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PRINCIPLES RELATED TO PRACTICAL CHEMISTRY (PART-1)

DETECTION OF NITROGEN, SULPHUR AND HALOGENS IN ORGANIC COMPOUNDS

DETECTION OF NITROGEN

  • Soda-lime test
organic compound + NaOH smell of NH3
(Not reliable since –NH2, –NO2 and – N = N-groups do not respond)

  • Lassaigne’s test
    (Common for N, S and X (halogens)
Lassaigne’s filtrate or sodium extract is prepared by fusing the organic compound with Na in ignition tube. Fused mass is dissolved in water, boiled and filtered. The filtrate is sodium extract which contains
Na + C + NNaCN sodium cyanide
2Na + SNa2S Sodium sulphide
Na + X NaX (Sodium halide)
Na + C + N + SNaCNS  (Sodium Sulphocyanide)

TEST FOR NITROGEN

Sod. extract + freshly prepared FeSO4 solution + FeCl3 solution + dil. H2SO4  green or blue colouration or sometimes blood red colour
2 NaCN + FeSO4 Fe(CN)2 + Na2SO4
4 NaCN + Fe(CN)2Na4 [Fe(CN)6]
3 Na4[Fe(CN)6] + 4 FeCl3Fe4[Fe(CN)6]3 + 12NaCl
                      ferric ferrocyanide
         (prussian blue)
FeCl3 + 3 NaCNS    Fe(CNS)3 + 3NaCl
Ferric sulphocyanide
 (Blood red)
(NH2.NH2 and diazo compounds do not give this test, diazo compounds are decomposed to give N2 and NH2.NH2 does  not contain C)

TEST FOR SULPHUR

TEST FOR HALOGENS

Sod. extract + Conc. HNO3 + AgNO3 Solution
  • If white precipitate soluble in NH4OH Cl present
       
  • If light yellow precipitate sparingly soluble in NH4OH Br present
  • If yellow precipitate insoluble in NH4OH  I present
    • Function of Conc. HNO3 : It decomposes NaCN and Na2S to avoid their interference

    • Layer test for bromine and iodine
Sod. extract

    • Beilstein's test
Organic compounds containing halogens when heated over Cu wire loop, give blue or green colour flame due to formation of volatile copper halides.
(Not reliable since thiourea, urea, pyridine and organic acids also give this test)

DETECTION OF THE FUNCTIONAL GROUPS

ALCOHOLIC GROUP (–OH)

  • Ceric ammonium nitrate lest
Treat 2 drops of the organic substance with 0.5 c.c. of ceric ammonium nitrate solution and dilute with 2 c.c. of water. A red colour indicates alcoholic group

  • Sodium test
2 cc of organic substance +a piece of anhydrous CaCl2 to absorb water if present. Transfer the organic substance to another test tube and add a dry piece of sodium. Effervescence indicate the presence of an alcohol

  • Evolution of HCl
Take 0.5 cc or 0.3 g of the original substance in a dry test tube and add to it about 6 drops of acetyl chloride or PCl5. Vigorous reaction takes place with the evolution of HCl fumes
The same test could be repeated with acetic anhydride when a characteristic. Fruity odour is obtained

  • Xanthate test
To 1 c.c. of the conc. aq solution of the original substance add 2 pallets of KOH. Heat and cool. Add 1 c.c. of the ether and 2-3 drops of CS2. Formation of yellow ppt. indicate the presence of alcoholic group
ROH + KOH     ROK + H2O

Distinction between 1º, 2º and 3º alcohol
2 ml of organic compound + 8 ml of Lucas reagent and shake.
Separation of an insoluble layer at once 3º alcohol Appearance of cloudiness within 4-5 minutes 2º alcohol solution remains clear 1º alcohol.

PHENOLIC GROUP (–OH)

  • Ferric Chloride Test
2 c.c. of aqs. or alcoholic solution + 2 drops of ferric chloride  deep colour change shows the presence of phenols.
  • Liebermann's Reaction
1 c.c. of conc.  small amount of O.S. + few crystals of  pour the contents of the test tube to a beaker half full of water
a deep red colour, then add NaOH solution
red colour changes to blue green

  • Phthalein Test
0.5g O.S. + 1.0 g phthalic anhydride + drop of conc.  add  dil. NaOH solution   characteristic colour

ALDEHYDIC GROUP (–CHO)

  • Schiff’s reagent  
5-6 drops or 0.1 g of O.S. + 2c.c. Schiff’s reagent
Shake and wait for 1-2 minutes Deep red or violet colour

  • Tollen’s reagent
5-6 drops or 0.1 g of O.S. + 2 c.c. Tollen’s reagent
warm and allow to stand the contents for 2-3 minutes
Formation of silver mirror or grey ppt.

  • Fehling solution
5-6 drops or 0.1 g of O.S. + 5 c.c. Fehling solution
boilRed ppt.
Red ppt
Note : Aromatic aldehydes do not respond to this test.

  • Benedict’s solution
0.1 g of O.S. + 2-3 c.c. of Benedict’s solution
BoilAppearance of red yellow ppt.

KETONIC GROUP

  • 2, 4–Dinitrophenylhydrazine test
1–2 drops or 0.1g of O.S. + 2c.c. of reagent
Shake vigorously and heat if necessary– A yellow or orange ppt.

  • Sodium bisulphite test
5-6 drops or 0.2 g of O.S. + 1 c.c. of saturated solution of sodium bisulphite-
Shake vigorously - A crystalline white ppt.

  • Sodium Nitroprusside test
5-6 drops or 0.1 g of O.S. + 2 c.c. of 0.5 % aqs sodium nitroprusside solution + 2–3 drops of NaOH–A red or purple colour
Note - Benzophenone does not give this test. First two tests are also given by aldehydes

CARBOXYLIC GROUP (–COOH)

  • Sodium bicarbonate test
Small amount of O.S. + 3 c.c. of saturated solution of sodium bicarbonate
Brisk effervescence

  • Litmus test
Place O.S. on moist blue litmus paper. If it changes to red the O.S. may be an acid
Note- Phenols also give this test

  • Ester test
Small quantity of O.S. + 4-5 drops of  drops of conc.  fruity smell
Ester (Fruity smell)

AMINO GROUP (–NH2, PRIMARY)

  • Carbylamine test
2 drops of O.S. + 2 drops of  of alcoholic caustic potashIntolerable offensive odour
Aromatic or aliphatic  amines give this test
  • Nitrous acid test
0.2 g of O.S. + 10 c.c. of  
Add 10% aqs Brisk effervescence (aliphaticamine)
A yellow oily layer is formed
No visible change

  • Azo-Dye test
This test is given by  aromatic amines.
0.2 g or 0.2 ml of O.S. + 2 ml conc. HCl (cool) + 10% aqs NaNO2(cold) pour into a beaker containing 10% alkaline Bright Orange/red dye

  • Distinguishing test for 1º, 2º and 3º amines (Hinsberg’s Test)
0.5 ml of O.S. + 2 ml of 25% NaOH + 1 ml of benzene sulphonyl chloride. Shake and Cool.
If ppt is obtained add 2 ml of conc. HCl
Precipitate dissolves in conc. HCl 3º amine
Precipitate does not dissolve in conc. HCl 2º amine
Insoluble mass is formed on addition of conc. HCl 1º amine

  • Solubility Test
Small amounts of O.S. + 2-3 ml dil HCl vigorously shake - If soluble, it may be an amine
RNH2 + HCl (RH3) Cl
Amines are basic in nature

  • Litmus Test
Place O.S. on moist red litmus paper. If it is changes to blue the O.S. may be an amine.

CHEMISTRY INVOLVED IN PREPARATION OF INORGANIC COMPOUNDS

FERROUS AMMONIUM SULPHATE (MOHR’S SALT)

Reagents - Iron fillings, ammonium sulphate and conc. sulphuric acid.
Procedure - Take 60 c.c. of distilled water in a conical flask and add to 8 c.c. of conc. sulphuric acid. Gradually add 10 gm of iron filings and heat to boiling. Now add 20 gm of ammonium sulphate and evaporate to one third of its volume. Cool the flask slowly and set aside to crystallise.
Formula -

POTASSIUM ALUMINIUM SULPHATE (POTASH ALUM)

Reagents - Potassium sulphate, Aluminium sulphate crystals
Procedure - 2.9 gm of potassium sulphate is dissolved in 30 c.c. of distilled water. 11.2 gm of aluminium sulphate is dissolved in 30 c.c. of distilled water. Mix the two solutions and leave overnight. Choose the well formed crystal and allow it to grow in the solution. Recrystallise by dissolving in minimum quantity of water.
Formula -

CHEMISTRY INVOLVED IN PREPARATION OF ORGANIC COMPOUNDS

ACETANILIDE

Reagents : Aniline 10 ml
Acetic anhydride 10 ml
Acetic acid 10 ml
Zinc dust 0.05 gm
Procedure : Take 10 ml of aniline in 150 ml of conical flask and add a mixture of 10 ml of acetic anhydride and 10 ml of acetic acid. Add 0.05 gm of Zinc dust. Reflux using water condenser for 15 minutes. Pour the hot mixture in 200 ml of ice cold water with constant stirring. Filter the crystals of acetanilide formed and wash with cold water. Recrystallise from dil. acetic acid (20 ml glacial acetic acid + 40 ml water).
Formula :
White shining flakes mpt 113ºC

IODOFORM

Reagents : Alcohol (5 gm); Sodium Carbonate (10 gm); Iodine (10 gm).
Procedure : 5 gm of ethyl alcohol and 10 gm of sodium carbonate in 50 ml of water are taken in round bottom flask. 10 gm of iodine is added in small fractions and the solution is stirred well and heated on water bath until the iodine has disappeared. Then cool, crystals of iodoform will separate. Wash them with distilled water
Formula :
Bright yellow

ANILINE YELLOW

Reagents:   Aniline 7ml
Conc. HCl 10ml
Sodium nitrite 2.5g
Sodium acetate 10.5g
Procedure: Take 7ml of aniline in 250ml of conical flask and add 10ml of Conc. HCl  followed by 40ml of water and shake. Flask is placed on a water bath containing crushed ice. 2.5g of sodium nitrite dissolved in 10ml of water added gradually and stirred. When sodium nitrite solution has been completely added, keep the solution for 5 minutes in the water bath. Dissolve 10.5g of sodium acetate in 20ml of water and slowly added to the contents in the flask. Keep the flask in water bath for 20min and stirred. A yellow ppt. of aniline yellow is obtained. This yellow solid is washed with water and then dry between folds of filter paper. The crude dye is recrystallise busing light petroleum.

p-NITRO ACETANILIDE

Reagents : Acetanilide 5 g
Glacial acetic acid 5 ml
Fuming 2 ml
Conc. 10 ml
Methylated spirit 20 ml
Procedure : Take 5 g of powdered acetanilide in 100 ml conical flask. Add 5 ml of glacial acetic acid and shake. Take 2 ml of fuming nitric acid in a test tube and cool it in a  freezing mixture. Add to it 2 ml of conc. sulphuric acid drop by drop with constant shaking and cooling. Add 8 ml of conc. drop by drop to the conical flask containing acetanilide and glacial acetic acid and place the flask in a freezing mixture. Add nitrating mixture to this flask drop by drop with constant stirring. Remove the conical flask from the freezing mixture and keep at room temperature for 15-20 minutes. Pour the contents of the flask on crushed ice in a beaker. Stir and filter the crude product. Crystallise the p-nitro acetanilide by dissolving it in methylated spirit and washing with cold water.
Formula :
Colourless crystalline solid mpt

CHEMISTRY INVOLVED IN THE TITRIMETRIC EXERCISES

Volumetric or titrimetric analysis are quantitative analytical techniques which employ a titration in comparing an unknown with a standard. In a titration, a volume of a standardized solution containing a known concentration of reactant ‘A’ is added incrementally to a sample containing an unknown concentration of reactant  ‘B’ till reactant ‘B’ is just consumed (stoichiometric completion). This is known as the equivalence point.
At this point we have N1V1 = N2V2.

Indicator : A compound added to the reacting solutions that undergo an abrupt change in a physical property usually a colour .

Types of indicators
  • Internal - They are added to the reacting solutions
  • External - Electrochemical devices such as pH meters

End point : The point at which a titration is stopped i.e. equivalence point.

Standard solution : The solution of known concentration of an acid, base or salt.
Examples of volumetric titrations

PREPARE A STANDARD SOLUTION M/100 OF OXALIC ACID. USING THIS SOLUTION DETERMINE THE MOLARITY OF THE GIVEN SOLUTION OF KMNO4

Theory - The reactions involved are
In this reaction oxalate ion is behaving as reducing agent and MnO4 ion as oxidising agent.
Oxidation half reaction
2CO2 + 2e
Reduction half reaction
MnO4 + 8H+ + 5e   Mn++ + 4H2O
by using molarity equation we can calculate the molarity of solution.
n1 × M1 × V1  = n2 × M2 × V2
(KMnO4 solution) (oxalic acid)
where n1 = 5 and n2 = 2

Indicator - KMnO4 is self indicator
End Point - Appearance of light pink colour

Observation
Molarity of oxalic acid solution = M/100
Volume of oxalic acid solution = 25 ml
Calculation -  n1 × M1 × V1 = n2 × M2 × V2
    (KMnO)             (oxalic acid)
Molarity of KMnO4 (M1) =

PREPARE A STANDARD SOLUTION OF M/100 FERROUS AMMONIUM SULPHATE. USING THIS SOLUTION DETERMINE THE MOLARITY OF THE GIVEN SOLUTION OF KMNO4

Theory - The reactions involved are
     
The reaction shown above is a redox reaction and hence the titration between ferrous ammonium sulphate and KMnO4 is a redox titration.
In this reaction ferrous sulphate from Mohr’s salt is behaving as reducing agent and KMnO4 is acting as oxidising agent.
Oxidation half reaction -
Fe++Fe+++ + e
Reduction half reaction
MnO4 + 8H+ + 5e Mn++ + 4H2O
by using molarity equation we can calculate the molarity of solution.
n1 × M1 × V1 = n2 × M2 × V2
(KMnO4) (FeSO4)

Indicator - KMnO4 is self indicator
End Point - Appearance of light pink colour

Observation
Molarity of ferrous ammonium sulphate = M/100
Volume of ferrous ammonium sulphate = 25 ml
Calculation :
Molarity of KMnO4 (M1) =

CHEMICAL PRINCIPLES INVOLVED IN SOME EXPERIMENTS

TO DETERMINE THE HEAT OF NEUTRALISATION OF A STRONG ACID WITH A STRONG BASE

APPARATUS
Calorimeter (or a thermos flask)
Thermometer, Glass stirrer,
200 c.c. N NaOH
200 c.c. N HCl

THEORY
The neutralisation reaction of a strong acid with a strong base is essentially the combination of one equivalent of H+ ions with one equivalent of OH- ions
Equal volumes of the acid and alkali of the same strength at the same temperature are mixed in a calorimeter and the rise in temperature is noted. Water equivalent of calorimeter is determined. The heat of neutralisation is calculated.

PROCEDURE
  1. Determination of the water equivalent of calorimeter :
Take a definite weight of cold water in the calorimeter and note its temperature. Add to it a definite amount of hot water of known temperature and note the temperature of the mixture. The water equivalent of calorimeter (W) is determined by the following equation.
(Weight of cold water + water equivalent) (Final temperature of mixture - temperature of cold water)
= (Weight of hot water) (Temperature of hot water - temperature of the mixture)
  1. Take 100 c.c. of N NaOH in the calorimeter and add to it 100 c.c. of N HCl. Mix them well with the stirrer. Note the final temperature of the mixture.

OBSERVATIONS
  • Water equivalent of the calorimeter
Weight of cold water = m1  
Weight of hot water = m2  
Temp. of cold water = t1°   C
Temp. of hot water = t2°   C
Temp. of mixture = t3°   C
Water equivalent W is given by
(m1 + W)(t3 – t1) = m2 (t2 – t3 )
  • Heat of Neutralisation
    Volume of acid = 100 c.c.
Volume of alkali = 100 c.c.
Initial temp. of acid = t° C
Initial temp. of alkali = t° C
Final temp. of the mixture = t1° C
Rise in temperature = (t1– t)ºC
The amount of heat produced on mixing acid and alkali
= (100 +100 + W) (t1 – t) = x calories
Here x is the heat of neutralisation

TO DETERMINE THE HEAT OF SOLUTION OF COPPER SULPHATE

THEORY
The amount of heat evolved or absorbed when one mole of substance (solid/liquid) is dissolved in large quantity of water such that further dilution does not make any heat. Change is known as heat of solution.

APPARATUS
Dewar’s vacuum flask, stirrer, thermometer physical balance, copper sulphate

PROCEDURE
  1. Determination of water equivalent of Dewar’s vacuum flask
Take a definite weight of cold water in a  Dewar’s vacuum flask and note its temperature. Add to it a definite amount of hot water of known  temperature and note the temperature of mixture. Determine the water equivalent by the following equation
(Weight of cold water+Water equivalent) (Final temperature of mixture - temperature of cold water)
= (Weight of hot water) (Temperature of hot water – temperature of mixture)
  1. In second aperture again take a definite weight of cold water in Dewar’s vacuum flask and note its temperature. Add to it definite amount of copper sulphate and stir the solution till the entire copper sulphate dissolves.
    Note down the final temperature of the solution
Initial temperature of water = T1º C
Initial weight of water = W1 g
Final temperature of copper sulphate solution = T2º C
Weight of copper sulphate = W2 g
Water equivalent of Dewar’s flask = W

CALCULATIONS
Total mass of solution = (W1 + W2)g
Rise in temperature = (T2 – T1)º C
Heat gained by flask = W(T2–T1) × 4.18 J
Specific heat of water = 4.18 J/gºC
Heat gained by solution = (W1 + W2)(T2–T1) × 4.18 J
Total heat change during dissolution of copper sulphate
(W2g)
Total heat change during dissolution of 1 mole of copper sulphate will be
Where M is molecular weight of copper sulphate.
Hence heat of solution of copper sulphate

PREPARATION OF LYOPHOBIC AND LYOPHILIC SOLS

Preparation of arsenious sulphide solution
Boil about 1 - 2 g pure  with about 500 ml of distilled water for about 10-12 minutes. Now cool the solution to room temperature and filter undissolved  if any. Dilute this solution and pass into it gas (free from HCl by passing through a tube containing soda lime and then through 3-4 wash bottles containing distilled water) till it is saturated. Then pass a stream of pure hydrogen washed through 3-4 wash bottles containing distilled water to remove the excess of H2S. Filter the solution to remove the coarse precipitate of  if necessary.

Preparation of Gelatin solution
Take about 400 ml of distilled water and heat it to 80-90℃. Add to it about 2g of powdered gelatin and dissolve it. Allow the solution to cool to room temperature. This gives a clear sol. of gelatin.

KINETIC STUDY OF IODIDE IONS WITH H2O2

Theory
In acidic medium, hydrogen peroxide (H2O2) oxidises iodide ions (I) to iodine (I2) which is detected by starch solution.
H2O2+ 2I + 2H+2H2O + I2
A known volume of sodium thiosulphate solution and starch solution is added to the reaction mixture for monitoring of reaction. I2 liberated react with sodium thiosulphate solution and reduced to iodide ions and when thiosulphate ions are totally consumed, the liberated iodine reacts with starch solution and gives blue colour, which detect the presence of iodine.
I2 +  2S2O23 S4O62– + 2IThiosulphate
I2+ starch solution   starch-iodine complex (Blue colour)
The time taken for the blue colour to first appear gives the idea about the rate of reaction.

Chemical required
0.1. M.KI soln., 2M HCl soln. Starch soln. 1M sodium thiosulphate sol. 4V H2O2 Soln.

Procedure
Take four different volume (apx. 25ml to 100ml ) of 0.1M KI soln. to four different flask and 50ml of 2M HCI  is added to each flask. By dilution volume of soln. made upto 200ml and then 5ml starch soln., 2ml of 1M thiosulphate soln. and 25ml of H2O2 soln. is added to each flask.

Observation
Time required for the blue colour to first appear in :
Flask A ................. seconds
Flask B ................. seconds
Flask C ................. seconds
Flask D ................. seconds

Conclusion
It has been observed that time required for the blue colour to appear  is decreases with increase in concentration of Iions. Thus. “The rate of reaction increases with increase in concentration of iodide ions.”

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Best Coaching Institute in Delhi for IIT JEE, NEET, AIIMS, PMT - Clear IIT Medical: Principles Related to Practical Chemistry (Part-1) | Chemistry Notes for IITJEE/NEET
Principles Related to Practical Chemistry (Part-1) | Chemistry Notes for IITJEE/NEET
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