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    University of Baghdad College of PharmacyDepartment 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

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    ii 

    Contents

    Tools and glass wares .....................................................................................................iii

     

    Qualitative analysis of organic compounds.....................................................................

    Determination of solubility class ......................................................................................

    Identification of alcohols .................................................................................................6

     

    Identification of aldehydes and ketones .........................................................................

    Identification of phenols ................................................................................................ 8 

    Identification of carboxylic acids ....................................................................................6

     

    Identification of carboxylic acids salts ............................................................................37

     

    Identification of alkyl and aryl halides ............................................................................39 

    Identification of amines .................................................................................................43 

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    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

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    lit

    P

    filt

    glass

    us paper

    tri dish

    r papers

    od (stirr

     

    r)

    gradu

    grad

    w

    iv

    spatula

    ted cyli

    ated pip

    ater bath

     

    ders

    ette burett

    spatul

    funnel

     and fun

    Hood

    el stand

     

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    Org. Chem. Lab. 

    1

    Qualitative Analysis of

    Organic Compounds 

    ualitative analysis of organic compounds helps identify andcharacterize 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 thatidentification 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 classificationaccording 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

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    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 thecompound (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 becauseof 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 isinsoluble 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. Predictionof 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

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    Org. Chem. Lab. 3

     I 

    H2O

    5% NaOH Ether

    5% HCl 5% NaHCO3

    96%

    cold H2SO

    4

    insol. sol.

    sol. insol.

    l   i    t   m u s  u

    n c h   a n  g e  d  

    l   i    t   m u

     s  b  l    u e 

    l   i    t   m u

     s r  e  d  insol. sol.

    insol. sol.

    insol. sol.

    insol. sol.

     N 

     I  I  I  I  A2

     B

     I  I  I  I  A1

     I  I  I  I  S 1

     I  I  I  I  S 2

     

    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.

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    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 differsfrom 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 theremaining 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 S 1class 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 S 2 class and includes ionic salts such as salts ofcarboxylic 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% sodiumhydroxide 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 aciditydegree of the compound. Note that testing solubility in 5% sodium

     bicarbonate solution is not needed if the compound is insoluble in 5%

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    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 phenolswith 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), thismeans that the compound is not an acid but, rather, is either a basiccompound 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 givenclass 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.

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    Org. Chem. Lab. 6

    H3C OH

    methanol

    cyclohexanol(cyclic)

    OH

    H2C

    OHH3C

    ethanol

    C

    H3C CH3H3C

    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 relatedto 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. Benzylalcohol 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

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    Org. Chem. Lab. 7

    Which alcohol has solubility class  S 2, what are the

    structural requirements present in this alcohol thatmade it under this solubility class?

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

    ceric ammonium

    nitrate

    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 watersolubility increases and vice versa. Besides, low molecular weight

    alcohols are soluble in water due to hydrogen bonding ability withwater 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.

    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 hydrogenhalides 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 ofalcohols by the Ce (IV) solution. This red colour disappears after a

    reasonable time due to completing the oxidation of this intermediate and

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    Org. Chem. Lab. 8

    (NH4)2Ce(NO3)6 + ROH

    ceric 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 ammoniumnitrate 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, andadd 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', whichalso causes the aldehyde or ketone to be tri-iodinated on the terminal

    methyl group producing iodoform as a yellow precipitate.

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    Org. Chem. Lab. 9

    ZnCl2 + HCl ZnCl3- + H+

    + HClC

    OH

    ZnCl2

    + H2OC

    Cl

    tertiary alcoholtertiary alkyl halide

    insoluble 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, inwhich 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-butanolis 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 andconcentrated 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.

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    Org. Chem. Lab. 10

    Procedure

    Mix 2-4 drops of the alcohol with few drops of Lucas reagent andobserve the results:

    •   benzyl alcohol gives immediate result as shown by the appearanceof 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.

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    Org. Chem. Lab. 11

    C O

    C

    O

    HR 

    C

    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 R Ŕ CO. 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 thiscarbonyl 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

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    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 yellowcolour (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  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.

    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. 

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    Org. Chem. Lab. 14

    The oxidation process requires an alkaline medium; thereforesodium 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 precipitatemaking it a water soluble complex. Note that since the medium isalkaline, 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% sodiumhydroxide 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 testtube 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. Ifthe 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 aketone.

     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.

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    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, andthen 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).

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    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

    heat

    OH

    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)5 NO].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 ofthe corresponding alcohol and the salt of the corresponding carboxylicacid (or the acid itself). Therefore, one molecule of the aldehyde serves as

    the oxidizing agent while the other serves as the reducing agent.

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    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 mLof 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 awhite 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.

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    Org. Chem. Lab. 18

     phenol

    OH

    hydroquinone(quinol, hydroquinol)

    OH

    OH

    resorcinol

    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

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    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.

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    Org. Chem. Lab. 21

    Procedure

    To a very dilute aqueous solution of the phenol (30-50 mg in 1-2 mLwater) 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,

    resorcinolviolet or blue

    o- and p-cresol greenish blue

    hydroquinone deep green

    α- and  β -naphthol no special colour

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    Org. Chem. Lab. 22

    OH

    OH

    resorcinol

    + 2

    C

    O

    C

    O

    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 phenolitself, 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 areinsoluble 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.

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    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 ofconcentrated 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 hydroxidesolution introduces an aldehyde group (-CHO) into the aromatic ring at

    the ortho- or para-positions: 

    compound colour

    α-naphthol green colour

     β -naphtholvery pale green with slightfluorescence

    phenol red to pink

    o-cresol red-violet

    m-cresol blue to pink

     p-cresol  no change

    resorcinol pale red colour with greenfluorescence

    hydroquinone violet colour

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    Org. Chem. Lab. 24

    Procedure

    To about 0.2 g of the phenol add 1mL of 30% sodium hydroxidesolution 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

    resorcinolred 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

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    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 roomtemperature. 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  A1. When they react with sodium

     bicarbonate, they evolve carbon dioxide gas. This is considered as

    a good simple indication of them.

    C

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    Org. Chem. Lab. 27

    C

    O

    OH

    R  C

    O

    HO

      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 flamewhereas 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 diluteammonia 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

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    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 excessinterferes 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

    lacticno special change

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    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.

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    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

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    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.

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    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 (seethe general test) add few drops of calcium chloride solution; a white

     precipitate of calcium tartrate is formed. This precipitate dissolves indilute hydrochloric acid but not in dilute acetic acid.

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    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 solutioncausing 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 acidFor procedure follow the same steps mentioned above for tartaricacid. 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.

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    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 identificationof 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 tartaricacid. 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 toyellow 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.

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    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.

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    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

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    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 dropsof 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

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    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. 

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    Org. Chem. Lab. 39

    CH3X

    metheyl halide

    C2H5 X

    ethyl halide

    Cl

     benzyl chloride chlorobenzene

    CHCl3 CCl4

    chloroform 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 (R   X, Ar   X) have boiling points higherthan 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 totheir 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 silvernitrate to give an insoluble silver halide (AgX), and the rate of this

    reaction indicates the degree of reactivity of the halogen atom in thecompound. 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 consistentwith results from elemental analysis (sodium fusion for detection of

    halogens).

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    Org. Chem. Lab. 40

    RX + AgNO3C2H5OH

    AgX + RONO2

    R XAg+

    R X Ag

    δ+

    δ+

    δ−

    AgX + R +

    R + NO3

    -

    R ONO2

    R 3CCl RI C C

    H

    R  CH2X

    H

    C C

    H

    R  H

    CH2X

    H2C

    Br

    CH2Br

    ArX C C

    H

    R  H

    X

    C C

    H

    R  X

    H

    HCCl3 CCl4

     

    It is obvious that this reaction follows S N1 mechanism. Generally thereactivity of alkyl halides towards this reagent is:

    R 3CX > R 2CHX > RCH2X

    Procedure

    Add one drop or a couple of crystals of the unknown to 2 mL of2% 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 andobserve 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 comparedwith the corresponding open-chain secondary halides. Cyclohexyl

    chloride is inactive, and cyclohexyl bromide is less reactive than 2-

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    Org. Chem. Lab. 41

    RCl + NaIacetone

    RI + NaCl

    RBr + NaI RI + NaBracetone

    R XI-

    I C X

    δ−

    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, ortertiary. This test depends on the fact that sodium chloride and sodium

     bromide are only very slightly soluble in acetone.

    The mechanism follows direct displacement (S N2) 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 andtertiary bromides react at 50°C, but the tertiary chlorides fail to react

    within the time specified. Tertiary chlorides will react if the test solutionsare allowed to stand for a day or two.

    These results are consistent with the following S N2 process:

    Procedure

    To 1 mL of the sodium iodide-acetone reagent in a test tube addtwo 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 reactionhas occurred. Occasionally, a precipitate forms immediately after

    combination of the reagents; this represents a positive test only if the

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    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 browncolour.

    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 ofchloroform 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.

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    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 groupsattached to the nitrogen atom: RNH2, R 2 NH, or R 3 N respectively where R

    is any alkyl or aryl group.

    Physical properties

    Like ammonia, amines are polar compounds and all of them canform 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

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    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 colouredcomplex. 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 saturatedaqueous 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 somecases 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 neededfor the colour to develop.

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    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 testModified

    Ramini test 

    Modified

    Simon test 

    tert -butylamine deep red red-brown

    dicyclohexylamine deep redviolet with

    precipitate

    diethylamine red-brown deep blue

    anilineturbid orange,

    changes to

    henna colour

    brown

    o-phenylenediamineturbid red

    brown

    turbid henna

    colour

     p,p-diaminodiphenylmethanelight brown

    precipitate

    light brown

    precipitate

     N,N-dimethylaniline

    brown

    changes to

    green

    brown changes

    to green (rapid

    change)

    diphenylamineorange- red to

    red brownpale orange

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    Org. Chem. Lab. 46

    R 2NH + HONO

    2° amine nitrous

    acid

     N -nitrosoamine

    (yellow oil or solid)

    N

    N O + H2O

    R 3N + H+

    3° aliphatic

    amine

    R 3NH+

    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 saltsdecompose 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.

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    Org. Chem. Lab. 47

    NaNO2 + HCl HONO + NaCl

    sodium nitrite nitrous acid

    NR 2 + HONO + HCl

    3° aromatic

    amine

    NHR 2C l + H2ONO

    hydrochloride salt of C -nitrosoamine

    (orange)

    NaOH

    NR 2 + 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, withshaking 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.

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    Org. Chem. Lab. 48

    ArN2Cl +

    ONa

    diazonium salt

    of primary

    aromatic aminessodium-2-naphthol

    + NaOH

    ONa

    NN Ar

    azo dye

    (red- orange)

    + NaCl + H2O

    R 2N H + C S2 + NH4OH + H2OCR 2N SNH2

    S

    carbon disulf ide

    dialkyldithiocarbamate

    NiCl2

    CR 2N 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 % sodiumhydroxide 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). 

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    Org. Chem. Lab. 49

    Procedure

    In a test tube dissolve 50 mg (1-2 drops) of the amine in 5 mLdistilled 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. Adefinite precipitation indicates that the unknown is a secondary amine. A

    slight turbidity is an indication of a trace of a secondary amine as animpurity.

    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 andmoisten a small area of newsprint paper with this solution. Place 2 dropsof 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.