qualitative organic analysis--sem 3

CONTENTS

I.Organic qualitative analysis (Scheme of organic

analysis)

II. Multistep synthesis

(a). Benzoic acid – m- nitrobenzoic acid - Methyl m-Nitrobenzoate

(b). Aniline -acetanilide- p-bromoacetanilide- p-bromoaniline

(c). Synthesis of dibenzal acetone by Aldol condensation

(d). Synthesis of methyl orange by coupling reaction.

III. Organic Estimations

(a). Estimation of Aniline/ Phenol

(b). Estimation of Ester

(c). Estimation of iodine value of Ester

(d). Estimation of Saphonification value of an oil/fat

(e). Estimation of Glucose using Fehling Solution

INSTRUCTIONS

Chemistry is a discipline based on observation. In lecture you will learn principles and theories

and in laboratory you have opportunity to experience these principles and theories in practice.

The following section presents some general guidelines. Making laboratory safety important.

Kindly follow the guidelines given below:

Laboratory aprons must be donned at all times in the lab and put up hair properly.

Sandals, open-toed shoes and high heels are not permitted in the laboratory.

Shorts or skirts cut above the knee are not permitted in the lab.

Never wear cloth that hangs.

Kindly follow the general behaviour listed:

Strictly avoid the use of regional languages in the lab.

No food or beverages will be permitted inside the lab.

Always read the upcoming experiments carefully and thoroughly, being used to

understand all of the directions before entering the lab.

Be in and ready, promptly when the lab begins.

Always read the labels of the reagents and never use a reagent from an unlabelled

bottle.

Never smell a chemical straight out of the container.

Never pour a waste chemical into drain or put in the garbage.

Never pick broken glassware with your bare hands, regardless of the size of the piece.

Please place all broken glassware in the appropriate broken glassware bucket.

Should any injury occur regardless of how minor it is report it immediately to the

Lab tutors.

Wash your hand frequently during the lab and definitely wash your hand twice at the

end of the lab.

QUALITATIVE ORGANIC ANALYSIS

SCHEME OF ORGANIC ANALYSIS

The Scheme of Analysis may be divided into five parts

1. Preliminary tests

2. Detection of elements

3. Detection of Characteristic groups

4. Confirmatory tests

5. Confirmation by preparing a solid derivative for identifying the organic compound.

Experiment Observation Inference

I Preliminary tests

1. Colour and appearance of

the substance are noted

2. Odour is noted

3. Solubility is noted

A little of the given

compound is shaken with the

following solvents

(a) Cold water

(b) Sodium hydroxide

(c) Dil.Hydrochloric acid

(a) Colourless

(b) Yellow

(c) Brown or black

(a) Pleasant fruity

(b) Fishy or

ammoniacal

(c) Kerosene like

smell

(d) Bitter almond

smell

(e) Pungent

(f) Carbolic

soluble

Insoluble

Soluble in sodium

hydroxide and reappears

as turbidity on adding

excess of dil. HCl.

Soluble and reappears as

turbidity on adding

excess of NaOH solution

Presence of hydrocarbon,

aldehydes, ketones, acids esters

etc.

Presence of aromatic

nitrocompounds

Presence of phenol or amine

Presence of ester

Presence of amines

Presence of hydrocarbon

Presence of benzaldehyde of

nitrobenzene

Presence of halogen compounds

Presence of phenol

Presence of sugars, lower

aliphatic alcohols, aldehydes,

ketones and esters.


hydrocarbons, amines, phenols,

higher aldehydes, ketones and

esters.

Presence of acids or phenols.

Presence of amines

II. Detection of Elements. Lassaigne’s Test. A small piece of metallic sodium is melted in an

ignition tube by gentle heating. Then small quantity of the substance is added. It is again heated

gently to complete the reaction and then strongly. When the ignition tube is red hot it is plunged

into distilled water taken in a china dish. The tube breaks and any residual sodium react with

water. The broken ignition tube is ground well with the bottom of a boiling tube. The mixture is

boiled well and filtered and the filterate is known as the sodium fusion extract. The following

tests are done with the extract.

1 To one portion of the sodium fusion

extract half of its volume of freshly

prepared ferrous sulphate solution is

added, boiled, few drops of ferric

chloride solution is added and

acidified with dil hydrochloric acid

A blue or green colouration

or precipitate is obtained

Nitrogen is present

2 Another portion of the extract is

acidified with dil. Nitric acid, boiled

well, cooled and silver nirate

solution is added

(a) White curdy precipitate

soluble in ammonia

(b) Yellowish white

precipitate sparingly soluble

in ammonia

(c) yellow precipitate

insoluble in ammonia

Chlorine is present

Bromine is present

Iodine is present.

3. To the third portion of the extract a

few drops of freshly prepared

sodium nitroprusside solution is

added

Violet colouration

Sulphur is present.

III DETECTION OF CHARACTERISTIC GROUPS

1 Test to find whether aliphatic or aromatic

(i)Ignition test. A small quantity of

the substance is ignited on nickel

spatula

(ii) Nitration test: A little of the

substance is added to a mixture

containing 2mL con. Sulphuric acid

and 1mL con. Nitric acid taken in

test tube. It is then heated on a

boiling water bath for about half an

hour and then poured into cold water

taken in the beaker

(a) Burnt with a non-smoky

flame

(b) Burnt with a smoky

luminous flame

Colourless solution

Yellow solution or

precipitate

Presence of aliphatic

substance


substance

Presence of alphatic

substance


substance.

2 Test to find out whether unsaturated or saturated

(i) Action of dilute potassium

permanganate: A little of the

substance is shaken with water and

(a) Immediate

decolourisation

Presence of

unsaturated

compound

one or two drops of dil. potassium

permanganate solution

(ii)Action of bromine water: A little

of the substance is dissolved in

suitable solvent( alcohol/ water) then

a little bromine water is added

(iii) Action of bromine in

carbontetrachloride: A little of the

substance is dissolved in carbon

tetrachloride and bromine in carbon

tetra chloride is added and shaken

(b) Slow decolourisation

Decolourisation without the

formation of a precipitate.

No decolurisation

(a) Decolourisation without

the evolution of

hydrogenbromide

(b)Decolourisation with the

evolution of hydrogen

bromide

(c) Decolourisation with

formation of a precipitate.

Presence of easily

oxidizable substance

like phenol, niro

phenol, amines,

belnzaldehyde, etc. Presence of unsaturated

compound.

Presence of saturated

compound.

Presence of

unsaturated substance

Presence of saturated

substances

Presence of easily

brominated

compounds like

phenols, aromatic

amines etc.

3 Action of con. Sulphuric acid: A

little of the substance is warmed

with con.H2SO4

(a) Charring with

effervescence due to the

liberation of sulphurdioxide,

carbondioxide,

carbonmonoxide and smell

of burnt sugar

(b) dissolves gradually on

heating

(c) White precipitate which

dissolved in excess of acid

Presence of

carbohydrate


hydrocarbon

Presence of basic

substance like

aromatic amines

4 (1) Action of sodium hydroxide

solution: A little of the

substance is boiled with dilute

sodium hydroxide solution

(a) Ammonia is evolved

(b) substance dissolved

(c)Separation of oil or

formation of an emulsion

(d)Solution turns deep

yellow in colour

Presence of amide

Presence of acidic

substances like acids

phenols and their

derivatives

Presence of anilides

Presence of

nitrophenols

(2)The given compound is boiled

with 20% sodium hydroxide for

half an hour then cooled and

acidified with dilHCl

White precipitate


esters and amides

5 Action of Sodalime: A little of the

substance is mixed with thrice its

mass of dry sodalime in a dry test

tube and heated. The smell of the

issuing gas is noted.

(a) Ammonia gas is evolved

(b) Kerosene lie smell

Presence of amides

and amines

Presence of acids

6 Action of sodium bicarbonate: to

one mL saturated solution of

sodium bicarbonate solution little of

the substance is added

Brisk effervescence of

carbondioxide

Presence of acids

7 Action of Metallic Sodium: To a

little of the substance ( if solid

dissolve in dry benzene) in a dry

test tube a small piece of metallic

sodium is added

Brisk effervescence

Presence of alcohols,

acids and phenols.

8 Action of ferric chloride solution:

To a little of the substance in Water

or alcohol a few drops of neutral

ferric chloride is added

(a) Violet colour

(b) A flocculent white

precipitate

(c) green colour changing to

a white precipitate

(d) Buff coloured precipitate

Presence of phenol

Presence of α-

naphthol

Presence of β-

naphthol

Presence of benzoic

acid, cinnamic acid or

phthalic acid

9 Action of Borsche’s reagent: To

one mL of Borsche’s reagent a little

of the substance is added and heated

over a water bath for five minutes.

Cooled and little water is added

A yellowish orange

precipitate is obtained

Presence of aldehydes

or ketones

10 Action of Schiff’s reagent: A little

of the substance is added to 1mL

Schiff’s reagent taken in test tube

and shaken well

Violet colour developed

within two minutes

Presence of aldehydes

11 Action of Tollen’s reagent: A little

of the substance is added to about

2ml tollen’s reagent in a clean test

tube and heated in a boiling water

bath

(a) Black or brown

precipitate

(b) Bright silver mirror is

formed

Presence of

polyhydric phenol

Presence of

aldehydes, reducing

sugars such glucose,

fructose, maltose etc.

12 Action of Fehling’s solution:

Fehling’s solution A and Fehling’s

solution B are mixed in equal

volumes. To 1 mL of this reagent a

little of the organic compound is

added and heated on a boling water

bath

Reddish brown precipitate is

formed

Presence of

aldehydes, polyhydric

phenols, and reducing

sugars.

13 Action of Molish’s reagent: To a

solution of substance in water added

a few drops of alcoholic solution of

α-naphthol. Then added about 1mL

of con.H2SO4 along the sides of the

test tube without disturbing

Violet ring is formed at the

junction

Presence of

Carbohydrates.

IV Confirmatory Tests

A If nitrogen is present is present, the following tests are conducted. Besides the following tests

for those groups for which indications are got are also done.

1

2

Action of sodium hydroxide

solution. A little of the substance

heated with sodium hydroxide

Action of Soda-lime: A little of the

organic substance is heated with

excess of dry soda-lime


(b)Separation of oil and

formation of an emulsion


(b) Amine is roduced

Presence of amides

Presence of anilides

Presence of amides

and amines

Presence of

aminoacids, toluidines

and anilides

3 Biuret test:A little of the substance

is gently heated in a dry test-tube

until it melts and then solidifies.

The residue is dissolved in a little

water and a dilute solution of

copper sulphate is added followed

by sodium hydroxide solution drop

by drop.

A violet colour is produced. Presence of diamide

like urea.

4 Action of nitrous acid: A little of

the substance is dissolved in dilute

hydrochloric acid, cooled in ice

water and a 10%solution of sodium

nitrite is added with shaking till it is

slightly in excess.

(a) Liberation of nitrogen

with the formation of

alcohol.

(b) Separation of yellow oil.

(c) Reddish brown solution

is obtained.

Presence of aliphatic

primary amines.

Presence of secondary

amines

Presence of tertiary

amines

5 With the solution obtained above

the following tests are done.

(i)To one portion of the solution an

alkaline solution of β-naphthol is

added.

(ii)A portion of the solution is

extracted with ether. The ether

extract is washed with sodium

hydroxide solution and then with

water. The ether is evaporated off

and Liebermann’s nitroso reaction

is conducted with the residual oil.

(iii)To another portion, dilute

A scarlet red precipitate is

formed.

Blue or green solution is

obtained.

Ether layer becomes deep


primary amines.

Presence of secondary

amines.

Presence of tertiary

sodium hydroxide solution is added

and then shaken with little ether.

green. amines.

6 Carbylamine reaction. To a little

of the substance few drops of

cholorofm and about 2ml of

alcoholic potash are added and

warmed.

Offensive smell is produced

Presence of primary

amine

7 Mulliken and Barker’s reaction.

A little of the substance is dissolved

in alcohol. A few drops of calcium

chloride solution is added and pinch

of zinc dust. Boiled for five

minutes, cooled and filtered into a

tube containing Tollens reagent.

Bright silver mirror or black

precipitate is obtained.

Presence of nitro

group.

8 Reduction of nitro group to amino

group. A little of the substance is

treated with few ml of dilute

hydrochloric acid and a pinch of

zinc dust. Heated for some time and

filtered. With the filtrate the

following tests are done.

(a) Carbylamine test is done with

one portion of the filtrate

(b) To another portion of the filtrate

dil. hydrochloric acid is added,

cooled in ice and sodium nitrite

solution is added in excess. Then

alkaline β-naphthol solution is

added

An offensive smell is

produced

A scarlet precipitate is

obtained

Presence of nitro

group


nitro group.

B If halogen is present, the following tests are conducted. Besides the following tests, tests for

those groups for which indications are got also done

1 Action with litmus. A little of the

substance is shaken with hot water

and tested with litmus

(a) soluble and acidic to

litmus

(b) Insoluble and acidic

(c) Insoluble and neutral

Aliphatic halogen

substituted acids


halogen substituted

acids

Presence of halogen

substituted

hydrocarbons, ketone

etc.

2 Action with silver nitrate solution.

A little of the substance is boiled

with sodium hydroxide solution for

15 minutes. Cooled, acidified with

dil. nitric acid and then added silver

(a) Precipitate of silver

halide is formed

(b) No precipitate of silver

Halogen is in the side

chain

Halogen is in the

nitrate solution. halide nucleus

3 Action of alcoholic silver nitrate.

To a little of the substance 2 ml of

alcoholic silver nitrate solution is

added and warmed gently.

(a) Precipitate of silver

halide is obtained

(b) No precipitate of silver

halide

Presence of halogen

in the side chain

Presence of halogen

in the nucleus.

C If sulphur is present, the following tests are conducted. Besides the following tests, the tests

for those groups which indications are got also done.

1 Action of alcoholic sodium

hydroxide To a little of the

substance 2 ml of alcoholic sodium

hydroxide solution is added and

warmed gently.

Ammonia is evolved

Presence of thiourea

or sulphonamide

2 Action of con. Hydrochloric acid.

To a little of the substance 2 ml

con. HCl is added and warmed

gently.

Pungent smell

Presence of

substituted thiourea.

3 Fusion with alkali. A little of the

substance is fused with sodium

hydroxide dissolved in water anf

hydrochloric acid is added

(a) Hydrogen sulphide is

evolved

(b) Sulphurdioxide is evolve

with the formation of phenol

(c) No phenol is formed but

precipitate of barium

sulphate when barium

chloride is added

(d) Ammonia is evolved

during fusion. No phenol is

formed. Sulhur dioxide is

evolved on adding acid

Presence of thio urea

Presence of sulphonic

acid

Presence of amino

sulphonic acid

Presence od

sulphonamide

D. If nitrogen, halogens and sulphur are absent, tests for the following groups for which

indications are got, are done.

I.Aldehydes

1. Schiff’s reagent test is

conducted

Violet colour is obtained Presence of aldehydes.

2 Borsche’s reagent test is

conducted.

Note: Ketones also answer this

test.

A yellow precipitate is

obtained.

Presence of aldehydes.

3 Tollen’s reagent test is

conducted.

Note: Other reducing reagents also

answer this test.

Bright mirror or black



4 Fehlling’s solution test is

conducted.

Note: Other reducing reagents also

Red precipitate is obtained. Presence of aldehydes.

answer this test

5 Sodium bisulphite test is

conducted: A little of the

substance is added to a saturated

solution of sodium bisulphite and

shaken well.

Note: ketones also answer this test.

White crystalline precipitate

is obtained.


6 Semicarbazide test: Dissolved

0.5g of semicarbazide

hydrochloride in 5ml of water and

added 0.5g of anhydrous sodium

acetate. It is warmed to get a

solution. Then added a small

quantity of the substance and

warmed on a water bath.

Note: ketones also answer this test.

White crystalline precipitate

is obtained.


II. Ketones.

1 Borsche’s reagent test is

conducted.

A yellow precipitate is

obtained.

Presence of ketones.

2 Semicarbazide test is conducted. White crystalline

precipitate is obtained

Presence of ketones.

3 Sodium bisulphate test is

conducted.

White crystalline

precipitate

Presence of ketones

4 Iodoform test is conducted: 0.25ml

of acetophenone is taken in a test

tube.Add 1.5ml of NaOH solution

followed by I2-KI solution until

the iodine colour persisted on

shaking. The solution is warmed.

Excess of iodine solution is

removed by adding dil.NaOH

solution drop by drop.The contents

is allowed to stand in the water

bath for 20minutes.

Yellow precipitate with

characteristic odour is

formed

Presence of ketones

containing the CH3-CO-

group

III.Acids

1 Tested with sodium bicarbonate

solution.

Effervescence Presence of acids.

2 Sodalime test is conducted Kerosene smell

obtained

Presence of acids.

3 Ester formation test is conducted.

About 0.5g of the substance is

heated gently with about 1 ml of

Pleasant fruity smell. Presence of acids.

ethanol and few a drops of

conc.sulphuric acid for about 1

minute. Cooled and poured into a

few ml of water in test-tube.

4 s-Benzylthiouronium salt test:

About 0.25 g of the acid is

dissolved in 2 ml of warm water.

The acid is neutralised by adding a

few drops of NaOH solution

(phenolphthalein can be used as an

indicator) .Then 2 drops of NH4Cl

are added followed by 0.5 g of s-

Benzylthiouronium chloride in

2ml water. It is cooled in ice.

White crystalline

precipitate.

Presence of acids

5 Fluorescein reaction. Fused

together in a dry test-tube a small

quantity of the substance with an

equal amount of resorcinol after

moistening the mixture with two

drops of conc. Sulphuric acid.

Cooled, dissolved in water and

then added excess of sodium

hydroxide solution.

A reddish solution

having an intense green

fluorescence is

produced.

Presence of dicarboxylic

acids.

IV.Phenols

1 Neutral ferric chloride solution test is

conducted. A little of the substance is

treated with neutral ferric chloride

solution.

(a) Violet blue or green colour.

(b) A flocculent white

precipitate.

Presence of

phenol.

Presence ofα-

naphthol.

2 Liebermann’s nitroso reaction: To

two drops of melted phenol, added

little solid NaNO2 . Heated gently for

1 minute.Cooled and added 4 drops

of conc.H2SO4. Diluted cautiously

with water.

Red solution which turned to

green or blue on adding

sodium hydroxide solution.

Presence of

phenol.

3 Phthalein fusion reaction: About 2

drops of melted phenol is mixed with

a small quantity of phthalic

anhydride in a dry test-tube. 2 drops

of conc.H2SO4 are added. The

mixture is heated at about 150oC for

2 min.Cooled and exess of 10 %

Red, bluish-purple, blur green

fluorescene, green or very faint

green colouration.

Presence of

phenol.

NaOH solution is added.

4 Benzoylation (Schotten Baumann

reaction) is conducted :Dissolved

about 0.25 g of phenol in about 5ml

of 10 % NaOH solution contained in

a boiling tube . About 1 ml of

benzoyl chloride is added. The

boiling tube is corked and shaken

vigorously for about 15 min.

Crystalline white precipitate. Presence of

phenol.

5 Azo-dye formation reaction:

Dissolved 2 drops of aniline in 1 ml

dil. HCl and well cooled in ice. A

few drops of saturated NaNO2

solution are added. The diazonium

solution thus obtained is added to a

well cooled solution of phenol in

aqueous NaOH solution.

Orange, scarlet,dark red,

brownish red solution or


Presence of

phenol.

V. Alcohols.

1 Test with metallic sodium is conducted. Brisk effervescence. Presence of alcohols.

2 Acetylation test: A little of the substance

is heated with glacial acetic acid and few

drops of conc. Sulphuric acid. Then

cooled and poured into excess of water

containing little sodium carbonate

solution.

Pleasant fruity smell

is produced.

Presence of alcohol.

VI. Esters.

1 Hydrolysis. A little of the substance is

refluxed with concentrated solution of

sodium hydroxide and then acidified with

conc. Hydrochloric acid.

White precipitate is

formed.

Presence of ester.

2 Hydroxamic acid formation. To a few

drops of the substance, added 0.2g of

hydroxylamine hydrochloride and about 5

ml of 10% sodium hydroxide solution and

the mixture gently boiled for 2 minutes.

Cooled and acidified with dilute

hydrochloric acid and then added a few

drops of ferric chloride solution.

A violet or a deep red-

brown colour

developed

immediately

Presence of ester

VII Carbohydrates.

1 Concentrated sulphuric acid test is

conducted. Warmed a little of the

substance with conc. Sulphuric acid.

Charring with smell of

burnt sugar

Presence of

carbohydrate.

2 Sodium hydroxide test is conducted.

A little of the substance is boiled with

sodium hydroxide solution

Solution turned yellow or

brown. Caramel smell is

emitted

Presence of

carbohydrate.

3 Molisch’s test is conducted. A deep violet ring is

formed.

Presence of

carbohydrate.

4 Treated with Tollen’s reagent Bright silver mirror or

black precipitate.

Presence of reducing

sugar.

5 Fehling’s solution test is conducted:

Warmed with Fehling’s solution.

Red precipitate is

formed.

Presence of reducing

sugar.

6 Osazone test is conducted: About 1 g

of sugar is dissolved in 15 ml water

and add 4 g of phenyl hydrazine

hydrochloride, 4 g of sodium acetate

and 1 ml glacial acetic acid. Heated

for 15 minutes in a water bath.

Yellow crystals are

formed.

Presence of

carbohydrate.

VIII. Hydrocarbons.

1 Odour is noted. Kerosene like smell

observed.

Presence of

hydrocarbons.

2 Sulphonation is conducted : To 1 ml of

fuming H2SO4 contained ina test tube, 2

drops of the substance are added and

shaken well for 3 min.

Substance has gone

into solution.

Presence of

hydrocarbon.

3 Nitration is conducted.

Note.(1) To nitrate naphthalene, about

0.5g of naphthalene is dissolved in 2 ml of

glacial acetic acid by gently warming,

cooled and heated to 800c after adding

conc. Nitric acid. It is then poured into

water when yellow crystals separate.

Yellow solid

obtained.

Presence of

hydrocarbon is

confirmed

4 Picrate test is conducted : Saturated

solutions of naphthalene and picric acid,

both in benzene are prepared separately.

These two solutions are mixed in a watch

glass and allowed to evaporate.

Red or yellow

precipitate.

Presence of

polynuclear

hydrocarbons.

5. Confirmation by preparing a solid derivative

The final step in the analysis of a sample organic compound is the preparatioof a

suitable solid derivative.

Preparation of Derivatives

Derivatives for Aromatic Hydrocarbons.

The main reactions carried out for the preparation of derivatives for aromatic hydrocarbons are

(a) nitration (b) side chain oxidation and (c) preparation of picrates for polynuclear

hydrocarbons.

(a) Nitration. Nitroderivatives can be prepared for benzene, toluene etc. About 1 ml of fuming

nitric acid and 1 ml of conc.sulphuric acid are mixed.About 0.25 ml of benzene or toluene is

added to the nitrating mixture. Then the mixture is heated on a boiling water bath for half an

hour,till a drop of mixture poured into water crystallizes immediately. The mixture is then

poured into cold water taken in beaker and stirred well. The crystals are filtered at the

pump,recrystallised from dilute alcohol, dried and then melting point is noted.

(b). Side chain oxidation. For aromatic hydrocarbons containing side chain like toluene or side

chain like xylenes, side chain oxidation can be effected for the preparation of their derivatives.

About 0.25 ml of the substance is mixed with about 12.5 ml of saturated potassium

permanganate solution and 1 g of anhydrous sodium carbonate. The mixtutre is then boiled for

half an hour under reflux. It is then transferred to a beaker, acidified with conc. Hydrochloric

acid and then added a saturated solution of sodium sulphite until the brown precipitate of

manganese dioxide has dissolved. It is cooled, filtered at the pump and recrystallised from hot

water. It is dried and melting point is noted.

(c) Picrates. Picrates can be easily prepared for polynuclear hydrocarbons like naphthalene

anthracene ctc About 0.25g of picric acid is also dissolved in hot benzene. About 0.25g of picric

acid also dissolved in hot benzene. These two solutions mixed well, poured into a watch glass

and kept for sometime. Coloured crystals of picrate separate. Melting point is noted.

Derivatives for Halogen compounds of Aromatic hydrocarbons.

(a) Nitration. For compounds having halogen in the nucleus like chlorobenzene, ortho-chloro

toluenes, para-dichlorobenzene etc. nitroderivatives are prepared. Nitration is carried out in

the same manner as aromatic hydrocarbos. Melting point is noted.

(b) Side chain oxidation. For compounds having halogen in the side chain like benzyl chloride

and for nuclear halogen compounds containing side chain oxidation can be adopted.Side

chain oxidation can be adopted exactly in the same manner as explained under aromatic

hydrocarbons. Melting point of the derivatives is found out.

Derivatives for alcohols.

The following derivatives can be prepared for alcohols.(a) benzoates and (b) oxidation products.

(a) Benzoylation(Schotten- Baumann reaction). About 0.25 g of the substance is dissolved in

about 4 ml of 10% sodium hydroxide taken in a boiling tube. About 0.5 ml benzoyl chloride is

added, corked the tube well and shaken vigorously for about 15 minutes.. (till the smell of

benzoyl chloride is no longer perceptible). Filtered, washed several times with water. Dried and

then recrystallised from alcohol. Melting point is determined.

(a) Oxidation. Side chain oxidation can be carried out in the case of alcohols like benzyl

alcohol. It is same as in the case of aromatic hydrocarbons

Derivatives for phenols.

The following derivatives can be prepared for phenols. (a) benzoyl derivatives (b) bromination

products (c) Nitration products and (d) picrates

(a) Benzoylation. Benzoylation can be easily carried out for phenols, cresols, α- naphthols, β-

naphthols and resorcinol. Details of benzoylation, refer under the derivatives of alcohols.

(b) Bromination. Bromination can be done in the case of phenols and cresols. A bout 0.25 g of

phenol is treated with saturated bromine water till the yellow colour due to excess of bromine

persists. The mixture should be shaken well after each addition of bromine water. The

crystallized bromo derivative is filtered at the pump, washed with water and dried. It is

recrystallised from alcohol, dried and melting point is determined.

(c) Nitration. Poly nitro derivatives can be prepared for certain phenols. About 0.25 g of phenol

is dissolved in about 1 ml of cold conc. Sulphuric acid and the solution poured slowly into about

6 ml of the nitrating mixture, containing equal volumes of concentrated nitric acid and sulphuric

acids. Then it is warmed for a few minutes on a water bath. If the reaction is violent and there is

tendency to form tarry matter, it has to be cooled in ice without warming on the water bath.

Cooled poured into ice water, filtered and recrystallised from dilute alcohol containing a few

drops of conc. Hydrochloric acid.

(d) Picrates. Picrates can be easily prepared for phenols. Details refer under derivatives of

hydrocarbons.

Derivatives for aldehydes and ketones.

The important derivatives for aldehydes and ketones are: (a) Phenyl hydrazones(b) 2,4- dinitro-

phenyl hydrazones (c) semicarbazone and (d) oximes.

(a) Phenylhydrazones. A solution of phenylhydrazine is prepared by dissolving 0.5g of

phenylhydrazine hydrochloride and 0.75 g of sodium acetate in 5 ml of water. About 0.25g of

aldehyde or ketone is dissolved in a little of alcohol and added to phenyl hydrazine solution. If a

clear solution is not obtained, more alcohol is added. The mixture is heated on a water bath for

about half an hour. The phenyl hydrazone is separated on cooling. It not a few drops of water are

added. The product is filtered off and crystallized from alcohol. The melting point is determined.

(a) 2,4- dinitrophenylhydrazones. Benzaldehyde acetophenone and benzophenone readily form

2,4- dinitrophenylhydrazones with 2,4- dinitrophenyl- hydrazine.(Borsche’s reagent). About 0.25

g of substance is diossolved in methanol. It is mixed with about 1 ml of Borsche’s reagent and

shaken vigorously for a few minutes, with scratching if necessary. If the yellowish orange

hydrazone does not separate, the solution is heated in a got water bath for about 10 minutes. It is

cooled, filtered at the pump, recrystallised from alcohol and melting point is determined.

(b) Semicarbazones. About 0.25 g of asemicarbazide hydrochloride is added to 2.5 ml of water

followed by 0.25g of anhydrous sodium acetate and warmed gently until a clear solution is

obtained. A solution of 0.25 g of the substance in 1 ml of methanol is added and warmed on a

water bath.It is cooled. Crystals of semicarbazone filtered and washed with water. It is

recrystallised from alcohol, dried and the melting point determined.

(c) Oximes. About 0.25 g of hydroxylamine hydrochloride is dissolved in about 2 ml of water.

About 0.25 g of sodium acetate and 0.1g of the compound are added into it. In case the

compound is water insoluble, sufficient amount of alcohol is added to the mixture to give a clear

solution. The mixture is then heated on a water bath for about 15 minutes and then cooled in

ice.Precipitation may be induced by adding a few drops of water. Filtered, washed with cold

water, recrystallised from dilute alcohol or benzene, dried and melting point is determined.

Derivatives for Acids. The following derivatives can be prepared for carboxylic acids (a) s- benzylthiouronium salts (b)

amides (c) anilides (d) bromo-derivatives (e) nitration and (f) acid anhydride.

(a) s- Benzylthiouronium salts. Dissolved about 0.2g of the acid in the minimum amount of hot

water, 5% aqueous sodium hydroxide solution is added until the solution is just alkaline to

methyl orange.Then one drop of dilute hydrochloric acid is added. The sodium salt of the acid

thus prepared is poured into a solution of 0.3g of s-benzylthiouronium chloride in 3ml of

water.The mixture is stirred and cooled in ice bath.Crystals are filtered at the pump,

recrystallised from ethanol containing 10% of water, dried and melting point determined.

(b) Amides. Amide derivatives can be easily prepared for benzoicacid, phthalic acid, cinnamic

acid and salicylic acids. About 0.5g of the acid is mixed with an equal quantity of phosphorous

pentachloride in a mortar. The mixture is ground well till the evolution of fumes ceased. Then

added a few ml of concentrated ammonia.Stirred well and some water is added. The amide

formed is filtered at the pump, washed with water and dried. It is recrystallised from dilute

alcohol and melting point is determined.

(c) Anilides. About 0.4g of pure aniline are taken in a dry test tube.The mixture is boiled under

reflux for about an hour,cooled and poured in an excess of dilute hydrochloric acid. It is filtered

at the pump, washed with water and dried.It is then recrystallised from dilute alcohol and melting

point determined.

(d) Bromo derivatives. Bromo derivatives can be easily prepared for cinnamic acid. About

0.25g of the acid is dissolved in boiling water. Excess of bromine water is added till brown

colour persisted. Crystals formed are filtered, washed with water and dried. Melting point

determined.

(e) Nitration. Nitro derivatives can be easily prepared for benzoic acid, salicylic acid etc.1ml of

nitrating mixture is prepared by mixing equal volumes of conc. nitric acid and conc.sulphuric

acid. About 0.25 g of the acid is added into the nitrating mixture in small portions at time with

shaking. It is then heated on a water bath for about 30 minutes. It is cooled and poured into

water. It is filtered at the pump, washed with water and dried. The melting point is determined.

(f) Acid anhydride. Anhydried can be prepared for ortho- carboxylic acid like phthalic acid.

About 0.25 g phthalic acid taken in a dry china dish and covered by means of an inverted

funnel.the stem of the funnel is closed by means of cotton wool. The china dish is gently heated.

Phthalic anhydride is formed which gets collected at the cooler side of the funnel. After cooling

the funnel is removed and the anhydride collected. The melting point of the anhydride is then

determined.

Derivatives for Esters.

The important method used for the preparation of derivatives of esters is hydrolysis to the

corresponding acid.

(a) Hydrolysis. About 1 ml or 1 g of the ester is mixed with about 10 ml of 20% solution of

sodium hydroxide in a R.B flask and boiled under reflux for about 45 minutes. It is then

transferred to abeaker, cooled and acidified with conc. Hydrochloric acid. The acid precipitated

is filtered at the pump. Washed with cold water and dried. Melting point is determined.

Derivatives of Amines.

The following derivatives may be prepared for primary and secondary amines.(a)acetyl

derivatives (b)benzoyl derivative and (c)picrates.In the case of tertiary amines, picrates are

commonly prepared.

(a) Acetylation. Since acetyl derivatives of aliphatic amines are usually soluble in cold

water,acetylation can be carried out in the case of aromatic amines like aniline ,toluidines,N-

methyl aniline etc.About 0.5 ml of the amine ,if liquid or 0.5g,if solid is taken in a small R B

flask or boiling test tube fitted with a reflux condenser.About 2.5ml of acetic anhydride and

acetic acid mixture (equal volumes) is added and refluxed gently for 15 minutes.It is then poured

into water.The solid anilide separated is filtered at the pump,washed with water and dried.It is

recrystallised from dilute alcohol and melting point is noted.

(b) Benzoylation.Benzoyl derivative can be prepared for primary amines like aniline, toluidines

and for secondary amines like N-methyl aniline.Details of benzoylation refer under preparation

of derivatives for phenols.

(c) Picrates.Picrate derivative can be prepared for primary,secondary(except diphenyl amine)

and tertiary amines.The given amine and picric acid(equal amounts)are dissolved separately in

cold ethanol to get saturated solutions.The two solutions are mixed and poured into a watch

glass.Coloured crystals of picrate separate.Melting point is determined.

(d) p-Nitroso derivative. p-Nitroso derivative can be prepared for the tertiary amine,N,N-

dimethylaniline.About 0.5 ml of N,N-dimethylaniline is dissolved in about 4ml of dilute

hydrochloric acid.It is cooled in ice and the added about 2ml of 20%sodium nitrite solution in

drops.It is kept in ice bath with stirring for 5 minutes.Then dilute sodium hydroxide solution is

added.A green precipitate of p-nitrosodimethylaniline is obtained.It is filtered at the pump ,dried

and melting point is determined.

Derivatives for Nitro Compounds.

The important derivatives for mononitro-compounds are: (a) The nitro group is reduced to

primary amino group.The primary amine obtained by reduction, can be diazotized and coupled

as explained under preparation of derivatives for phenols.If aromatic primary amine is obtained

by reduction,it can be diazotized and coupled with β-naphthol in alkaline solution (b) Further

nitration to get solid dinitro compounds (c) In the case polynitro compounds, they can be

partially reduced to solid nitroanilines and hence partial reduction serves a method for the

preparation of derivative for polynitro hydrocarbons.

(a) Reduction of mono-nitro compounds. As already explained, mononitro compounds are

reduced to the corresponding primary amino compounds and with the amino compound

benzoylation and azodye formation conducte.

(b) Nitration. Nitration of benzene to solid meta-dinitrobenzene can be easily carried out.1ml of

conc.nitric acid and 1ml of conc.sulphuric acid are mixed together in a boiling test-tube.About

0.25ml of nitrobenzene is added with shaking. The mixture is heated in a boiling water bath for

about 15 minutes.It is then poured into cold water. It is filtered at the pump,washed with water

and dried.It is recrystallised from alcohol and melting point is noted.

(c) Reduction of polynitro hydrocarbons to aminonitro hydrocarbons. This method is used

for the preparation of derivative for meta-dinitrobenzene.About 0.5g of powdered sulphur is

added to a solution of 1.5g of sodium sulphide in about 7ml of water. The mixture is boiled until

a clear solution is obtained.

About 1g of meta-dinitrobenzene is boiled with about 50ml of water in a beaker. To the

boiling solution is added the sodium sulphide solution prepared above, in a thin stream with

stirring.When the addition is over, the mixture is boiled for about 30 minutes more and filtered

hot.The filtrate is cooled when meta-nitroaniline separates. It is filtered at the pump, washed with

cold water and dried. It is then recrystallised from hot water, dried and melting point is

determined.

Note:(i)For nitrophenols,benzoylation does not proceed smoothly and hence nitrophenols are

reduced to aminophenols and then benzoylation is conducted (methods of reduction and

benzoylation already explained)to obtain dibenzoyl derivative.(ii) For nitroaniline, benzoyl

derivatives can be prepared.

Derivatives for Amides. For amides other than urea,hydrolysis can be effected for the preparation of derivative. If the

original compound is an aromatic amide, alkaline hydrolysis followed by acidification with

hydrochloric acid gives a solid organic acid with definite melting point.In the case of aliphatic

amides, the acid obtained after hydrolysis will remain in solution. In such case, the cold solution,

when carefully neutralized and treated with s-benzylthiouronium chloride, deposits the

thiouronium salt.

(a) Hydrolysis. About 1 g of aromatic amide is taken in a R.B flask fitted with a reflux

condenser. About 10 ml of 10% sodium hydroxide solution is added. It is heated for about 30

minutes. It is cooled and acidified with conc. Hydrochloric acid. The precipitated acid is filtered

at the pump, washed, recrystallised from hot water, dried and melting point determined.

Derivatives for Urea

(a) Urea nitrate. A concentrated solution of urea in about 1 ml of water is prepared. Then a few

drops of conc. Nitric acid are added with shaking. White crystalline precipitate of urea nitrate

separates. It is filtered at the pump, dried and melting point is determined.

(b) Urea Oxalate. A concentrated solution of urea in about 1 ml of water is prepared. Then add a

concentrated aqueous solution of oxalic acid in drop with shaking. White crystalline precipitate

of urea oxalate separates. Filtered at the pump, dried and melting point is determined.

Derivative for Thiourea.

s- Benzyl thiouronium chloride. About 0.5 g powdered thiourea and 0.8 ml of benzyl chloride

are added to one ml of 95% ethanol in a small R.B. flask or boiling test tube fitted with reflux

condenser. The mixture is warmed on a water bath with gentle shaking until effervescence

subside. Then the mixture is boiled under reflux for 30 minutes. The solution is cooled in ice

bath when crystals of s-Benzylthiouronium chloride separate. Crystals are filtered at the pump,

dried and melting point is determined.

11. Derivatives for Anilides.

The following derivatives can hbe prepared: (a) Hydrolysis to the corresponding acid and amine

(b) bromo derivative and nitration. (c) Nitration.

(a) Hydrolysis. Anilides undergo hydrolysis very slowly by alkalies and hence acid hydrolysis

is usually employed. A bout 0.5 g of anilide is mixed with 5 ml of 70% sulphuric acid in a R.B

flask or boiling test- tube fitted with a reflux water condenser. The mixture is gently boiled for

about 15 minutes. Then the solution is cooled and diluted with about 5 ml of water. By

hydrolysis, acetanilide gives liquid acetic acid and liquid aniline. With the aniline obtained,

solid derivatives can be prepared and their melting points determined.In the case of benzanilide,

solid benzoic acid is obtained by hydrolysis. The solid is filtered, dried and melting point is

determined.

(b) Bromination. Little of the anilide is dissolved in acetic acid. Then bromine in acetic acid is

added with shaking until brown colour remained. It is then poured into water. The precipitated

p-bromo derivative is filtered at the pump, washed with water and dried. It is recrystallised from

alcohol, dried and melting point determined.

(c) Nitration. Anilides are nitrated by using 80% nitric acid at 00

c and then poured into ice cold

water. Nitration leads to a mixture of o- nitroderivative and p-nitroderivative. Ortho- derivative

is soluble in cold alcohol while para- derivative is insoluble.

12 Derivative for carbohydrates.

Osazone. About 1 g of sugar is dissolved in 15 ml water and add 4 g of phenyl hydrazine

hydrochloride, 4 g of sodium acetate and 1 ml glacial acetic acid. Heated for 15 minutes in a

water bath. The osazone formed is filtered, washed with water and dried. It is then recrystallized

from alcohol, dried and melting point is determined.

Multistep synthesis

1. Benzoic acid – m- nitrobenzoic acid - Methyl m-Nitrobenzoate

Step1:

Electrophilic Aromatic Substitution of Benzoic Acid to Produce m-Nitrobenzoic Acid

Recall from your lecture class that a carboxylic acid would be a meta-director in an

electrophilic aromatic substitution reaction. In practice, this nitration reaction can result in the

production of quite a bit of the ortho product as well, unless the temperature is kept very cold

throughout the reaction. All of the materials that will be used in the experiment are in proportion

to the amount of benzoic acid that is being reacted. Not more than 3 g of PhCOOH is used, and

record its mass carefully

Overall Reaction:

COOH

Con. HNO3

Con.H2SO4COLD

NO2

COOH

Benzoic acid m-Nitrobenzoic acid

Mechanism:

HNO3 + 2 H2SO4NO2

+ HSO4 H3O+

NO2

HSO4

COOH COOH

NO2H

H2SO4

NO2

COOH

First, prepare a nitrating mixture (NM) by slowly adding con. H2SO4 to con. HNO3 while

cooling it in a small Erlenmeyer flask in an ice/water/salt bath to 0oC or less. Make this NM in

proportion to the amount of benzoic acid that will be reacting, although the benzoic acid will not

be in this mixture. For each g of benzoic acid, use 1 mL of concentrated H2SO4 and 0.67 mL of

concentrated HNO3 to prepare this NM. Keep it cold.

Second, prepare the reaction mixture (RM) in a large Erlenmeyer flask; this container

will maximize cooling during the reaction. Add concentrated H2SO4 to the Erlenmeyer and cool

it to 00C or less. 2.5 mL of H2SO4 for each gram of benzoic acid is used. Add the benzoic acid

slowly to the H2SO4, keeping the temperature below 00C. During the course of this mixing and

the reaction to follow, the RM should stay below 00C and never exceed 5

0C. When all of the

benzoic acid has been added to the H2SO4, it will be rather paste-like.

Now, double-check that the RM is colder than 00C and slowly add the COLD NM to the

COLD RM, mixing carefully and keeping it cold. Use a short disposable pipette to transfer it and

be sure that the rate of addition allows for efficient cooling; remember that the RM should stay

below 0oC and never exceed 5

0C. Add the NM very slowly at first, but the rate can be speeded

up as the reaction proceeds. Use the temperature as a guide. After all of the NM has been added,

keep the mixture cold for another 10-15 minutes with occasional stirring.

Finally, pour the mixture over ice/water slurry of about 150 g of ice and 200 mL of water.

Stir vigorously and the product precipitates. Filter the product from the mixture, wash well with

cold water, and allow it to dry. When the product is completely dry, obtains its mass and

calculates the theoretical and percentage of yield for the reaction. Check its purity by mp, IR.

The product is usually of sufficient purity to use for the next step, but if that is not the case,

recrystallize it from hot water.

Step 2:

Fischer Esterification of m-Nitrobenzoic Acid to Produce Methyl m-Nitrobenzoate

Overall Reaction:

Con.H2SO4

NO2

COOH

m-Nitrobenzoic acid

+ CH3OHHEAT

NO2

COOCH3

Methyl m-Nitrobenzoate

Mechanism:

NO2

CO

O

H

SO O

OO

H

H

SO O

OOHNO2

CO

O

H

H

NO2

CO

O

H

H

NO2

CO

O

H

H

NO2

C

O

O

H

H

NO2

CO

O

H

H

CH3

O H

NO2

C O

O

H

H

O

CH3

H

transfer of proton

NO2

C O

O

H

H

O

CH3

H

NO2

C

O

OH

H

O

CH3

H

SO O

OOH

NO2

C

O

O CH3

As in the previous step, the amounts of reagents used for this procedure will depend on

the mass of m-nitrobenzoic acid that is used in the reaction. Use no more than 3 g of it. It is also

critical that the m-nitrobenzoic acid is completely dry, since this reaction is an equilibrium

process and water in a wet sample will drive the reaction in the reverse direction, reducing the

yield of the product.

For each gram of m-nitrobenzoic acid 8 mL of methanol is required and for each 20 mL

of methanol, 1 mL of concentrated H2SO4is required. Consider the total volume of this mixture,

and choose a round bottom flask that holds about twice that volume; in other words, choose a

flask so that it is about half full. Put the three materials in the proportions described above into

the round bottom flask with a couple of boiling chips, and attach a reflux condenser to form a

reflux apparatus. Heat to reflux for 1 hour.

Pour the reaction mixture into ice /water slurry (use a total volume of slurry of about 5

times the volume of methanol used) and stir. Once the ice is melted, use suction filtration to

isolate the product and wash it with water. The crude product should be recrystallized from

methanol or methanol/water. Once it is completely dry, determine its mass and calculate its

theoretical and percent yield. Also, determine its purity by mp and IR.

Multi-Step Synthesis Yield Calculation.

When you carry out a series of reaction steps, you usually want to know the efficiency of the

whole process. To do so, you can use the percent yields for each step to compute the overall

percent yield. This is easiest to explain with an example. Suppose you carried out four reactions

in sequence with the percent yields given below.

Step 1: 87.5%; Step 2: 91.2%; Step 3: 79.3%; and Step 4: 81.9%

The overall percent yield is computed as shown, here.

Overall Percent Yield: 0.875 x 0.912 x 0.793 x 0.819 x 100 = 51.8% overall.

Be sure to compute the overall percent yield for the two steps of you synthesis of methyl m-

nitrobenzoate.

2. Aniline -acetanilide- p-bromoacetanilide- p-bromoaniline

Step 1: Acetylation of Aniline

In the first step we need to put the removable acetyl protecting group on the nitrogen of aniline.

The acetyl group is electron withdrawing and it therefore makes the lone pair on the nitrogen less

reactive either in an oxidation reaction or a protonation reaction. Bromination of aniline suffers

from lack of control. The electron donating amino group activates the ring to such a great extent

that usually tribromoaniline is isolated. However, if aniline is converted to acetanilide the

monosustitution is easily achieved because the acetamino group cannot activate benzene ring

towards electrophilic attack as well as simple amino group does. The acetamino group is less

effective in donating electron density to the benzene ring, because the electron pair on the

nitrogen atom is delocalized by both the carbonyl group and the phenyl ring. It should be kept in

mind, however, that the acetamino group is still an activating group.

2. The acetic acid used in the brominating step causes protonation. Protonation of the nitrogen of

aniline makes it a very strong deactivating group, making the aromatic ring less susceptible to

reaction and would be a meta director. Another value of the acetyl protecting group is that it is

bulky group and preferentially directs the bromination to the para position rather than the ortho

position. The full mechanism for the reaction is givenbelow. Acetic anhydride ispartially

protonated by the acetic acid. This makes the anhydride an even better electrophile for the

nucleophilic nitrogen of aniline. This attacks to form the tetrahedral intermediate, which, after

proton transfer, loses acetic acid.

NH2

C

O

OH3C

CH3C

O

CH3COOH

NHCOCH3

Procedure:

Dissolve 4.0 mL of aniline in 10 mL of acetic acid in a 100 mL round bottom flask. To this

solution, add 5 mL of acetic anhydride and mix well by swirling. CAUTION: the reaction is

exothermic and the flask becomes warm. Add two boiling chips, attach a condensing column and

attach the hoses for water cooling. Heat at a gentle reflux for fifteen minutes. After fifteen

minutes, allow the flask to cool slightly. Cautiously add 5 mL of cold water through the top of

the condenser into the reaction mixture. Boil the solution for an additional five minutes so as

to hydrolyze any unreacted acetic anhydride. After boiling for five minutes, allow the reaction

mixture to cool slightly and then pour it slowly with stirring into 30 mL of ice cold water. After

allowing the mixture to stand for 15 minutes with occasional stirring, collect the precipitate by

suction filtration using your Buchner funnel. Be sure to wet the filter paper before you filter.

Disconnect from the vacuum, wash the solid crystals with 10 mL of cold water and

reconnect the vacuum tube for a couple of minutes more so as to dry the product as much

as possible. Transfer the crystals to a watch glass and leave them to dry. Weigh the product and

record the yield.

Mechanism

C

O

O CH3

C

O

H3C

C

O

CH3O

H+C

O

O CH3

C

O

H3C

H

C

O

CH3O

+

C

O

O CH3

C

O

H3C

H

H2N

C

O

H3C OH

N HH

C

O

O CH3

H

C

O

H3C

C

O

CH3O

NH

C

O

O CH3

H

C

O

H3C

C

O

CH3O

H

NHCOCH3

+ C

O

CH3HO

Bromination of acetanilide

NHCOCH3

Br2

Glacial acetic acid

NHCOCH3

Br

Acetanilide the starting reagent in this synthesis is a mild analgesic and a mild antipyretic. The

antipyretic properties of acetanilide was discovered by accident when a sample, improperly

labeled and thought to be naphthalene was inadvertently administered to a patient in 1886.

N

H

C

H3C

O

and

NH

C

H3C

O

N

H

C

H3C

O

Place 2.0g of acetanilide in 50mL Erlenmeyer flask. Add 7.5mL of glacial acetic acid and swirl

the flask until the solid dissolves. Fill 1:4 (V/V) bromine/glacial acetic acid in a burette and

clamp it on the stand. Add about 4.4 mL of the bromine acetic acid mixture drop by drop over a

10 minute period with stirring. Stir the reaction mixture for about 15 minutes to complete the

reaction. Pour the reaction mixture into a 250mL beaker containing 60mL water. Rinse the flask

with 15mL of water and add this rinse to the beaker. Stir the precipitated solid well with a

stirring rod to break up any chunks. The solution is in orange colour due excess bromine.

Destroy the excess bromine by adding small portion of sodium bisulphate. Collect the product by

vacuum filtration and wash the residue four times with 15mL portions of water. Flow air through

the residue for drying. Collect the solid, spread it on a watch glass to dry out. Once dried weigh

the crude and recrystallize a small portion from dilute alcohol. Record the melting point, and IR

spectrum.

Alternative Green Procedure:

NHCOCH3NHCOCH3

Br

CAN, KBr

H2O, EtOH

Chemicals Required: Acetanilide - 1 g

Potassium bromide - 1 g

Ceric ammonium nitrate - 6 g

Ethanol - 15 mL

Water - 15 mL

In a 250 ml conical flask acetanilide (1 g) was dissolved in ethanol (15 ml). Then

potassium bromide (1 g) and ceric ammonium nitrate (6 g) were dissolved in water (15 ml). This

solution was transferred into an addition funnel. This solution was added drop wise to the conical

flask containing acetanilide solution. After the addition was over, the reaction mixture was

stirred for 10 minutes in room temperature (white crystals appeared). Then this solution was

poured into ice-cold water. The white crystals were filtered through Buchner funnel and the solid

was dried.

Yield: 1.34 g (85 %)

M.p. of p-bromoacetanilide = 165 o

C

Mechanism:

Ce(IV) + BrH2O

[Br ]Ce(IV)

[Br ]

NHCOCH3NHCOCH3

[Br ]

HBr

-H

NHCOCH3

Br

Preparation of p-bromoaniline

NHBr

H3C

O(1) H

+/H2O

(2) OHNH2Br

Transfer all of the crude p-bromoacetanilide that was prepared above to a 100 mL round bottom

flask. Add 10 mL of water and 10 mL of concentrated hydrochloric acid. Add two boiling chips

and reflux the mixture at a gentle boil for 15- 20 minutes using your heating mantle. Since we

are not using a flammable organic solvent, it is also safe in this step to use a low flame with your

Bunsen burner, using your tripod and wire gauze. As you heat, swirl the flask to ensure mixing

and to dissolve any remaining solid. All of the solid will dissolve and the solution will become

orange in color.After 20 minutes of reflux at a gentle boil, remove the heat source and add 15 mL

of water. Cool the flask to room temperature. Crystals of p-bromoaniline may separate. Prepare a

solution containing 10 mL concentrated aqueous 40 mL water and 25 g ice in a 400 mL beaker.

Pour the solution of p-bromoaniline from above into this solution. Stir the suspension and test the

pH of thesolution by placing a drop on the test strip. It must be strongly basic (blue to litmus).

Ifnot, add more concentrated ammonia.Collect the orange precipitate of p-boromoaniline by

suction filtration using Buchner funnel and wash the solid filter cake with 10 mL of cold water.

Recrystallize the rest of the material from water. Take about 0.5 gram of your product and

recrystallize it from approximately 15mL of water. Use your 50 mL Erlenmeyer. Put ~ 0.5 g in

the Erlenmeyer and add 10 mLof water. Use your stirring rod to disperse the compound as much

as possible in the water to aid in dissolution. Heat the suspension to boiling. (It is safe to use

your Bunsen burner here since we are using water). Once the water is boiling, stir the solution

add more water 1-2 mL at a time until all of the material has dissolved. Once the solution

becomes clear, remove it from the heat and allow it to cool slowly to room temperature and then

immerse it in an ice bath. Collect the crystals by suction filtration using your Hirsh funnel. Leave

the crystals to dry; then determine the melting point.

Record the weight and calculate the percent yield. Run a TLC of the crude and the recrystallized

sample using dichloromethane as the solvent

Mechanism

NHBr

H3C

O

+ HBr NH

C

H3C

OHBr NH

C

H3C

OH

Br NH

C

H3C

OH

O

H

H

Br NH

C

H3C

OH

O

H

HBr NH

C

H3C

O

HHSO4

OH

Br NH2+ H3C

O

OH

3. Synthesis of Dibenzalacetone by Aldol Condensation The reaction of an aldehyde with a ketone employing sodium hydroxide as the base is an

example of a mixed aldol condensation reaction, the Claisen-Schmidt reaction. Dibenzalacetone

is readily prepared by condensation of acetone with two equivalents of benzaldehyde. The

aldehyde carbonyl is more reactive than that of the ketone and therefore reacts rapidly with the

anion of the ketone to give a α-hydroxyketone, which easily undergoes base-catalyzed

dehydration. Depending on the relative quantities of the reactants, the reaction can give either

mono- or dibenzalacetone.

H O

+ H3C

C

CH3

O

O

Benzaldehyde Acetone Dibenzalacetone

1,5-diphenyl-1,4-pentadiene-3-one

Melting point 1100C, max 320nm, max=34,300 In the present experiment, sufficient ethanol is present as solvent to readily dissolve the starting

material, benzaldehyde, and also the intermediate, benzalacetone. The benzalacetone, once

formed, can then easily react with another mole of benzaldehyde to give the product,

dibenzalacetone.

Procedure:

In a 125 mL Erlenmeyer flask, dissolve 0.020 moles sodium hydroxide (pellets) in

4.0 mL of water. In a 50 ml Erlenmeyer flask weigh out accurately 0.0160 moles benzaldehye

andweigh into the same flask 0.0080 moles acetone. Add 10 ml of 95% ethanol and pourthis

mixture into the prepared solution of sodium hydroxide. Mix and swirl occasionallyfor fifteen

minutes. A yellow, flocculent precipitate should form. Filter the solid product by vacuum using

your spatula to transfer as much of the solid as possible. After no more liquid is coming through

the filter paper, disconnect the filter flask from the vacuum line, wash the solid with 10 mL water

and, after about one minute, reconnect to the vacuum. Repeat the wash in the same way using 2-

3 mL chilled 95% ethanol. Allow air to be sucked around the crystals for about 2 minutes.

Recrystallize the product from ethyl acetate using a water bath and hot plate to heat the solvent.

Ethyl acetate is flammable. Use approximately 2.5 mL of solvent per gram of product. Add about

1/2 the expected amount of ethyl acetate, stir with your spatula and heat thesuspension to boiling.

Add more ethyl acetate in 1 mL portions, reheating to boiling each time, until all solid material

dissolves (solution becomes clear). Allow the solution to cool slowly to room temperature and

then cool in an ice bath. Collect the final product on the Buchner funnel by suction filtration.

Record the weight of your compound and calculate the percent yield. Take the melting point.

Mechanism

C

O

CH2H3C

HOH

C

O

CH2H3C

CH O

C

O

CH2H3C

C

OH

O

H H

C

O

CHH3C

C

OH

H

+

O HH

- H2O

CO CH

H2C

C

H

HOH

CO CH

CH2

C

H

C

H

O

C OHC

H2C

C

H

C

H

O

O

H H

OH

C OHC

HC

C

H

C

H

OH

+

HOH

C OHC

HC

C

H

C

H

-H2O

OH

Acetone enolizes in the strongly basic conditions. Note that benzaldehye cannot

enolize and so it must act as the electrophile. The nucleophilic alpha carbon then attacks the

carbonyl of benzaldehyde. After proton transfer there is loss of water to give the a,b-unsaturated

carbonyl that is stabilized by conjugation with the phenyl substituent. Notice how the π-electrons

of the phenyl ring are delocalized all the way onto the carbonyl and onto the other carbonyl in

the final dibenzylacetone product.

4. Preparation of Methyl Orange

Methyl orange is synthesized from sulfanilic acid and N,N-dimethylaniline using a diazonium

coupling reaction. The first step is called “diazotization.” Sodium sulfanilate reacts with sodium

nitrite in hydrochloric acid (i.e., nitroso cation) to form an unstable “diazonium salt.” The second

step is the “diazonium coupling reaction.” The diazonium ion is used in situ, and reacts with N,

N-dimethylaniline to form the acidic azo dye.

+

SO3H

NH2

SO3

NH3

-

+ Na2CO32 2

SO3- +Na

NH2

+ CO2 + H2O

Sulfanilic Acid (zwitterion) Sodium Sulfanilate

SO3- +Na

NH2

SO3- +Na

N N+

HCl / NO2-

Cl-

Sodium SulfanilateDiazonium Chloride

SO3- +Na

N N+ Cl-

+

NCH3H3C

CH3COOH N NO3SNa N

CH3

CH3

N,N-Dimethylaniline

Methylorange

Methyl orange forms beautiful orange crystals and is used as an acid-base indicator The anion

form is yellow and the acid form is red.

N N N

CH3

CH3

S

O

O

OXH

N N N

CH3

CH3

S

O

O

HO

Yellow pH> 4.4 Red pH< 3.1 Inner salt form Mechanism

The first step is simply an acid base reaction. In order to dissolve the sulfanilic acid in the

aqueous solution we add sodium carbonate.. When we add sodium nitrite and HCl, the nitroso

ion is formed from sodium nitrite and this reacts with the amine to form a nitrosoammonium

adduct that loses water under the acidic conditions after proton transfer. This gives the

diazonium salt. Aromatic diazonium salts are stable at low temperature. The terminal nitrogen of

the diazonium salt is very electron deficient. It can be attacked by good nucleophiles. We

dissolve the dimethylaniline in acetic acid. This forms the dimethylaniline acetate salt. Neutralize

this in situ and the dimethylaniline becomes a good nucleophile due to the activating effect of the

dimethylamine substituent. Attack is in the para position due to hindrance at the ortho position

by the bulky dimethylamine substituent.

NH2S

O

O

O

Na O

N

O

+ HCl

HO

N

OHCl O

N

O

H

H

H2O + O N

H

Na O

C O Na

O

HO S NH2

O

O

ON

HClHO S N

O

O

N O

H ClH

H

H2O

HO S N

O

O

N O H

H H ClH2O

HO S N

O

O

N O H

H

HO S N

O

O

N Cl NH3C

CH3

H

C

O

OH3C

OHN

H3C CH3

HO S N

O

O

N N

CH3

CH3H

HO S N

O

O

N N

CH3

CH3

Procedure:

In order to avoid any excess of a reagent that could decompose or cause decomposition and

produce tar to contaminate our dye, you need to weigh the quantities of solid reagent very

carefully to the accuracy of 0.05 g or better. In this experiment you will have to calculate for

yourself some of the amounts of needed reagents. After you have calculated them, confirm your

results with the instructor before proceeding.

Dissolve 0.010 mole of sulfanilic acid (anhydride) in about 50 ml of a solution of sodium

carbonate containing 0.010 to 0.0125 moles of sodium carbonate in a 125 ml Erlenmeyer flask.

The solution is prepared by the stockroom and its strength is indicated on the bottle, but you

must calculate the exact amount needed. Warm the mixture slightly to speed up dissolution. Test

one drop of the solution to make sure it is alkaline. If not, add a small amount (1-2 mL) sodium

carbonate solution and check the pH again. Then add 0.010 moles sodium nitrite and cool to

25°C (room temperature). Put 40 g of ice in a 400 mL beaker and add enough hydrochloric acid

of a 6M or a 12 M solution in order to provide a total of 0.030 mol HCl in the beaker. Add the

sulfanilate solution prepared above in a fine stream while stirring continuously. Keep this

solution cold in the ice bath at all times. It now contains your diazonium salt, which will

decompose if it becomes warm. It is only partially soluble in the aqueous solution and will

precipitate as a bluish-greenish solid. Prepare a solution of N,N-dimethylaniline (0.010 mol) in

0.010 mol of acetic acid in a 25 ml Erlenmeyer flask.

Now add the dimethylaniline acetate solution slowly with constant stirring to the suspension of

the diazonium salt. A dull, reddish-purple mass should appear. Now, very slowly add about 30

mL of 1.0 M sodium hydroxide solution with constant stirring. Add the NaOH a few mL at a

time. The addition should take 10 -15 minutes. The actual coupling reaction does not occur until

you add the NaOH. The reaction takes place best at about pH 7. Keep adding the NaOH until the

solution becomes basic (blue to litmus). If the sodium hydroxide is added too quickly, then free

dimethylaniline will separate out as an oily phase. This then leaves an equivalent amount of the

diazonium salt unreacted. This excess salt decomposes to brown tar on warming to room

temperature and contaminates the otherwise beautiful crystalline orange dye. At the end of the

coupling reaction a yellow-orange or golden color should be observed. The product will now be

recrystallized from the reaction mixture. Heat the reaction mixture to boiling using your tripod

and Bunsen burner. Everything should dissolve and the solution should be clear (though it will

be highly colored). If all the material does not dissolve when the solution is heated to boiling,

add more water as needed. Then, allow it to cool slowly to room temperature to crystallize and

then place the flask in an ice bath to get it as cold as possible. Remember: do not stir or shake the

solution when it is cooling. Allow the crystals to form in an undisturbed flask. They will be

much purer and larger if they form slowly in a motionless flask.

ORGANIC ESTIMATIONS

1. ESTIMATION OF ANILINE/PHENOL.

The reaction to be studied in this experiment is between bromate and bromide ions in the

presence of acid and occurs according to the equation,

KBrO3

+ 5KBr + 3H2SO

4 3K

2SO

4 + 3H

2O + 3Br

2

In this reaction, the potassium and sulphate ions are “spectator” ions in that they are not

themselves materially affected, so, in ionic terms the reaction may be summarised as,

5Br-

+ BrO3

-

+ 6H+

+ 3Br2

+ 3H2O

The free bromine generated reacts with phenol /aniline forming tribromophenol/aniline.

OH

+ 3Br2

OH

Br

Br

Br

+ 3HBr

Equivalent mass of phenol/aniline =

= 15.5 (aniline)

A bromate-bromide mixture which easily liberate bromine in presence of an acid is used so as

keep the concentration of bromine a constant.

Requirments:

1. Approximately N/10sodium thiosulphate.

2. Approximately N/10 brominating mixture.

3. 10% potassium iodide solution.

4. starch solution.

Procedure:

(a) standardisation of sodium thiosulphate solution. About 1.25 g of A.R. potassium

dichromate is weighed out into a 250 ml standard flask. It is dissolved in water and made up to

the mark. 20 ml of the made up solution is pippeted out into a conical flask. About 3 ml of conc.

HCl is added ,followed by 5 ml of 10% KI solution. It is titrated against sodium thiosulphate

solution using starch as the indicator. Titration is repeated till concordant results are obtained.

(b)Estimation of aniline/phenol. The given aniline solution is made up to 100 ml. 20 ml of

aniline and 40 ml of brominating mixture are pippeted out into a stoppered conical flask and

diluted with 25 ml of water. 5ml of conc. HCl is added, and the flask is shaken for a minute to

mix the reactants. It is allowed to stand for 30 minutes with occasional shaking of the contents of

the flask. Flask is cooled under tap and 20 ml of 10% KI solution is added in the cup around the

stopper. The stopper is dislodged whereupon the iodide solution is drawn into the flask with no

loss of bromine. The flask is shaken for 30 seconds and allowed to stand for 10 minutes.the

stopper is removed and the neck of the flask and stopper are washed with a little water. The free

iodine is titrated against sodium thiosulphate using starch as the indicator. The volume of

thiosulphate will be equivalent to the excess of bromine.

A blank analysis is carried out using 20 ml of brominating mixture and 20 ml of water, the

procedure being otherwise identical with the analysis of aniline.

Calculation: Let the strength of sodium thiosulphate be = N1

Let 20 ml of brominating mixture = V ml of Na2S2O3

Amount of brominating mixture used in the estimation = 40 ml

40 ml of brominating mixture = 2V ml of Na2S2O3

20 ml of aniline solution+40 ml of brominating

mixture after the reaction = V2 ml of Na2S2O3

Amount of sodium thiosulphate equivalent to aniline = (2V-V2) ml

Normality of Aniline =

Result:

Mass of aniline in the whole of the given solution = ………….. g

2. Estimation of Ester

Principle: Ester is hydrolyzed quantitatively with known volume of standard alkali. The unreacted alkali is

then titrated against standard acid. The amount of reacted alkali can be found out. From this, the

amount of ester can be calculated.

CH3-COOC2H5 + NaOH CH3COONa+ C2H5OH

Procedure:

About 1 g of given ester is weighed out into a 250 ml round-bottomed flask. 50 ml of standard

N/2 sodium hydroxide solution is added a reflux condenser is fitted into the flask. The contents

into the flask are refluxed on a stand bath for 2 hours. The completion of hydrolysis is indicated

by the disappearance of pleasant smell of ester. The contents of the flask are quantitatively

transferred into 250 ml standard flask and made up to mark. 25 ml of the solution is titrated

against N/2 HCl. From the titre value, percentage of ester in the given sample is calculated.

Calculation:

Weight of ester = W g

Normality of NaOH = N1

Normality of HCl = N2

50x N1 = Volume of 1 N NaOH

Volume of HCl Unreacted NaOH = V2 ml

V2x N2 = Volume of unreacted NaOH x Normality of NaOH

Volume of unreacted NaOH =

= V3ml

V3XN1 = Volume of 1 N NaOH

Volume of 1N NaOH unreacted NaOH = V3XN1 = V4

Volume of 1N NaOH that has reacted =V1-V4

1000 ml 1 N NaOH 1000 ml 1N ester = 88 g of ester

(where 88 is the molecular weight of CH3-COOC2H5)

1 ml 1 N NaOH =

g of ester

(V1-V4) ml 1 N NaOH = = W1 g

Percentage composition of ester =

Result:

The percentage composition of the ester =…………g

3. Estimation of iodine value of an ester

The iodine value is expressed in grams of iodine for the amount of halogens linked with 100g

test sample, and is used as degree of unsaturated bond of fats and oils. Iodine value is a measure

of the total number of double bonds present in fats and oils. It is expressed as the «number of

grams of iodine that will react with the double bonds in 100 grams of fats or oils». The

determination is conducted by dissolving a weighed sample in a non-polar solvent such as

cyclohexane, then adding glacial acetic acid. The double bonds are reacted with an excess of a

solution of iodine monochloride in glacial acetic acid (“Wijs’ solution”). Mercuric ions are added

to hasten the reaction. After completion of the reaction, the excess iodine monochloride is

decomposed to iodine by the addition of aqueous potassium iodide solution, which is then

titrated with standard sodium thiosulfate solution.

Principle: A solution of a definite mass of oil in a suitable solvent such as carbon tetrachloride is treated

with a known excess of iodine monochloride in glacial acetic acid (Wij’s solution). The excess

iodine monochloride is treated with excess of potassium iodide and the liberated iodine is

estimated by titration with standard thiosulphate solution. From the results the iodine value is

calculated.

Requirements:

1.Wij’s solution(Iodine monochloride)

2. Standard sodium thiosulphate solution N/10

3. Approximately 10% solution of potassium iodide.

4. Carbon tetrachloride.

5. Freshly prepared 1% starch solution.

Procedure:

(a) Preparation of wij’s solution: About 6.5 g of pure finely powdered iodine is accurately

weighed and dissolved in 500 ml of pure glacial acetic acid contained in a round bottim flask.

The flask is warmed to facilitate the dissolution of iodine. When cooled, 50 ml of the solution is

transferred into another flask and pure dry chlorine is passed through it till the colour changes

from dark brown to clear orange. The remaining iodine solution is then added, when the colour

of the solution turns to light brown. The solution is next heated on a water bath for 20 minutes.

ICl + KI I2 + KI

ICl 2I

(b)Estimation: About 0.2 g of oil is weighed out into a clean dry stoppered bottle of 500 ml capacity. It

is then dissolved in about i0 ml of carbon tetrachloride. 25 ml of iodine monochloride solution is

then run in from a burette. The resulting mixture, if turbid, is cleared by adding more carbon

tetrachloride. The bottle is gently rotated to mix the contents thoroughly. The bottle is then kept

aside for about half an hour. Then 20 ml of 10% KI solution are added and the mixture diluted by

adding 200 ml of water. The mixture is then titrated with standard thiosulphate solution using

starch as indictor.

A blank determination is carried out without the oil using exactly the same quantity of

carbon tetrachloride and the same pipette for delivering the wij’s solution.

Calculation

If V1 ml of thiosulphate is required for the blank and V2 ml for reacting with the excess of iodine

monochloride in the actual experiment,

Then the iodine value.

S = The strength of thiosulphate

W = Mass of oil taken

Result:

Iodine value of the given oil = ……….g

4. Estimation of saponification value of an oil or fat

Fats and oils are the principle stored forms of energy in many organisms. They are highly

reduced compounds and are derivatives of fatty acids. Fatty acids are carboxylic acids with

hydrocarbon chains of 4 to 36 carbons; they can be saturated or unsaturated. The simplest lipids

constructed from fatty acids are triacylglycerols or triglycerides. Triacylglycerols are composed

of three fatty acids each in ester linkage with a single glycerol. Since the polar hydroxyls of

glycerol and the polar carboxylates of the fatty acids are bound in ester linkages, triacyl glycerols

are non polar, hydrophobic molecules, which are insoluble in water. Saponification is the

hydrolysis of fats or oils under basic conditions to afford glycerol and the salt of the

corresponding fatty acid.

It is important to the industrial user to know the amount of free fatty acid present, since this

determines in large measure the refining loss. The amount of free fatty acid is estimated by

determining the quantity of alkali that must be added to the fat to render it neutral. This is done

by warming a known amount of the fat with strong aqueous caustic soda solution, which

converts the free fatty acid into soap. This soap is then removed and the amount of fat remaining

is then determined. The loss is estimated by subtracting this amount from the amount of fat

originally taken for the test.

The saponification number is the number of milligrams of potassium hydroxide required

to neutralize the fatty acids resulting from the complete hydrolysis of 1g of fat. It gives

information concerning the character of the fatty acids of the fat- the longer the carbon chain; the

less acid is liberated per gram of fat hydrolysed. It is also considered as a measure of the average

molecular weight (or chain length) of all the fatty acids present. The long chain fatty acids found

in fats have low saponification value because they have a relatively fewer number of carboxylic

functional groups per unit mass of the fat and therefore high molecular weight.

Requirements:

1. N/2Alcoholic potash.

2. N/2 Hydrochloric acid

Procedure: About 1 to 2 g of ester (oil or fat) is weighed out accurately into a round bottomed flask. The

flask is fitted with reflux condenser. 25ml of N/2 alcoholic potash are added and the flask is

heated on water bath for about one hour. When reaction is completed, the liquid becomes clear.

A blank experiment is performed simultaneously with the same quantity of alcoholic

potash. Both flasks are cooled and the alkali in both is estimated by titration with N/2

hydrocholric acid using phenolphthalein as indicator. From the results the saponification value is

calculated

Calculation

Let V1 and V2 be volumes of standard acid required for the estimation and the blank and W be

the mass of ester/oil taken

Then the alkali used up by the ester = (V2-V1) ml

1 ml of Normal alkali = 56.1 mg of KOH

Hence saponification value =

Result Saponification value of the given ester = …………..g

5. ESTIMATION OF GLUCOSE

Glucose is a very important monosaccharide in biology. It is one of the major products of

photosynthesis. The living cell uses it as a source of energy and metabolic intermediate. The

name comes from the Greek word glykys, which means "sweet", plus the suffix "-ose" which

denotes a sugar. Two stereoisomers of the aldohexose sugars are known as glucose, only one of

which (D-glucose) is biologically active. This form (D-glucose) is often referred to as dextrose

monohydrate, or, especially in the food industry, simply dextrose (from dextrorotatory glucose).

Fehling's solution is always prepared fresh in the laboratory. It is made initially as two separate

solutions, known as Fehling's A and Fehling's B. Fehling's A is a blue aqueous solution of copper

(II) sulfate pentahydrate crystals, while Fehling's B is a clear solution of aqueous potassium

sodium tartrate (also known as Rochelle salt) and a strong alkali (commonly sodium

hydroxide).Equal volumes of the two mixtures are mixed together to get the final Fehling's

solution, which is a deep blue colour. In this final mixture, aqueous tartrate ions from the

dissolved Rochelle salt chelate to Cu2+

(aq) ions from the dissolved copper sulphate crystals, as

bidentate ligands giving the bistartratocuprate(II) complex as shown below.

Methylene Blue:

Methylene blue is a heterocyclic aromatic chemical compound with the molecular formula

C16H18N3SCl. It has many uses in a range of different fields, such as biology and chemistry. At

room temperature it appears as a solid, odorless, dark green powder, which yields a blue solution

when dissolved in water. The hydrated form has 3 molecules of water per molecule of methylene

blue. Methylene blue should not be confused with methyl blue, another histology stain, new

methylene blue, or with the methyl violets often used as pH indicators. The International

Nonproprietary Name (INN) of methylene blue is methylthioninium chloride.

Methylene blue is widely used as redox indicator in analytical chemistry. Solutions of this

substance are blue when in an oxidizing environment, but will turn colorless if exposed to a

reducing agent. The redox properties can be seen in a classical demonstration of chemical

kinetics in general chemistry, the "blue bottle" experiment. Typically, a solution is made of

dextrose, methylene blue, and sodium hydroxide. Upon shaking the bottle, oxygen oxidizes

methylene blue, and the solution turns blue. The dextrose will gradually reduce the methylene

blue to its colorless, reduced form. Hence, when the dissolved oxygen is entirely consumed, the

solution will turn colorless.

Theory of Estimation of Glucose: A freshly prepared Fehling’s solution is first standardized by titration against a standard solution

of pure glucose A.R. The standardized Fehling’s solution is then used to determine the amount of

glucose in an unknown sample or solution by direct titration.

The Fehling’s solution being a solution of cupric ions is blue in colour and at the end point

changes to red colour precipitate of cuprous oxide. As the supernatant liquid is blue and the

precipitate is red in colour, there may be some difficulty in determination of end point

accurately. Hence sometimes a methylene-blue indicator is employed for accurate determination

of the end point.

C6H12O6 + 2CuO → C6H11O5.COOH + Cu2O

Glucose CupricOxide Gluconic Acid Cuprous oxide

(Fehling’s solution)

Procedure: (a) Standardisation of fehlings solution. About 1.25 g of glucose is accurately weighed out into

a 250 ml standard flask. It is dissolved in water and made up to 250 ml. 20 ml of freshly prepared

fehling solution (10 ml each of I and II) is pipette out into a conical flask. It is diluted with equal

volume water and boiled. To the boiling solution standard solution of glucose is added from the

buretteuntill the blue colour just disappeared. This gives an approximate value of volume of the

glucose required. The exact value is obtained by repeating the titration by adding so much of

glucose solution that 0.5 ml to 1 ml will be required to complete the titration to another sample

of fehling solution , the solution is kept boiling 3 to 5 drops of 1% aq.solution of methylene blue

is added to it give a blue colour. The titration is completed within a minute. The end point will be

the disappearance of blue colour with red ppt of Cu2O. The titration is repeated to get concordant

values.

(b)Estimation of glucose: Make up the given solution to 250 ml. pipette out 20 ml of the

Fehling solution to a 250 ml conical flask diluted with an equal volume of water, heat to boiling

add glucose solution, from a burette until the blue colour just disappears. This gives the

approximate value of the glucose solution required. To obtain the exact value, repeat the titration

and add so much of the glucose solution. So that 0.5 to 1 ml more is required to complete the

reduction. Heat the solution to boiling for 2 minutes. Then without the removal of the flame

beneath the flask add 3-5 drops of 1% aq methylene blue indicator. Complete the titration in 1

minute by adding glucose solution drop wise until the colour of methylene blue just disappears.

Repeat the experiment till the concordant value (+ 0.1 ml) is obtained.

Calculation:

Weight of glucose in 250 mL = W1g

Weight of glucose /mL of the solution =

g

20mL of Fehling solution V1mL of glucose (standard) solution

Weight of glucose/ mL of Fehling solution

g

20 mL of the Fehling solution = V2 mL of glucose (estimation) solution

Weight of glucose in the whole of given solution =

g

Result:

Weight of glucose in the whole of the given solution = ……………g

qualitative organic analysis--sem 3

Documents