University of Baghdad College of Pharmacy Department of Pharmaceutical Chemistry

A laboratory manual on

for second year students

Azhar M. Jaism (M. Sc. Ph. Chem.) Duraid H. Mohammad (M. Sc. Ph. Chem.) Oct., 2012

HO O OH

O

O

ii

Contents

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 ......................................................................... 11 Identification of phenols ................................................................................................ 18 Identification of carboxylic acids .................................................................................... 26 Identification of carboxylic acids salts ............................................................................ 37 Identification of alkyl and aryl halides ............................................................................ 39 Identification of amines ................................................................................................. 43

iii

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

litm

Pe

filte

glass r

mus paper

etri dish

er papers

rod (stirre

r

er)

gradua

gradu

w

iv

spatula

ated cylin

uated pip

ater bath

nders

ette

h

burette

spatula

funnel

e and funn

Hood

a

nel stands

s

Org. Chem. Lab. 1

Qualitative Analysis of Organic Compounds

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. On 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. • Color, taste, and odor of the compound. • Determination of the solubility group (solubility classification

according to the general families). 2. Chemical 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.

Q

Org. Chem. Lab. 2

Determination of Solubility Class

olubility 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:

S

Org. Chem. Lab.

3

I

H2O

5% NaOH Ether

5% HCl 5% NaHCO3

96%cold H2SO4

insol. sol.

sol. insol.litm

usunchanged

litmusblue

litmusred

insol. sol.

insol. sol.

insol. sol.

insol. sol.

NIIIIA2

B

IIIIA1

IIIIS1

IIIIS2

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 (RNH4+) or organic acids salts with

alkali metal cations (RCOO-). • Ionic compounds. • Polar compounds “like dissolves like”. • Organic 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.

Org. Chem. Lab.

4

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 water-soluble compounds only and no further solubility tests using the remaining solvents are to be done.

Accordingly two probabilities are there:

1. Compounds 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 S1class and includes, e.g., aldehydes, ketenes, and aliphatic acids.

2. Compounds 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 S2 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% NaOH & 5% NaHCO3 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%

Org. Chem. Lab.

5

• 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.

sodium hydroxide solution, but rather, 5% hydrochloric acid solution should be used. Two probabilities are there:

1. Compounds soluble in both bases. This group is given class A1. This class includes strong acids that

have the ability to react with weak bases (carboxylic acids) and phenols with electron withdrawing groups (e.g., –NO2). Protons are weakly attached and can be given easily.

2. Compounds soluble in 5% sodium hydroxide solution only. This group is given class A2 and it includes phenols, amides, and

amino acids (weak acids).

5% HCl 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 (RNH2), 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.

Cold concentrated H2SO4 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. On 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.

Org. Chem. Lab.

6

H3C OH

methanol

cyclohexanol(cyclic)

OH

H2C

OHH3C

ethanol

CH3C CH3

H3C

tert-butanol

CH2

benzyl alcohol(aromatic)

H2C

CHH3C

sec-butanol

CH3

OH

OH

OH

R OH

the generalformula

Identification of Alcohols

lcohols are organic compounds that may be considered as derivatives of water in which one of the hydrogen atoms of water molecule (H-O-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.

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.

A

Org. Chem. Lab.

7

Which alcohol has solubility class S2, what are the structural requirements present in this alcohol that made it under this solubility class?

2[(NH4)2Ce(NO3)6] + RCH2OH

ceric ammoniumnitrate

RC

O

H + 2[(NH4)2Ce(NO3)5] + 2HNO3

alcohol aldehyde

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 S1 such as ethanol and methanol. Alcohols that are insoluble in water are related to class N such as benzyl alcohol, sec-butanol, and cyclohexanol.

Chemical 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 (Ceric ammonium nitrate reagent) Ceric 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.

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 Ce (IV) solution. This red colour disappears after a reasonable time due to completing the oxidation of this intermediate and

Org. Chem. Lab.

8

(NH4)2Ce(NO3)6 + ROHceric ammonium

nitrate

(yellow)

(NH4)2Ce(NO3)5 + HNO3

alcohol

(red)

OR

C

OH

H

H3C

the reduction to the colourless Ce (III) solution producing the corresponding aldehyde or ketone.

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.

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.

Org. Chem. Lab.

9

ZnCl2 + HCl ZnCl3- + H+

+ HClC

R

R

R

OHZnCl2

+ H2OC

R

R

R

Cl

tertiary alcohol tertiary alkyl halideinsoluble in water

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 ˚C) 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.

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.

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.

Org. Chem. Lab.

10

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 15-20

minutes (giving a cloudy solution). • in primary alcohols one layer appears.

Org. Chem. Lab.

11

C O

C

O

HRC

O

HH3C

CH

O

CH

O

OH

C

O

CH3H3CC

CH3

O

CCH3

O

CH2

C

O

formaldehyde acetaldehyde benzaldehyde

salicylaldehydeacetone acetophenon

benzyl methyl ketone benzophenone

Identification of Aldehydes and Ketones

ldehydes are compounds of the general formula RCHO; ketones are compounds of the general formula RŔCO. 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.

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

A

Org. Chem. Lab.

12

compounds and alcohols is, therefore, oxidation-reduction relation. Tertiary alcohols can’t undergo this reaction.

Physical properties

• All aldehydes and ketones are liquids except formaldehyde, which is gas (boiling point -21˚C), and benzophenone, which is solid (melting point 48˚C). 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, (CH2O)n, or trioxane, (CH2O)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˚C while its oxidation product, acetone, boils at 56˚C, ethanol boils at 78˚C while its oxidation product, acetaldehyde, boils at 21˚C.

• 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 S1(e.g., formaldehyde and acetone).Aldehydes and ketones that are insoluble in water are classified under class N such as benzaldehyde and benzophenone.

Chemical 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.

Org. Chem. Lab.

13

C O + HH

RY C

H

R OH

Y

NHNH2 +O2N

NO2

C OR

R

δ+ δ−

NO2N

NO2

HN C

R

R+ H2O

2,4-dinitrophenyhydrazine 2,4-dinitrophenyhydrazone(imine)

yellow-orange ppt.

• 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.

Procedure 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. Only 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.

Org. Chem. Lab.

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 (Ag2O), 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. Only aldehydes can reduce Fehling’s reagent (a deep blue solution) to give a red cuprous oxide precipitate.

Org. Chem. Lab.

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 (CuSO4.5H2O) with few drops of concentrated sulfuric acid whereas Fehling’s B solution is an aqueous solution of potassium sodium tartrate (C4H4KNaO6,4H2O) 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. On 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).

Org. Chem. Lab.

16

OH

C

O

HH

OH

+ 2

resorcinol formaldehyde

HO

OH

OH

OH

OH

HO

OH

conc. H2SO4

- H2O

polymer

CH

O

benzaldehyde

CH

O

benzaldehyde

+NaOH

heatOH

benzyl alcohol

CONa

O

sodium benzoate

+

b) Sodium nitroprusside test To few drops of the compound add 1 mL of sodium nitroprusside

(Na2[Fe(CN)5NO].2H2O) solution and excess of 30% sodium hydroxide solution, a red color complex is a positive test. 4. 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.

5. Cannizzaro reaction

Benzaldehyde, salicylaldehyde, and formaldehyde can undergo Cannizzaro reaction because they do not have an alpha hydrogen atom.

In this type of reactions the aldehyde undergoes a self oxidation-

reduction 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.

Org. Chem. Lab.

17

What is paraformaldehyde and from what aldehyde is it from? Write down its molecular formula.

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.

Org. Chem. Lab.

18

phenol

OH

hydroquinone(quinol, hydroquinol)

OH

OHresorcinol

OH

OH

OH

CH3

OH

CH3

m-cresolo-cresol

OH

CH3

p-cresol

HO

β -naphthol

OH

α-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.

Identification of Phenols

henols are organic compounds with a hydroxyl group attached directly to benzene or substituted benzene. They have the general formula Ar-OH. Examples of them include

phenol (also known as carbolic acid), hydroquinone, resorcinol, o-cresol, m-cresol, p-cresol, β-naphthol, and α-naphthol.

P

Org. Chem. Lab.

19

OH

+ C2H5I

OC2H5

+ HI

phenol ethyl phenyl ether(Phenetole)

aq. NaOH

heat

Physical properties

• 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.

Chemical 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 A2. Presence of electron withdrawing group at the phenyl ring strengthens the acidity of the phenol making it of solubility class A1 (e. g, nitrophenol). However, phenol itself is of solubility class S1since it is water soluble.

Types of phenols reactions

1. Reactions at the phenolic hydroxyl group (-OH); 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.

Org. Chem. Lab.

20

2. Substitution at the aromatic ring; e.g., bromination and nitration reactions:

• reaction with bromine water.

• reaction with dilute nitric acid.

Chemical 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.

Org. Chem. Lab.

21

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 (50-100 mg) dissolved in water add 1 drop of ferric chloride solution and observe the resulting 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:

compound colour phenol, m-cresol, resorcinol violet or blue

o- and p-cresol greenish blue

hydroquinone deep green

α- and β-naphthol no special colour

Org. Chem. Lab.

22

OH

OH

resorcinol

+ 2

CO

CO

O

phthalic anhydride

HO O OH

O

O

fluorescein(pale red with green fluorescence

in alkaline medium)

+ 2H2O

2. Reaction with bromine water

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. On 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.

Org. Chem. Lab.

23

The fluorescence of resorcinol is due to the presence of an oxygen linkage between the two phenolic nuclei (in basic medium).

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: 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 (-CHO) into the aromatic ring at the ortho- or para-positions:

compound colour

α-naphthol green colour

β-naphthol very pale green with slight fluorescence

phenol red to pink

o-cresol red-violet

m-cresol blue to pink

p-cresol no change

resorcinol pale red colour with green fluorescence

hydroquinone violet colour

Org. Chem. Lab.

24

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:

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 0.5-1 minutes, allow the mixture to stand for 5 minutes with occasional vigorous shaking. The

compound colour

phenol yellow or no colour

resorcinol red colour with weak fluorescence

α-naphthol dark green

β-naphthol deep blue that turns to green

o-cresol deep orange

m-cresol pale orange

p-cresol yellow

hydroquinone deep brown

Org. Chem. Lab.

25

1. Why phenols are stronger than alcohols as acidic compounds? 2. Give the difference in water solubility among resorcinol, hydroquinone, and catechol.

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.

Org. Chem. Lab.

26

C

O

OH

carboxyl group

C

O

H OH C

O

H3C OH

formic acid acetic acid

COOH

COOH

oxalic acid

CH2COOH

CH2COOH

succinic acid

COOH

CH

HO CH3

lactic acid

CHCOOH

CHCOOH

HO

HO

tartaric acid

CH2COOH

CCOOHHO

CH2COOH

citric acid

COOH

benzoic acid

COOH

OH

salicylic acid

Identification of Carboxylic Acids

arboxylic acids are organic compounds that have a carboxyl group attached to an alkyl group(RCOOH) or to an aryl group (ArCOOH). 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

Physical properties

• Only 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 S1. Water insoluble acids dissolve in both sodium hydroxide solution and sodium bicarbonate solution, being classified under class A1. When they react with sodium bicarbonate, they evolve carbon dioxide gas. This is considered as a good simple indication of them.

C

Org. Chem. Lab.

27

C

O

OH

R C

O

HO

R

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.

• 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.

• Aromatic carboxylic acids burn with a yellow smoky flame whereas aliphatic ones burn with a blue flame without smoke.

Chemical 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. On 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.

Chemical 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

Org. Chem. Lab.

28

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). Cool the solution and then add few drops of ferric chloride solution to get different colours (solutions or precipitates) as follows:

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.

acid result

formic and acetic red solution

succinic and benzoic light brown precipitate

salicylic violet solution oxalic, tartaric, citric, and lactic no special change

Org. Chem. Lab.

29

If the solution is still acidic (little ammonia is added), colourless complexes are formed between the acid and ferric ions, a false negative result.

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. On the other hand benzoic acid is liberated as a white precipitate because it is insoluble in water since it is aromatic.

Org. Chem. Lab.

30

Salicylic acid gives a violet colour:

2. Special tests for formic acid

Since formic acid has a hydrogen attached to the carbonyl group (HC=O) 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

Org. Chem. Lab.

31

CH3COOC2H5 + H2Oconc. H2SO4

ethyl acetate(ester)

CH3COOH +C2H5OH

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.

3. 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:

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. Cool 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.

Org. Chem. Lab.

32

5. Special tests for tartaric acid

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.

Org. Chem. Lab.

33

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.

6. Special tests for oxalic acid

a) Reaction with potassium permanganate solution Oxalic 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. Oxalic acid gives negative result with this reagent.

Org. Chem. Lab.

34

7. Special tests for lactic acid

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.

Org. Chem. Lab.

35

b) Reaction with calcium chloride solution For procedure follow the same steps mentioned above for tartaric acid. Citric acid gives the same results.

c) Reaction with Fenton's reagent For procedure follow the same steps mentioned above for tartaric acid. Citric 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.

Org. Chem. Lab.

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:

acid

tests results with description

FeCl3 test H2SO4 test CaCl2 test Fenton's test

citric

oxalic

tartaric

lactic

Org. Chem. Lab.

37

Can you skip the neutralization step in the reaction between carboxylic acid salts and ferric chloride solution? Explain.

Identification of Carboxylic Acids Salts

arboxylic acids salts are organic compounds with the general formula (RCOOM) where (RCOO-) 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.).

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.

C

Org. Chem. Lab.

38

What is the proper technique for smelling chemicals?

• 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:

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.

Org. Chem. Lab.

39

CH3Xmetheyl halide

C2H5 X

ethyl halide

Cl

benzyl chloride chlorobenzene

CHCl3 CCl4chloroform carbon tetrachloride

CH2Cl

Identification of Alkyl and Aryl Halides

Physical properties

All alkyl halides and chlorobenzene are colourless liquids when pure except iodoform, CHI3, 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. Chlorobenzene possesses aromatic odour.

Alkyl and aryl halides (RـــX, ArـــX) 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.

Chemical 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).

Org. Chem. Lab.

40

RX + AgNO3C2H5OH

AgX + RONO2

R XAg+

R X Agδ+δ+ δ−

AgX + R+

R+NO3

-

R ONO2

R3CCl RI C CH

R CH2X

H

C CH

R H

CH2XH2C

Br

CH2Br

ArX C CH

R H

XC C

H

R X

H

HCCl3 CCl4

It is obvious that this reaction follows SN1 mechanism. Generally the reactivity of alkyl halides towards this reagent is:

R3CX > R2CHX > RCH2X

Procedure 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:

pri- and sec- chlorides, benzyl chloride, and 1-chloro-2,4-

dinitrobenzene give result only on warming: Chlorobenzene, chloroform, iodoform, carbon tetrachloride, and

vinylic chlorides don't give any positive result even on heating: Cyclohexyl halides exhibit a decreased reactivity when compared

with the corresponding open-chain secondary halides. Cyclohexyl chloride is inactive, and cyclohexyl bromide is less reactive than 2-

Org. Chem. Lab.

41

RCl + NaIacetone

RI + NaCl

RBr + NaI RI + NaBracetone

R XI-

I C Xδ−δ−

I R

bromohexane, although it will give a precipitate with' alcoholic silver nitrate.

2. Sodium iodide in acetone test.

This test, complementing the alcoholic silver nitrate test, is used to classify aliphatic chlorides and bromides as primary, secondary, or tertiary. This test depends on the fact that sodium chloride and sodium bromide are only very slightly soluble in acetone.

The mechanism follows direct displacement (SN2) process; therefore, the order of reactivity of simple halides is:

primary> secondary> tertiary

With sodium iodide, primary bromides give a precipitate of sodium bromide within 3 min at 25°C, whereas the chlorides give no precipitate and must be heated to 50°C in order to effect a reaction. Secondary and tertiary bromides react at 50°C, but the tertiary chlorides fail to react within the time specified. Tertiary chlorides will react if the test solutions are allowed to stand for a day or two.

These results are consistent with the following SN2 process:

Procedure

To 1 mL of the sodium iodide-acetone reagent in a test tube add two drops of the compound. If the compound is a solid, dissolve about 0.1 g in the smallest possible volume of acetone, and add the solution to the reagent. Shake the test tube, and allow the solution to stand at room temperature for 3 min. Note whether a precipitate is formed and also whether the solution turns reddish brown, because of the liberation of free iodine. If no change occurs at room temperature, place the test tube in water bath at 50°C.Excessive heating causes loss of acetone and precipitation of sodium iodide, which can lead to false-positive results. At the end of 6 min, cool to room temperature and note whether a reaction has occurred. Occasionally, a precipitate forms immediately after combination of the reagents; this represents a positive test only if the

Org. Chem. Lab.

42

precipitate remains after the mixture is shaken and allowed to stand for 3 minutes.

3. Differentiation between alkyl and aryl halides (Formaldehyde- sulfuric acid test)

With this test aryl halides (chlorobenzene) produce pink, red, or bluish red colour whereas alkyl halides produce yellow, amber, or brown colour.

Procedure This reagent is prepared at the time of use by adding 1 drop of

formaldehyde to a test tube containing 1 mL concentrated sulfuric acid. In another test tube add 1 drop of the compound to be tested to 1 mL of hexane. From this solution take 1-2 drops and add them to 1 mL of the reagent. Shake well and observe the colour.

4. Special tests for chloroform a) Riemer- Tiemann reaction

For procedure and chemical equations refer to "Identification of Phenols" experiment. Resorcinol results in a red colour with slight fluorescence in the aqueous layer while α- or β- naphthol results in a deep blue aqueous layer fading to green.

b) Reduction of Fehling's reagent

For preparation of Fehling's reagent and chemical equations refer to "Identification of Aldehydes and Ketones" experiment. Boil 1 mL of chloroform gently (water bath) with 3 mL of Fehling's reagent with constant shaking for 3- 4 minutes. Reduction occurs and reddish cuprous oxide slightly separates.

Org. Chem. Lab.

43

H3C N H

H

N

CH3

H

H N CH3

H

H

Identification of Amines

mines are basic organic compounds that are considered as derivatives of ammonia. They are classified as primary, secondary, or tertiary according to the number of groups

attached to the nitrogen atom: RNH2, R2NH, or R3N respectively where R is any alkyl or aryl group.

Physical properties

Like ammonia, amines are polar compounds and all of them can form intermolecular hydrogen bonds except tertiary amines.

They have lower boiling points than alcohols or carboxylic acids of

the same molecular weight but higher boiling points than non polar

A

Org. Chem. Lab.

44

compounds. Methylamine is gas while o-phenylenediamine and p,p-diaminodiphenylmethane are solids. The others are liquids.

All amines are capable of forming hydrogen bonds with water, thus those with six carbon atoms or less are quite soluble in water. They are soluble in organic solvents as ether, alcohol and benzene.

All of them have fish like odour except the methylamines and ethylamines which smell just like ammonia.

Aromatic amines are colourless when pure, but they are easily oxidized by air becoming coloured. They are generally very toxic and can be absorbed through the skin.

Chemical reactions

All classes of amines (primary, secondary, and tertiary) have an unshared pair of electrons on the nitrogen atom, just like ammonia. That is why they are similar to ammonia in their chemical behavior (mainly basicity and neocleophilicity).

1. Ramini and Simon tests (Sodium nitroprusside tests). (conventional Ramini and Simon tests) In Ramini test the amine reacts with acetone and the product

interacts with sodium nitroprusside (Na2[Fe(NO)(CN)5].2H2O) that is dissolved in 50 % aqueous methanolic solution to produce a coloured complex. In Simon test acetone is replaced by 2.5 M acetaldehyde solution. These two tests distinguish between primary and secondary aliphatic amines.

To distinguish between aromatic amines (primary, secondary and tertiary) the modified Ramini and Simon tests are applied. These tests use the same reagents and procedure of the conventional tests but the modifications are the replacement of the methanolic solution of sodium nitroprusside by a solution of sodium nitroprusside in dimethylsulfoxide (modified sodium nitroprusside reagent) and the use of a saturated aqueous solution of zinc chloride instead of water.

Procedure • Ramini test

To 1 mL of methanolic sodium nitroprusside solution add 1 mL of distilled water, 5 drops of acetone, and about 30 mg of the amine. In most cases the characteristic colour appears in a few seconds, although in some cases 2 minutes may be necessary.

• Simon test Follow the above procedure exactly but use 5 drops of 2.5 M

acetaldehyde solution instead of acetone. Up to 2 minutes may be needed for the colour to develop.

Org. Chem. Lab.

45

• Modified Ramini test To 1 mL of the modified sodium nitroprusside reagent add 1 mL of

saturated aqueous zinc chloride solution, 5 drops of acetone, and about 30 mg of the amine. Primary and secondary aromatic amines produce orange-red to red-brown colours within a period of few seconds to 5 minutes. Tertiary aromatic amines give a colour that changes from orange-red to green over a period of about 5 minutes.

• Modified Simon test Follow the above procedure exactly but use 5 drops of 2.5 M

acetaldehyde solution instead of acetone. Primary aromatic amines give an orange-red to red-brown colour within 5 minutes; secondary aromatic amines give a colour changing from red to purple within 5 minutes; tertiary aromatic amines give a colour that changes from orange-red to green over a period of about 5 minutes.

Examples are outlined in the following table:

Amine Ramini test Simon test Modified Ramini test

Modified Simon test

tert-butylamine deep red red-brown

dicyclohexylamine deep red violet with precipitate

diethylamine red-brown deep blue

aniline turbid orange,

changes to henna colour

brown

o-phenylenediamine turbid red brown

turbid henna colour

p,p-diaminodiphenylmethane light brown precipitate

light brown precipitate

N,N-dimethylaniline brown

changes to green

brown changes to green (rapid

change)

diphenylamine orange- red to red brown pale orange

Org. Chem. Lab.

46

R2NH + HONO

2° amine nitrousacid

N -nitrosoamine(yellow oil or solid)

NR

RN O + H2O

R3N + H+

3° aliphaticamine

R3NH+

soluble

2. Nitrous acid test.

The reaction of amines with nitrous acid (HNO2) is another test that classifies the amine not only as primary, secondary, or tertiary, but also as aliphatic or aromatic.

Primary aromatic and aliphatic amines react with nitrous acid to form an intermediate diazonium salt. The aliphatic diazonium salts decompose spontaneously by rapid loss of nitrogen, particularly when the original amino group is attached to a secondary or tertiary carbon, while most aromatic diazonium salts are stable at 0°C but lose nitrogen slowly on warming to room temperature.

Secondary amines undergo a reaction with nitrous acid to form N-

nitrosoamines, which are usually yellow oils or solids. These are carcinogenic compounds; therefore, students are advised not to perform nitrous acid test for secondary amines.

Tertiary aliphatic amines do not react with nitrous acid, but they

form a soluble salt. Tertiary aromatic amines react with nitrous acid to form the orange-

coloured hydrochloride salt of the C-nitrosoamine. Treating the solution with base liberates the blue or green C-nitrosoamine.

Org. Chem. Lab.

47

NaNO2 + HCl HONO + NaClsodium nitrite nitrous acid

NR2 + HONO + HCl

3° aromaticamine

NHR2Cl + H2ONO

hydrochloride salt of C-nitrosoamine(orange)

NaOH

NR2 + NaCl + H2ONO

C-nitrosoamine(green)

Procedure Nitrous acid is prepared instantaneously by the reaction of sodium

nitrite and hydrochloric acid: In a test tube dissolve 0.5 mL or 0.5 g of the amine in a mixture of

1.5 mL of concentrated hydrochloric acid and 2.5 mL of water, and cool the solution to 0°C in a beaker of ice. In another test tube dissolve 0.5 g of sodium nitrite in 2.5 mL of water, and add this solution drop wise, with shaking to the cold solution of the amine hydrochloride. Move 2 mL of the final solution to another test tube, warm gently, and examine for evolution of gas. Results

• The observation of rapid bubbling or foaming as the aqueous sodium nitrite solution is added at 0°C indicates the presence of a primary aliphatic amine.

• The evolution of gas (bubbling) upon warming to room temperature indicates that the amine is a primary aromatic amine, and the solution should be subjected to the coupling reaction (test 3).

• If a pale yellow oil (heavier than water) or low-melting solid, which is the N-nitrosoamine, is formed with no evolution of gas, the original amine is a secondary amine.

• If a dark-orange solution or an orange crystalline solid is formed, which is the hydrochloride salt of the C-nitrosoamine, the amine is tertiary aromatic. Treating 2 mL of this solution with few drops of 10 % sodium hydroxide or sodium carbonate solution produces the bright-green or -blue nitrosoamine base.

Org. Chem. Lab.

48

ArN2Cl +

ONa

diazonium saltof primary

aromatic aminessodium-2-naphthol

+ NaOH

ONaN

N Ar

azo dye(red- orange)

+ NaCl + H2O

R2NH + CS2 + NH4OH + H2OCR2N SNH2

S

carbon disulfidedialkyldithiocarbamate

NiCl2

CR2N S

S

2Ni

• If only solubilization of the amine is obtained with no other results, the amine is tertiary aliphatic.

3. Coupling reaction (a test for primary aromatic amines).

Procedure Dissolve 0.1 g of 2-naphthol in a mixture of 2 mL of 10 % sodium

hydroxide solution and 5 mL distilled water. Add 2 mL of the cold diazonium solution and observe the result. The formation of a red- orange dye (red precipitate in case of phenol) with evolution of gas only upon warming indicates that the compound is a primary aromatic amine.

4. Carbon disulfide reagent test (for secondary aliphatic amines).

Org. Chem. Lab.

49

Procedure In a test tube dissolve 50 mg (1-2 drops) of the amine in 5 mL

distilled water (or 1-2 drops of concentrated hydrochloric acid if necessary). In another test tube mix o.5- 1 mL of concentrated ammonia solution with 1 mL of nickel chloride in carbon disulfide reagent (NiCl2/CS2). Add 0.5- 1 mL from the first test tube to the second one. A definite precipitation indicates that the unknown is a secondary amine. A slight turbidity is an indication of a trace of a secondary amine as an impurity.

5. Lignin test (for primary and secondary aromatic amines).

This test depends on the action of lignin in the newsprint paper.

Procedure Dissolve 10- 20 mg of the amine in a few drops of ethanol and

moisten a small area of newsprint paper with this solution. Place 2 drops of 6 N hydrochloric acid on the moistened spot. The immediate development of a yellow or an orange colour is a positive test for a primary or secondary aromatic amine.