Vnsgu Sf Sem 2 Organic Practical

ORGANIC PREPARATION SEM-II

Preparation Of β-Resorcylic Acid From Resorcinol

Aim:- Organic Preparation Of β-Resorcylic Acid

Requirement:-Resorcinol, K2CO3, dil. HCl, NaHCO3, Conc. HCl, H2O, Reflux Condenser,Gas Inlet Tube

Procedure:-

Place A Solution Containing 10 gms of Resorcinol, 50 gms of NaHCO3 & 100 ml of H2O ina 250 ml R.B.F fitted with a reflux condenser & gas inlet tube heat it on a water bath forabout one hour then reflux vigorously over a flam on wire gauge for 30 min. passing arapid stream of CO2 through the soln.[about 15 to 20 min]

Acidify the soln. while still hot by adding conc. HCl. Allow to cool to room temp. chill in anice bath and collect the crude β-Resorcylic Acid by filtration.

Recrystallize the product by boiling water in present of a little decolorizing carbon filterand cool in an ice salt mixture with stirring collect and dry the pure β-Resorcylic Acid

Theoretical Yield=14.0gms

M.P.:-216-217 oC

Reaction:-

OH

OH

COOH

OH

OH

NaHCO3

-Resorcylic Acid

Resorcinol

Calculation:-

110 gms Resorcinol gives 154 gms β-Resorcylic Acid

10 gms Resorcinol gives (?) gms β-Resorcylic Acid

= = 14.0 gms β-Resorcylic Acid

Theoretical Yield=14.0gms

Practical Yield = _________ gms

% Yield =


Preparation Of Resacetophenone From Resorcinol

Aim:- Organic Preparation Of Resacetophenone

Requirement:-Resorcinol, Anhydrous ZnCl2, Glacial HAc, 50% HCl

Procedure:-

Take 10 gms resorcinol & heat the mixture on a sand bath until it just begins to boil. Nowallow the reaction to complete itself on sand bath without further heating for half hour.

Cool the reaction mixture and dilute it with 50% HCl[about 2 to 3 T.T] keep it for sometimes & shake it well Resacetophenone slowly separate from the solution in orange needfilters it & wash water & dry it. Recrystallized it from the boiling water.

Theoretical Yield=13.8 gms

M.P.:-144-147 oC

Reaction:-

OH

OH

OH

OH

COCH3

CH3COOH

ZnCl2-H2O

Calculation:-

110 gms Resorcinol gives 152 gms Resacetophenone

10 gms Resorcinol gives (?) gms Resacetophenone

= = 13.8 gms Resacetophenone



% Yield =


Preparation Of Benzimidazole From o-phenylene diamine

Aim:- Organic Preparation Of Benzimidazole

Requirement:- o-phenylene diamine, 90% Formic acid, 10% NaOH, Ice

Procedure:-

Take 10gms of o-phenylene diamine in 250 ml R.B.F & 6 ml of 90% Formic acid. Heat themixture on a water bath at 100 oC for 2 hours cool the flask & then add 10% NaOH soln.slowly with constant rotation of the flask until the mixture is just alkaline [check on litmuspaper] filter off the crude Benzimidazole at the pump & wash with ice cold water forrecrystallization, dissolve the crude product in about 150 ml of water boil it & add 75 gmsof decolorizing carbon & digest for 15 minutes. Filter rapidly at the pump cool the filtrateto about 10 oC filter off the Benzimidazole wash with cold water & dry at 100 oC.


M.P.:-171-172 oC

Reaction:-

H-COOH

NH2

NH2

2 Hrs

-2H2O N

CH

HN

O-phenylene diamine Formic acid Benzimidazole

Calculation:-

108 gms O-phenylene diamine gives 118 gms Benzimidazole

10 gms O-phenylene diamine (?) gms Benzimidazole

= = 10.93 gms Benzimidazole



% Yield =


Preparation Of 4-Methyl-7-Hydroxy Coumarin From Resorcinol

Aim:- Organic Preparation Of 4-Methyl-7-Hydroxy Coumarin

Requirement:- 15 ml Conc. H2SO4, 3.7 gms Resorcinol, 5 ml Ethyl Acetoacetate, NaOH &HCl

Procedure:-

15 ml Conc. H2SO4 in a wide necked 50 ml flask cool it. Externally till the temp. of the acidis about 5 oC mean while prepare a solution of Resorcinol [3.7 gms to 4.0 gms] in 4.4 ml ofEthyl acetoacetate.

Now add the solution of Resorcinol in Ethyl Acetoacetate to Conc H2SO4 slowly. So that thetemp of mixture does not rise about 10 oC continue the stirring for 30 minutes. Pour themixture into crushed ice when the solid separates. Filter of the Coumarin at the pump. .Recrystallized it from Alcohol


M.P.:-188-190 oC

Reaction:-

CH2

C

C

C2H5O O

H3C OOH

OH

Conc. H2SO4

OHO

CH3

O

Resorcinol Ethyl Acetoacetate 4-Methyl-7-Hydroxy Coumarin

Calculation:-

110 gms Resorcinol gives 176 gms 4-Methyl-7-Hydroxy Coumarin

4.0 gms Resorcinol (?) gms 4-Methyl-7-Hydroxy Coumarin

=.

= 10.93 gms 4-Methyl-7-Hydroxy Coumarin


Practical Yield = __________ gms

% Yield =


Preparation Of Methyl Orange From Sulphanilic Acid

Aim:- Organic Preparation Of Methyl Orange

Requirement:- Sulphanilic Acid, Anhydrous Na2CO3, NaNO3, NaCl, Dimethyl Aniline,20% NaOH, ice etc.

Procedure:-

in a 250ml beaker take 5.0 gms of Sulphanilic acid, 1.5 gm of Anhydrous Na2CO3 & 50 mlof water and warm until a clear solution is obtained cool the soln. under the tap to about 15oC & then place it in a ice bath. Now add this cold soln. of NaNO3 into a cold solution ofSulphanilic Acid. Slowly with constant stirring.

Now pour this resulting solution slowly & with stirring into a 500 ml beaker containing 5ml of Conc. HCl & 30 gms Crushed ice.

Stir well this mixture for 10 min &maintain the temp. Below 10 oC using the ice bath.Diazobenzene sulphonate will soon separate out. Do not filter this off as they will dissolveduring the next stage of preparation.

Now dissolve 3.0 ml of Dimethyl Aniline into cold soln. of diazotized Sulphanilic acid withconstant stirring. Allow the mixt. To stand for 10 min the red or acid form of methylOrange will gradually separated.

After 10 min add slowly & with stirring 25 ml of 20% NaOH soln. the mixture will assumea uniform Orange color due to separation of the sodium salt of Methyl Orange in fineparticles.[add NaOH till Alkaline]

Now heat the contents almost to boiling stage & add 5.0gms of NaCl & again warm it at 80-90 oC. Until the salt has dissolved completely.

Allow the mixture to cool undisturbed for 15 min & then cool it in ice bath. Filter theproduct wash with water & recrystallized it from hot water.


M.P.:-Not well defined

Reaction:-

NH2

SO3H

Diazotization

NaNO2+HCl

0-5 oC

N=N

SO3H

N(CH3)2

coupling Rean.

Dimethyl aniline

Methyl Orange

SO3Na

N=N N(CH3)2


Calculation:-

173 gms Sulphanilic Acid gives 327 gms Methyl Orange

5.0 gms Sulphanilic Acid (?) gms Methyl Orange

=.

= 9.45 gms Methyl Orange



% Yield =

Preparation Of P-Nitro Aniline From Acetanilide

Aim:- Organic Preparation Of P-Nitro Aniline

Step 1:- Acetanilide to P-Nitro Acetanilide [Nitration]

Requirement:- powdered Acetanilide (dry), Glacial HAc, Conc. H2SO4, ice, Fuming HNO3,Alcohol

Procedure:-

Take 10 gms of powdered Acetanilide (dry) & 15 ml Glacial HAc in a 250ml beaker & add20 ml of Conc. H2SO4 with stirring the mixture becomes warm and clear solution isobtained.

Surround the beaker with freezing mixture of ice & salt and stir the solution mechanically

Now, take 5 ml fuming HNO3 & 3 ml Conc. H2SO4 in a 100 ml beaker cool the acid solution& transfer it in a separating funnel.

When the temp. of the solution from separating funnel drop by drop and maintain the tembelow 10 oC.

When all the mixed acid has been added remove the beaker from the freezing mixture andallow it to stand at room temperature for at least about 1 hour and then pour the reactionmixture into an ice containing water.

Crude P-Nitro Acetanilide is at once precipitated allow it to stand for 15 min. filter theproduct, wash it with cold water till free from acids[checks on litmus paper] dry it.Recrystallized the product with alcohol.


M.P.:- 214 oC


Reaction:-

NHCOCH3NHCOCH3

NO2

Cold Mixture

Fuming HNO3 + Conc. H2SO4

Glacial HAc0 - 10 oC

Acetanilide

P-Nitro Acetanilide

Calculation:-

135 gms Acetanilide gives 180 gms P-Nitro Acetanilide

10 gms Acetanilide (?) gms P-Nitro Acetanilide

= = 13.3 gms P-Nitro Acetanilide



% Yield =

Step 2:- P-Nitro Acetanilide to P-Nitro Aniline [Acid Hydrolysis]

Requirement:- P-Nitro Acetanilide, 70% Conc. H2SO4[60ml Conc. H2SO4+45 ml H2O],10% NaOH or Conc. Ammonia Solution.

Procedure:-

Take 10 gm P-Nitro Acetanilide & 50 ml of 70% Conc. H2SO4 in a 250ml R.B.F attach areflux condenser & reflux it for about half hour [or until a test sample remains clear upondilution with 2 – 3 times its volume with water]

The P-Nitro Aniline is present in the liquid as sulphate. Pour the clear hot solution intocold water and precipitate the P-Nitro Aniline by adding excess of Conc. Ammoniasolution. Cool the mixture in ice water.

Filter the yellow crystalline P-Nitro Aniline. Wash with water & dry it. Recrystallize theproduct with hot water.


M.P.:- 147 oC


Reaction:-

NHCOCH3

NO2

P-Nitro Acetanilide

NH2

NO2

CH3COOH

P-Nitro Aniline

Acid Hydrolysis

Calculation:-

180 gms P-Nitro Acetanilide gives 138 gms P-Nitro Aniline

10 gms P-Nitro Acetanilide (?) gms P-Nitro Aniline

= = 13.3 gms P-Nitro Aniline



% Yield =

Preparation Of P-Bromo Aniline From Acetanilide

Aim:- Organic Preparation Of P-Bromo Aniline

Step 1:- Acetanilide to P-Bromo Acetanilide [Bromination]

Requirement:- powdered Acetanilide (dry), Glacial HAc, Bromine, ice, NaHSO3

Procedure:-

Take 5 gms of finely powdered Acetanilide in 15 ml of glacial HAc in R.B.F & cool it in icebath in another 100 ml beaker take 3 ml of bromine & dissolve it in 12 ml of HAc &transfer the solution slowly and with constant shake to cold solution of acetanilide which ispreserved in ice bath when all the bromine has been added the solution will turn an orangedue to the slight excess of bromine. All the reaction mixture into ice cold water in a 500 mlbeaker stir the mixture and add just sufficient NaHSO3 solution to remove the orangecolor. Filter the product, wash with ice cold water and dry it. Recrystallized the productwith Alcohol.


M.P.:- 167 oC


Reaction:-

NHCOCH3

P-Bromo Acetanilide

NHCOCH3

BrAcetanilide

Bromination

Br2 in Glacial HAc

Calculation:-

135 gms Acetanilide gives 214 gms P-Bromo Acetanilide

5 gms Acetanilide (?) gms P-Bromo Acetanilide

= = 7.92 gms P-Bromo Acetanilide



% Yield =

Step 2:- P-Bromo Acetanilide to P-Bromo Aniline

Requirement:- P-Bromo Acetanilide, Ethanol, KOH(solid), ice

Procedure:-

Take 7 gms P-Bromo Acetanilide & 15 ml Ethanol in a 250 ml R.B.F attached a refluxcondenser & heat the mixture till a clear solution is obtained [if necessary add morealcohol]

Meanwhile prepare a solution of 4 gms KOH in 10 ml of waters. Add this solution in a clearsoln. of R.B.F & reflux the contents for about 2 hours. Cool & add the contents of the flaskin a 250 ml beaker containing 100 ml ice cold water. The precipitate of P-Bromo Aniline isobtained. Filter the product and dry it. Recrystallized the product with rectified sprite.


M.P.:- 66.6 oC


Reaction:-

P-Bromo Acetanilide

NHCOCH3

Br

P-Bromo Aniline

NH2

Br

Alkiline Hydrolysis

KOH

Calculation:-

214 gms P-Bromo Acetanilide gives 172 gms P-Bromo Aniline

7 gms P-Bromo Acetanilide (?) gms P-Bromo Aniline

= = 5.6 gms P-Bromo Aniline



% Yield =

Result:-

Step Product TheoraticalYield

PracticalYield

% ofYield

Structural Formula M.P

β Resorcylic Acid

Reaction:

Mechanism:

Resacetophenone

Reaction:

Mechanism:

OH

O H-

OH

O

OCH3C H

OH

OH

H

OH

OH

COCH3

CH3CO

o-Phenylene Diamine

Reaction:

Mechanism:

4-Methyl-7-Hydroxy Coumarin

Reaction:

Mechanism:

Methyl Orange

Reaction:

Mechanism:

N HH

SO3H

N HH

SO3Na

N NH

SO3Na

O N

SO3Na

NH O H

N

SO3Na

N N

SO3Na

N Cl

Na2CO3

-NaHCO3

Diazotization

NaNO2+2HCl0 - 5 oC+NO

ProtonShift

-H2O

+Cl

N N Cl

SO3Na

N

H

CH3

CH3N N

SO3Na

N

CH3

CH3

HCl

N N

SO3Na

N

CH3

CH3

Methyl Orange

P-Nitro Aniline

Step 1:- Acetanilide to P-Nitro Acetanilide [Nitration]

Reaction:

Mechanism:

NHCOCH3

H

NHCOCH3

H

NHCOCH3

NO2

NHCOCH3

NO2

Step 2:- P-Nitro Acetanilide to P-Nitro Aniline [Acid Hydrolysis]

Reaction:-

NHCOCH3

NO2

P-Nitro Acetanilide

NH2

NO2

CH3COOH

P-Nitro Aniline

Acid Hydrolysis

Mechanism:

P-Bromo Aniline

Step 1:- Acetanilide to P-Bromo Acetanilide [Bromination]

Reaction:

Mechanism:

Step 2:- P-Bromo Acetanilide to P-Bromo Aniline

Reaction:

Mechanism:

ORGANIC ESTIMATION SEMESTER-II

1

Determine the % purity of Formaldehyde

Aim:- to determine the % purity of Formaldehyde by Hydroxyl Amine Hydrochloride MethodRequirement:- Formaldehyde, 10 % Hydroxyl Amine Hydrochloride soln., Exact 0.1N NaOHsoln., Bromo phenol blue IndicatorProcess:-

1. Blank Experiment:-Take 25 ml of distilled H2O in a conical flask. Add about 1 ml Bromo phenol blueIndicator & then add 10 ml soln. of 10 % Hydroxyl Amine Hydrochloride & titrate itagainst 0.1N NaOH soln. from the burrete yellow soln. changes to blue violet is the endpoint. Take 3 to 4 readings. Let the constant burrete reading “V2” ml.

2. Actual Experiment:-Weight out accurate about 0.270 to 0.300 gms of the given formaldehyde soln. in aweighting bottle. Transfer it to a conical flask by 25 ml dist. H2O. then add about 1 mlBromo phenol blue Indicator & then add 10 ml soln. of 10 % Hydroxyl AmineHydrochloride &allow to stand for 10 min & shake frequently. Then titrate the liberatedHydrochloride acid against 0.1N NaOH soln. from the burrete yellow soln. changes toblue violet is the end point. Take 3 to 4 readings.let the constant burrete reading “V1” ml.

Reaction:-

Observation:-Blank Estimation:-

Burette:- 0.1N NaOH Soln.

Flask:- 10 ml 10 % Hydroxyl Amine Hydrochloride + 25 ml dist. H2O

Indicator:- Bromo phenol blue Indicator

End point:- yellow to blue colour

No. I.B.R F.B.R Difference Mean1 0.0 ml2 0.0 ml3 0.0 ml

Observation:-Actual Estimation:-Burette:- 0.1N NaOH Soln.Flask:- weighted formaldehyde soln. + 10 ml 10% Hydroxyl Amine Hydrochloride + 25 ml dist.H2OIndicator:- Bromo phenol blue IndicatorEnd point:- yellow to blue colour


2

No. Weighted F.B.R % of purity1 gms2 gms3 gms

Calculation:-

% of HCHO =( ).

N=0.104M=30Result:-% purity of Formaldehyde:-___________.

Determine the Amount of Acid + Amide

Aim:- you are given a mixed solution of acid & amide. Determine the amount of acid & amide inthe given solution.

Requirement:- 0.1N HCl (exact), 0.1N NaOH (exact), 1.5N NaOH (approx.), phenolphthaleinindicator

Process:-Blank estimation:-[standardization of NaOH]Take 25 ml of 1.5N NaOH in a 250 ml S.M.F & diluted it to 250 ml with dist. H2O. take 25 ml ofthe diluted soln. in a conical flask & titrate it against 0.1N HCl soln. from burette usingphenolphthalein as an indicator disappearance of pink color is the end point. Take 3 or 4 reading.Let the constant burette reading “A” ml.

Estimation of Acid:-Take 25 ml of given mixed soln of acid+amide in a 250 ml S.M.F & dilute it to 250 ml dist. H2O.Take 25 ml of this diluted soln.in a conical flask and titrate it against 0.1N NaOH soln. fromburette using phenolphthalein as an indicator. Appearance of pink color is the End point. Take 3to 4 burette reading. Let the constant burette reading “D” ml.

Actual Estimation [Estimation of Acid + Amide]:-Take 25ml of given mixed soln. of acid + amide in 250 ml R.B.F. add exact 25 ml of 1.5 NNaOH by pipette & 2 to 3 T.T dist. H2O attach on air condenser & refluxed the content on a sandbath. Cool the content & transfer the Hydrolyzed.Take 25 ml this diluted Hydrolyzed soln. in a conical flask & titrate it against 0.1N HCl soln.from burette using phenolphthalein as an indicator. Disappearance of pink color is the end point.Take 3 to 4 reading. Let the constant burette reading. Let the burette reading “B” ml.


3

Observation: Blank estimation [standardization of NaOH soln.]Burette:- 0.1N HCl Soln.Flask:- 25 ml dil. Soln. of NaOHIndicator:- phenolphthalein IndicatorEnd point:- disappearance of pink color


25ml diluted NaOH soln. required (A)=_____ml 0.1N HCl.

Normality of given NaOH soln.=. 10

=__________ N

Estimation of Acid:-Burette:- 0.1N NaOH Soln.Flask:- 25 ml dil. Soln. of [Acid + Amide]Indicator:- phenolphthalein IndicatorEnd point:- Appearance of pink color


25 ml diluted solution Reqn. D =_________ ml 0.1 N NaOH

Reaction:-

Calculation:-

1000 ml 1 N NaOH = 59 gms Succinic Acid1 ml 0.1 N NaOH = 0.0059 gms Succinic AcidD ml 1 N NaOH = 0.0059 X D gms Succinic Acid

E = _______ gms Succinic Acid

Amount of succinic acid in 25 ml diluted solution E = ________ gms

Amount of succinic acid in given solution = E X 10

X = ________ gms


4

Actual estimation [Estimation of Acid + Amide]:-

Burette:- 0.1N HCl Soln.Flask:- 25 ml dil. Refluxed [Hydrolyzed Soln.]Indicator:- phenolphthalein IndicatorEnd point:- Disappearance of pink color


25 ml diluted refluxed [Hydrolyzed Soln.] required B = _______ ml 0.1 N HCl

Reaction:-

Now,

1. The amt. of NaOH soln. added in terms of 0.1 N HCl A = ______ ml2. The amt. of NaOH soln. unused in terms of 0.1 N HCl B = ______ ml3. The amt. of NaOH soln. used up for acid + amide in terms of 0.1 N HCl C = A – B ml

C = ________ ml4. The amt. of NaOH soln. used up for amide only in terms of 0.1 N HCl F = C – D ml

F = _______ ml

Now,

1000 ml 1 N NaOH = 1000 ml 1 N HCl = 59 gms Acetamide1000 ml 1 N HCl = 59 gms Acetamide1 ml 0.1 N HCl = 0.0059 gms AcetamideF ml 0.1 N HCl = 0.0059 X F gms Acetamide

G = _______ gms AcetamideAmt. of Acetamide in the given solution = G X 10

Y = ________ gms.

RESULT:-1. Normality of given NaOH solution = _______ N2. Amt. of Acid in the given solution = ________ (X) gms3. Amt. of Amide in the given solution = ________ (Y) gms


5

Determine the Amount of Acid + Ester

Aim:- you are given a mixed solution of acid + Ester. Determine the amount of acid & E in thegiven solution.

Requirement:- 0.1N HCl (exact), 0.1N NaOH (exact), 1.5N NaOH (approx.), phenolphthaleinindicator

Process:-Blank estimation:-[standardization of NaOH]Take 25 ml of 1.5N NaOH in a 250 ml S.M.F & diluted it to 250 ml with dist. H2O. take 25 ml ofthe diluted soln. in a conical flask & titrate it against 0.1N HCl soln. from burette usingphenolphthalein as an indicator disappearance of pink color is the end point. Take 3 or 4 reading.Let the constant burette reading “A” ml.

Estimation of Acid:-Take 25 ml of given mixed soln of acid+Ester in a 250 ml S.M.F & dilute it to 250 ml dist. H2O.Take 25 ml of this diluted soln.in a conical flask and titrate it against 0.1N NaOH soln. fromburette using phenolphthalein as an indicator. Appearance of pink color is the End point. Take 3to 4 burette reading. Let the constant burette reading “D” ml.

Actual Estimation [Estimation of Acid + Amide]:-Take 25ml of given mixed soln. of acid + ester in 250 ml R.B.F. add exact 25 ml of 1.5 N NaOHby pipette attach water condenser & reflux it on water bath till hydrolysis is complete [ 1 and halfhour hour].Cool it and transfer the reflux solution in a 250 ml S.M.F & dilute it to250 ml with dist. H2O.Take 25 ml this diluted refluxed soln. in a conical flask & titrate it against 0.1N HCl soln. fromburette using phenolphthalein as an indicator. Disappearance of pink color is the end point. Take3 to 4 reading. Let the constant burette reading. Let the burette reading “B” ml.

Observation: Blank estimation [standardization of NaOH soln.]Burette:- 0.1N HCl Soln.Flask:- 25 ml dil. Soln. of NaOHIndicator:- phenolphthalein IndicatorEnd point:- disappearance of pink color


25ml diluted NaOH soln. required (A)=_____ml 0.1N HCl.

Normality of given NaOH soln.=. 10

=__________ N


6

Estimation of Acid:-Burette:- 0.1N NaOH Soln.Flask:- 25 ml dil. Soln. of [Acid + Ester]Indicator:- phenolphthalein IndicatorEnd point:- Appearance of pink color


25 ml diluted solution Reqn. D =_________ ml 0.1 N NaOH

Reaction:-

Calculation:-

1000 ml 1 N NaOH = 60 gms HAc1 ml 0.1 N NaOH = 0.0060 gms HAcD ml 1 N NaOH = 0.0060 X D gms HAc

E = _______ gms HAc

Amount of HAc d in 25 ml diluted solution E = ________ gms

Amount of HAc in given solution = E X 10

X = ________ gms

Actual estimation [Estimation of Acid + Amide]:-

Burette:- 0.1N HCl Soln.Flask:- 25 ml dil. Refluxed Soln.Indicator:- phenolphthalein IndicatorEnd point:- Disappearance of pink color



7

25 ml diluted refluxed Soln. required B = _______ ml 0.1 N HCl

Reaction:-

Now,

1. The amt. of NaOH soln. added in terms of 0.1 N HCl A = ______ ml2. The amt. of NaOH soln. unused in terms of 0.1 N HCl B = ______ ml3. The amt. of NaOH soln. used up for acid + ester in terms of 0.1 N HCl C = A – B ml

C = ________ ml4. The amt. of NaOH soln. used up for ester only in terms of 0.1 N HCl F = C – D ml

F = _______ ml

Now,

1000 ml 1 N NaOH = 1000 ml 1 N HCl = 88 gms Ethyl acetate1000 ml 1 N HCl = 88 gms Ethyl acetate1 ml 0.1 N HCl = 0.0088 gms Ethyl acetateF ml 0.1 N HCl = 0.0088 X F gms Ethyl acetate

G = _______ gms Ethyl acetateAmt of Ethyl acetate in the given solution = G X 10

Y = ________ gms.

RESULT:-5. Normality of given NaOH solution = _______ N6. Amt. of Acid in the given solution = ________ (X) gms7. Amt. of Ester in the given solution = ________ (Y) gms

Vnsgu Sf Sem 2 Organic Practical

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