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DETAILED STUDY OF VITAMIN B1,B2,&B6 1 Vitamin B: (Thiamine or aneurin): Source: Occurs abundantly in rice and wheat. It is also found in yeast ,egg, meat, beans, Peas,etc. in small qunatity it is also present in liver and kidney. Isolation: Source is extracted with acidulated water at pH Adorption on fullers earth which adsorbs only vitamin B1. Eluted with quinine sulphate solution. Excess of quinine is then precipitated by Ba(OH)2 and add AgNO3 On treatment with HCl the silver salt is converted to chloride hydrochloride of vitamin B1 which is crystalline and removed easily. Deficiency disease: deficiency in human causes loss of appetite, gastrointestinal disturbances, 2 muscular weakness, pain in arms and legs and decrease in blood Pressure.In severe cases leads to Beriberi. CONSTITUTION: 1.Molecular formula of thiamine chloride hydrochloride is C12H18N4Cl2OS 2. with sodium sulphite saturated with SO2 at room temperature cleaved into two compounds A and B. C12H18N4Cl2OS + Na2SO3 SO2 C6H9NOS + C6H9N3O3S A + 2NaCl B CONSTITUTION OF COMPOUND A; 1.The molecular formula is C6H9NOS . 2.Compound is basic but does not react nitrous acid the nitrogen atom is in teritiary form. indicates that 3.It gives monochloro compound when reacted with HCl showing the presence of one hydroxyl group . U.V spectrum of the chloro compound is similar to that of the parent hydroxyl compound, the later is present in side chain. . . 4. the unreactivity of the sulphur atom and the uv spectrum of compound A suggested that the sulphur atom is present in a thiazole ring, 3 Confirmation: sulphur atom is lost when A is treated with alkaline plumbite solution. 5.Oxidation of the compound A with HNO3 gives an acid ( C5H5N02S) which was identified as 4-methyl thiazole -5-carboxylic acid II. The structure of this has been proved by synthesis. CH3 HO NH 1.condensation C + HC 2.hydrolysis C SH Thioformamide N CH3 Br S COOH COOC2H5 alpha-Bromo ethylacetoacetate II From the above points it is clear that the compoud A has the structure as II with the only difference that one of the side chains of A has been replaced by –COOH group in II. 6. From the above it is clear that the side chain is C2H50 which may be either -CH2CH2OH or –CHOH.CH3 and hence the compound A III or IV. CH3 CH3 N N S III CH2CH2OH S IV CHOHCH3 4 On oxidation , either of these compounds could be converted to II. But the compound A is optically inactive . So, III is the structure of A because the compound IV is the structure of A because the compound IV is optically active. 7. The structure III for A has been confirmed by synthesis , in which thioformide is condensed with 3-chloro 5-ethoxy -2pentanone .the later compound is synthesised from ethylacetoacetate and β-bromo diethyl ether as shown below OC CH3 BrCH2CH2OC2H5 + OC NaOC2H5 COOC2H5 CH C2H5OOC C H2 CH2CH2OC2H5 SO2Cl2 CH3 CH3 OC ketonic hydrolysis Cl OC CH Cl C CH2CH2OC2H5 Keto form C2H5OOC CH2CH2OC2H5 CH3 HO C Cl C enol form CH2CH2OC2H5 3-chloro 5-ethoxy2-pentanone 5 The hydrochloride of this compound is identical with that of the product obtained from thiamine. It also gives acid II on oxidation with nitric acid. Londergan have synthesised compound A 2-methyl furan. HO NH CH3 1.Condensation C + CH3 N 2.hydrolysis HC C SH Cl CH2CH2OC2H5 Thioformamide CH2CH2OH S compound A ( III) Synthesis of compound A H2-Pd/c CH2 H2C COCH3 HCl CH2OH CH3 O -H20 CH3 N S CH2CH2OH 1.Cl2 2.HCSNH2 in HCOOH O CH3 6 Constitution of compound B: 1.The molecular formula is C6H9N3O3S. 2. On heating with NaOH it yields sodium sulphate indicating the presence of a sulphonic acid group in compound B . 3. The compound B was found to contain one –NH2 group as it formed monohydroxy derivatives. Along with the evolution of nitrogen when treated nitrous acid. 4. the compound B on heating with HCl under pressure yields ammonia and a compound C. the U.V spectrum is very similar to that of 6-hydroxy pyrimidine. So, the compound B is probably a 6-amino pyrimidine derivative . C6H9N3O3S + H20 HCl 1500c C6H8N2O4S + NH3 7 5.Reduction of B with sodium in liquid ammonia yields 6-amino -2,5 dimethyl-pyrimidine . This structure is confirmed by its synthesis. O CH3 + H2N H5C2OOC HN C.CH3 NH ACETAMIDINE C2H5ONa HOHC H3C N 1.POCi3 formyl propionic ester 2.NH3-C2H5OH H3C H2N N N 4-Amino-2,5-dimethylpyrimidine CH3 8 The position of sulphonic acid group: This can be present on either side of the two methyl groups since amino group was found to be free in compound B. 6. When thiamine is reduced with sodium in liquid ammonia. a compound D is formed which is Identified as 6-amino 5-aminomethyl -2- methyl pyrdimidine .proved by its synthesis. NH2 CN N + H2N NH CH3 C2H5O C2H5ONa C C H H3C N CN 1.CH3COOH 2.H2-Pd-C NH2 CH2NH2 N H3C N D Thus in compound D, there is an amino group instead of the sulphonoic acid 9 in B is attached to the methyl group at position 5 and thus proved by its synthesis. NH2 NH2 NH2 CH2NH2 N CH2Br 1.HONO H3C SO3 H3C D N NaHSO3 2.HBr N Na2SO3H N N H3C N POINT OF ATTACHMENT OF A AND B: H3C NH2 H2 C Cl- N N+ HOH2CH2C CH2NH2 Na/C2H5OH N NH3 S ClHH2N N CH3 H3C Na2SO3 NH2 CH3 N CH2SO3H N + CH2CH2OH S H3C N B A 10 Synthesis: structure is proved by its synthesis(williams): NH2HBr CH2Br N H3C CH3 + N N CH2CH2OH S NH2HBr Br CH2 CH3 N N CH2CH2OH S H3C N AgCl in CH3OH NH2HCl Cl CH2 CH3 N 11 N S H3C N CH2CH2OH thiaminechloride hydrochloride VITAMIN B2(RIBOFLAVIN): SOURCE: The best sources of the vitamin are yeast, green vegetables , liver, wheat germ, egg yolk,k milk, meat, fish etc. Isolation: the material is acidified with con.HCl and the vitamin is adsorbed on fullers earth. The adorbate is treated with basic solvents (NH3,causticsoda etc). After several precipitations and resolutions thallium salt of vitamin is formed. Lactoflavin is crystallised from alcohol. DEFICIENCY: It is necessary for growth and health deficiency in riboflavin results in a.inflammation of the tongue, b. chelosis. 12 CONSTITUTION; 1.‘the molecular formula of riboflavin is C17H20 N4O6. 2. Silver salt of riboflavin on acetylation gives a tetra-acetate , indicating the presence of four hydroxyl groups. 3.Oxidation of riboflavin with lead tetra-acetate yields formaldehyde. it indicates the presence of Primary alcoholic group. 4.There is no primary amino group since riboflavin does not react with nitrous aci . 5.Alkaline hydrolysis of the vitamin yields urea, indicating the presence of the fragment –NH-CO-NH. The other 2 nitrogens are teritiary. 6. Irradiation of an alkaline solution of riboflavin yields a new compound lumniflavin or photoflavin. Similar treatment of the vitamin in acid or neutral solution gives lumichrome. 13 7. STRUCTURE OF LUMIFLVIN: a. Molecular formula of lumiflavin is C13H13N4O2 b. The photolysis reaction of riboflavin to produce lumiflavin can be represented as follows. C17H20N4O6 hv C13H22N4O2 C4H8O4 + c. Lumiflavin can niether be acetylated nor oxidised by lead tetra-acetate showing there by that The C4 fragment is a tetrahydroxy butyl side chain. R-CHOH-CHOH-CHOH-CH2OH Lumiflavin was shown to contain a methyl –imine group which was not present in riboflavin . Therefore the methyl imino group must have replaced the missing side chain. e. Alkaline hydrolysis of lumiflavin yields urea and amino carboxylic acid A. d. C13H22N4O2 alkali + 2H2O C12H12N2O3 A + CH4N2O urea 14 Since two molecules of water are required in the reaction, urea must come from a ring system, and not from a side chain ureide or guanidino group which would require only one molecule of water for hydrolysis . the acid A easily eliminates a carbondioxide molecule to give a substance B. This suggests that A is a β- keto carboxylic acid. -CO2 C12H12N2O3 C11H12N2O B A The compound B shows the properties of a lactam and gives one molecule of glyoxalic acid and a compound C on boiling with NaOH solution. C11H12N2O NaOH +2H2O B CHOCOOH + C9H14N2 GLYOXALIC ACID C Structure of the compound C: By the usual tests, compound C was found to be an aromatic diamino compound. Since it gave a blue precipitate with ferric chloride, it must have N-methyl-o- phenylenediamine nucleus. NHCH3 N-Methyl-o-phenylene diamine NH2 . 15 The molecular formulae of the compound C (C9H14N2) and the above mentioned nucleus (C7H10N2) shows that C2H4 should be accounted for. This can be done by assuming the Presence of either an ethyl group or two methyl groups in the benzene ring But Khun showed the presence of two methyl groups by a series of synthetic reactions and Thus the compoud C was identified as N-methyl-4,5 diamino –o –xylene. H3C NHCH3 H3C NH2 The above structure for compound C is proved by its synthesis. Therefore, the structure of the compound B can be represented as follows which also explains the required products of hydrolysis. CH3 H3C H3C N CO NHCH3 NaOH + COOH CHO H3C N B H3C NH2 C 16 Since A is a beta –keto carboxylic acid of B, it can be represented as below. CH3 CH3 H3C H3C N CO H3C N N CO -CO2 H3C COOH N B Since A and one moleculae of urea are obtaine from lumiflavin, the latter would be 6,7,9Trimethyl isoalloxazine (6,7,9 trimethyl flavin). CH3 H3C N CH3 N CO NH H3C N lumiflavin H3C N alkali CO Ba(OH)2 H3C N COOH O + NH2 CONH2 17 Finally the structure of the lumiflavin has been confirmed by synthesis which involves the condensation of N-methyl -4,5-diamino-o –xylene with alloxan.the former was obtained From o-xylene as below. 18 Structure of lumichrome: Analytical work similar to that described for lumiflavin showed that the structure of lumichrome is I (6,7 –dimethyl alloxazine) Synthesis of lumichrome: H3C NH2 H N O O + NH H3C NH2 O O H3C N H3C N H N O NH 19 O SIDE CHAIN OF RIBOFLAVIN: 1. The reaction mixture from which lumichrome was isolated gave positive reaction for a pentose sugar, so the side chain is a sugar having five carbon atoms. 2. Zerewitnoff method shows that riboflavin has five active hydrogens atoms. Since one active Hydrogen atom is present on the N3 of lumiflavin .the rest four active hydrogen atoms must Be present in the side chain in the form of four –OH groups .must be present as a terminal CH2OH group. Further more, riboflavin forms a diisopropylidene derivative with acetone two 1,2 –glycolic systems must be present in the side chain (riboflavin). The above points lead to the following structure of the side chain of riboflavin. -CH2-CHOH-CHOH-CHOH-CH2OH CH2(CHOH)3CH2OH 10.Thus riboflavin can be written as below. H3C N N CO NH H3C N O Lactoflavin The side chain contains three asymmetric carbons and hence there are 8 optically active forms For riboflavin. The actual nature of the side chain (D- ribityl) at position 9 was proved by its synthesis 20 Structure is proved by its synthesis: Khun synthesis: H3C NH2 1. condensation + H3C CHO (CHOH)3CH2OH 2. reduction D(-) ribose NO2 CH2(CHOH)3CH2OH H3C HO NH H N O + NH H3C O NO2 O 2-amino-4,5 -dimethyl -N-D-ribityl aniline. alloxan CH2(CHOH)3CH2OH H3C N N CO NH H3C N 21 O Vitamin B6 ,Pyridoxine or Adermin : It refers to a group of three compounds namely ,pyridoxine, or adermin, pyridoxal and pyridoxamine which are interconvertible in the form of their phosphates. Occurrence: it is found in the plant as well as in animal eg: yeast and fish muscle. Most abundant sources for the vitamin are yeast and rice polishings followed by seeds and cereals. ISOLATION: The vitamin splits up, by heating ,from protein . extracted with water, and acidified with dilute H2SO4 impurities are removed by silver nitrate with suitable pH . The filterate is acidified to pH 2. and the vitamin is absorbed on fullers earth. The elution is carried out with Ba(OH)2 and the vitamin is precipitated by 22 any of the acids, H2SO4, phosphotungstic acid. This precipitation is repeated for many times and lastly the vitamin is obtained as hydrochloride. (M.P: 2090C) Deficiency diseases: In rats causes a specific dermatistis (acrodynia) ,inhibition of growth and reduction in accessory reproductive organs . In humans deficiency develops the symptoms of pellagra or anemia . Constitution : 1.Molecular formula is C8H11NO3. 2. Pyridoxine behaves as a weak base, and the usual tests showed the absence of methoxyl, Methylamino and primary amino groups. 3.Zerewitnoff method showed the presence of three active hydrogen atoms. 4. With FeCl3 deep red colour diazomethane forms monomethyl ether indicates that one hydroxyl group is present as phenolic. Confirmation: Uv spectrum of pyridoxine is quite similar with that of β- hydroxy pyridine nucleus. Also indicates the presence of a pyridine nucleus in pyridoxine. 23 5. The mono methyl ether of pyridoxine does not give colour reaction with FeCl3 it forms diacetate with acetic anhydride and With HBr it forms hydrobromide of a dibromide. Indicates : that the hydroxyl groups are alcoholic in nature. 6. Mono methyl ether of pyridoxine doesnot reacts with lead tetra acetate indicates the absence of two adjacent hydoxyl groups in a side chain. 7. Khun-Roth method of oxidation CH3 group in Pyridoxine. points out the presence of one –C- 8. MME of pyridoxine , with alkaline KMnO4 tricarboxylic acid(C9H7NO7). gives a methoxy pyridine The tricarboxylic acid with ferrous sulphate gives a blood colour is a characteristic reaction to pyridine 2 carboxylic acid. Indicates one of the three acidic groups in the tricarboxylic acid must be in the 2- position. 24 8. MME of pyridoxine with alk KMnO4 gives a molecule of CO2 and anhydride of a Carboxylic acid. indicates that in the parent dicarboxylic acid, the two carboxyl groups are in the ortho positions. Further confirmation: The ortho dicarboxylic acid with FeSO4 does not give a red colour , I: it doesnot have a carboxylic group in the second position. Therefore it follows that ,on decarboxylation, the tricarboxylic acid eliminates the 2- carboxylic group to form the anhydride, thus the tricarboxylic acid may have either of the following structures COOH COOH HOOC OCH3 OR N HOOC OCH3 COOH HOOC N Now we have that pyridoxine methyl ether has one –OCH3 and two alcohollic groups thus the two –COOH groups in the tricarboxylic acid could arise from the two alcoholic groups (-CH3OH) . The third thus must have arrived from a methyl group,Thus the following 25 structures are possible for the pyridoxine and its corresponding methyl ether. CH2OH CH2OH HOH2C OH HOH2C OR N OH CH3 H3C N two possible structures for pyridoxine CH2OH CH2OH HOH2C HOH2C OCH3 OCH3 OR N H3C CH3 N Two possible structures for pyridoxine methtl ether 9. MME of pyridoxine with barium permanganate gives a dicarboxylic acid I. The dicarboxylic acid forms anhydride with acethic anhydride and gives a florescent dye with resorcinol . IN: the 2 –COOH groups are in o-positions to each other , moreover, the dicarboxylic acid 1 does not give color with FeSO4 +showing that neither of the two –COOH groups is in α -position to the nitrogen. Hence the structure of the dicarboxylic acid ,1, may be either A or B COOH HOOC COOH OCH3 HOOC OCH3 OR N CH3 26 H3C N The position of the two –COOH groups in 1 corresponds to the 2 alcoholic groups in pyridoxine. Khun compared these two dicarboxylic acids with that of the synthetic compounds and found that the dicarboxylic acid obtained from pyridoxine is A as it resembles with the synthetic Dicarboxylic acid from 4-methoxy-3-methyl isoquinoline. CH3 COOH N KMnO4 CH3 OCH3 H3CO N HOOC OCH3 HOOC N CH3 COOH 27 Hence the correct structures for 1 is A and thus pyridoxine will be 2 which explains all the reactions. CH2OH HOH2C CH2OH OH N acetylation HOH2C CH3 PyridoxineII OAc N CH3 CH2N2 CH2OAc CH2OH HOH2C HOH2C OCH3 OCH3 acetylation N N CH3 CH3 Barium permanganate HBr COOH HOOC OCH3 CH2Br BrH2C OH N CH3 Dicarboxylicacid (I) N CH3 28 Synthesis: structure is confirmed by synthesis , e.g., that of Harris and Folkers CH2OC2H5 CH2OC2H5 CN CN C O CH2 CH2 C C Pyridine CH3OH H3C NH2 H3C O N H O ethoxy acetyl acetone CH2OC2H5 O2N CH2OC2H5 CN CN HNO3 (CH3CO)2 O H3C N OH H3C N OH 1.PCl5 2.H2+Pd -c 29 CH2OC2H5 CH2OC2H5 H2N H2N CN CN H2 +Pd C H3C H3C N Cl OH 1.HNO2 2.HBr CH2Br CH2OH HO N CH2OH BrH2C OH 1.H2 O AgCl H3C N H3C N Biological function: these are involved in a number of important metabolic reactions of the α –amino acids,e.g Transamination, racemization, decarboxylation and elimination reactions. 30 REFERENCES: 1.ORGANIC CHEMISTRY NATURAL PRODUCTS VOLUME -2---------O.P AGARWAL 2. ORGANIC CHEMISTRY NATURAL PRODUCTS VOLUME-1------------GURUDEEP CHATWAL 31 THANK U 32