Chapter 3 Tests for Siling Labuyo Capsaicin, a kathamine alkaloid derivative (Phenylalkylamine), does not precipitate like most alkaloids since there’s only a 0.14% yield from the chillii plant. (Arnold Brossi, 1990)
N- (4- hydroxyl- 3 methoxybenzyl)- 8- methyl- non- 6- enamide General Test for Alkaloid: General Test Positive Result Mayer’s Cream or pale yellow precipitate Wagner’s Brown or reddish brown precipitate Hager’s Tannic Acid Vitali-Morin
Yellow precipitate Amorphous or Crystalline Precipitate Violet precipitate
Procedure for Preparing the Reagents: Reagent Mayer’s (the reagent is available in the laboratory) Wagner’s ( Iodo-potassium iodide) Hager’s (Picric acid) (the reagent is available in the laboratory) Tannic Acid (the reagent is from the extract of the tannin group) Vitali-Morin
Experimental Result Pale yellow solution Brown or reddish precipitate Yellow solution Amorphous precipitate No reaction
brown
Procedure Dissolve 1.358g of HgCl2 in 60ml water and pour into a solution of 5g KI in 10ml of water. Add sufficient water to make 100ml. Dissolve 2g of iodine and 6g of KI in 100ml water. Dissolve 1g of picric acid in 100ml of water. Dissolve 10g of tannic acid in 10ml of ethanol and dilute with water to 100ml. See procedure for general test.
Procedure for General Testing of Alkaloid: Test Procedure Mayer’s To 1mL of the extract, add 2-3 drops of the Mayer’s reagent. Wagner’s To 1mL of the extract, add 2-3 drops of the Wagner’s reagent. Hager’s To 1mL of the extract, add 2-3 drops of the Hager’s reagent. Tannic Acid To1mL of the extract, add 2-3 drops of the Tannic acid solution reagent. Vitali-Morin To 1mL of the extract, add 5-10 drops nitric acid. Evaporate to dryness. Dissolve the residue in acetone. Add methanolic solution of potassium hydroxide. Principles Behind the Tests: Test Principle
Mayer’s Wagner’s Hager’s Tannic Acid Vitali-Morin
Halogenation followed by Complex reaction mechanism. Halogenation followed by Complex reaction mechanism. Nitration followed by Complex reaction mechanism (Bronsted-Lowry theory) Complexation reaction mechanism (Bronsted-Lowry theory) Positive for Tropane alkaloid. (the group’s alkaloid is not a Tropane alkaloid)
Discussion: The alkaloids are a class of vegetable compounds which are chemically similar to ammonia. Like this, the free alkaloids are of a basic nature and they form salts by uniting with acids without the setting free of hydrogen. Only a few of the alkaloids are liquid, most of them, as well as their salts, being crystallize-able solids. Their salts are, as a rule, much more easily soluble in water than the alkaloids themselves. They are decomposed by alkalis, the alkaloid being set free, just as ammonia is freed from its salts by the same reagents. When the solution is concentrated this process often results in their precipitation, as the alkaloid is less soluble than its salts. There is a great difference in the comparative solubility of the alkaloids in many of the organic liquids like chloroform, benzene, ether, petroleum ether, amyl alcohol, etc., and the methods devised for their separation are based upon this. They all have a bitter taste, as do their salts. The separation and identification of the alkaloids is not as simple an operation as that of the metallic compounds. It should be borne in mind that a very small amount of impurity may conceal or to a great degree, modify the reactions. (Arnold Brossi, 1990) In the Bronsted-Lowry theory defines an acid as any chemical species (molecule or ion) that is able to lose, or donate a hydrogen ion (proton) while a base is any species with the ability to gain or accept a hydrogen ion (proton). It follows that if a compound is to behave as an acid, donating a proton, there must be a base to accept the proton. So the Bronsted-Lowry concept can be defined by the reaction acid + base
conjugate base + conjugate acid.
The conjugate base is the ion or molecule remaining after the acid has lost a proton, and the conjugate acid is the species created when the base accepts the proton. The reaction can proceed in either forward or backward direction; in each case the acid donates a proton to the base. Mayer’s Test gives a negative result for Ephedrine and Purine alkaloids. It may also give a false negative result for the impurities of the extracted alkaloid may have interfered with the structure of the capsaicin. Wagner’s test is similar in principle to the Mayer’s test where the reagent changes the degree or angle of the bonds leading to a complex reaction mechanism. In Hager’s test, the nitro groups are introduced and sulfonic acid groups are displaced. The reaction of the picric acid is highly exothermic and over time, the picric acid slowly sublimes even at room temperature making the test results give a false negative result. A salt is formed with the alkaloid since picric acid is an acid and was combined with the alkaloid which is basic. Tannic acid is a polyphenol. Its weak acidity is due to the numerous phenol groups in the structure which can precipitate many organic and inorganic compounds. Tannic acid complexes with and precipitates alkaloids such as nicotine, apomorphine, strychnine; and it also precipitates, to a lesser extent, metals like aluminium, lead, mercury, and silver. Thus, the addition of tannic acid in the extracted alkaloid gives an amorphous precipitate since tannic is an acid and alkaloids are basic, they would form a salt (A salt, in chemistry, is defined as the product formed from the neutralization reaction of acids and bases).
Specific Test for Alkaloid:
Specific Test 2,4- DNP (test for aldehydes and ketones) Baeyer’s (test for active unsaturation) Nitration (test for aromacity) Bromine (test for unsaturation) Jones’ (Chromic Acid test)
Positive Result Experimental Result Orange precipitate (positive for Turbid orange solution ketones) Brown solid Brown solution with brown precipitate Yellow oily layer Yellow oily layer Colorless Colorless Further addition turns the solution to reddish color Blue-green solution Blue-green solution
Procedure for Specific Testing of the Alkaloid: Test Procedure 2,4-DNP Add 2 drops of the plant extract in 2mL 95% Ethanol to 2,4- Dinitrophenylhydrazine reagent (available in the laboratory) in a test tube. Shake vigorously. Baeyer’s Addition of distilled water and solid potassium permanganate into the extract. Nitration Place 5 drops of plant extract in a test tube. Add 8 drops of the Nitrating mixture (2mL conc. Sulfuric acid and 2 ml Nitric acid in an Erlenmeyer flask immersed in water bath). Then shake the test tube. Bromine To 1mL of the extract, add 1-2 drops of pure bromine. Jones’ Dissolve 10 mg or 2 drops of the extract in 1 ml of pure acetone in a test tube and add to the solution 1 small drop of Jones reagent (chromic acid in sulfuric acid). Principles behind the Tests: Test Principle 2,4- DNP Addition-Elimination (condensation reaction) Baeyer’s Oxidation with reduction (Redox reaction) Nitration Electrophilic aromatic substitution Bromine Electrophilic addition Jones’ Oxidation reaction Discussion: 2,4-DNP: The chemistry of the reaction: The overall reaction is given by the equation:
R and R' can be any combination of hydrogen or hydrocarbon groups (such as alkyl groups). If at least one of them is a hydrogen, then the original compound is an aldehyde. If both are hydrocarbon groups, then it is a ketone. (Ralph Shriner, 2004) Look carefully at what has happened.
The product is known as a "2,4-dinitrophenylhydrazone". The reaction is known as a condensation reaction. A condensation reaction is one in which two molecules join together with the loss of a small molecule in the process. In this case, that small molecule is water. (Ralph Shriner, 2004) In terms of mechanisms, this is a nucleophilic addition-elimination reaction. The 2,4dinitrophenylhydrazine first adds across the carbon-oxygen double bond (the addition stage) to give an intermediate compound which then loses a molecule of water (the elimination stage). (Ralph Shriner, 2004) You can just use it to test for the presence of the carbon-oxygen double bond. The positive result is an orange or yellow precipitate from a carbon-oxygen double bond which indicates the presence of an aldehyde or ketone. (Ralph Shriner, 2004) Baeyer’s: The Baeyer’s test is similar to that of the Bromine test which tests for active unsaturation, the target of which is the double bonds. In oxidation with reduction, the double bonds get broken and the substrate oxidizes and becomes a diol compound. The reagent gets reduced in the process. The KMnO4 (Mn+7) reagent is reduced to MnO2 (Mn+4). (Ralph Shriner, 2004) Potassium permanganate is purple. Since it is reduced, thus the purple color disappears. MnO2 on the other hand is a muddy brown precipitate which will appear for positive results. (Ralph Shriner, 2004)
Nitration: Nitration test is a test for aromacity. The target of which is the benzene ring. The principle involved is the electrophilic aromatic substitution because it substitutes hydrogen with NO2. The reagent is the concentrated nitric acid and the catalyst is the concentrated sulfuric acid. When the two solutions react, it releases NO2+ which is called nitronium ion. To put it simply, this nitronium ion attaches to the benzene ring and to retain the aromatic
stability, will in return release the hydrogen attached to where the NO2 was attached. Thus, the product is a nitrocompound which is responsible for the yellow oily layer in the specific test result. (Ralph Shriner, 2004)
Bromine: Bromine test is a test for unsaturation and the target is the double bonds. The principle involved is electrophilic addition because the double bonds are broken and the electrophile is added to the substrate. Br2 reagent separates to Br+ and Br-. The electrophile Br+ will react with the electron-rich portion of the substrate which is the double bond. It gets the electron, thus destroying the double bond and making the other side positive, meaning electron-deficient. To compensate for it, it will get electrons from the nucleophile which is Br-. Thus, the final product is a dibromo-compound. Br2 is red because it was combined with the substrate in the reaction so that there won’t be any free Br2. As a result, the red color of the free Br2 disappears. Thus, the positive result was the disappearance of the red color. But further addition of pure bromine would make the solution saturated. (Ralph Shriner, 2004)
Jones’: A positive test is marked by the formation of a green color within 5 seconds upon addition of the orangeyellow reagent to a primary or secondary alcohol. A positive test for aldehydes and primary or secondary alcohols consists in the production of an opaque suspension with a green to blue color. (Ralph Shriner, 2004)
