group: e.g. morphine 19  Isoquinoline group Papaverine (C20H21NO4) Solid alkaloid melting at 1470C, and optically inactive. *Determination of papaverine structure: 1- Since papaverine adds on one molecule of methyl iodide to form a quaternary iodide, the nitrogen atom is in the tertiary state. 2- The application of the Zeisel method shows the presence of four methoxy groups, the demethylated product is know as papaveroline. C20H21NO4 + 4HI 4CH3I + C16H13NO4 3- When oxidized with cold dilute permanganate, papaverine is converted into the secondary alcohol papaverinol, which on more vigorous oxidation with hot dilute permanganate is oxidized to the ketone papaveraldine. The foregoing reactions led to the conclusion that papaverine contains a methylene group Oxid. Oxid. (C19H19NO4)CH2 (C19H19NO4)CHOH (C19H19NO4)CO cold hot 4- When papaverine oxidized with hot concentrated permanganate, it broken down into small fragments, veratric acid which indicate the presence of group (a) in papaverine, 6,7- dimethoxyisoquinoline-1-carboxylic acid which indicate the presence of group (b) in papaverine, the presence of the latter 20  group also accounts for the isolation of other two fragments; metahemipinic acid (c) and pyridine-2,3,4-tricarboxylic acid (d) Cx MeO N MeO OMe Cy OMe (a) (b) COOH MeO COOH HOOC MeO COOH HOOC N (c) (d) 5- The total carbon content of (a), 9 carbon atoms, plus that of (b) 12 carbon atoms is 21 carbon atoms. But papaverine contains only 20 carbon atoms, there is however a -CH2- group present and if we assume that –Cx and –Cy are one, we can draw papaverine structure as follows MeO N MeO CH2 OMe OMe 21  *Synthesis of papaverine: MeO MeO CH2Cl MeO CH2CN HCHO KCN HCl MeO MeO MeO H2/Ni i) HCl ii) PCl5 MeO CH2CH2NH2 MeO CH2COCl MeO MeO (I) (II) MeO MeO Heat P2O5 (I) + (II) - HCl N NH -H2O MeO MeO OC CH2 CH2 MeO MeO OMe OMe MeO N MeO Pd on asbestos CH2 2000C MeO OMe 22  Phenanthrene group Morphine (C17H19NO3) HO O NMe H HO Morphine is the chief of alkaloid in opium and was the first alkaloid to be isolated, and melts at 2540C. *Determination of morphine structure: 1- The usual tests show that the nitrogen atom is in the tertiary state, and since morphine forms a diacetate (heroin) and a dibenzoate, two hydroxyl groups are therefore present in the molecule Acet. Morphine Heroin (diacetate) 2- Morphine gives the ferric chloride test for phenols, thus it contains one hydroxyl phenolic group. The second hydroxyl group is secondary alcoholic by the fact that morphine can be converted to monohalogeno derivative by the action of halogen acid 23  Indol group Gramine (C11H14N2) Solid alkaloid, melts at 1460C , it has been found in barley mutants, it raises the blood pressure in dogs when administered in small doses. *Synthesis of gramine: Synder et al. have synthesized gramine by Mannish reaction from indol, formaldehyde and dimethyl amine in aqueous solution CH2N(Me)2 aq. + HCHO + Me2NH solution N N H H Cocaine (C17H21NO4) CO2CH3 NCH3 OOCC6H5 Solid coca alkaloid, melting at 980C, occurs in coca leaves, sparingly soluble in water, but its hydrochloride is quite soluble and it used as a local anesthetic. 24 *Determination of cocaine structure: 1- When heated with water, cocaine is hydrolyzed to methanol and benzoylecgonine, Thus cocaine contains a carbomethoxy group and benzoylecgonine a carboxyl group. Heat C17H21NO4 + H2O MeOH + C16H19NO4 2- When benzoylecgonine is heated with barium hydroxide solution, further hydrolysis occurs, the products obtained being benzoic acid and ecgonine COOH Ba(OH)2 C16H19NO4 + H2O C9H15NO3 + 3- Later, Willsttter et al. synthesized ecgonine by means of the Robinson method, using succinaldehyde, methyl amine and β- oxomomoethyl glutrate as starting materials. CO2C2H5 CO2C2H5 H CHO KOH + NCH3 + O NCH3 O CHO H CO2H CO2H CO2C2H5 CO2H Heat i)Na/Hg NCH3 O NCH3 OH ii)Hyd. 25  CO2CH3 CO2H i)CH3OH/HCl NCH3 OOCC6H5 NCH3 OH ii)C6H5COCl *Synthesis of cocaine: 26  Some questions on alkaloids 1- Three alkaloids A (C8H15NO), B (C17H19NO3) and C (C20H21NO4), they have pyrrolidine, piperidine and isoquinoline as main nuclei, respectively. a) Explain what is meant by the term "alkaloid" and show what kinds of physical methods are now being used to elucidate alkaloid structure. b) Draw the structure of A, B and C, try to determine the structure of one and how can you prepare other one. 2- Just by equations, show how can you convert; a) Piperonal into piperic acid. b) Ephedrine into 1,2- methyl phenyl ethylene dioxide. c) Nicotine into nicotinic acid. 3- Suggest a synthesis for the following compounds; a) Adrenaline. b) Piperine from piperic acid. c) Hygrine. 4- Explain by equations, how can you determine; a) The methylene group in papaverine. b) The structure of one of the coca alkaloids. c) The methylene ether group in piperine. d) The presence of methoxy group in cusparine. e) Ephedrine structure. f) Hygrine structure. 5- Illustrate the relation between each of the following pairs; a) Veratric acid - papaverine. b) Nicotine - hygrinic acid. c) Ephedrine - propiophenone. d) Ecgonine - cocaine. e) Cusparine - protocatechuic. f) Morphine - heroin. 6- a- Oxidation of papaverine with hot concentrated KMnO4. b- Controlled oxidation of cusparine with CrO3. c- Subjection of ephedrine to Hofmann exhaustive Methylation. 27  The above three experiments help in the elucidation of papaverine, cusparine and ephedrine, how?. 7- Using equations, illustrate the following; a) Careful oxidation of piperic acid. b) Synthesis of nicotine from nicotinic acid c) Preparation of coniine. 8- Write with examples the different classes of alkaloids and show briefly the general methods of their isolation. 9- a- Show by equations how can you prove the presence of; 1) Methoxy groups in papaverine. 2) Methylene ether group in piperine. b- Explain the effect of oxidation on; nicotine and hygrine. c- Give a method for preparation of coniine or piperine. 10- Three Alkaloids A (C10H15NO), B (C10H14N2) and C (C17H19NO3), they have phenyl ethyl amine, pyrrolidine - pyridine and Phenanthrene as main nuclei, respectively; draw the structure of A, B and C, try to determine the structure of one of them and give a method for preparation of anther one. 28