3
EXPERMENTS
1. ESTIMATION OF PHENOL
2. ESTIMATION OF ANILINE
3. ESTIMATION OF KETONE
4. SAPONIFICATION VALUE OF OIL
5. IODINE VALUE OF AN OIL (HANUS METHOS)
6. ESTIMATION OF GLUCOSE
7. PREPARATION OF TRIBROMOBENZENE FROM ANILINE 8. PREPARATION OF m-NITRO BENZOIC ACID FROM METHYL BENZOATE
9. PREPARATION OF 2, 4-DINITROPHENYLHYDRAZINE FROM CHLOROBENZENE
10. PREPARATION OF BENZANILIDE FROM BENZOPHENONE
5
ESTIMATION OF PHENOL
Ex.No:1
Date:
AIM:
To estimate the amount of phenol present in the whole of the given solution.
PRINCIPLE:
Phenol reacts with bromine to give tribromophenol. Phenol is readily brominated by a known excess of acidified bromide - bromate mixture to give 2,4,6-tribromophenol. The excess of unreacted bromide is calculated by titrating it against sodium thiosulphate solution along with KI solution. From the quantity of bromine consumed, the amount of phenol is calculated. For bromination, bromide- bromate mixture is used instead of bromine because the later losses its strength readily due to volatility of bromine.
2KBrO3 + 4KBr + 6HCl → 6KCl + 6H2O + 3Br2
Since one mole of phenol consumes six equivalence of bromine, the equivalent weight of phenol is 1/6 of its molecular weight.
Equivalent mass of phenol = Molecular mass of phenol 6
= 94/6=15.66g
6
Preparation of standard potassium dichromate solution:
Weight of potassium dichromate in 250 ml =
Strength of potassium dichromate =
Titration-I
Standardization of sodium thiosulphate:
Std. K2 Cr2 O7 Vs Na2S2 O3
S.No | Volume of K2 Cr2 O7 (ml) | Burette Reading (ml) | Volume of thiosulphate ( ml) | Concordant value (ml) | |
Initial | Final | ||||
1 | 20 | ||||
2 | 20 | ||||
3 | 20 | ||||
Calculation:
Volume of potassium dichromate (V1) = Strength of potassium dichromate (N1) = Volume of sodium thiosulphate (V2) = Strength of sodium thiosulphate (N2) = V1 N1/V2
7
CHEMICALS REQUIRED:
∙ N/10 Sodium thoisulphate
∙ N/10 Bromide – bromate mixture
∙ 10% KI
∙ Starch
∙ Sample solution of phenol
PROCEDURE:
1. Preparation of Winkler’s solution:
The brominating mixture (winkler’s solution) is prepared by dissolving 15 g of potassium bromide and 3g of KBrO3 in one litre.
2. Standardisation of Sodium thiosulphate solution:
About 1.2 g of AR potassium dichromate is weighed accurately. It is dissolved in water and made upto 250ml. 20ml of the made up solution is pipetted out into a clean conical flask. About 20ml of dil.H2SO4 and followed by 10ml of 10% KI are added to it. The librated iodine is titrated against sodium thiosulphate solution till it becomes straw yellow colour and then added 1 ml of starch as indicator. The end point is the change of colour from blue to green. Titration is repeated for concordant values and the strength of thiosulphate is calculated.
8
Titration-II
Standardization of brominating mixture
Brominating mixture vs Std Na2S2O3; Indicator: Starch
S.No | Volume of brominating mixture (ml) | Burette Reading (ml) | Volume of thiosulphate (ml) | Concordant value (ml) | |
Initial | Final | ||||
1. | 20 | ||||
2. | 20 | ||||
3. | 20 | ||||
Calculation:
Volume of brominating mixture (V3) = 20 ml
Volume of sodium thiosulphate required for
20 ml of Brominating Mixture =
Volume of sodium thiosulphate required for
40 ml of brominating mixture (V4)
3. Standardisation of Bromate-Bromide Mixture:
Exactly 20 ml of bromate-bromide mixture is pipetted out into a clean conical flash. About 10ml of 10% KI solution is added followed by 5ml of con. HCl. The liberated iodine is titrated against standard thio sulphate solution till it becomes straw yellow colour and then added 1 ml of starch as indicator. The end point is the disappearance of blue colour. Titration is repeated for concordant values. From this value the amount of thio sulphate required for the brominating mixture is calculated.Estimation of phenol
Titration-II
Volume of sodium thiosulphate required for
20ml of phenol +40ml brominting mixture after reaction (V5) = Volume of sodium thiosulphate required for 40ml
brominting mixture (V4) = Volume of sodium thiosulphate required for
20ml phenol alone (V4-V5) = Strength of sodium thiosulphate (N2) = Normality of phenol = (V4 - V5) X (N2)/ 20 Amount of phenol present in the whole
of the given solution = Normality of phenol X 15.6/10 Duplicate:
Estimation of phenol:
Titration-III
Volume of sodium thiosulphate required for
20ml of phenol +40ml brominting mixture after reaction V5 = Volume of sodium thiosulphate required for
40ml brominting mixture (V4) = Volume of sodium thiosulphate required for
20ml phenol alone (V4 -V5) = Strength of sodium thiosulphate (N2) = Normality of phenol = (V4 - V5) X (N2) / 20 Amount of phenol present in the
whole of the given solution = Normality of phenol X 15.6/10 =
11
4. Estimation of phenol:
The given phenol solution is made up to 100ml. 20ml of made up phenol solution and 40ml of brominating mixture are pipetted out into a stoppered conical flask and diluted with 25 ml of water, 5 ml of con. 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 10ml of 10% KI solution is added. The liberated iodine is titrated against standard thiosulphate solution till it becomes straw yellow colour and then added 1 ml of starch as indicator.The end point is the disappearance of blue to colourless. The volume of thiosulphate consumed will be equivalent to the excess of bromine. From the titre value, the strength of phenol and hence the amount of phenol are calculated. A duplicate is also conducted.
REPORT:
The amount of phenol present in the whole of the given solution =
ESTIMATION OF ANILINE
Ex: No: 2
Date:
AIM:
To determine the amount of aniline present in the whole of the given solution.
PRINCIPLE:
Aniline reacts with bromine to give tribromoaniline.
Aniline is readily brominated by a known excess of acidified bromide – bromate mixture to give 2,4,6-tribromoaniline; the excess of unreacted bromide liberate I2 from KI and the liberated I2 is titrated against standard thiosulphate solution .From the quantity of bromine consumed, the amount of aniline is calculated . For bromination, bromide -bromate mixture is used instead of bromine because the later loses its strength due to volatility of bromine.
2KBrO3 + 4KBr + 6HCl → 6KCl + 6H2O + 3Br2
Since one mole of aniline consumes six equivalence of bromine, the equivalent weight of aniline is 1/6 of its molecular weight.
Equivalent weight of aniline = Molecular weight of aniline 6
= 93/6 = 15.5g
Preparation of standard potassium dichromate solution: Weight of potassium dichromate in 250 ml =
Strength of potassium dichromate =
Titration-I
Standardization of Sodium thiosulphate:
Std. K2Cr2O7 Vs Na2S2O3
S.No | Volume of K2Cr2O7 (ml) | Burette Reading (ml) | Volume of Na2S2O3 (ml) | Concordant value (ml) | |
Initial | Final | ||||
1 | 20 | ||||
2 | 20 | ||||
3 | 20 | ||||
Calculation:
Volume of potassium dichromate (V1) =
Strength of potassium dichromate (N1) =
Volume of sodium thiosulphate (V2) =
Strength of sodium thiosulphate (N2) = V1 N1/V2 =
15
CHEMICALS REQUIRED:
∙ N/10 Sodium thoisulphate
∙ N/10 Brominating mixture
∙ 10% KI
∙ Starch
∙ Sample solution of aniline
PROCEDURE:
1. Preparation of winkler’s solution:
The brominating mixture (winkler’s solution) is prepared by dissolving 3g of KBrO3 and 15 g of potassium bromide in one litre water.
2. Standardisation of Sodium thiosulphate solution:
About 1.2 g of AR potassium dichromate is weighed accurately. It is dissolved in water and made upto 250ml. 20ml of the made up solution is pipetted out into a clean conical flask. About 20ml of dil. H2SO4 and followed by 10 ml of 10% KI are added to it. The librated iodine is titrated against sodium thiosulphate solution till it becomes straw yellow colour and then added 1 ml of starch as indicator. The end point is the change of colour from blue to green. Titration is repeated for concordant values and the strength of thiosulphate is calculate.
Titration-II
Standardization of Brominating Mixture:
Brominating mixture vs Std Na2S2O3; Indicator: Starch
S.No | Volume of Brominating Mixture (ml) | Burette Reading (ml) | Volume of thiosulphate (ml) | Concordant value (ml) | |
Initial | Final | ||||
1 | 20 | ||||
2 | 20 | ||||
3. | 20 | ||||
Calculation:
Volume of brominating mixture (V3) =
Volume of sodium thiosulphate required
for 20ml brominating mixture =
Volume of sodium thiosulphate required
for 40ml brominating mixture (V4) =
3. Standardization of Bromate-Bromide Mixture:
Exactly 20 ml of bromate-bromide mixture is pipetted out into a clean conical flash. About 10 ml of 10% KI solution is added followed by 5 ml of con. HCl. The liberated iodine is titrated against standard thio sulphate solution till it becomes straw yellow colour and then added 1 ml of starch as indicator. The end point is the disappearance of blue colour. Titration is repeated for concordant values. From this value the amount of thio sulphate required for the brominating mixture is calculated.
18
Estimation of aniline:
Titration-II
Volume of sodium thiosulphate required for20ml
of aniline + 40ml brominating mixture after reaction (V5) = Volume of sodium thiosulphate required for
40ml brominating mixture (V4) = Volume of sodium thiosulphate required for
20 ml Aniline alone ( V4 - V5) = Strength of sodium thiosulphate (N2) = Normality of Aniline = (V4 -V5) X N2/2 0 Amount of Aniline present in the whole of the
given solution =
Duplicate:
Titration-III
Volume of sodium thiosulphate required for 20ml of
aniline + 40ml brominating mixture after reaction (V5) = Volume of sodium thiosulphate required for
40ml brominating mixture (V4) = Volume of sodium thiosulphate required for
20ml aniline alone (V4 - V5) = Strength of sodium thiosulphate (N2) = Normality of aniline = (V4 - V5) X N2 / 2 0
Amount of aniline present in the whole
of thegiven solution = St. of phenol X 15.6/10 =
19
4. Estimation of Aniline:
The given aniline solution is made up to 100ml. 20ml of made up aniline solution and 40ml of brominating mixture are pipetted out into an iodine flask and diluted with 25 ml of water, 5 ml of con. HCl is added and the flask is shaken for a minute to mix the reactant. It is allowed to stand for 30 minutes with occasional shaking of the contents of the flask. Flask is cooled under tap water and 10ml of 10% KI solution is added. The end point is the disappearance of blue to dirty white precipitate. The volume of thiosulphate cosumed will be equivalent to the excess of bromine. From the titre value, the strength of aniline and hence the amount of aniline are calculated. A duplicate also conducted.
REPORT:
The amount of aniline present in the whole of the given solution =
20
Preparation of standard potassium dichromate solution:
Weight of potassium dichromate in 250 ml =
Strength of potassium dichromate =
Titration-I
Standardisation of Sodium thiosulphate:
Std. K2 Cr2 O7 Vs Na2S2 O3
S.No | Volume of K2 Cr2 O7 (ml) | Burette Reading (ml) | Volume of Na2 S2 O3 (ml) | Concordant value (ml) | |
Initial | Final | ||||
1 | 20 | ||||
2 | 20 | ||||
3. | 20 | ||||
CALCULATION :
Volume of potassium dichromate (V1) =
Strength of potassium dichromate (N1) =
Volume of sodium thiosulphate (V2) =
Strength of sodium thiosulphate (N2) = V1 N1/V2
21
ESTIMATION OF KETONE
EX. No: 3
Date:
AIM:
To estimate the amount of ketone present in the whole of the given solution.
PRINCIPLE:
Acetone or any ketone containing group forms iodoform when shaken well with the mixture of iodine and aqueous KOH solution according to the following equation.
I2 + 2 KOH → KI + KOI + H2O
CH3-CO - CH3 + 3KOI → CH3-CO- CI3 + 3KOH
Triiodo dimethyl ketone
CH3-CO- CH3 + KOH → CHI3 + CH3-COOK
Iodoform
3 molecules of Iodine = 1 molecule of dimethyl ketone 3 molecules of Iodine = 58 g of dimethyl ketone
6 litres of 1 N Iodine = 58 g of dimetyl ketone
1 ml of 1 N Iodine = 58/6000
= 0.0097 g of dimethyl ketone
The excess of iodine decomposed by the following equation and the liberated iodine is titrated against thiosulphate using starch as indicator.
KI + KOI + H2SO4 → K2SO4 + I2 + H2O
(excess)
Na2S2O3 + I2 → 2NaI + Na2S4O6
22
TITRATION-II:
Standardisation of Iodine solution
Iodine solution vs std Na2S2 O3; Indicator: Starch
S.No | Volume of Iodine solution (ml) | Burette Reading (ml) | Volume of thiosulphate (ml) | Concordant value (ml) | |
Initial | Final | ||||
1 | 20 | ||||
2 | 20 | ||||
3. | 20 | ||||
CALCULATION:
Volume of Iodine solution (V3) = 20 ml
Volume of sodium thiosulphate required for
20 ml of Iodine solution =
Volume of sodium thiosulphate required for
40 ml of Iodine solution (V4) =
23
CHEMICALS REQUIRED:
* AR K2Cr2O7 - 1.2 g in 250 ml (0.lN)
* Sodium thiosulphate - 6 g in 250 ml (N/10)
* Potassium iodide - 10 g in 100 ml (10%)
* Concentrated HCl - 20ml
* Starch solution - 10 ml freshly prerpared
* Aqueous iodine - 2.5 g in 200 ml
* KOH - 11 g in 250 ml (lN)
*Acetone - 5 g in one litre water (about N/2)
PROCEDURE
1. Sandardisation of Thiosulphate:
A standard solution of N/10 K2Cr2O7 is prepared by weighing about 1.2g of AR crystals accurately in a chemical balance, dissolved in distilled water and made upto 250 ml in a standard flask. 20 ml of standard K2Cr2O7 solution is pipetted out to clean conical flask, about 5 ml of con. HCl is added followed by 10 ml of 10% KI solution. The liberated iodine is immediately titrated against sodium thiosulphate taken in the burette. When the solution acquires straw yellow colour, 1 ml of starch is added and titration is continued (in drops) with constant shaking. The end point is the change in colour from blue to green. Titrations are repeated for concordant value. From the end point, the strength of thiosulphate is calculated.
24
Titration-II
Volume of thiosulphate required by (ketone + I2) solution (A ml) = Blank titre value (20ml I2+30mlKOH) (Bml) = B ml of thiosulphate is required for 50 ml of iodine solution = 20X A / B
=… x ml of iodine
Therefore, reacted iodine = 20 – unreacted iodine =
1 ml of 1N iodine solution = 0.0097g of dimethyl ketone (50 – x) ml of ----------- iodine = 0.0097 X (50–x) X strength iodine
=
Amount of ketone present in the whole
of the given solution = x0.0097/10 = Z g
Duplicate:
Volume of thiosulphate required by (ketone + I2) solution (A ml) = Blank titre value (20 ml I2+ 30ml KOH) (B ml) = B ml of thiosulphate is required for 50 ml of iodine solution A ml of thiosulphate is required for
Unreacted iodine = 50X A / B = x ml of iodine Therefore, reacted iodine = 50 – unreacted iodine 1 ml of1N iodine solution = 0.0097g of dimethyl ketone (50 – x) ml of ______ iodine = 0.0097 X (50–x) X strength of iodine Amount of ketone present in the
whole of the given solution =…..x 0.0097/10
25
2. Standardisation of Iodine:
Exactly 20 ml of iodine solution pipetted out into a conical flask, about 1 ml of starch is added and titrated against thio solution, end point is the disappearance of blue colour. The titration is repeated for concordant titer value.
3. Estimation of Acetone:
The given ketone is made upto 100ml in a standard flask. 20 ml of made up solution is pipetted out into clean conical flask. 30 ml of I N KOH is added with constant shaking. 40 ml of iodine solution is added into the conical flask and the mixture is shaken well for about 15 minutes. It is then acidified with about 20 ml of con HCl and the liberated iodine is titrated against sodium thiosulphate using starch as indicator. The end point is disappearance of blue colour.
A blank titration is carried out without ketone using the same volume (30 ml) of KOH and (50 ml) iodine against sodium thiosulphate solution using starch as indicator.
REPORT:
The amount of acetone present in the whole of
the given solution =
26
Titration I
Standardisation of HCl
Std Na2CO3 vs HCl; Indicator: methyl orange
S.No | Volume of sodium carbonate (ml) | Burette Reading (ml) | Volume of HCl (ml) | Concordant value (ml) | |
Initial | Final | ||||
1 | 25 | ||||
2 | 25 | ||||
3. | 25 | ||||
Titration II
Determination of Saponification value of an oil.
KOH + oil vs Std. HCl; Indicator: phenolphthaelin
S.No | Volume of KOH + Oil (ml) | Burette Reading (ml) | Volume of HCl (ml) | Concordant value (ml) | |
Initial | Final | ||||
1 | 25 | ||||
2 | 25 | ||||
3. | 25 | ||||
27
SAPONIFICATION VALUE OF OIL
EX. No: 4
Date:
Aim :
To determine the saponification value of an oil.
Principle:
The Saponification value is defined as the number of milligrams of KOH required for the hydrolysis of 1g of an oil or fat. To determine the Saponification value, a weighed quantity of the given oil is refluxed with a known volume of alcoholic potassium hydroxide solution. The unused alkali is then back-titrated against standard acid.
Saponification value of some common oils are given below: (1)Mustard oil = 174 (5) Linseed oil = 188-195 (2)Castor oil = 175-183 (6) Gingelly oil = 188-193 (3)Olive oil = 185-196 (7) Cotton seed oil = 194-196 (4)Groundnut oil = 186-194 (8) Coconut oil = 253-262
Chemicals required
AR sodium carbonate crystals = 7g/250 ml water Alcoholic potash(KOH) = 6g dissolved in 200ml rec.spirit (0.5N) Phenolphthalein = 5ml
HCl (about N/2) = 10ml conc. HCl in 200ml water
28
Titration III
Blank Titration .
KOH vs Std. HCl; Indicator: phenolphthaelin
S.No | Volume of KOH (ml) | Burette Reading (ml) | Volume of HCl (ml) | Concordant value (ml) | |
Initial | Final | ||||
1 | 25 | ||||
2 | 25 | ||||
3. | 25 | ||||
Calculation
Calculate the strength of HCl solution (Titration 1)
Let w g be the weight of oil taken.
Let A ml be the volume of HCl required for excess alcoholic potash Let B ml be the blank titre value.
1000 ml 1N HCl = 56.1 g of KOH
1ml,of 1N HCl = 56.1 mg of KOH
Saponification value = 56.1 X (B-A) X strength Of HCl /w
29
Procedure
1. Standardisation of HCl
About 7g of analar sodium carbonate crystals (eq. wt = 53) are weighed accurately in a chemical balance, dissolved in water and the solution in made upto 250ml in a standard flask. 20ml is pipette out into a claen conical flask. About 2 drops methyl orange indicator are added and titrated against hydrochloric acid taken in the burette (till the light pink colour appears). The titration is repeated to get concordant value. The strength of HCl solution. is calculated.
2. Determination of Saponification value of an oil.
About 1g of the given oil is weighed accurately in a chemical balance and transferred into 250ml round-bottom flak. 25ml of alcoholic potash (about N/2) is added from the burette slowly into the flask. It is then fitted with an air-condenser and heated on a water-bath for about 30 minutes. A
blank is run simultaneously with the same quantity of alcoholic potash, but without the oil. Both the flask are cooled and titrated against standard HCl solution using about 1ml of phenolphthalein indicator. The end point is the disappearance of pink colour.
Report:
The Saponification value of given oil is =
30
31
IODINE VALUE OF AN OIL (HANUS METHOS)
EX. No: 5
Date:
Aim :
To determine the iodine value of an oil.
Principle
Iodine value is defined as the number of parts by weight of iodine reacting with 100 parts by weight of an oil or fat. The drying power of an oil is generally proportional to its iodine value. Coconut oil ( a non-drying oil) has low iodine value of 6-10; linseed oil (a drying oil) has high iodine of 77- 88 value. Iodine value indicates the degree of unsaturation of the fatty acids present in the oil or fat. The determination of the iodine value is great importance in characterizing an oil and also in finding the proportion of an adulterant in a sample of the oil.
In Hanus mothod of determination of the iodine value, a known weight of oil is dissolved in CCl4 and treated with a known excess volume of iodine monobromide solution. The unused IBr is back titrated against standard sodium thiosulphate solution.
Iodine values of some common oils are given below:
(1)Coconut oil = 6-10 (5) Cotton seed oil = 103-111. (2)Olive oil = 79-88 (6) Gingelly oil = 103-117. (3)Castor oil = 84. (7) Mustard oil = 105-110. (4)Groundnut oil = 83-100. (8) Linseed oil = 175-202.
32
Titration-I
Standardization of sodium thio sulphate:
Std. K2 Cr2 O7 Vs Na2S2 O3
S.No | Volume of K2 Cr2 O7 (ml) | Burette Reading (ml) | Volume of thiosulphate ( ml) | Concordant value (ml) | |
Initial | Final | ||||
1 | 20 | ||||
2 | 20 | ||||
3 | 20 | ||||
Calculation:
Volume of potassium dichromate (V1) =
Strength of potassium dichromate (N1) =
Volume of sodium thiosulphate (V2) =
Strength of sodium thiosulphate (N2) = V1 N1/V2
33
Chemical required
AR Potassium dichromate = 1.2 g/250 ml water. Sodium thiosulphate = 6g/250 ml water (about N/10). Potassium iodine = 10g/100ml.water.
Starch solution (freshly prepared) = 10ml.
IBr = Iodine (13.2 g/1 litre HOAc) + 3 ml. liquid bromine. (100ml.required for each student)
Carbon tetrachloride = 20ml.
Procedure
Titration 1: Standardisation of Sodium thiosulphate.
A standard solution of N/10 K2Cr2O7 is prepared by weighing about 1.2 g of analar crystals (eq.wt. = 49) accurately in a chemical balance, dissolving in water and making upto 250 ml in a standard flask. 20ml of standard K2Cr2O7 solution is pipette out into a clean 250ml conical flask. About 5ml con. HCl is added, followed by 10ml of 10% aq. KI solution. The liberated iodine is immediately titrated against thiosulphate solution taken in the burette. When the solution acquires as straw- yellow colour, 1ml of freshly prepared starch is added and the titration is continued (in drops) with constant shaking. The end point is the change in colour from blue to green (due t0 Cr3+). The titration is repeated to get concordant values. The strength of thiosulphate solution is calculated.
Cr2O72-+ 14 H++ 6 I- → 2 Cr3+ + 7 H2O + 3 I2
2S2O32-+ I2 → 2I-+ S4O62-
34
Calculation
Let A ml be the volume of Na2S2O3 required for the excess IBr solution. Let B ml be the blank titre value.
Iodine value = (B-A) X 127/1000 X 100/w X strength of thiosulphate. Where w is the weight of oil taken, and 127 is the atomic weight of iodine.
35
TITRATION 2:
Determination of Iodine value.
About 0.5 to 1g of the given oil is weighed accurately in a chemical balance and dissolved in 10ml of CCl4 in a 500ml iodine flask and 25ml of iodine monobromide solution is added to the flask by using a burette and the time is noted. The resulting mixture, if turbid, is cleared by adding a small addition known volume of CCl4. The stopper is moistened with a few drops of aq. KI solution and then inserted into the bottle. The bottle is kept aside for about 40-60 minutes, with occasional shaking. Then, the reaction mixture is diluted with 200ml of water, followed by the addition of 20ml of 10% aq. KI solution. The mixture is titrated against sodium thiosulphate taken in the burette nearing the end point starch is added as indicator. The end point is disappearance of blue colour. A duplicate is also conducted.
A blank titration is carried out without oil, using exactly the same quantity of CCl4 (10ml plus any additional amount used) and 25ml of iodine monobromide solution using starch as indicator.
Report:
The iodine value of given oil =
36
Preparation of standard oxalic acid solution Weight of oxalic acid in 250 ml =
Strength of oxalic acid =
37
ESTIMATION OF GLUCOSE
Ex.No:6
Date:
AIM:
To estimate the amount of glucose present in the whole of the given solution
PRINCIPLE:
Glucose is a reducing sugar, when glucose is boiled with an excess of alkaline Copper hydroxide it is oxidized to gluconic acid.
The precipitate is red colour cuprous oxide dissolved in a warm acidic solution of ferric alum, Cu2O reduces ferric sulphate (in ferric alum) to ferrous sulphate, and the reduced ferrous sulphate is titrated against KMnO4.
Cu2O + Fe2 (SO4) 3 + H2SO4 → 2 CuSO4 + 2FeSO4 + H2O 10Fe SO4 +8 H2SO4 +2KMnO4 → K2SO4 + 2MnSO4 + 5 Fe2 (SO4)3 + 8H2O 1000 ml of 1N KMnO4 = 63.6g of Cu
1 ml of N/10 KMnO4 = 6.36 g of Cu
The amount of Cu equivalent of sugar is calculated from the conversion table.
1 mg of Cu = 0.5441 mg of glucose
=
38
Titration – I
Standardisation of KMnO4
Std.Oxalic acid vs KMNO4
S.No | Volume of Oxalic acid (ml) | Burette Reading (ml) | Volume of KMnO4 (ml) | Concordant value (ml) | |
Initial | Final | ||||
1 | 20 | ||||
2 | 20 | ||||
3. | 20 | ||||
Calculation:
Volume of Oxalic acid (V1) =
Strength of Oxalic acid (N1) =
Volume of KMnO4 (V5) =
Strength of KMnO4 (N2) = V1 N1/V2
1 ml of 0.1N KMnO4 contains 6.36g of Cu
Therefore, 1ml of 0. N KMnO4 contains = 6.36 X / 0.1 =
39
CHEMICALS REQURED:
o Analar oxalic acid = 1.6 gin 250 ml
o KMnO4 = 0.1N
o Felhings “A” solution
o Felhings “A” solution
o Acidified Ferric alum solution = 24 g dissolved in 20 ml dil.H2SO4 and diluted to 200
ml.
o Dilute sulphuric acid
PROCEDURE:
1. Standardisation of KMnO4:
A standard solution of N/10 oxalic acid is prepared by weighing about 1.6 g of AR oxalic acid and made upto 250 ml in a standard flask. 20 ml of standard oxlic acid is pipetted out in a conical flask, about 20 ml of con. Sulphuric acid is added, then the solution is heated to 60˚C, and the hot solution is titrated against KMnO4 taken in a burette. The end point is the appearance of pale permanent pink colour. Titrations are repeated for concordant value. From the end point, the strength of KMnO4 is calculated.
40
Titration –II
Volume of glucose solution = 20.0ml
Volume of KMnO4 =
1 ml of 0.1 N KMnO4 contains = 6.354mg of Cu 1 ml of 0. KMnO4 contains =
ml of standard KMnO4 =
1 mg of Cu contains =
__ mg of Cu contains =
Amount of glucose present in the whole of the given solution =
Titration –II (Duplicate)
Volume of glucose solution = 20.0ml
Volume of KMnO4 =
1 ml of 0.1 N KMnO4 contains = 6.354mg of Cu 1 ml of 0. KMnO4 contains =
ml of standard KMnO4 =
1 mg of Cu contains =
____ mg of Cu contains =
Amount of glucose present in the whole of the given solution =
41
2. Estimation of glucose:
The given glucose solution is made upto 100 ml. 20 ml of the made up solution is pipetted out into clean conical flask. About 20ml of Felhings ‘A’ solution and 20 ml of Felhing ‘B’ solution are added. Then the solution is heated to boiling for 2 to 3 minutes. The red cuprous oxide precipitate is formed and allowed to settle. The precipitate is filtered using filter paper and the precipitate is washed with water to remove the excess of copper solution.
The filter paper along with precipitate is transferred into a clean conical flask and acidified ferric alum is added slowly till the red precipitate is dissolved. About 20 ml of dil H2SO4 is added and titrated against KMnO4. The end point is the appearance of pale permanent pink colour. A duplicate titration is also carried out.
From the titre value, the amount of copper and hence the amount of glucose can be determined.
REPORT:
The amount of glucose present in the whole of the given solution =
42
43
PREPARATION OF TRIBROMOBENZENE
AIM:
To prepare tribromobenzene from tribromoaniline
PROCEDURE:
STAGE I: Preparation of 2, 4, 6-tribromoaniline from aniline
CHEMICALS REQUIRED:
∙ Aniline _ 3ml
∙ Bromine _ 4ml
∙ Glacial acetic acid _ 15ml
∙ Aqueous Ethanol (1: 1) _ 10ml
3 ml Aniline is dissolved in 5 ml of glacial acetic acid in a conical flask fitted with a cork. A solution of Br2 in glacial acetic acid, (4 ml Bromine 10 ml glacial acetic acid) is taken in the boiling tube using a dropper; about 1 ml of bromine in acetic acid is added at a time to aniline in the conical flask and the mixture is shaken well. The addition of bromine is continued till the pink colour of the solution in the flask becomes light yellow. The flask is corked well and shaken vigorously for about 10-15 minutes. About 50 ml of cold water is then added to the conical flask and the mixture is shaken vigorously for 2 minutes. Colourless crystals of tribromoaniline is separated and filtered. It is washed with water, dried and weighed. About 1 g of the crude sample is recrystallised from aqueous alcohol and the melting point is determined.
STAGE- 11: Preparation of tribromobenzene from tribromoaniline: About 5g of 2, 4, 6 tribromoaniline is dissolved with 10 ml of rectified sprit and 7 ml of benzene are mixed in a round bottom flask and it is fitted with reflux condenser and heated on a water bath till the solution starts boiling. About 2 g of NaNO2 is added through the top of the condenser and mixture is shaken vigorously and it is then heated (75 minutes) on a boiling water bath till no more gas is evolved. The solution is then cooled in ice, mixture of tribromobenzene and sodium sulphate crystallized out. Filter with suction on a Buckner funnel. The crystal is washed with the small quantity of alcohol and then repeatedly with water to remove all the Na2SO4. About 1 g of the sample is recrystallised from hot rectified spirit and melting point of the recrystallised sample is noted.
REPORT:
The yield of 2,4,6-tribromoaniline =
44
45
PREPARATION OF m-NITRO BENZOIC ACID FROM METHYL BENZOATE
AIM:
To prepare m- nitrobenzoic acid from methyl benzoate
CHEMICAL REQUIRED:
∙ Methyl benzoate = 5ml
∙ Con H2SO4 = 20 ml
∙ Nitrating mixture and Ethanol
PROCEDURE:
STAGE: 1 Preparation of m- nitromethylbenzoate:
10 ml of Con H2SO4 and 5 ml of methyl benzoate are taken in a conical flask and cooled to 0◦ C. In another beaker, 7ml of Con H2SO4 and 7ml of Con HNO3 are taken and cooled to 00 C. The nitrating mixture is added to the conical flask with vigorous stirring. The addition is carefully done. The mixture is always kept with in the range of 5-100 C. The mixture is stirred well for about 15 minutes and poured into ice. The crude m nitromethylbenzoate is separated and filtered and washed with water. The precipitate is then taken in a conical flask and agitated with about 50 ml of ethanol in order to remove small amount of o-nitromethylbenzoate, ester and other impurities. The cooled mixture is filtered, washed with ethanol and the solid is dried. It is recrystallised from ethanol.
Result:The yield of m- nitromethylbenzoate =
STAGE: 2 Preparation of of m- nitrobenzoic acid:
In a 250 ml round bottom flask fitted with a reflux condenser 9g of m nitro methyl benzoate and 5g of NaOH in 20 ml of water is taken. The mixture is heated to boiling for 30 minutes then the reaction mixture is diluted with equal volume of water and poured into ice. To this mixture, 2 ml of Con. HCl is added so that m-nitrobenzoic acid is precipitated. The crude solid is filtered, washed and dried. It is recrystallised from 1:1 HCl and melting point is determined.
REPORT:
The yield of m-nitrobenzoic acid =
46
47
PREPARATION OF 2, 4-DINITROPHENYLHYDRAZINE FROM CHLOROBENZENE
AIM:
To prepare 2, 4-dinitrophenylhydrazine from chlorobenzene
CHEMICAL PRQUIRED:
▪ Chlorobenzene
▪ Con HNO3
▪ Conc.H2SO4
▪ Aqueous hydrazine
▪ Ethanol
PROCEDURE:
STAGE: 1
Preparation of 2, 4-dinitrochlorobenzoic acid from chlorobenzene
In 250 ml round bottom flask fitted with condenser, 5 ml of chlorobenzene and 8 ml of con.H2SO4 are taken. The mixture is heated on a water bath and to this mixture, the nitrating mixture (10ml of con. HNO3 and 7 ml of con H2SO4) is added in small portions. The mixture is stirred during addition and temperature is maintained below 100C, after the completion of addition, the mixture is heated on a water bath for 2 hours. After cooling the contents of the flask it is poured on crushed ice taken in a 500ml beaker. The solid is recrystalised from alcohol.
STAGE: 2
Preparation of 2, 4-dinitrophenylhydrazine:
About 5g of pure 2, 4-dinitrochlorobenzene is dissolved in 10 ml of ethylene glycol in a 100 ml round bottom flask. The solution is warmed to get the clear solution. The flask is cooled on ice bath to 10◦ C, 1-4 ml of 64%aqueous hydrazine solution is added drop wise with constant stirring, care must be taken during the addition and the temperature is not raised above 15◦ C. When the addition is completed, 5 ml of methanol is added and the flask is heated on a water bath for 20 minutes. The yield is cooled, washed with ethanol. It is filtered and dried. The recrystallisation is done using ethanol.
REPORT:
The yield of 2, 4-dinitrophenylhydrazine =
48
49
PREPARATION OF BENZANILIDE FROM BENZOPHENONE
AIM:
To prepare Benzanilide from Benzophenone.
CHEMICALS REQUIRED:
▪ Benzophenone - 5 ml
▪ Hydroxylamine hydrochloride - 3 g
▪ Rectified spirit - 10 ml
▪ NaOH - 6 g
PROCEDURE:
STAGE I: Preparation of Benzoxine from Benzophenone: A mixture of 5 ml benzophenone and 3 g of hydroxyl amine hydrochloride is placed in a 100 ml of round bottom flask. 10 ml of rectified spirit and 10 ml of distilled H2O are added; 6 g of sodium hydroxide pellets is added to the same flask in small portions with constant stirring. The reaction becomes vigorous. The round bottom flask is cooled in running water after all the addition of sodium hydroxide is over. The above content is refluxed for 15 minutes. The round bottom flask is then cooled and the contents are poured into a beaker which already contains 5 ml of hydrochloric acid in 100 ml of H2O. The oximes get precipitated. It is then filtered and washed with cold water. It is recrystallized from methanol.
STAGE II: Preparation Benzanilide from Benzoxime:
CHEMICALS REQUIRED:
Benzoxime - 3 g
▪ Ether - 20 ml
▪ PCl5 (or) SOCl2 - 3 g or ml
▪ Ethanol - 10 ml
PROCEDURE:
3 g of benzoxime is dissolved in 20 ml of anhydrous ether in a conical flask. 3 g of powered thionyl chloride is added to the flask. Shake the conical flask for thorough mixing. The excess of solvent ether is distilled by keeping it in hot water bath. 20 ml of water is added to the same conical flask and it is boiled for 3 minutes. The benzanilide obtained is filtered and dried. It is recrystallized from ethanol.
REPORT:
The yield of benzanilide =
50
51
PREPARATION OF PHTHALIDE FROM PHTHALIC ANHYDRIDE
AIM:
To prepare phthalide from phthalic anhydride
CHEMICALS REQUIRED:
Phthalic anhydride - 24.75g, Urea - 5g, Rectified spirit - 10ml PROCEDURE:
STAGE I: Preparation of Phthalide from Phthalic anhydride: Initimately mix 24.75g of pure phthalic anhydride and 5g of urea and place the mixture in a 1-litre long necked round bottomed flask . Heat the flask in an oil bath at 130-135oc. When the contents have melted, effervescence commence and gradually increases in vigrour. After 10-20 minutes the mixture suddenly froths up to about three times in the original volume (raise in temp 150-160o) and becomes almost solid. Remove flame from the beneath of the bath and cool. Add about 20 ml of water to disintegrate the solid in the flask. Filter in the pump, wash with little water and then dry at 100o c .the yield is phthalimide and melting point is 233. It is recrystalized from rectified sprit.
STAGE II: Preparation Phthalic acid from Phthalide:
CHEMICALS REQUIRED:
Phtahlide - 3g, Zinc Powder - 24.75g, NaOH (20%) - 20 ml PROCEDURE:
In a 1 litre three necked flask stirred 24.75g of high quality zinc powder to a thick paste with a solution of 0.125g of crystallised copper (II) sulphate in 20ml of water and then add 41ml of 20% NaOH solution. Cool the flask in an ice bath to 5oc stir the contents mechanically and add 18.4 g 0f phthalamide in small portions at such a rate that the temperature does not rise above 8oc (about 30 minutes are required) for the addition . Continue the stirring for half an hour dilute with 50ml of water, warm on a water bath until the evolution of ammonia ceases (about 3 hrs) and concentrate to a volume of about 50ml by distillation under reduced pressure. Filter cool in ice and render the filterate acid to congored paper with concentrated HCl much of the phthalide separates as an oil , but in order to complete the lactonisation of hydroxymethylbenzoic acid , boil for an hour , transfer while hot to the beaker. The oil solidifies on cooling to a hard red brown cake. Recrystalise from water.
REPORT:
The yield of Phthalic acid =