A laboratory manual on

Transcription

1 University of Baghdad ollege of Pharmacy Department of Pharmaceutical hemistry A laboratory manual on for second year students H Azhar M. Jaism (M. Sc. Ph. hem.) Duraid H. Mohammad (M. Sc. Ph. hem.) ct., 2012

2 ontents Tools and glass wares... iii Qualitative analysis of organic compounds... 1 Determination of solubility class... 2 Identification of alcohols... 6 Identification of aldehydes and ketones Identification of phenols Identification of carboxylic acids Identification of carboxylic acids salts Identification of alkyl and aryl halides Identification of amines ii

3 Tools and Glass Wares Below are the mostly used tools and glass wares in organic chemistry laboratory (courses II and III): reagent bottles glass stoppers dropper Bunsen burner washing bottle beaker test tubes test tube holder test tube holder test tube rack tripod stand wire gauze iii

4 litmus paper spatula spatulaa Petri dish graduated cylinders funnel filter papers graduated pipette burettee and funnel stands glass rod (stirrer) water bath Hood iv

5 1 Q Qualitative Analysis of rganic ompounds ualitative analysis of organic compounds helps identify and characterize unknown organic compounds. Many organic compounds are usually a component of a mixture of several compounds that might be considered as impurities. These impurities may be side products resulted during the preparation of the organic compound or may be decomposition products of the original pure organic compound and this occurs during storage under unsuitable conditions. n the other hand, some compounds may be obtained and stored pure because of their high degree of stability. In most cases a good separation and purification should precede qualitative analysis of organic compounds so that identification will be successful. The qualitative analysis of any organic compound should follow these steps: 1. Physical properties studying. State of the organic compound (solid, liquid, gas) Determination of the melting point or boiling point. olor, taste, and odor of the compound. Determination of the solubility group (solubility classification according to the general families). 2. hemical properties studying. Effect of the compound or its solution on litmus paper. Determination of elements in the organic compound (nitrogen, sulfur, or halogens). Detection of the organic groups, i.e. group classification to get more specific families. Specific classification tests. Preparation of derivatives.

6 2 Determination of Solubility lass Solubility class determination gives an idea about the type of the functional group present in the compound, the polarity and molecular weight of the compound, and the nature of the compound (acidic, basic, neutral). This is accomplished by testing the solubility of the compound in either of the following sets of solvents: distilled water, 5% sodium hydroxide solution, 5% sodium bicarbonate solution, 5% hydrochloric acid solution, and cold concentrated sulfuric acid, or distilled water and ether. It is well known that hydrocarbons are insoluble in water because of their non polar nature. If an unknown compound is partially soluble in water, then this indicates that a polar functional group is present. Additionally, solubility in certain solvents often leads to more specific information about the functional group. For example, benzoic acid is insoluble in water, but is converted by 5% sodium hydroxide solution to a salt, sodium benzoate, which is readily water soluble. In this case, then, the solubility in 5% sodium hydroxide solution of a water insoluble unknown is a strong indication of an acidic functional group. Prediction of the molecular weight and size may sometimes be obtained from the result of solubility tests. For example, in many homologous series of monofunctional compounds, the members with fewer than about five carbon atoms are water soluble, whereas the higher homologs are insoluble. The first step to follow is to test the solubility of the compound in water. Generally and for solubility classification purposes, the compound is said to be soluble in any solvent if it dissolves to the extent of about 3 % (0.1 gm/3 ml or 0.2 ml/3 ml). This is achieved by dissolving about 0.1 gm of the solid compound or 3-4 drops of the liquid compound in gradually increasing volumes of the solvent up to 3 ml (max. allowed volume is 3 ml) with shaking. This technique is the one that should be followed in solubility classification to determine whether the compound is soluble or insoluble in that solvent. When solubility in dilute acid or dilute base is being considered, the significant observation to be made is whether it is significantly more soluble in aqueous acid or aqueous base than in water. Such increased solubility is the desired positive test for acidic or basic functional groups. Below is a very useful scheme for solubility classification:

7 3 H 2 insol. sol. 5% Na Ether insol. sol. sol. insol. 5% Hl 5% NaH 3 96% cold H 2 S 4 insol. insol. sol. sol. insol. sol. litmus unchanged litmus blue litmus red IS 2 B I N IA 2 IA 1 IS 1 Discussion on solvents Water Water is a polar solvent with a dielectric constant equals to 80. It has the ability to form hydrogen bonding and can act either as an acid or a base. Therefore it can dissolve: Salts of ammonium ion (RNH 4 + ) or organic acids salts with alkali metal cations (R - ). Ionic compounds. Polar compounds like dissolves like. rganic compounds with low molecular weight (less than 5 carbon atoms) such as alcohols, aldehydes, ketones, and carboxylic acids. Water is useful to determine the degree of acidity of a compound, even if the compound is insoluble in water, using litmus paper (acidic, basic, or neutral). Water is the first solvent used to determine the solubility class of a compound. If the compound is water soluble, the next step is to test its solubility in ether.

8 Ether Ether is a non-polar solvent having a dielectric constant of 4.3. It cannot form hydrogen bonding (unassociated liquid). Therefore, it differs from water in that it cannot dissolve ionic compounds such as salts. It dissolves most water insoluble compounds; therefore, in the determination of solubility class, the importance of ether is for watersoluble compounds only and no further solubility tests using the remaining solvents are to be done. 4 Accordingly two probabilities are there: 1. ompounds soluble in both water and ether. These compounds: are non-ionic. contain five or less carbon atoms. contain an active group that is polar and can form hydrogen bonding. contain only one strong polar group. This division of compounds is given S 1 class and includes, e.g., aldehydes, ketenes, and aliphatic acids. 2. ompounds soluble in water only (but not in ether). These compounds: are ionic. contain two or more polar groups with no more than four carbon atoms per each polar group. This group is classified as S 2 class and includes ionic salts such as salts of carboxylic acids and amines and compounds with more than one active group such as poly hydroxylated compounds and carbohydrates. Note that solubility in ether is tested only for water-soluble compounds. For water insoluble compounds use the left side of the solubility classification scheme, i.e. test solubility in sodium hydroxide rather than ether. 5% Na & 5% NaH 3 Water insoluble compounds must be tested first in 5% sodium hydroxide solution which is a basic solvent. It reacts with water insoluble compounds that are capable of donating protons such as strong and weak acids. The stronger the acid, the weaker the base it can react with. Water insoluble compounds that dissolve in 5% sodium hydroxide solution must also be tested for solubility in 5% sodium bicarbonate solution. Therefore, for water insoluble acidic compounds sodium hydroxide solution is considered as a detecting solvent whereas sodium bicarbonate solution is called as a sub classifying solvent since it can react with strong acids only. That is, these two solvents give an idea about the acidity degree of the compound. Note that testing solubility in 5% sodium bicarbonate solution is not needed if the compound is insoluble in 5%

9 sodium hydroxide solution, but rather, 5% hydrochloric acid solution should be used. 5 Two probabilities are there: 1. ompounds soluble in both bases. This group is given class A 1. This class includes strong acids that have the ability to react with weak bases (carboxylic acids) and phenols with electron withdrawing groups (e.g., N 2 ). Protons are weakly attached and can be given easily. 2. ompounds soluble in 5% sodium hydroxide solution only. This group is given class A 2 and it includes phenols, amides, and amino acids (weak acids). 5% Hl If the compound is insoluble in water and sodium hydroxide solution (and, hence, insoluble in sodium bicarbonate solution too), this means that the compound is not an acid but, rather, is either a basic compound or a neutral compound. 5% hydrochloric acid solution, which can dissolve basic compounds such as amines (RNH 2 ), is used for such a compound. If the compound is soluble in this solvent, then it is given class B. This class includes primary, secondary, and tertiary amines. old concentrated H 2 S 4 If the compound is insoluble in water, 5% sodium hydroxide solution, and 5% hydrochloric acid solution, solubility in cold concentrated sulfuric acid should be tested. If the compound is soluble in this acid, it belongs to class N which includes neutral compounds such as high molecular weight alcohols, aldehydes, ketones, esters, and ethers (more than four carbon atoms), and unsaturated hydrocarbons. n the other hand, compounds that are insoluble in cold concentrated sulfuric acid belong to class I which includes inert aliphatic (saturated) hydrocarbons, aromatic hydrocarbons, haloalkanes, and aryl halides. Show by a chemical equation only how can concentrated sulfuric acid dissolve oxygen containing neutral compounds. Depending on the chemical structure discuss your results of solubility class of the compounds given to you.

10 Identification of Alcohols 6 Alcohols are organic compounds that may be considered as derivatives of water in which one of the hydrogen atoms of water molecule (H--H) has been replaced by an alkyl or substituted alkyl group. Therefore, properties of alcohols may be related to properties of both water and hydrocarbons. The alkyl group could be primary, secondary, or tertiary, and may be open chain or cyclic. Accordingly, alcohols may be defined as organic compounds that contain hydroxyl groups attached to alkyl, substituted alkyl, or cyclic alkyl group. R the general formula H 3 methanol H 2 H 3 ethanol H 2 H 3 H 3 H sec-butanol H 3 H 3 H 3 H 2 tert-butanol cyclohexanol (cyclic) benzyl alcohol (aromatic) Physical properties Alcohols are colourless liquids with a special faint odour. Benzyl alcohol and cyclohexanol have characteristic odours. Aliphatic alcohols burn with blue flame (without smoke) while aromatic alcohols burn with yellow smoky flame. Boiling points of alcohols are considerably high (being associated liquids); they increase as the molecular weight (number of carbons) increases. Alcohols are miscible with water except benzyl alcohol, cyclohexanol, and sec-butanol (which is very slightly soluble in water.

11 Solubility classification Alcohols are polar compounds because of the presence of the hydroxyl group which is also responsible for their ability to form hydrogen bonding. The degree of the polarity depends on the size of the alkyl side chain; the polarity decreases as the size of the alkyl side chain increases, or in another word, as the hydroxyl group /hydrocarbon ratio of alcohols increases, their water solubility increases and vice versa. Besides, low molecular weight alcohols are soluble in water due to hydrogen bonding ability with water molecules. Therefore, alcohols that are soluble in water and ether are classified under class S 1 such as ethanol and methanol. Alcohols that are insoluble in water are related to class N such as benzyl alcohol, sec-butanol, and cyclohexanol. 7 Which alcohol has solubility class S 2, what are the structural requirements present in this alcohol that made it under this solubility class? hemical properties Alcohols are neutral compounds that don t change the colour of litmus paper. All reactions of alcohols are related to its active hydroxyl group and are of two types: a) removal of the hydroxyl itself as in the reaction with hydrogen halides to form alkyl halides or in the dehydration reaction to form a double bond. b) removal of the proton only from the active hydroxyl as in the formation of esters or in the reaction with active metals such as sodium. 1. General test (eric ammonium nitrate reagent) eric ammonium nitrate (yellow solution) is an oxidizing agent that reacts with alcohols to give a red complex and with phenols to give a brown to greenish brown precipitate. 2[(NH 4 ) 2 e(n 3 ) 6 ]+RH 2 ceric ammonium nitrate alcohol R H +2[(NH 4 ) 2 e(n 3 ) 5 ]+2HN 3 aldehyde Each mole of the alcohol requires two moles of the reagent. The red coloured complex is an intermediate for the oxidation of alcohols by the e (IV) solution. This red colour disappears after a reasonable time due to completing the oxidation of this intermediate and

12 the reduction to the colourless e (III) solution producing the corresponding aldehyde or ketone. 8 (NH 4 ) 2 e(n 3 ) 6 +R ceric ammonium nitrate (yellow) alcohol R (NH 4 ) 2 e(n 3 ) 5 +HN 3 (red) Procedure Water soluble (miscible) alcohols; mix two drops of the alcohol with one drop of ceric ammonium nitrate solution. A red complex indicates a positive test. Water insoluble (immiscible) alcohols; mix two drops of the alcohol with 0.5 ml dioxane, shake well, and add one drop of the reagent to get a positive red complex. This test gives positive results with primary, secondary, and tertiary alcohols (up to 10 carbons), poly hydroxylated compounds such as carbohydrates, and hydroxylated carboxylic acids, aldehydes and ketones. 2. Specific tests a) Iodoform (Haloform) test This test is specific for alcohols which have a free methyl group and a hydrogen attached to the carbon bearing the hydroxyl group such as ethanol and sec-butanol. H 3 H The overall reaction is: The alcohol is oxidized to the corresponding aldehyde or ketone by the action of the produced oxidizing agent 'sodium hypoiodite', which also causes the aldehyde or ketone to be tri-iodinated on the terminal methyl group producing iodoform as a yellow precipitate.

13 Procedure (Note: read the procedure carefully and completely before starting) Dissolve about 3 drops of the alcohol in about 2 ml of distilled water (or 2 ml of dioxane for water insoluble compounds), add about 1 ml of 10% sodium hydroxide solution, then add iodine solution drop wise with shaking until either a yellow iodoform precipitate is formed, in which case the test is positive and is completed, or the dark colour of the iodine solution is present. In the latter case allow the solution to stand for 3 minutes during which period check for the appearance of the yellow precipitate at the bottom of the test tube. If there is no precipitate, warm the solution in water bath (60 ) for about 3 minutes with shaking from time to time and check for the yellow precipitate. During warming, if the colour of iodine disappears, add few additional drops of iodine solution with shaking until either the yellow precipitate is formed or the dark iodine colour persists, and then complete warming. Then get rid of the excess iodine by the addition of 10% sodium hydroxide solution drop wise with shaking to obtain the yellow precipitate. If the precipitate is not formed, allow the solution to stand for 10 minutes to get the positive result. Finally if no precipitate is formed after the 10 minutes- standing period, dilute the solution with an equal volume of distilled water to obtain the iodoform precipitate. It is important to proceed through all these steps so that only at the final step you can say that the test is negative. Both ethanol and sec-butanol give positive iodoform test and they can be differentiated only by testing their solubility in water; sec-butanol is less soluble in water than ethanol. 9 b) Lucas test This test often provides classification informations on alcohols and is used to distinguish between the different types of alcohols (primary, secondary, or tertiary). It depends on the formation of alkyl chloride as a second liquid phase. Lucas reagent is prepared from anhydrous zinc chloride and concentrated hydrochloric acid. Zinc chloride is added to increase the ionization of hydrochloric acid. R R R Znl 2 +Hl Znl 3 - +H + Benzyl alcohol shows the fastest positive result. Tertiary alcohols are faster in the formation of conjugated halides than secondary alcohols. Primary alcohols and methanol don t react and don t form two layers. +Hl Znl 2 R R R l +H 2 tertiary alcohol tertiary alkyl halide insoluble in water

14 Procedure Mix 2-4 drops of the alcohol with few drops of Lucas reagent and observe the results: benzyl alcohol gives immediate result as shown by the appearance of two phases. tertiary alcohols give two phases that separate within 2-3 minutes. secondary alcohols give two phases that separate after minutes (giving a cloudy solution). in primary alcohols one layer appears. 10

15 Identification of Aldehydes and Ketones 11 Aldehydes are compounds of the general formula RH; ketones are compounds of the general formula RŔ. The groups R and Ŕ may be aliphatic or aromatic, and in one aldehyde, formaldehyde, R is hydrogen. Both aldehydes and ketones contain the carbonyl group,, and are often referred to collectively as carbonyl compounds. It is this carbonyl group that largely determines the chief chemical and physical properties of aldehydes and ketones. R H H 3 H H f ormaldehyde acetaldehyde benzaldehyde H 3 H 3 H H 3 acetone salicylaldehyde acetophenon H 2 H3 benzyl methyl ketone benzophenone Aldehydes and ketones differ from alcohols in having two less hydrogen atoms. Removal of these two hydrogens from a primary alcohol as a result of oxidation yields an aldehyde; where as their removal from a secondary alcohol gives a ketone. The relation between these carbonyl

16 compounds and alcohols is, therefore, oxidation-reduction relation. Tertiary alcohols can t undergo this reaction. 12 Physical properties All aldehydes and ketones are liquids except formaldehyde, which is gas (boiling point -21 ), and benzophenone, which is solid (melting point 48 ). Formaldehyde is handled either as an aqueous solution (formalin, an aqueous solution of 40% formaldehyde and 15% methanol.) or as one of its solid polymers: paraformaldehyde, (H 2 ) n, or trioxane, (H 2 ) 3. Low molecular weight aldehydes and ketones (less than 5 carbons) are appreciably soluble in water, although they do not have the ability to form hydrogen bonds (unlike alcohols), aromatic ones are insoluble in water, and all of them are soluble in organic solvents. They are colorless except benzaldehyde, which has a pale yellow colour (due to oxidation) with a characteristic odour. The boiling points of aldehydes and ketones are lower than those of the alcohols from which they are derived; isopropyl alcohol boils at 82.5 while its oxidation product, acetone, boils at 56, ethanol boils at 78 while its oxidation product, acetaldehyde, boils at 21. Aliphatic aldehydes and ketones burn with a blue flame (without smoke) while aromatic ones burn with a yellow smoky flame. Solubility classification Aldehydes and ketones that are soluble in water are soluble in ether too and are classified under class S 1 (e.g., formaldehyde and acetone).aldehydes and ketones that are insoluble in water are classified under class N such as benzaldehyde and benzophenone. hemical properties All reactions of aldehydes and ketones are related to the carbonyl group (the active group). Aldehydes contain a hydrogen atom attached to its carbonyl while ketones don t. This difference in the chemical structure affects their chemical properties in two ways: a) aldehydes are easily oxidized to the corresponding acids and have reducing properties while ketones are not oxidized under similar conditions and do not show reducing properties. b) aldehydes are usually more reactive than ketones towards nucleophilic addition, the characteristic reaction of carbonyl group.

17 R H + H Y R H Y 13 Both aldehydes and ketones are neutral compounds that don t change the color of litmus paper. 1. General test (2,4-Dinitrophenylhydrazine reagent) Both aldehydes and ketones give yellow or orange precipitate with 2,4-dinitrophenylhydrazine reagent. 2 N N 2 NHNH 2 + R R N 2 δ + δ 2 N N H N R R +H 2 2,4-dinitrophenyhydrazine Procedure 2,4-dinitrophenyhydrazone (imine) yellow-orange ppt. To 2 drops of the compound add 3 drops of the reagent, a yellow or orange precipitate will be formed. If the compound is insoluble in water, dissolve it in 1 ml of methanol and then add the reagent. 2. Differentiation between aldehydes and ketones Differentiation between aldehydes and ketones is achieved by taking the advantage of the fact that aldehydes can be easily oxidized while ketones cannot (they need stronger oxidizing agents).two reagents can be used for this purpose, Tollen s reagent or Fehling s reagent. nly aldehydes give positive results with these two reagents. a) Tollen's test (Reduction of ammoniacal silver nitrate) Tollen s reagent is the combination of silver nitrate solution with ammonium hydroxide in the presence of sodium hydroxide solution. Aldehydes show positive result with this reagent because the reaction between them involves the oxidation of the aldehyde to the corresponding carboxylic acid with an accompanying reduction of the silver ions from this reagent to silver element in the form of silver mirror on the inner side of the test tube.

18 14 The oxidation process requires an alkaline medium; therefore sodium hydroxide solution is used, and in order to overcome the formation of the brown silver oxide precipitate (Ag 2 ), ammonium hydroxide is used to serve as a complexing agent for this precipitate making it a water soluble complex. Note that since the medium is alkaline, salts of the produced carboxylic acid are formed rather than the acid itself. Procedure Preparation of Tollen s reagent To 3mL of silver nitrate solution add 2-3 drops of 10% sodium hydroxide solution, and then add drop wise very dilute ammonia solution with continuous shaking until all the brown precipitate of silver oxide is dissolved. This reagent should be freshly prepared prior before use. Add 2-3 drops of the compound to 2-3 ml of Tollen s reagent, a silver mirror will be formed. If no reaction occurs, warm the test tube in water bath for few minutes (note that excessive heating will cause the appearance of a false positive test by decomposition of the reagent). The formed silver mirror can be washed using dilute nitric acid. If the test tube is not very clean, silver metal forms merely as a granular gray or black precipitate. False-negative tests are common with water insoluble aldehydes. A negative result indicates that the compound is a ketone. b) Reduction of Fehling's reagent This test, like Tollen s test, is used to distinguish aldehydes from ketones. nly aldehydes can reduce Fehling s reagent (a deep blue solution) to give a red cuprous oxide precipitate.

19 15 Procedure Preparation of Fehling's reagent Fehling s reagent is prepared by mixing exactly equal volumes of Fehling s A and Fehling s B solution in a 1:1 ratio immediately before use (usually 1 ml of each).fehling s A solution is an aqueous solution of copper sulfate pentahydrate (us 4.5H 2 ) with few drops of concentrated sulfuric acid whereas Fehling s B solution is an aqueous solution of potassium sodium tartrate ( 4 H 4 KNa 6,4H 2 ) and sodium hydroxide. Add 5 drops of the compound to 1 ml of Fehling s solution, and then heat in water bath for 5 minutes (with shaking for water insoluble compounds). Aldehydes change the color of Fehling s solution from blue to green, orange precipitate, and then red precipitate or copper mirror. Ketones don t change the color of this reagent. n the other hand, this test does not give a sharp result with aromatic aldehydes. 3. Special tests for aldehydes and ketones containing a terminal methyl group These compounds include acetaldehyde, acetone, acetophenone, and benzyl methyl ketone. All of them have a methyl group attached to the carbonyl group: a) Iodoform (Haloform) test Follow the same procedure of iodoform test mentioned earlier (identification of alcohols).

20 b) Sodium nitroprusside test To few drops of the compound add 1 ml of sodium nitroprusside (Na 2 [Fe(N) 5 N].2H 2 ) solution and excess of 30% sodium hydroxide solution, a red color complex is a positive test Polymerization reaction To 0.5 ml of formaldehyde or salicylaldehyde add 0.2 gm of resorcinol and drop-by-drop concentrated sulfuric acid to get a red or reddish violet color, or a white ring that changes to a reddish violet ring. + 2 H H conc. H 2 S 4 -H 2 H H resorcinol formaldehyde polymer 5. annizzaro reaction Benzaldehyde, salicylaldehyde, and formaldehyde can undergo annizzaro reaction because they do not have an alpha hydrogen atom. In this type of reactions the aldehyde undergoes a self oxidationreduction in the presence of a strong basic medium to yield a mixture of the corresponding alcohol and the salt of the corresponding carboxylic acid (or the acid itself). Therefore, one molecule of the aldehyde serves as the oxidizing agent while the other serves as the reducing agent. H + H Na heat + Na benzaldehyde benzaldehyde benzyl alcohol sodium benzoate

21 Procedure To few drops of benzaldehyde (or the other aldehydes) add 0.5 ml of 30% sodium hydroxide solution and heat gently on a water bath with shaking for five minutes. A precipitate of sodium benzoate is produced. Dissolve this precipitate by adding few drops of distilled water, and then add drops of concentrated hydrochloric acid to liberate benzoic acid as a white precipitate. As mentioned earlier formaldehyde can undergo this reaction ; however, this reaction can't be relied on for testing formaldehyde since the acid produced, formic acid, is liquid that can't be observed separately as compared to the solid benzoic acid resulted from benzaldehyde. 17 What is paraformaldehyde and from what aldehyde is it from? Write down its molecular formula.

22 Identification of Phenols Phenols are organic compounds with a hydroxyl group attached directly to benzene or substituted benzene. They have the general formula Ar-. Examples of them include phenol (also known as carbolic acid), hydroquinone, resorcinol, o-cresol, m-cresol, p-cresol, β-naphthol, and α-naphthol. 18 phenol hydroquinone (quinol, hydroquinol) resorcinol H 3 o-cresol m-cresol H 3 H 3 p-cresol H α-naphthol β -naphthol 'Phenols' is the term used to call all the members of this class of organic compounds. The simplest member is called phenol. Try to find out from where the term 'phenol' was derived.

23 Physical properties 19 Phenols are liquids (e. g., o- and m-cresol) or solid crystalline compounds(e. g., phenol and resorcinol). They are coloured due to air oxidation and have a special odour. Pure compounds are colourless. They have high boiling points because of intermolecular hydrogen bonding. Phenol itself is soluble in water due to its ability to form hydrogen bonding with water. For other phenols the solubility in water decreases by increasing the molecular weight. (cresols and naphthols are insoluble in water). Phenols burn with a yellow smoky flame due to the presence of aromatic ring. hemical properties Phenols are weak acidic compounds, so they are soluble in strong alkaline solutions only (e. g., sodium hydroxide solution). For this reason they fall into solubility class A 2. Presence of electron withdrawing group at the phenyl ring strengthens the acidity of the phenol making it of solubility class A 1 (e. g, nitrophenol). However, phenol itself is of solubility class S 1 since it is water soluble. Types of phenols reactions 1. Reactions at the phenolic hydroxyl group (-); e.g., ether formation and salt formation: phenol reacts with sodium hydroxide to form sodium phenoxide. ethyl phenyl ether is formed when phenol reacts with ethyl iodide. 2 H H 5 I aq. Na heat +HI phenol ethyl phenyl ether (Phenetole)

24 2. Substitution at the aromatic ring; e.g., bromination and nitration reactions: 20 reaction with bromine water. reaction with dilute nitric acid. hemical reactions 1. Reaction with ferric chloride Phenols react with ferric chloride to give coloured compounds due to the presence of the enol group. Actually this reaction is considered as a test for any compound with enol group.

25 Procedure To a very dilute aqueous solution of the phenol (30-50 mg in 1-2 ml water) or to a few crystals of the solid phenol ( mg) dissolved in water add 1 drop of ferric chloride solution and observe the resulting colour: 21 compound phenol, m-cresol, resorcinol violet or blue colour o- and p-cresol greenish blue hydroquinone deep green α- and β-naphthol no special colour Hydroquinone undergoes oxidation in the presence of ferric chloride resulting in a deep green solution (crystals may separate) and, on further addition of ferric chloride solution, a yellow solution of p-benzoquinone is produced:

26 2. Reaction with bromine water 22 This reaction is an example of substitution reaction at the phenyl ring (mentioned earlier). Procedure To a concentrated aqueous solution of the phenol or to the phenol itself, add bromine water gradually. At first the bromine is decolourized and then, on adding an excess, a white or yellowish-white precipitate of a poly bromo-derivative is produced with all except hydroquinone and α- and β-naphthol. n gradually adding bromine water to a solution of hydroquinone, a deep red coloration is produced, followed by the separation of deep green crystals which then dissolve giving a yellow solution. The naphthols decolourize bromine water, but usually no precipitate of the bromo compound can be obtained. This test is not very satisfactory with those phenols which are insoluble in water, owing to the difficulty of distinguishing the bromo compound from the original phenol. 3. Phthalein test Many phenols yield phthaleins which give special colours (sometimes with fluorescence) in alkaline solutions when reacted with phthalic anhydride and a little amount of concentrated sulphuric acid. Phenol and resorcinol are examples. H H 2 phthalic anhydride resorcinol fluorescein (pale red with green fluorescence in alkaline medium)

27 The fluorescence of resorcinol is due to the presence of an oxygen linkage between the two phenolic nuclei (in basic medium). 23 Procedure In a dry test tube put about 0.1 g of the phenol and an equal amount of phthalic anhydride or phthalic acid, mix well, and add 1-2 drops of concentrated sulphuric acid. Heat gently on a direct flame for 1 minute until the crystals of the mixture melt and fuse. Then cool the test tube and add excess of 10% sodium hydroxide solution. Results should be as follows: compound α-naphthol β-naphthol phenol o-cresol m-cresol p-cresol resorcinol hydroquinone green colour colour very pale green with slight fluorescence red to pink red-violet blue to pink no change pale red colour with green fluorescence violet colour If the resultant colour is not so clear you can dilute with water. 4. Reimer-Tiemann reaction Treatment of phenol with chloroform and aqueous sodium hydroxide solution introduces an aldehyde group (-H) into the aromatic ring at the ortho- or para-positions:

28 Procedure To about 0.2 g of the phenol add 1mL of 30% sodium hydroxide solution and 1 ml of chloroform, heat on water bath, and observe the colour of the aqueous layer: 24 compound phenol resorcinol α-naphthol β-naphthol o-cresol m-cresol p-cresol hydroquinone colour yellow or no colour red colour with weak fluorescence dark green deep blue that turns to green deep orange pale orange yellow deep brown 5. Reduction of potassium permanganate Phenols reduce potassium permanganate solutions and undergo oxidation to quinones. The manganese is reduced from +7, which gives a purple solution, to +4, which is brown. This test is highly successful with dihydroxylated phenols than phenol itself. Procedure Add 0.1 g or 0.2 ml (3-4 drops) of the compound to 2 ml of water or ethanol. Add 2% aqueous potassium permanganate solution drop by drop with shaking until the purple colour of the permanganate persists. If the permanganate color is not changed in minutes, allow the mixture to stand for 5 minutes with occasional vigorous shaking. The

29 disappearance of the purple color and the formation of a brown suspension, which is manganese (II) oxide, at the bottom of the test tube is a positive result for the presence of phenols Why phenols are stronger than alcohols as acidic compounds? 2. Give the difference in water solubility among resorcinol, hydroquinone, and catechol.

30 Identification of arboxylic Acids 26 arboxylic acids are organic compounds that have a carboxyl group attached to an alkyl group(r) or to an aryl group (Ar). The 'R' may be a hydrogen and the result is formic acid. They may be mono carboxylated, multi carboxylated, substituted (e. g., hydroxyl groups), or they may be aromatic H H 3 carboxyl group formic acid acetic acid H 2 H 2 oxalic acid succinic acid H H 3 H H H H H H H 2 H 2 lactic acid tartaric acid citric acid benzoic acid salicylic acid Physical properties nly formic acid, acetic acid, and lactic acid are liquids at room temperature. The others are solids. Low molecular weight carboxylic acids are soluble in water and, therefore, lie under class S 1. Water insoluble acids dissolve in both sodium hydroxide solution and sodium bicarbonate solution, being classified under class A 1. When they react with sodium bicarbonate, they evolve carbon dioxide gas. This is considered as a good simple indication of them.

31 Their boiling points are generally high due to the association through hydrogen bonds: two molecules of the carboxylic acid are held together by two hydrogen bonds rather than one. 27 H R R Aromatic carboxylic acids burn with a yellow smoky flame whereas aliphatic ones burn with a blue flame without smoke. hemical properties The acidic properties of carboxylic acids are attributed to the proton of the carboxyl group. Mono carboxylic acids are weak acids except formic acid, which is the strongest. The tendency of the alkyl group to release electrons weakens the acid; thus formic acid is the strongest. n the other hand presence of electron withdrawing groups (such as halogens) especially on the alpha carbon increases the acidity. Reactions of carboxylic acids are related to: the proton as in salt formation reactions. removal of the hydroxyl group as in conversion to derivatives such as esters, amides, or acid chlorides. substitution either in the alpha position of aliphatic acids or in the meta position of aromatic ones. Give an example of a carboxylic acid with α- halogenation (name and chemical structure). Which parameter will you look for to compare its acidity with other acids? Give the general formula of esters, amides, and acid chlorides. hemical reactions 1. General test (Ferric chloride test) The acid solution should be made neutral before performing the test with ferric chloride solution. This is achieved by adding very dilute ammonia solution drop by drop with shaking to a solution of about 0.5 g of the solid acid or 2 drops of the liquid acid in 1 ml water until the medium becomes basic as indicated by changing the colour of litmus paper to blue or changing the colour of phenolphthalein indicator from colorless to pink, in which case the characteristic odour of ammonia is predominant. At this stage the solution is slightly basic. To make the

32 solution neutral the excess ammonia should be removed by gently heating the test tube in a water bath with shaking from time to time until both the odour of ammonia and the pink colour disappears. (In case of oxalic, tartaric, citric and lactic acids keep a portion of their neutral solution for use in calcium chloride test). ool the solution and then add few drops of ferric chloride solution to get different colours (solutions or precipitates) as follows: 28 acid formic and acetic succinic and benzoic salicylic oxalic, tartaric, citric, and lactic result red solution light brown precipitate violet solution no special change The steps of this test are: When the solution is basic (excess ammonia): Therefore elimination of the excess ammonia is important since the brown orange precipitate of ferric hydroxide formed by this excess interferes with the colour of the ferric salt of the acid resulting in a false result.

33 If the solution is still acidic (little ammonia is added), colourless complexes are formed between the acid and ferric ions, a false negative result. 29 As mentioned in the above table formic acid and acetic acid form a red coloured solution in this test: Succinic acid and benzoic acid give a light brown precipitate: To distinguish between these two acids add few drops of dilute sulphuric acid to this light brown precipitate with shaking thereby liberating the free carboxylic acid back. If the liberated acid is water soluble then it is succinic acid which is aliphatic. n the other hand benzoic acid is liberated as a white precipitate because it is insoluble in water since it is aromatic.

34 Salicylic acid gives a violet colour: Special tests for formic acid Since formic acid has a hydrogen attached to the carbonyl group (H=) it can reduce certain compounds while being oxidized: a) Reduction of mercuric chloride Formic acid reduces mercuric chloride to mercurous chloride in the form of white precipitate and, in the presence of excess acid, mercurous chloride is reduced to mercury element as a gray precipitate. To few drops of the acid add few drops of mercuric chloride solution, and then heat to get a white precipitate. Add excess of the acid with heating to get the gray precipitate of elemental mercury. b) Tollen's test Refer to the experiment of identification of aldehydes and ketones for preparation of Tollen's reagent and procedure of this test. c) Alkaline potassium permanganate test Formic acid reacts with potassium permanganate solution, a strong oxidizing agent, in alkaline medium causing decolourization of this violet reagent. Mix 2 3 drops of the acid with 5 ml of sodium bicarbonate solution, and then add 1% potassium permanganate solution drop by

35 drop and observe the disappearance of the original violet colour of the reagent which will be followed by the appearance of a brown precipitate of manganese dioxide Special test for acetic acid (ester formation) Acetic acid, on contrary to formic acid, neither can be oxidized by, nor can reduce any of the reagents applied to formic acid. Instead, it undergoes ester formation reaction: H H 5 conc. H 2 S 4 H 3 2 H 5 +H 2 ethyl acetate (ester) Mix 1 ml of acetic acid with 2 ml of ethanol in a test tube and add to this mixture 2 3 drops of concentrated sulphuric acid. Heat the test tube in a water bath for 10 minutes, and then pour the mixture into another test tube containing 5 ml sodium bicarbonate solution; the characteristic fruity odour of ethyl acetate can be smelt, which indicates the formation of this ester. 4. Special test for succinic acid (Fluorescence test) In a dry test tube mix equal quantities of succinic acid and resorcinol with 2 drops of concentrated sulphuric acid. Heat the mixture on direct flame for 1 minute until the mixture melts. ool and add excess of 10% sodium hydroxide solution to get a red colour with green fluorescence. If the colour is not so clear dilute with water.

36 5. Special tests for tartaric acid 32 a) Reaction with concentrated sulphuric acid When a mixture of about 0.5 g of tartaric acid and 1 ml of concentrated sulphuric acid is heated gently on a flame with shaking heavy charring takes place and carbon monoxide, carbon dioxide, sulphur dioxide gases are evolved. b) Reaction with calcium chloride solution To about 1 ml of the cold neutral solution of the tartaric acid (see the general test) add few drops of calcium chloride solution; a white precipitate of calcium tartrate is formed. This precipitate dissolves in dilute hydrochloric acid but not in dilute acetic acid.

37 c) Reaction with Fenton's reagent Dissolve about 0.5 g of tartaric acid in 1 ml of water and then add 1 drop of ferrous sulphate solution followed by 2 drops of hydrogen peroxide solution. Then add excess of 10% sodium hydroxide solution until an intense violet colour is observed. In this reaction the components of Fenton's reagent (hydrogen peroxide and iron) undergo an oxidation-reduction reaction that results in the generation of ferric ions which form ferric salt of dihydroxyfumaric acid that is responsible for the violet colour Special tests for oxalic acid a) Reaction with potassium permanganate solution xalic acid reacts with acidic potassium permanganate solution causing decolourization of this reagent. It doesn't react with this reagent under alkaline medium. Dissolve 0.5 gm of the acid in 2 3 ml of distilled water and add 2 3 ml of dilute sulfuric acid. Heat gently (water bath), and then add potassium permanganate solution drop by drop and observe the disappearance of the violet color of the reagent. b) Reaction with calcium chloride solution For procedure follow the same steps mentioned above for tartaric acid. The same results are obtained. c) Reaction with concentrated sulphuric acid For procedure follow the same steps mentioned above for tartaric acid. The same gases are bubbled out with a little darkening. d) Reaction with Fenton's reagent For procedure follow the same steps mentioned above for tartaric acid. xalic acid gives negative result with this reagent.

38 7. Special tests for lactic acid 34 a) Iodoform test Lactic acid can undergo iodoform formation reaction since it contains a free methyl group and a hydrogen attached to the carbon bearing the hydroxyl group: For procedure follow the same steps mentioned in the identification of alcohols experiment. b) Reaction with concentrated sulphuric acid For procedure follow the same steps mentioned above for tartaric acid. The same gases are bubbled out with a considerable blackening but without a marked charring. c) Reaction with calcium chloride solution For procedure follow the same steps mentioned above for tartaric acid. Lactic acid gives negative result. d) Reaction with Fenton's reagent For procedure follow the same steps mentioned above for tartaric acid. Lactic acid gives negative result with this reagent. 8. Special tests for citric acid a) Reaction with concentrated sulphuric acid For procedure follow the same steps mentioned above for tartaric acid. The same gases are bubbled out and the mixture turns to yellow but does not char. Acetone dicarboxylic acid is also formed, and its presence is tested by heating the mixture for 1 minute, cool, add a few milliliters of water and make alkaline with 30% sodium hydroxide solution. Add a few milliliters of sodium nitroprusside solution and observe the intense red colouration of the medium.

39 b) Reaction with calcium chloride solution For procedure follow the same steps mentioned above for tartaric acid. itric acid gives the same results. 35 c) Reaction with Fenton's reagent For procedure follow the same steps mentioned above for tartaric acid. itric acid gives negative result with this reagent. 9. Special test for salicylic acid (ester formation) In addition to the characteristic violet colour obtained with ferric chloride, salicylic acid can also be detected by ester formation test. In this test methyl salicylate ester separates out as an organic phase having a characteristic odour. Follow the same procedure and conditions used for esterification of acetic acid but use methanol instead of ethanol. Not that methanol is toxic internally so never withdraw it by mouth to avoid accidental ingestion.

40 36 Methyl salicylate, also known as wintergreen oil, is used widely in pharmaceutical topical preparations, give its main use with the name of a popular topical preparation. Both ethyl acetate and methyl salicylate separate as an organic phase during ester formation test, how can you detect the organic layer practically and theoretically? Fill the following table: tests results with description acid Fel 3 test H 2 S 4 test al 2 test Fenton's test citric oxalic tartaric lactic

41 Identification of arboxylic Acids Salts 37 arboxylic acids salts are organic compounds with the general formula (RM) where (R - ) refers to the carboxylic acid part and (M + ) is the alkali part which, in this experiment, may be either a metal cation (Na + or K + ) or ammonium (NH4 + ). These salts are colourless or white crystalline solids and are soluble in cold or hot water. Identification of the carboxylic acid part (anionic part) The carboxylic acid part can be identified by the usual steps for identification of carboxylic acids starting with ferric chloride test and, according to the result observed; the proper special test should be performed then to conclude the carboxylate name (formate, lactate, salicylate, etc.). an you skip the neutralization step in the reaction between carboxylic acid salts and ferric chloride solution? Explain. Identification of the alkali part (cationic part) Identification of sodium or potassium cations Place about 0.1 g of the salt on the edge of a metal spatula and start heating it gently on a flame with gradual increase in the heat strength. Sodium and potassium salts leave a residual amount of solid on the spatula in addition to the carbon coming from decomposition of the organic part. This residual solid may be sodium carbonate or potassium carbonate and can be detected, after cooling, by the addition of few drops of dilute hydrochloric acid solution which results in a strong effervescence within the residual solid due to liberation of carbon dioxide gas: During ignition observe the colour of the flame. Sodium salts burn with a golden yellow flame whereas potassium salts burn with a purple flame.

42 Identification of ammonium cation Repeat the ignition procedure mentioned above and note that ammonium salts don't leave any residual solid except the carbon coming from decomposition of the organic part. After cooling, addition of few drops of dilute hydrochloric acid does not result in any effervescence. Ammonium cation can be detected as follows. Place few crystals of the salt in a test tube and add 0.5 ml of 10% sodium hydroxide solution. At this stage free ammonia is liberated and can be smelt easily: 38 Place a small filter paper over the top of the tube and fold it down around the tube. Add 2 drops of 10% copper sulphate solution on the filter paper covering the mouth of the test tube. Heat the test tube mildly on a flame to boil the mixture. The liberated ammonia will react with the copper ions present on the filter paper resulting in a blue colour. What is the proper technique for smelling chemicals?

43 Identification of Alkyl and Aryl Halides 39 Physical properties All alkyl halides and chlorobenzene are colourless liquids when pure except iodoform, HI 3, which is a yellow crystalline solid with a characteristic odour. Methyl iodide, ethyl iodide and bromide, chloroform, and carbon tetrachloride have sweetish odours. Benzyl chloride has a sharp irritating odour and is lachrymatory. hlorobenzene possesses aromatic odour. Alkyl and aryl halides, XR ) ( XAr have boiling points higher than the parent hydrocarbon because of the heavier molecular weight. Accordingly, for a given compound, iodides have the higher boiling point than bromides and chlorides. In spite of their polarity alkyl halides are insoluble in water due to their inability to form hydrogen bonds. They are soluble in most organic solvents. Iodo-, bromo-, and polychloro- compounds are denser than water. H 3 X metheyl halide 2 H 5 X ethyl halide H 2 l Hl 3 l 4 chloroform carbon tetrachloride l benzyl chloride chlorobenzene hemical reactions 1. Reaction with alcoholic silver nitrate. Alcoholic silver nitrate reagent is useful in classifying halogen compounds. Many halogen containing compounds react with silver nitrate to give an insoluble silver halide (AgX), and the rate of this reaction indicates the degree of reactivity of the halogen atom in the compound. Besides, the identity of the halogen can sometimes be determined from the colour of the silver halide produced; silver chloride is white (turns to purple on exposure to light), silver bromide is pale yellow, and silver iodide is yellow. These should, of course, be consistent with results from elemental analysis (sodium fusion for detection of halogens).

44 40 RX + AgN 3 2 H 5 AgX + RN 2 R X Ag+ δ + δ δ + R X Ag AgX + R + R + N - 3 R N 2 It is obvious that this reaction follows S N 1 mechanism. Generally the reactivity of alkyl halides towards this reagent is: Procedure R 3 X > R 2 HX > RH 2 X Add one drop or a couple of crystals of the unknown to 2 ml of 2% ethanolic silver nitrate solution. If no immediate reaction is observed, stand for 5 minutes at room temperature and observe the result. If no reaction takes place, warm the mixture in water bath for 30 seconds and observe any change. If there is any precipitate (AgX) add several drops of 1 M nitric acid solution to it; silver halides are insoluble in this acid. tert- chlorides, methyl and ethyl iodides, allylic chlorides, and ethyl bromides give fast result at room temperature: R H 2 X R H Br R 3 l RI H 2 H H H H 2 X H 2 Br pri- and sec- chlorides, benzyl chloride, and 1-chloro-2,4- dinitrobenzene give result only on warming: hlorobenzene, chloroform, iodoform, carbon tetrachloride, and vinylic chlorides don't give any positive result even on heating: R H R X ArX Hl 3 l 4 H X H H yclohexyl halides exhibit a decreased reactivity when compared with the corresponding open-chain secondary halides. yclohexyl chloride is inactive, and cyclohexyl bromide is less reactive than 2-