Notes On Purification Of Organic Compound 3r1i44

Methods of Purification of Organic Compounds : Organic compounds obtained either from natural source (or) synthesized in laboratory contaminated with impurities. Various methods are used for removal of impurities from an organic compound depends on the nature of compound and type of impurities present in it. The following methods are commonly used for purification 1) Sublimation 2) Crystallisation 3) Distillation 4) Solvent extraction (differential extraction) 5) Chromatography. Note: Most of the pure compounds contains sharp Melting point & Boiling points.

Sublimation : “The process of conversion of solid sublimate to vapour state directly by heating without ing through liquid state is called sublimation”. solid vapour  This method is used for purification of solids  Sublimation process is used for separation of sublimable volatile compounds from non sublimable impurities.  Sublimation is generally used for purification of camphor, napthalene, Anthracene, Benzoic acid, phthalic anhydride, Anthraquinone, Indigo, Iodine, HgCl2, solid SO2.

Crystallisation :  It is used for purification of solid organic compounds. GENERAL ORGANIC CHEMISTRY

 Crystallisation is based on the difference in solubilities of the compound and impurities in a suitable solvent.  The principle involved in this method is impure compound dissolved in a solvent is sparingly soluble at low temperature, but appreciably soluble at high temperature.  Insoluble impurities are removed by filtration in hot condition.  If a compound is highly soluble in one solvent and very little soluble in another solvent, then crystallisation is carried out by using mixture of these solvents.  Impurities, which impart colour are removed by adsorbing over activated charcoal.  Repeated crystallisation is required if organic compound contains impurities of comparable solubilities.  The process of separation of different components of a mixture by repeated crystallisation is called fractional crystallisation.  Fractional crystallisation is used for separation of two or more soluble substances which have different solubilities in the same solvent.  Most commonly used solvents for crystallisation are water, alcohol, ether, chloroform, carbontetrachloride, acetone, benzene, petroleum ether.  Sugar having an impurity of common salt can be crystallised from hot ethanol, since sugar dissolves in hot ethanol but common salt does not.  Fractional crystallisation can be used to separate a mixutre of KClO3 (less soluble) and KCl (more soluble).

185

Distillation :  Distillation in an important method used to separate i) Volatile liquids from non volatile impurities. ii) Liquids having sufficient difference in boiling points.

Simple Distillation : This process is used for purification of liquids which does not undergo decomposition at their boiling points.  The vapourisation of a liquid by heating and subsequent condensation of vapours by cooling is known as distillation.  Liquid mixture is taken in a round bottom flask and heated carefully, the vapour component with lower boiling point distills first, the vapour formed is condensed by using condenser and the liquid is collected in a receiver. The vapours of component with higher boiling point distills latter.  The liquids that have boiling point difference greater than 400C can be purified by this method. Eg: i) Chloroform (B.P. 334 K) & Aniline (B.P. 457 K) ii) Ether (B.P. 308 K)& Toluene (B.P. 384 K) iii) Benzene (B.P. 353 K)& Aniline(B.P. 475 K)

Fractional Distillation :  Fractional distillation is used if the difference in boiling point of two liquids is less than 400 C .  Vapours of liquid mixture are ed through fractionating column before condensation,which is fitted over mouth of the round bottom flask.  Vapours of liquid with higher boiling point condense before the vapours of liquid with lower boiling point, the vapours rai in the fractionating column is richer in more volatile component.  Fractionating column provides many surfaces for heat exchange between ascending vapours and descending condensed liquid.  Each successive condensation and vapourisation unit in the fractionating column is called a theoritical plate.  Liquids forming a constant boiling mixture (azeotropic mixture) can not be separated by this method.  Fractional distillation is used to separate different fractions of crude oil in petroleum industry.  This method is used for separation of mixture of acetone (B.P. 330K) and methyl alcohol (B.P. 338K)  Mixture of benzene and toluene can be separated by fractional distillation.

EX.1: How is ethyl alcohol purified from methylated spirit. Sol. Methylated spirit is ethyl alcohol contaminated mainly with methyl alcohol. Ethyl alcohol is purified by fractional distillation since the difference in boiling point is less.

Distillation Under Reduced Pressure (Vacuum Distillation) :  This method is used to purify liquids having very high boiling points, which decompose at or below their boiling points.  These liquids are made to boil at a temperature lower than their normal boiling point by reducing pressure on their surface with the help of vacuum pump.  Glycerine H2O2, formaldehyde are purified by vacuum distillation.  Glycerol can be separated from spent-lye in soap industry by using vacuum distillation.  Sugar cane juice is concentrated in sugar industry by evaporation under reduced pressure which saves lot of fuel.

Steam Distillation :  This method is used for separation and purification of organic compounds (solids or liquids) which i) are steam volatile ii) are insoluble in water. iii) Posess high vapour pressure (10-15 mm of Hg at 373 K) iv) Contains non valatile impurities.  Steam distillation is based on Dalton’s law of partial pressure i.e., P = P1 + P2. where P = Atmospheric pressure P1 = Vapour pressure of organic liquid P2 = Vapour pressure due to water.  Compounds which can be purified by steam distillation are aniline, nitrobenzene, bromobenzene, o-nitrophenol, o-hydroxy benzaldehyde (salicylaldehyde), o-hydroxy acetophenone, turpentine oil, essential oils. EX.2: Mention about the purification of (a) aniline and (b) naphthalene Sol. a)Aniline can be purified by steam distillation because it is immiscible with water and steam volatile. b) Naphthalene can be purified by sublimation because it changes on heating directly to vapour state and on cooling, it changes back into solid form.

186

GENERAL ORGANIC CHEMISTRY

Solvent Extraction (Differential Extraction) :  Definition: The process of isolating an organic compound from its aqueous solution by shaking with a suitable solvent is called differential extraction. It is also called solvent extraction  When an organic compound is present in an aqueous medium, then it is separated by shaking it with an organic solvent in which it is more soluble than in water.  Solvent should be immiscible with water and organic compound to be separated should be highly soluble in it.  Organic solvent and aqueous solution are immiscible with each other, so they can form two distinct layers which can be separated by separatory funnel.  Organic solvent is distilled or evaporated to get organic compound.  If organic compound is less soluble in organic solvent then large quantity of solvent is required to extract small quantity of compound, which is said to be continuous extraction.  Benzoic acid can be extracted from its aqueous solution using benzene as solvent.  Ether is a better solvent in differential extraction due to : i) its less polarity ii) least reactivity iii) less boiling point iv) higher solubility of organic compounds

  

 





Separation by Chemical Methods :  It is used for mixture of substances which are chemically different. Eg: 1) Separation of acidic and basic compounds of coal-tar. ammonical dil. CuC CCu HCl HC CH Cu C l 2 2 (impure) (red ppt.) (pure)

2) HC CH

3) Pyroligneous Acid

Ca(OH)2

Ca(OH) Calcium

acetate

conc. HCl

CH3OOH

(from wood distillation industry)

4) CH3OH (Impure)

Methyl oxalate NaOH(aq) CH3OH (Crystalline)

(pure)

Chromatography :  This method is used for separation of mixtures into their components, purification of compounds and also to test the purity of compounds.  Chromatography is obtaind from the greek word GENERAL ORGANIC CHEMISTRY

187

“Chroma’’ means colour and “graphy’’ means writing. This method was first used for separation of coloured substances found in plants. This method was described by Tswett. This Technique consists of two phases one is stationary phase of large surface area while the second is moving phase which is allowed to move slowly over the stationary phase. Stationary phase is either liquid or solid, while moving phase may be liquid or gas. The technique of chromatography is based on the rates at which the components of the mixture moves through a porous medium (called stationary phase) under the influence of some solvent (or) gas (called mobile phase). Mixture of substances is applied on a stationary phase which may be solid or liquid. A pure solvent, a mixture of solvents or a gas is allowed to move slowly over the stationary phase, the components of the mixture get gradually separated from one another. Recovery of separated substances by using suitable solvent is known as elution. The solvent used is known as eluant. S. Chromatography Stationary Mobile Phase No. Process Phase 1. Column chromatography (Adsorption)

Solid

Liquid

2. Liquid-liquid partition chromatography

Liquid

Liquid

3. Paper chromatography

Liquid

Liquid

4. Thin layer chromatography (TLC)

Liquid(or) Liqid solid

5. Gas-liquid chromatography (GLC)

Liquid

Gas

Solid

Gas

Solid

Liquid

6. Gas-solid chromatography (GSC) 7. Ionic change chromatography

 Based on the principle involved chromatography is classified in to a) Adsorption chromatography and b) partition chromatography.

a. Adsorption Chromatography :  Adsorption chromatography is based on the fact that different compounds are adsorbed on an adsorbent to different degrees.  Commonly used adsorbents are sillica gel, alumina, magnesium oxide, cellulose powder, activated animal charcoal.  When a mobile phase is allowed to move over stationary phase, the components of the mixture move by varying distances over stationary phase.  There are two main types of chromatographic techniques based on principle of differential adsorption (i) Column chromatography and (ii) Thin layer chromatography (TLC) i) Column Chromatography  It involves separation of a mixture over a column of adsorbent packed in a glass tube, which is fitted with stop cock at its lower end.  The mixture to be adsorbed on the adsorbent is placed at the top of the stationary phase.  An appropriate eluant, which is a liquid or a mixture of liquids is allowed to flow down the column slowly.  The most readily adsorbed substances are retained near the top and others come down to various distances in the column. ii) Thin Layer Chromatography (TLC)  It involves separation of substances of a mixture over a thin layer of an adsorbent coated on glass plate.  The glass plate is coated with adsorbent (ex: silica gel, alumina) as a thin layer (about 0.2mm thick) is called chromatography plate or chroma plate .  The solution of mixture to be separated is applied as small spot about 2cm above from one end of the TLC plate.  The glass plate is placed in a closed jar containing the eluant, as the eluant rises up, the components of the mixture move up along the eluant to different distances depending on their degree of adsorption and separation takes place.  The relative adsorption of each component of the mixture is expressed in of its retardation factor i.e, R f value.

Y X

(base line)

Rf 

Distance moved by the substance from base line (x) Distance moved by the solvent from base line (y)

 The spots of coloured compounds are visible on TLC plate due to their original colour.  The colourless compound which fluroscene are detected with ultraviolet light  Spots of compounds are even detected by allowing them to adsorb iodine, will show up as brown spots.  Some times an appropriate reagent is sprayed on the plate. Eg: Amino acids are detected by spraying the plate with ninhydrin solution.

Partition Chromatography :  Partition chromatography is based on continuous differential partitioning of components of a mixture between stationary and mobile phases.  Paper chromatography is a type of partition chromatography.  In paper chromatography a special quality paper known as chromatography paper is used.  In Chromatography paper, cellulose helps as inert , and water absorbed from air on to hydroxyl groups of cellulose acts as stationary phase.  The chromatography paper spotted with the solution of mixture at the base is suspended in a suitable solvent or mixture of solvents, this solvent (s) acts as mobile phase.  The solvent rises up the paper by capillary action and flows over the spot.  The paper selectively retains different components according to their differing partition in the two phases. The paper strip so developed is called chromatogram.

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GENERAL ORGANIC CHEMISTRY

 The spots of the separated coloured compounds are visible at different heights from the position of initial spot on the chromatogram.  The spots of the separated colourless compounds may be observed either under ultraviolet light or by the use of appropriate spraying agent.

Additional Information : Applications of Chromatography : I)

II)

III)

IV)

V)

i) In Chemical Industry column chromatography is used for separation of required components obtained after synthesis. ii) TLC is useful for monitoring large scale column chromatography. Pharmaceutical industry : Chromatography is used for separation of chiral compounds to obtained pharmaceutically active optical isomer. Food Industry : Chromatography techniques are used for quality control in food industry. It is used to determine presence and to separate additives, flavours etc. It is also used to detect presence of contaminents like mould, bacteria in food. Environment - Testing Lab : Presence and quality of pollutants in air and drinking water can be determined by chromatography technique. Diagnostic Technique : Presence of certain drugs and the marker compounds for medical diagnosis in blood and urine are determined.

 Water vapours turn anhydrous copper sulphate in to blue. CuSO  5 H 2O  CuSO4 .5H 2O

4 anhydrous  colour less 

Detection of Nitrogen, Sulphur Halogens & Phosphorus :  Nitrogen, sulphur, halogens and phosphorus present in an organic compound are detected by Lassaigne’s test.  Organic compounds are fused with dry sodium in fusion tube and fused mass after extraction with water is boiled and filtered, the filtrate is called sodium fusion extract.  During prepartion of sodium fusion extract covalent compound is converted into ionic compound.  The following reactions takes place  Na  C  N   NaCN  2Na  S   Na2 S  2 Na  X 2   2 NaX (X=Cl,Br or I)

Test for Nitrogen : Sodiumfusion extract is boiled with freshly prepared ferrous sulphate ( FeSO4 )solution and then acidified with concentrated sulphuric acid. The formation of prussian blue colour confirms the presence of nitrogen.

Qualitative Analysis of Organic Compounds (Detection of Elements) :

2 NaCN  FeSO4  Fe(CN )2  Na2 SO4

 The qualitative analysis of an organic compound involves detection of all elements present in it.

Fe(CN )2  4 NaCN  Na4 [ Fe(CN )6 ] sodium hexacyano ferrate(II) On heating with concentrated sulphuric acid some Iron(II) ions are oxidised to Iron(III) ion, which reacts with sodium hexacyano ferrate(II) to produce Iron (III) hexacyanoferrate(II) (ferric ferrocyanide) which is prussian blue in colour.

Detection of Carbon and Hydrogen :  Carbon and hydrogen are detected by heating the compound with cupric oxide (CuO).  Carbon present in the compound is oxidised to carbondioxide, which turns lime water milky.  Hydrogen present in the compound is converted in to water, which turns anhydrous copper sulphate into blue. C  2CuO  2Cu  CO2 

H 2  CuO  Cu  H 2O 

 Carbondioxide turns lime water milky. Ca(OH )2  CO2  CaCO3  H 2O

3 Na4 [ Fe(CN )6 ]  2 Fe2 ( SO4 )3  Fe4 [ Fe(CN )6 ]3  6 Na2 SO4 ferricferrocyanide (prussianblue)  This test fails in case of diazo compounds.  If the amount of nitrogen present is less, then prussian blue is present in collodial form and the solution looks green.

(milky)

GENERAL ORGANIC CHEMISTRY

hydrated

 blue 

189

EX.3: Hydrazine does not give Lassaigne’s test. Why ? Sol. In the Lassaigne’s test, nitrogen is converted to cyanide by combining with carbon of the compound. Hydrazine does not contain carbon and hence cyanide cannot form. EX.4: Why diazonium salts do not show positive Lassaigne’s test for nitrogen ? Sol. Diazonium salts are unstable and lose nitrogen as

NaCl  AgNO3  AgCl   NaNO3 (white ppt) AgCl  2 NH 4OH  [ Ag ( NH3 )2 ]Cl  2H 2O (soluble complex) ii) Yellowish precipitate, sparingly soluble in ammonium hydroxide indicates presence of bromine NaBr  AgNO3  AgBr   NaNO3 yellowish ppt iii) Yellow precipitate, insoluble in ammonium hydroxide indicates presence of Iodine.

N 2 gas on heating. Hence during fusion, no sodium cyanide is formed in Lassaigne’s extract due to the loss of nitrogen.

Test for Sulphur : a)

Sodium fusion extract is acidified with acetic acid and lead acetate is added to it, a black precipitate of lead sulphide is formed, which indicates presence of sulphur. Na2 S  (CH 3COO ) 2 Pb  PbS  2CH 3COONa

b)

black Sodium fusion extract is treated with freshly prepared sodium nitroprusside, appearance of violet colour (purple) indicates presence of sulphur. Na2 S  Na2 [ Fe(CN )5 NO ]  Na4 [ Fe(CN )5 NOS ]

(sodium nitro prusside) (violet)  In case both nitrogen and sulphur are present in an organic compound sodium thiocyanate is formed,

NaI  AgNO 3  AgI   NaNO3 yellow precipitate  Nitrogen and sulphur are also present in the compound, the sodiumfusion extract is boiled with concentrated nitric acid to decompose sodium cyanide & sodium sulphide formed during Lassaigne’s test , otherwise they interfere with silver nitrate test for halogens. NaCN  HNO3  NaNO3  HCN 

Na2 S  2 HNO3  2 NaNO3  H 2 S   If NaCN and Na2 S are not decomposed, then white and black precipitates of AgCN and Ag 2 S are formed respectively with silver nitrate solution.

Beilstein’s Test :

which gives blood red colour with neutral FeCl3 solution. Na  C  N  S  NaSCN

A copper wire flattened at one end is heated in an oxidising flame of Bunsen burner. The heating is continued till it does not impart blue colour flame. The hot end of copper wire is now touched with the organic substance and is once again kept in flame, the appearance of green or blue colour indicates the presence of halogens in the organic compound. Limitations : a) Substances such as urea, thiourea do not contain halogens but gives this test. b) It does not tells which halogen is present in organic compound.

3NaSCN  FeCl3  Fe( SCN )3  3NaCl (blood red) (or) Fe 3  SCN   [ Fe( SCN )]2 (blood red)  If sodium fusion is carried out with excess of sodium, the thiocyanate decomposes to yield cyanide and sulphide, these ions gives their usual tests.

NaSCN  2 Na  NaCN  Na2 S

Test for Halogens :  Sodiumfusion extract is acidified with nitric acid and then treated with silver nitrate solution. i) White precipitate, soluble in ammonium hydroxide indicates presence of chlorine.

Chlorine Water Test for Bromine and Iodine :  Both AgBr and AgI are yellow precipitates, it is a littile bit difficult to identify given halogen in bromine or iodine, to confirm it chlorine water test is used.

190

GENERAL ORGANIC CHEMISTRY

 Sodium fusion extract is acidified with dilute

H 2 SO4 (or) HNO3 , to this 1(or) 2ml of chloroform (or) Carbon tetrachloride is added and then excess of chlorine water is added with constant shaking. i) If chloroform (or) carbon tetrachloride layer becomes yellow (or) brown indicates presence of bromine. 2 NaBr  Cl2  2 NaCl  Br2 Br2 dissolves in chloroform (or) carbontetrachloride gives yellow (or) brown colour. ii) If chloroform (or) carbon tetra chloride layer becomes violet indicates presence of iodine. 2 NaI  Cl2  2 NaCl  I 2

I 2 dissolves in chloroform (or) carbontetrachloride gives violet colour.  Presence of NaCN (or) Na2 S in sodium fusion extract does not interfere in this test.

Quantitative Analysis : It involves the estimation of percentage composition of various elements by suitable methods.

Estimation of Carbon and Hydrogen :  Carbon and hydrogen are estimated by Liebig’s combustion method.  A known mass of an organic compound is burnt in the presence of excess of oxygen (free from CO2) and Cupric oxide (CuO)  Carbon and hydrogen present in the compound are oxidised to CO2 and H2O respectively y y  C x H y   x   O2   xCO2  H 2O  4 2   CO2 and H2O produced are weighed by absorbing in concentrated solution of potassium hydroxide and anhydrous calcium chloride (or) magnesium perchlorate respectively %C 

12 weight of CO2 formed   100 44 weight of organic Compound

%H 

2 weight of H 2O formed   100 18 weight of organic Compound

Test for Phosphorus : The compound is heated with oxidising agent (sodium peroxide) or with fusion mixture (sodium carbonate and potassium nitrate) phosphorus present in the compound is oxidised to sodium phosphate. The solution is boiled with nitric acid and then treated with ammonium molybdate, a canary yellow (ammonium phospho molybdate) precipitate formation (or) yellow colouration indicates presence of phosphorus.

EX.5: On complete combustion, 0.246g of an organic compound gave 0.198g of carbon dioxide and 0.1014g of water. Determine the percentage composition of carbon and hydrogen in the compound. Sol. % of carbon =

12  0.198  100  21.95 44  0.246

2 P  5 Na2O2  2 Na3 PO4  2 Na2O 

% of hydrogen =

Na3 PO4  3HNO3  H 3 PO4  3NaNO3

2  0.1014  100  4.58 18  0.246

H 3 PO4  12( NH 4 ) 2 MoO4  21HNO3 

Estimation of Nitrogen :

( NH 4 )3 PO4 .12MoO3  21NH 4 NO3  12 H 2O (ammonium phospho molybdate)

Nitrogen present in organic compound is estimated by a) Dumas method b) Kjeldahl’s method

Dumas Method :

Test for Oxygen :  There is no direct test for oxygen.  If organic compound is heated in a dry test tube in nitrogen atmosphere, if water droplets are formed on the walls of the test tube indicates presence of oxygen.  Presence of oxygen can be known by testing functional groups containing oxygen. Eg: –OH, –COOH, –CHO, –NO2 etc GENERAL ORGANIC CHEMISTRY

 In this method nitrogen present in the organic compound is converted in to N2(molecular nitrogen)  A weighed amount of organic compound is heated with cupric oxide in an atmosphere of carbondioxide  Carbon and hydrogen present in the compound are

191

oxidised to CO2 and H 2O , while N 2 is set free.

 Some oxides of nitrogen formed are reduced to free nitrogen by ing over heated copper gauze

% of nitrogen = 28 vol.of N 2 at STP   100 22400 wt.of organic compound

y  C x H y N z   2 x   CuO  2 

xCO2 

28 41.9   100  17.46 22400 0.3 22400 ml of dinitrogen at STP weigh = 28g

=

y z y  H 2O  N 2   2 x   Cu 2 2 2 

 Oxides of nitrogen + Cu  N 2  CuO  The mixture of gases produced is collected over caustic potash solution (KOH solution) which absorbs CO2.  Nitrogen is collected in the upper part of nitro meter.

41.9 ml dinitrogen at STP weight =

0.3 g of organic compound contains

V2 

PV T 732.3  30  273 1 1  2   27.4cm3 T1 P2 288  760

Percentage of nitrogen in organic compound =

28 vol.of N2 at STP  100 22400 wt.of organiccompound

PV T 700  50  273 1 1  2   41.9ml T1 P2 760  300

28  41.9 100   17.46 22400 0.3 percentage of nitrogen = 17.46

Kjeldahl’s Method :  In this method nitrogen present in the organic compound is converted in to ammonia (NH3)  A known mass of organic compound containing nitrogen is heated with concentrated sulphuric acid in presence of K2SO4 and CuSO4 then nitrogen present in the compound is converted in to ammonium sulphate.  Organic compound + H 2 SO4   NH 4 2 SO4  The resulting solution is distilled with excess of sodium hydroxide

 NH 4 2 SO4  2 NaOH  Na2 SO4  2 NH 3  2H 2O

28 27.4   100  13.6 22400 0.25 EX.7: 0.3g of an organic compound gave 50ml. nitrogen at 270C and 715mm pressure. If the aqueous tension at 270C is 15mm, calculate the percentage compoistion of nitrogen in the compound. Sol. Mass of the substance = 0.3g Volume of the moist dinitrogen=50ml Temperature = 27 0 C = 27+273 = 300K Pressure = 715 - 15 = 700mm Volume of dintrogen at STP :

v2 

28  41.9 g 22400

of N 2 100 g of organic compound contain

28 Volume of nitrogen in ml at STP %N    100 22400 Weight of organic compound

EX.6: 0.25g of an organic compound gave 30cm 3 of moist dinitrogen at 288 K and 745 mm presure. Calculate the percentage of nitrogen. (Aqueous tension at 288K=12.7mm) Sol. Mass of the substance = 0.25g Volume of moist dinitrogen = 30cm 3 Temperature = 288K Pressure = 745 -12.7 = 732.3mm Volume of dinitrogen at STP :

28  41.9 g 22400

 Ammonia evolved is absorbed in a known but excess volume of standard HCl (or) H2SO4solution.  The acid left unreacted is estimated by titration against standard solution of sodium hydroxide. Percentage of Nitrogen = 14 VN   100 1000 wt. of organic compound

%N 

1.4  V  N Weight of organic compound

Where V = Volume of acid in ml neutralised by ammonia N = Normality of acid.  This method is simpler and more convenient  It is mainly used to find percentage of nitrogen

192

GENERAL ORGANIC CHEMISTRY

present in food stuffs, soils, fertilizers and various agricultural products.  This method is not applicable to compounds containing nitro (-NO2), Nitroso (NO), azo group

m. eq of H 2 SO4 unused = 20 - 15.4 = 4.6 % of nitrogen = 1.4  m.eq. of H 2 SO4 unused 1.4  4.6   12.88 wt.of organic compound 0.5

O

      N  N    N  N   , azoxy compounds    

and nitrogen present in the ring (pyridine

,

N

quinoline

) because nitrogen N present in these compounds is not quantitatively converted in to ammonium sulphate. EX.8: In Kjeldahl’s estimation of nitrogen, the ammonia evolved from 0.5g of an organic compound neutralised 10ml of 1M H 2 SO4 . Calculate the percentage of nitrogen in the compound.

Estimation of Halogens Carius Method :  A weighed amount of an organic compound is heated with fuming nitric acid in the presence of silver nitrate contained in a hard glass tube known as carius tube.  Carbon and hydrogen present in the compound is converted in to CO2 and H2O.  Halogen present in the organic compound is converted in to silver halide.  The precipitate is washed, dried and weighed. Percentage of halogen = Atomicweightof halogen Weightof silverhalide formed  100 Mwtof . silverhalide Weightof organiccompound %Cl 

% Br 

Sol. 10 ml of 1M H 2 SO4 = 20ml of M NH 3 1000 ml of 1M ammonia contains 14g nitrogen

%I 

14  20 g N2 20 ml of 1M ammonia contains 1000

14  20 100  56.0 1000  0.5 EX.9: In Kjeldahl’s estimation of nitrogen, the ammonia obtained from 0.5g of an organic substance was ed into 100cm 3 of M H 2 SO4 . The excess of acid required 10 M NaOH for neutralisation. 154 cm3 of 10 calculate the percentage of nitrogen in the compound.

% of nitrogen =

Sol. m eq. of H 2 SO4 taken = molarity x basicity x volume (ml) =

1  2  100  20 10

m.eq of NaOH =

80 wt. of AgBr formed  100 188 Wt. of organic compound

127 Wt. of AgI formed  100 235 Wt. of organic compound

EX.10: In Carius method, 0.1890g of an organic compound gave 0.2870g of silver chloride. Calculate the percentage of chlorine in the compound. Sol. Weight of subtance = 0.1890g Weight of silver chloride = 0.2870g weigh of AgCl  35.5  100 % of chlorine = weight of subtance 143.5 0.2870  35.5  100  37.8 0.1890  143.5 EX.11: One gram of a bromoalkane on heating with excess silver nitrate in Carius tube method gave 0.94g of yellow precipitate. What is the percent weight of halogen ? Sol. Weight of subtance = 1g Yellow precipitate is AgBr ; Weight of AgBr = 0.94g

1  1154  15.4 10

GENERAL ORGANIC CHEMISTRY

35.5 wt. of AgCl formed   100 143.5 Wt. of organic compound



% of Bromine = 193

0.94  80  100  40 1188

EX.12: In carius method of estimation of halogen. 0.15g of an organic compound gave 0.12g of AgBr. Find out the percentage of bromine in the compound. Sol. % of bromine =

80  0.12  100  34.04 188  0.15

 Phosphoric acid is precipitated as magnesium ammonium phosphate (Mg NH4 PO4), by addition of magnesia mixture (MgCl2 + NH4OH + NH4Cl)  Magnesium ammonium phosphate is washed, dried and it is heated strongly to get magnesium pyrophosphate  Mg2 P2O7  .

Estimation of Sulphur Carius method:  A weighed amount of organic compound is heated in a carius tube with sodium peroxide or fuming nitric acid.  Sulphur present in the compound is oxidised in to sulphuric acid, which is treated with BaCl2 solution gives precipitate of BaSO4.  It is filtered, the precipitate is washed, dried and weighed.

H PO

3 4  Phosphoric acid 

0.4813g BaSO4 contain

%P 

32  0.4813 100  42.10 233  0.157 EX.14: On heating 0.2g of an organic compound with a mixture of barium chloirde and nitric acid, 0.466g of barium sulphate was obtained. Calculate the percentage of sulphur. Sol. Weight of substance = 0.2g Weight of barium sulphate = 0.466g Weight percentage of sulphur =

% of suphur =

wt. of Mg 2 P2O7 formed 62   100 222 wt. of organic compound

 Phosphoric acid is precipitated as ammonium phospho molybdate  NH 4 3 PO4 .12MoO3 by adding ammonia and ammonium molybdate. (Molecular mass of (NH4)3PO4.12MoO3 = 1877) %P 

32  0.4813 g sulphur 233

ammonium phosphate 

(Magnesium Pyrophosphate)

wt of BaSO4 formed 32 %S    100 233 wt. of organic compound

Sol. 233g BaSO4 contain 32g sulphur

Mg NH 4 PO4

 Magnesium

2MgNH 4 PO4   Mg 2 P2O7  2 NH 3  H 2O 

Atomic wt. of sulphur wt of BaSO4 formed %S    100 Mol. wt. of BaSO4 wt. of organic compound

EX.13: In sulphur estimation, 0.157 g of an organic compound gave 0.4813 g of barium sulphate. What is the percentage of sulphur in the compound ?

Magnesia m i x ture  

31 wt of  NH 4 3 PO4 .12 MoO3 formed  100 1877 wt. of organic compound

Estimation of Oxygen :  Usually percentage of oxygen in organic compound is determined by method of difference % of oxygen = 100 - (sum of the percentages of all other elements)  Oxygen present in the organic compound is estimated by Aluise’s method.  A known amount of organic compound is subjected to pyrolysis in a stream of nitrogen.  The mixture of gaseous products containg oxygen is ed over red-hot coke, then all the oxygen is converted in to carbon monoxide.

compound   O2  other gaseous products  1373 K 2C  O2   2CO  CO formed is quantitatively converted in to CO2

by ing over warm Iodine pentoxide  I 2O5  5CO  I 2O5  I 2  5CO2

0.466  32 100  32 0.2  233

The resulting gaseous mixture  CO2 and I 2  is

Estimation of Phosphorus Carius Method : A weighed amount of organic compound is heated with fuming nitric acid, then phosphorus present in the compound is oxidised to phosphoric acid. 194

ed through potassium iodide solution, which absorbs iodine, and then ed over KOH to absorb CO2. % of oxygen 

16 wt . of CO2 formed   100 44 wt . of organic compound

GENERAL ORGANIC CHEMISTRY

EX.15: 0.2g of an organic compound on analysis gave 0.147g of carbondioxide, 0.12g of water and 74.6 c.c of nitrogen at S.T.P. Calculate the weight percentages of constituents. Sol. Weight of compound = W=0.2g Weight of CO2  W1  0.147 g

Platinic Chloride Method for Bases :  Organic bases combines with chloroplatinic acid

 H 2 PtCl6  to form insoluble platinichloride, which on ignition gives metallic platinum.  If ‘B’ is mono acidic base then formula of saltwill

Weight of H 2O  W2  0.12 g

be B2 H 2 Pt Cl6 .

Volume of N 2 at STP = 74.6 c.c. % of carbon =

B2 H 2 Pt Cl6   Pt  Molecular mass of platinum salt Mass of platinum salt  Atomic mass of platinum Mass of platinum

W1  12  100 0.147  12  100   20.04% W  44 0.2  44 % of Hydrogen =

If E in equivalent weight of base then 2 E  410 Mass of platinum salt  195 Mass of platinum

W 2  2  100 0.12  2  100   6.66% W  18 0.2  18 % of Nitrogen =

E

V2  28  100 7 4 .6   4 6 .6 3 % W  22400 8  0 .2 Remaining is oxygen.

Molecular mass of base = Equivalent mass of base x acidity

Emperical Formula :

% of Oxygen = 100   %C  % H  % N  = 100  73.33  26.67% Note: Presently estimation of elements (C, H and N) in organic compound is estimated by using CHN elemental analyser by taking very small amount of substance (1-3 mg), results are displayed on screen with in short time.

Chemical Methods Used to Find Molecular Mass Silver Salt Method for Acids : Organic acid form insoluble silver salts, which on heating undergoes decomposition to leave a residue of metallic silver.

RCOO Ag    Silver salt

 The simplest whole number ratio between the atoms of various elements present in one molecule of a substance is called emperical formula.

Calculation of Emperical Formula :   



Ag Silver  residue 

The steps involved are Divide mass percentage of each element by its atomic mass, gives relative number of atoms. Simplest ratio is obtained when the figures obtained is divided by lowest number. If the simplest ratio obtained is not a whole number ratio, then multiply all the figures with suitable integer to get simplest whole number ratio Write symbols of various elements side by side with above numbers at the lower right corner of each, which is emperical formula of compound.

Molecular Formula :

Equivalent wt of silver salt mass of silver salt  Equivalent wt. of silver Mass of silver

The actual number of atoms present in one molecule of a substance is called molecular formula. Molecular formula = (Emperical formula)n where

E  108  1 Mass of silver salt  108 Mass of silver

 Mass of silver salt  E 108  107  Mass of silver  Molecular weight of acid = Equivalent weight of acid(E) x basicity. GENERAL ORGANIC CHEMISTRY

 1  Mass of platinum salt 195  410   2  Mass of platinum 

n

Molecular wt. of thecompound Emperical formula weight of thecompound

 If vapour density is given then molecular weight = 2 x vapour density.

195











Determination of Molecular Formula of Gaseous Hydrocarbons (Eudiometry) : Eudiometry is a direct method used to find molecular formula of gaseous hydrocarbon without finding percentage composition of elements and its molecular weight. A known volume of gaseous hydrocarbon is mixed with excess of pure and dry oxygen in eudiometer tube placed inverted in a trough of mercury. The mixture is exploded by ing an electric spark between platinum electrodes. As a result carbon and hydrogen of hydrocarbon are oxidised to CO2 and H2O vapour respectively. The tube is allowed to cool to room temperature, then water vapour is condensed in to liquid water, whose volume occupied is almost negligible. Thus the gaseous mixture left in eudiometer tube is CO2 and O2. Caustic potash solution is then introduced in to eudiometer tube which absorbs CO2 completely. The gas left is unused O2.

2NaOH  CO2  Na2CO3  H 2O  Decrease in volume on introducing KOH solution gives volume of CO2 used.  Some times volume of O2 left unused is found by introducing pyrogallol and noting the decreasein volume.  Let molecular formula of gaseous hydrocarbon is C x H y . On combustion one volume of it forms

 The volume of air displaced is calculated at STP PV PV 1 1 2 2 condition by using T  T 1 2 Mol. mass of volatile substance = Mass of substance taken  22400 Volume of air displaced in ml at STP

Methods of PUrification of Organic Compounds 1.

2.

3.

4.

‘x’volumes of CO2 & ‘y/2 ’volumes of water vapour. y y  Cx H y   x   O2  xCO2  H 2O 4 2  

5.

y  1 vol  x   vol x vol y/2 vol 4   For 1 volume of hydro carbon y  Volume of O2 used =  x   vol 4  Volume of CO2 produced = x vol.

Contraction on explosion and cooling = 1 

6. y 4

(volume of liquid water is neglected)

7.

Determination of Molecular Mass by Victor Meyer’s Method :  A known mass of the volatile substance is vapourised in victor meyer’s method.  The vapours obtained displaces an equal volume of air in to graduated tube.

8.

196

A mixture of camphor and KCl can be separated by 1) Evaporation 2) Sublimation 3) Filtration 4) Decantation Impure Naphthalene is purified by 1) Fractional crystallisation 2) Fractional distillation 3) solvent extraction 4) Sublimation Which of the following method is used for the purification of solids 1) Distillation under reduced pressure 2) Distillation 3) Steam distillation 4) Sublimation Simple distillation can be used to separate liquids which differ in their boiling points at least by 1) 50 C 2) 100 C 3) 400 C-500 C 4) 1000C Simple distillation of liquids involves simultaneously 1) Vapourisation and condensation 2) Condensation and vapourisation 3) Vapourisation and sublimation 4) Sublimation and condensation A mixture of benzene and toluene can be separated by 1) Crystallization 2) Solubility 3) Separating funnel 4) Fractional distillation The best and latest technique for isolation, purification and separation of organic compounds is 1) Crystallization 2) distillation 3) sublimation 4) chromatography Chromatographic technique is used for the separation is GENERAL ORGANIC CHEMISTRY

9.

1) Camphor 2) Alcohol & Water 3) Acetone and Methanol 4) Plant pigments Chromatography is a technique based on 1) Solubilities of solute 2) Adsorption of solute 3) Chemical adsorption followed by dispersion 4) Differential adsorption of different constitents of a mixture A very common adsorbent used in column chromatography is 1) Powderd charcoal 2) Silica gel 3) Chalk 4) Sodium carbonate In column chromatography, the moving phase consists 1) A substance which is to be separated 2) Mixture of eluent and substance to be separated 3) eluent 4) Absorbent The relative adsorption of each component of the mixture is expressed in of 1) Adsorption factor 2) Retardation factor 3) Co-factor 4) Sorption factor Paper chromatography is 1) Adsorption chromatography 2) Partition chromatography 3) Ion exchange chromatography 4) all of these In paper chromatogrpahy 1) Moving phase is liquid and stationary phase is solid 2) Moving phase is liquid and stationary phase is liquid 3) Moving phase is solid and stationary phase is solid 4) Moving phase is solid and stationary phase is liquid

19. For which of the following compounds the Lassaigne’s test for the detection of nitrogen will fail 1) H 2 N  CO  NH 2 2) C6 H 5  NH  NH 2 HCl 3) NH 2  NH 2 HCl

Qualitative analysis of Organic Compounds :

4) H 2 N  CO  NH  NH 2 .HCl Organic compound is fused with metallic sodium for testing nitrogen, sulphur and halogens because 1) To make the solution alkaline 2) To convert into elemental state of nitrogen, sulphur and halogens 3) To convert covalent compound in to ionic compound 4) To decrease fusion temperature In Lassaigne’s extract, nitrogen in organic compound is converted to 1) N2 2) NH3 3) NO 4) CN– The compound not formed in the positive test for nitrogen with the Lassaigne’s solution of an organic compound is a) Fe4[Fe(CN)6]3 b) Na3[Fe(CN)6] c) Fe(CN)3 d) Na3[Fe(CN)5NOS] 1) b, c, d 2) a, b 3) a, b, c 4) a only The Lassaigne’s solution when heated with ferrous sulphate and acidified with sulphuric acid gave intense blue colour indicating the presence of nitrogen. The blue colour is due to the formation of 1) Na4  Fe  CN 6  2) Fe3  Fe  CN 6 

15. In detection of CO2, lime water turns milky due to formation of 1) CaO 2) CaCl2 3) CaCO3 4) Ca(HCO3)2 16. H2O vapours on ing through anhydrous CuSO4 turns it to 1) Green 2) Blue 3) Violet 4) White 17. Lassaigne’s test is used in qualitative analysis to detect 1) Nitrogen 2) Sulphur 3) Chlorine 4) All of these 18. Which of the following elements in an organic compound cannot be detected by Lassaigne’s test? 1) N 2) S 3) Cl 4) H

3) Fe2  Fe  CN 6  4) Fe4  Fe  CN 6  3 24. In the Lassaigne’s test the Sulphur present in the organic compound first changes into 1) Na2SO3 2) CS2 3) Na2SO4 4) Na2S 25. When lassaigne’s extract (Na2S) is acidified with acetic acid and then lead acetate solution is added to it, the colour of the precipitate is 1) Blue 2) Black 3) Red 4) White 26. Sodium extract gives blood red colour when treated with FeCl3. FOrmation of blood red colour confirms the presence of 1) Only nitrogen 2) Only sulphur 3) Only halogens 4) Both Nitrogen and Sulphur

10.

11.

12.

13.

14.

GENERAL ORGANIC CHEMISTRY

20.

21.

22.

23.

2

197

27. The presence of halogen in an organic compound is detected by 1) Iodoform test 2) Molisch’s test 3) Layer test 4) Million’s test 28. ClCH2COOH is heated with fuming HNO3 in the presence of AgNO3 in carius tube. After fibration and washing a white precipitate is obtained. The precipitate is of 1) Ag2SO4 2) ClCH2COOAg 3) AgCl 4) AgCN 29. The Blistein’s test is a rapid test used for organic compounds to detect 1) Phosphorous 2) Sulphur 3) Halogens 4) Nitrogen

Quantitative Analysis of Organic Compounds

37. In organic compounds, Sulphur is estimated as 1) BaSO4 2) BaCl2 3) Ba3(PO4)2 4) H2SO4 heat 38. Organic compound    O2 + other gases 2C+O2  2CO 5CO+I2O5  I2+5CO2 the reactions given above form is basis of direct estimation of 1) Nitrogen in organic compound 2) Oxygen in organic compound 3) Phosphorus in organic compound 4) Iodine in organic compound 39. 1.4 g of hydrocarbon on combustion gasve 1.8g water. The empirical formula of hydrocarbon is 1) CH 2) CH2 3) CH3 4) CH4 40. Molecular weight of an organic acid is given by Equivalent weight Basicity 2) Equivalent weight Basicity 3) (Equivalent weight) x (Basicity) 4) (Equivalent weight) x (valency)

30. Percentage of carbon in an organic compound is determined by 1) Duma’s method 2) Kjeldahl’s method 3) Carius method 4) Liebig’s method 31. In the Liebig’s method for the estimation of C and H, the organic compound is fused with 1) CuO pellets 2) Copper turnings 3) Iron fillings 4) Zinc-copper couple 32. In Duma’s method nitrogen in organic compound is estimated in the form of 1) N2 2) NO 3) NH3 4) N2O5 33. In Kjeldhl’s method of estimation of nitrogen, copper sulphate act as 1) Oxidizing agent 2) Reducing agent 3) Catalytic agent 4) Hydrolysing agent 34. Kjeldhal’s method cannot be used for the estimation of nitrogen in 1) C6 H5  N  N  C6 H5

01) 2 07) 4 13) 2 19) 3 25) 2 31) 1 37) 1

02) 4 08) 4 14) 2 20) 3 26) 4 32) 1 38) 2

03) 4 09) 4 15) 3 21) 4 27) 3 33) 3 39) 2

04) 3 10) 2 16) 2 22) 1 28) 3 34) 4 40) 3

05) 1 11) 2 17) 4 23) 4 29) 3 35) 2

06) 4 12) 2 18) 4 24) 4 30) 4 36) 2

Methods of PUrification of Organic Compounds :

2) N

NO2 3)

1)

1. 4) All of these

COOH 35. In Kjeldahl’s method, nitrogen present in the organic compound is first converted into 1) NH3 2) (NH4)2SO4 3) N2 4) NO 36. Halogen can be estimated by 1) Duma’s method 2) Carius method 3) Leibig’s method 4) All of these 198

Organic liquid vapourises at a temperature below its boiling point in steam distillation because 1) Mixture boils when sum of vapour pressure of water and organic liquid becomes equal to atmospheric pressure. 2) Steam distillation is actually distillation under increased pressure. 3) Water vapour does not contribute to its boiling point 4) Atmospheric pressure is reduced GENERAL ORGANIC CHEMISTRY

2.

3.

A liquid which decomposes at or below its boiling point can be purified by 1) steam distillation 2) simple distillation 3) fractional distillation 4) distillation under reduced pressure Which of the following statement is incorrect? 1) Fixed melting point can be used to test the purity of the solid organic compound 2) Hydrogen peroxide is purified by steam distillation. 3) Impurities cause a decrease in the melting point of the compound 4) Crystallisation is based on sparingly solubility of compound at low temperature

Assertion & Reason

4.

5.

6.

7.

1) If both A and B are true and the R is the correct explanation of the A. 2) If both A and R are true but R is not the correct explanation of the A. 3) If A is true but R is false 4) If A is false but R is true Assertion (A) : A mixture of plant pigments can be separated by chromatogrphy. Reason (R) : Chromatography is used for the separation of coloured substances into individual components. Assertion (A) : Moving phase is liquid and stationary phase is solid in paper chromatography. Reason (R) : Papaer chromatography is used for analysis of polar organic compounds. Assertion (A) : Thiphene present in commercial benzene as an impurity can be removed by shaking the mixture with cold concentrated H2SO4. Reason (R) : Thipohene is a heterocyclic aromatinc compound. Assertion (A) : Refining of petroleum involves fractional distillation Reason (R) : Fractional distillation involves repeated distillation.

Qualitative Analysis of Organic Compounds 8.

9.

halogens because 1) To neutralise alkaline solution of sodium fusion extract. 2) To convert sodium cyanide and sodium sulphide into HCN and H2S which are volatile. 3) To convert sodium cyanide and sodium sulphide in to sodium thiocyanate. 4) To get white precipitate of AgCN and black 10. Assertion (A) : Potassium can be used in lassigned test. Reason (R) : Potassium reacts vigorously. 11. Assertion (A) : During test for nitrogen with Lassaigne extract on adding FeCl3 solution sometimes a red precipitate is obtained. Reason (R) : Sulphur is also present.

Comprehension An organic compound was fused with sodium metal and extracted with distilled water. On adding fresh prepared FeSO4 solution followed by the addition of FeCl2 and dil. HCl produced greenish blue solution. 0.30 g fo the organic compound after kjeldahlisation evolved a gas (X) which was ed on 100ml of 0.1 M H2SO4. The excess of acid required 20 ml of 0.5 M NaOH for neutralisation. 12. The blue colour of solution is due to 1) K 4  Fe  CN 6 

13.

14. 15.

16.

The covalent compound which does not give positive test in Lassaigne’s test is 1) 1, 3-Dinitrobenzene 2) Glycine 3) Urea 4) Hydrazine Nitric acid is added to sodium extract and boiled before adding silver nitrate to test

GENERAL ORGANIC CHEMISTRY

199

2) Na3  Fe  CN 6 

3) Na2  Fe  CN 5 NO  4) Fe4  Fe  CN 6  3 Dil. HCl is added to the solution in the above test to 1) make the solution acidic 2) to dissolve FeSO4 3) to dissolve Fe(OH)2 4) to change FeSO4 to FeCl2 The gas X obtained during kjeldahlisation is 1) N2 2) N2O 3) NH3 5) N2H4 The amount of acid left after neutralisation with base is 1) 10 ml 2) 90 ml 3) 50 ml 5) 80 ml The organic compound which suitably fits into the above data is 1) CH3CONH2 2) C6H5CONH2

S 3)

H 2 N-C-NH 2

4) H2N-CO-NH2

Qualitative Analysis of Organic Compounds 17. 0.246g of an organic compound on complete combustion gave 0.198g oc carbondioxide and 0.1014g of water, then the percentage composition of carbon and hydrogen in the compound respectively. 1) 4.58, 21.95 2) 21.95, 4.58 3) 45.8, 2.195 4) 2.195, 45.8 18. 0.2033g of an organic compound in Dumas method gave 31.7 mL of moist N2 at 140C and 758 mm pressure. Percentage of N2 in the compound is (Aq. Tension at 140C = 14mm) 1) 18.44% 2) 16.89% 3) 15.60% 4) 16.00% 19. 0.5g of an organic compound containing nitrogen on kjeldahlising required 29 ml of N/5 H2SO4 for complete neutralisation of ammonia. The percentage of nitrogen in the compound is 1) 33.34 2) 16.24 4) 21.64 4) 14.84 20. 29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl’s method and the evolved ammonia was absorbed in 2omL of 0.1M HCl solution. The excess of the acid required 15mL of 0.1M NaOH solution for complete neutralization. The percentage of nitrogen in the compound is 1) 23.7 2) 29.5 3) 59.0 4) 47.4 21. 0.197g of a substance when heated with strong nitric acid and silver nitrate gave 0.3525g os silver iodide. Percentage of iodine is 1) 95% 2) 96.70% 3) 95.50% 4) 98.05% 22. If 0.32g of an organic compound containing sulphur produces 0.233 g of BaSO4. Then teh percentage of sulphur in it is. 1) 10 2) 15 3) 20 4) 25 23. If 0.1g of an organic compound containing phosphorus gave 0.222g of Mg2P2O7, then the % of phosphorus in the compound is 1) 31 2) 0.2 3) 76 4) 62 24. 0.4g of an organic compound gave 0.188g of solver bromide by a halogen estimation method. The percentage of bromine in the compound is (at. wt. of Ag=108, Br=80) 1) 20 2) 40 3) 46 4) 60 25. The molecular formula of an organic compound is C4H9N. The volume of N2 that will be given by 0.2g of the above compound at STP is .... (ml) 1) 31.5 2) 50 3) 63 4) 93

26. 0.73g of organic compound on oxidation gave 1.32g of carbondioxide. The percentage of carbon in the given compound will be 1) 49.32 2) 59.32 2) 29.32 4) 98.64 27. An organic compound has C and H percentage in the ratio 6:1 by mass and C and O percentage in the ratio 3:4 by mass the compound is 1) HCHO 2) CH3OH 3) CH3CH2OH 4) (COOH)2 28. Assertion (A) : During digestion with concentrated H2SO4 nitrogen of organic compound is converted into (NH4)2SO4 Reason (R) : (NH4)2SO4 on heating with alkali liberates NH3. 29. Assertion (A) : Oils are purified by steam distillation. Reason (R) : The compounds which decompose at their boiling points can be purified by steam distillation 30. Assertion (A) : In Duma’s method when an organic compound is heated with cupric oxide, ‘N’ is converted to N2 gas. Reason (R) : Cupric oxide oxidizes carbon and hydrogen to CO2 and water vapour

01) 1 07) 2 13) 1 19) 2 25) 1

1.

2.

200

02) 4 08) 4 14) 3 20) 1 26) 1

03) 2 09) 2 15) 3 21) 2 27) 1

04) 2 10) 1 16) 4 22) 1 28) 2

05) 4 11) 1 17) 2 23) 4 29) 3

06) 2 12) 4 18) 1 24) 1 30) 2

The ammonia evolved from the treatment of 0.3g of an organic compound for the estimation of nitrogen was ed in to 100 ml of 0.1M sulphuric aicd. The excess of acid requried 20 ml of 0.5M sodium hydroxide solution for complete neutralisation. The organic compound is 1) Acetamide 2) benzamide 3) Urea 4) thiourea 0.2g of an organic compound on analysis gave 0.147g of carbondioxide, 0.12g of water and 74.6 c.c of nitrogen at S.T.P. Calculate the weight percentage of oxygen? 1) 20.04 2) 6.66 3) 46.63 4) 26.67 GENERAL ORGANIC CHEMISTRY

3.

If a compound on analysis was found to contain C=18.5%, H=1.55%, Cl=55.04% and O=24.81%, then its empirical formula is 1) CHClO 2) CH2ClO 3) C2H2OCl 4) ClCH2O 4. An organic compound having molecular mass 60 is found to contain C=20%, H=6.67%, and N=46.67% while rest si oxygen on heating it gives NH3 along with a solid residue. The solid residue gives violet colour with alkaline copper sulphate solution. The compound is (AIEEE-2005) 1) CH3NCO 2) CH3CONH2 3) (NH2)2CO 4) CH3CH2CONH2 5. In a compound C, H and N atoms are present in 9:1:3:5 by weight. Molecular weight of compound is 108. Molecular formula of compound is(AIEEE-2002) 1) C2H6N2 2) C3H4N 3) C6H8N2 4) C9H12N3 6. Certain organic compounds contains carbon atoms twice the number of nitrogen atoms and hydrogen atoms are seven times the number of nitrogen atoms. The compound is mono acidic base and 0.45g of the compound neutralises 0.01 mole of HCl. The compound is 1) (CH3)2NH 2) C2H5NH2 3) N2H4 4) Both 1 and 2 7. 0.302g of organic compound gave 0.268g of silver bromide. The percentage of bromine in the sample is 1) 20 2) 50 3) 37.75 4) 75 8. In an estimation of S by Carius method 0.2175g of the compound gave 0.5825g of BaSO4 Percentage of S is 1) 36.78% 2) 45.50% 3) 39.48% 4) 30.69% 9. If 0.75g of an organic compound in Kjeldahl’s method neutralized 30ml of 0.25 NH2SO4, the percentage of nitrogen in the compound is 1) 20 2) 50 3) 80 4) 14 10. 0.28 g of an organic compound in Dumas method liberated 24ml of nitrogen at STP. The percentage of nitrogen in the compound is 1) 20 2) 10.71 3) 80 4) 50 11. One gram of X-N2Cl on strong heating gave 160ml of N2 gas of STP what could be X 1) C6H5 2) CH3 3) C2H5 4) C6H13 GENERAL ORGANIC CHEMISTRY

12. 0.1092g of certain diabasic Organic acid neutralised 21ml of decinormal solution of NaOH. The molar mass of acid is 1) 75gmol-1 2) 52gmol-1 3) 208gmol-1 4) 104gmol-1 13. Tyrosine, (an amino acid) is one of the constitutent of certain protein and is present to the extent of 0.22% by mass. If molar mass of tyrosine id 181gmol -1. The minimum molecular mass of protein is 1) 7 x 104 U 2) 82273U 3) 92200U 4) Above 105U 14. The Silver Salt of a Monobasic acid on ignition gave 60% of Ag. The molecular mass of the acid is 1) 37 2) 33 3) 73 4) 77 15. Each mole of Haemoglobin contains four moles of iron. If the percentage of iron in Haemoglobin is 0.35% by mass. The molar mass of Haemoglobin is 1) 6.4 x 104 gmol-1 2) 5.6 x 104 gmol-1 3) 6.4 x 105 gmol-1 4) 5.6 x105 gmol-1 16. In Carius tube, the compound ClCH2COOAg was heated with fuming HNO3 & AgNO3. After filtration and washing, a white precipitate was formed. The precipitate is 1) Ag2SO4 2) AgNO3 3) AgCl 4) ClCH2COOAg 17. In Victor Meyer’s method 0.2g of an organic substance displaced 56ml of air at STP the molecular weight of the compound is 1) 56 2) 112 2) 80 4) 28 18. 116mg of a compound on vapourisation in a Victor Meyer’s apparatus displaces 44.8mL of air measured at STP. The molecular mass of the compound is 1) 58g 2) 48g 3) 116g 4) 44.8g 19. 10.0ml of a mixture of methane and ethylene was exploded with 30ml (excess) of oxygen. After cooling, the volume was 21.0ml. Further treatment with caustic potash reduced the volume to 7.0ml. The amounts of methane and ethylene in the mixture respectively are 1) 5ml of CH4 + 5ml of C2H4 2) 6ml of CH4 + 4ml of C2H4 3) 3ml of CH4 + 7ml of C2H4 4) 4ml of CH4 + 6ml of C2H4 20. 121g of an amide obtained from a carboxylic

201

21.

22.

23.

24.

25.

ascid, RCOOH upon heating with alkali liberated 17g of ammonia. The acid is 1) Acetic acid 2) Propanoic acid 3) Benzoic acid 4) Butanoic acid An organic liquid has atomic ratio C:H:N:S as 2:3:1:1. Each mole of this basic liquid can be neutralised by 2 mole of HCl. Also 1.02g of chloroplatinate of this base, on ignition gave 0.4g of platinum. The ratio of molecular mass to emperical mass of liquid is (at. wt of Pt is 195) 1) 1 2) 2 3) 3 4) 4 Silver salt of certain Organic acid with atomic ratio C:H:O as 2:3:2 contains 65.06% of silver. If each molecule of acid has two ionisable H atoms, the formula of the acid 1) C3H6O3 2) C4H6O4 3) C2H3O2 4) C3H6O2 9.9g of an amide with molecular formula C4H5NxOy on heating with alkali liberated 1.7g of ammonia. If the percentage of oxygen is 32.33%, then the ratio of N & O atoms in the compound is 1) 2:1 2) 1:2 3) 2:5 4) 2:3 A complex compound of cobalt with the composition Co=22.58%, H=5.79%, N=32.2%, O=12.20% and Cl=27.17% on heating, loses ammonia to the extent of 32.63% of its mass. The number of molecules of ammonia present in one molecule of the cobalt complex is (At. mass of Co=58.9) 1) 7 2) 5 3) 8 4) 12 5.0g of certain metal, X (atomic mass = 27) is converted into 61.7% crystalline sulphate containing 48.6% by mass of water of crystallization. The simplest formula of the compound is 1) X 2  SO4 3 18 H 2O

2) X 2  SO4 3 14 H 2O

3) X 2  SO4 3 16 H 2O

4) X 2  SO4 3 12 H 2O

26. Match column-I with their characteristic reaction in column-II Column-I A) NH 2  NH 3Cl B) HO

NH 3 I

COOH

C) HO O2 N

D)

NH 3Cl NH

NH 3 Br

NO2

Column-II I) Sodium fusion extract of the compound gives prussian blue colour with FeSO4 II) Gives yellow precipitate with AgNO3 solution which is sparingly soluble in NH4OH III) Gives white precipitate with AgNO3 solution IV) Gives yellow precipitate with AgNO3 solution whcih is insoluble in NH4OH 1) A-III; B-I,IV; C-I, III; D-I, II 2) A-I; B-IV; C-II, III; D-I, II 3) A-III; B-I, IV; C-I, II; D-II, III 4) A-III; B-IV; C-I, II; D-III, IV 27. List-I (Chromatographyprocess) A) Ion exchange chromatography B) GSC C) GLC D) Paper chromatography List-II (Stationary phase-Mobile phase) 1) Liquid - liquid 2) solid - liquid 3) solid - gas 4) liquid - gas 4) solid - solid The correct match is 1) A-2, B-3, C-4, D-1 2) A-3, B-2, C-4, D-1 3) A-4, B-3, C-2, D-1 4) A-1, B-3, C-4, D-2 28. List-I (Process of purification) A) Crystallization B) Sublimation C) Fractional distillation D) Steamdistillation List-II (Principle involved the process) 1) Liquids which are immiscible in water possessing high boiling point, steam volatile. 2) The compound should be soluble in the solvent at its boiling temperature 3) The compound should have high vapour pressure below its melting point 4) Liquids which has B.Pt difference less than 400C The correct match is 1) A-2, B-3, C-4, D-1 2) A-2, B-3, C-1, D-4 3) A-4, B-2, C-1, D-3 4) A-3, B-2, C-4, D-1

202

GENERAL ORGANIC CHEMISTRY

29. List-I (Colour) A) blood red B) Prussian blue C) Violet D) White

List-II (Element) 1) Cl 2) S 3) N and S 4) P 5) N

The correct match is 1) A-1, B-4, C-2, D-3 3) A-3, B-5, C-2, D-1 30. List-I (Colour) A) Prussian blue B) Violet C) Blood red D) Colourless The correct match is 1) A-4, B-1, C-3, D-2 3) A-1, B-2, C-3, D-4

01) 3 07) 3 13) 2 19) 2 25) 1

1.

2.

02) 4 08) 1 14) 3 20) 3 26) 1

03) 1 09) 4 15) 1 21) 1 27) 1



% of Hydrogen = 

2) A-2, B-4, C-3, D-1 4) A-4, B-3, C-2, D-1 List-II (Element) 1) [Fe(CN)5NOS]4– 2) Fe4[Fe(CN)6]3 3) [Fe(SCN)]2+ 4) AgCl 5) Na4[Fe(CN)6] 2) A-3, B-1, C-2, D-4 4) A-2, B-1, C-3, D-4

04) 3 10) 2 16) 3 22) 2 28) 1

05) 3 11) 1 17) 3 23) 2 29) 3

W2  2  100 W  18

0.12  2  100  6.66% 0.2  18

% of Nitrogen =

W2  28  100 W  22400

74.6  46.63% 8  0.2 Remaining is oxygen % of Oxygen = 100 - (%C + %H + %N) = 100 - 73.33 = 23.67 % Element % Relative no. Simplest composition of atoms ratio C 20 1.67 I H 6.67 6.67 4 N 46.67 3.33 2 O 26.66 1.67 1 Emperical formula (CH4N2O) molecular weight of organic compound = 60 n x 60 = 60, n=1  Molecular formula = CH4N2O 

4.

06) 1 12) 4 18) 1 24) 2 30) 4

Number of milli equivalents of H2SO4 = N x V(ml) = 20 number of milli equivalents of NaOH used to neutralise excess of H2SO4 = 10 milli equivalents of acid used to neutralise ammonia = 20 - 10 = 10 % of nitrogen = 1.4  N  V  ml  1.4  10   46.66 0.3 0.3 % of nitrogen in urea 28    NH 2CONH 2   100  46.6  60   Weight of compound = W = 0.2g weight of CO2 = W1 = 0.147g Weight of H2O = W2 = 0.12g Volume of N2 at STP = 74.6 c.c

O i.e., NH  C  NH 2 2 urea 

O

O

O

NH 2  C  NH  C  NH 2  NH 3 When gently heated, urea loss ammonia to form biuret. When an aqueous biuret solution is treated with NaOH solution and a drop of CuSO4 solution, a violet colour is produced. This is known as biuret test, which is characteristic of all compounds containing. || O

203

O

NH 2  C  NH 2  NH 2  C  NH 2   

 C  NH

% of Carbon = 7 GENERAL ORGANIC CHEMISTRY

0.147  12  100  20.04% 0.2  44

5.

Element

weight ratio

C

9

H

1

11.

13.



14.

3.5

0.45  0.01 mole GMWt

15.

0.45  45 gmol 1 compound = 0.01 Now as per given condition, atomic ratio C:H:N is 2:7:1 Emperical formula is Mol. mass 45    1 . Thus, both the Emperical mass 45 formulae are correct. N2 produced by 1g of XN2Cl = 160ml

100  181  82272.7 g 0.22

 108W   107  Molar mass of an acid = n   x   108  100   1  107   180  107  73u 60   No.of iron atoms per molecule = 4 Mass of iron per molecule = 4 x 56 = 224u 0.35 parts by mass of iron is present in haemoglobin = 100parts 224 parts mass of iron is present in haemoglobin 100  224  64000 (or) 6.4  104 gmol 1 0.35 In Carius method, Cl is converted into AgCl. 44.8mL of air displaced by 116  116mg  g , So, 22400mL of air 1000 116 22400   58 g displaced by, 1000 44.8 Let the vol.of CH 4 = xml vol.of C2H4 = (10-x)ml Equations for combuston of CH4 and C2H4 respectively are : 

or Molar mass of 16. 18.

19.

160 mole 22400 1mole of N2 will be produced from salt 

i) CH 4  2O2  CO2  2 H 2O xml 2xml xml

22400  140 g 160 Molar mas of XN2Cl=140gmol-1 Molar mass of X = 140-28-35.5=76.5gmol-1 This pertains to C6H5 group.

ii) C2 H 4  3O2  2CO2  2 H 2O (10-x) 3(10-x) 2(10-x)ml Vol. after contraction and cooling i.e., Vol. of CO2 produces +O2 left unused = 21ml; Vol.of CO2 produced = 21-7 = 14ml...(iii) But from eq.(i) and (ii), total vol.of CO2 produced = xml + 2(10-x)ml....(iv) Equating (iii) & (iv), x+2(10-x)=14 (or) x=6 Vol. of CH4 = 6ml and C2H4 = 10-6=4ml



12.

0.1092  104 gmol 1 3 1.05  10 0.22 of tyrosin is present in protein = 100g; 181g of tyrosin is present in protein M 

3.5 1  1 14 4 Emperical formula = C3H4N molecular weight of the compound = 108 54n = 108 n=2 molecular formula = (Emperical formula)n = (C3H4N)2 = C6H8N2 0.45g of base = 0.01 mole of HCl

N

6.

Relative no. Simplest of atoms ratio 9 3  3 12 4 1 1 4 1

H 2 X  2 NaOH  Na2 X  2 H 2O milli moles of NaOH consumed = 21 x 0.1 = 2.1 1 milli moles of acid = (milli moles of base) 2 1   2.1  1.05 2 moles of acid = 1.05 x 10–3; 1.05 x 10-3 = 0.1092 mwt.

20.

204

RCONH 2  NaOH  RCOONa  NH 3 i.e. one mole of amide on reaction with NaOH give one mole of ammonia (NH3) Molecular mass of RCONH2 is A + 12 + 16 + 14 +2 = 121 GENERAL ORGANIC CHEMISTRY

21.

A + 44 = 121  A = 77 Hence, R group with mol. Wt.(A) is C6H5 Thus aicd is C6H5COOH Emp. formula is C2H3NS, acidity of base = 2,

= 0.383 : 5.74 : 2.3 : 0.766 : 0.766 : 1.94 = 1 : 15 : 6 : 2 : 2 : 5 Thus, the complex contains six N-atoms out of which 5 are present in form of NH3.

2  1.02  195   410   87 gmol 1  0.4 

Molecular mass = 2 

22.

25.

% of M in crystalline sulphate =

5  100  82% 61.7

% of water = 48.6% (given)

Molar mass 87  1 Emp mass 87 Emperical formula is C2H3O2

%of SO 24 = 100 - (48.6 + 8.2) = 43.2% 8.2 43.2 48.6 : : 27 96 18 = 0.3 : 0.45 : 2.7 = 1 : 1.5 : 9 = 2 : 3 : 18. X : SO42  : H 2O

 108W   107  Mol. mass = Basicity   x 

 108 100   2  107   118.0  65.06 

23.

Mol mass 118  2 Emperical mass 59 Mol. formula = (C2H3O2)2 = C4H6O4 1 mole of NH3 (17g) will be obtained from 1mole

Method of Purification of Organic Compounds : 1.

9.9  17  99 g 1.7 Mol.wt of amide = 99

of amide 

%of C =

12  4 100  48.48 99

5 1  100  5.05 %of H = 99

% of N =

14 x 1400 x  100  99 99

16 y 1600 y 100  99 99 But % of O = 32.33 (given)

2.

3.

% of C =

4.

1600 y 32.33  99  32.33 or y  2 99 1600 Further % of N = 100-(48.48+5.05+32.33)

But % N =

24.

1400 x (as calculated above) 99

1400 x  14.14  x  1 thus 99 x : y that is N : O = 1 : 2 Ratio of CO : H : N : O : Cl : NH3



22.58 5.79 32.2 12.26 27.17 32.63 : : : : : 58.9 1 14 16 35.5 17

GENERAL ORGANIC CHEMISTRY

5.

6. 205

A mixture contains four solid organic compounds A, B, C and D. On heating only C changes from solid state to vapour state directly. C can be separated from the rest in the mixture by 1) Distillation 2) Sublimation 3) Fractional distillation 4) Crystallisation Anthracene is purified by 1) Filtration 2) Crystallisation 3) Distillation 4) Sublimation Separation of two substances by fractional crystallisation depends upon their difference in 1) Densities 2) Solubilities 3) melting point 4) Boiling points Simple distillation can be used to separate 1) A mixture of benzene (b.p. 800C) and thiophene (b.p. 840C) 2) A mixture of ethanol (b.p. 78.16C) and water (b.p. 1000C) 3) A mixture of ether (b.p. 350C) adn toluene (b.p. 1100C) 4) None of the above Ortho and para nitrophenols can be separated by 1) crystallization 2) Steam distillation 3) sublimation 4) solvent extraction Two volatile liquids A and B differ in their

7.

8.

9.

10.

11.

12.

13.

14.

boiling points by 150C. The process which can be used to separated them is 1) Fractional distillation 2) Steam distillation 3) Distillation under reduced pressure 4) Simple distillation In steam distillation, the sum of the vapour pressure of the volatile compound and that of water is 1) Equal to atmospheric pressure 2) Less than atmospheric pressure 3) More than atmospheric pressure 4) Exactly half of the atmospheric pressure Turpentine oil can be purified by 1) Steam distillation 2) Vacuum distillation 3) Fractional distillation 4) Sublimation Elution is the process used for 1) Crystallization of a compound 2) Sublimation of a compound 3) Extraction of a compound 4) Distillation of a compound Chromatographic techniques of purification can be used for 1) Coloured compounds 2) Liquids 3) Solids 4) All of these Chromatography is a valuable method for the separation, isolation, purification and identification of the constituents of a mixture and it is based on general principle of 1) Rates at which components moves under the influence of mobile phase 2) Phase distribution 3) Interphase separation 4) Phase operation In adsorption chromatography mobile phase will be 1) Only solid 2) Only liquid 3) Only gas 4) Liquid as well as gas Which of the following can be used as adsorbent in adsorption chromo\atography 1) Silica gel 2) Alumina 3) Cellulose powder 4) All of these In column chromatography stationary phase is 1) Only solid 2) Only liquid

3) Only gas 4) All of these 15. Components present in the mixture separated over the adsorbent column, is called 1) Chromatography 2) Band specturm 3) Line spectrum 4) Distribution 16. Two substances when separated on the basis of partion coefficient between two liquid phase, then the technique is known as 1) column chromatography 2) paper chromatography 3) GLC 4) TLC 17. Fixed melting point of an organic compound informs 1) Purity of an organic compund 2) Conductivity of compound 3) Chemical nature of compound 4) Whether the compound is liquid or gas.

Qualitative Analysis of Organic Compounds : 18. The presence of carbon in an organic compound is detected by heating it with 1) Sodium metal to convet it into NaCN 2) CaO to convert it into CO which burns with blue flame 3) CuO to convert it into CO2 which turns lime water milky 4) Cu wire to give a bluish - green flame 19. In Lassaigne’s method organic compound fused with 1) Sodium metal 2) Zinc dust 3) Sodium carbonate and Zinc dust 4) Calcium metal 20. Which of the following statements is not applicable to Lassaigne’s test? 1) The extra elements (N, S and halogens) are tested as their inorganic salts. 2) During fusion of organic compound with sodium, Na2S, NaCN and NaX are formed. 3) It can even be used for the detection of hydrogen in the organic compound. 4) It is possible to differentiate between the different halogens. 21. A compoudn which does not give a positive

206

GENERAL ORGANIC CHEMISTRY

22.

23.

24.

25.

26.

27.

result in the Lassaigne’s test for nitrogen is 1) Urea 2) Hydroxyl amine 3) Glycine 4) Phenylhydrazine Medium of sodium extract is 1) Neutral 2) Basic 3) Acidic 4) Depends on organic compound Organic compound containing nitrogen and sulphur are preent in Lassaigne’s extract as 1) NaCNS 2) NaSCN 3) Na2S 4) NaCN The compound that doe not give a blue colour in Lassaigne’s test 1) Aniline 2) Glycine 3) Pyridine 4) Urea In lassaigne’s test for nitrogen, the blue colour is due to the formation of 1) Sodium cyanide 2) Sodium ferrocyanide 3) Ferric Ferrocyanide 4) Potassium ferrocyanide Lassaigne’s test gives a violet colouration with sodium nitroprusside, it indicates presence of 1) N 2) S 3) O 4) Cl For detection of sulphur in an organic compound, sodium nitroprusside is added to the sodium extract. A violet colour is obtained which is due to the formation of

3) Pure and dry O2 4) Pure and dry He 31. In the Liebig’s method, if ‘w’ is the mass of compound taken and ‘x’ is amount of CO2 formed then 1) %C =

32.

33.

34. 35.

36.

37.

3) Na4  Fe  CN 5 NOS  4) Na2  Fe  CN 5 NOS  28. The function of boiling the sodium extract with conc. H NO 3 before testing the halogens is 1) To convert Fe++ into Fe+++ 2) To make solution clear 3) To make solution acidic 4) To destroy cyanide and sulphide ions 29. Liebig’s method is used for the estimation of 1) Nitrogen 2) Sulphur 3) Carbon and hydrogen 4) Halogens 30. Which gas is introduced into the combustion tube in Liebig’s method? 1) Pure and dry CO2 2) Pure and dry N2 GENERAL ORGANIC CHEMISTRY

2) %C =

12 w  100 44 x

12 x 12 x   100 4) %C =  44 w 44 w In kjeldahl’s method to estimate nitrogen, compound is heated with conc. H2SO4 in presence of 1) CaSO4 2) (NH4)2SO4 3) CuSO4 4) P2O5 In Kjeldahl’s method, potassium sulphate acts as 1) Catalytic agent 2) Dehydrating agent 3) Boiling Point elevator 4) Reducing agent In Carius method halogens are estimated as 1) X2 2) BaX2 3) PbX2 4) AgX In organic compounds, phosphorus is finally estimated as 1) Mg2P2O7 2) H3OP4 3) Mg3(PO4)2 4) P2O5 An alkane has Carbon and Hydrogen ratio (by mass) is 5.1428 : 1. Its molecular formula is 1) C7H19 2) C5H12 3) C6H14 4) C8H18 If two compounds have the same emperical formula but different molecular formulae they must have 1) Different percentage composition 2) Different molecular weight 3) same viscosity 4) same vapourdensity

3) %C =

1) Fe4  Fe  CN 6  3 2) Fe  SCN 3

12  x 16  w

207

01) 2 07) 1 13) 4 19) 1 25) 3 31) 3 37) 2

02) 4 08) 1 14) 1 20) 3 26) 2 32) 3

03) 2 09) 3 15) 1 21) 2 27) 3 33) 3

04) 3 10) 4 16) 2 22) 2 28) 4 34) 4

05) 2 11) 1 17) 1 23) 2 29) 3 35) 1

06) 1 12) 4 18) 3 24) 3 30) 3 36) 3

7.

Method of Purification of Organic Compounds : 1.

2.

3.

Which of the following statements is not correct regarding purification of liquids by steam distillation? 1) Impurities must be non-volatile 2) The liquid must be completely immiscible with water 3) The liquid must possess high boiling point 4) The liquid must be miscible in water Distillation under reduced pressure method is used to purity the liquids in which the liquids 1) have high boiling points 2) have low boiling points 3) do not decompose at their boiling points 4) highly volatile Components of rectified spirit cannot be separated by distillation because 1) it forms an Azetropic mixture 2) it has components with same boiling points 3) it contains immiscible liquids 4) it has high vapour pressure

8.

9.

10.

11.

12.

Qualitative Analysis 4.

5.

6.

Which of the following compounds will give blood red colour while doing the Lassaigne’s test for N? 1)  NH 2  2 CO

2) C6 H 5 SO3 H

3)  NH 2  2 CS

4) CHCl3

13.

CCl4 contains four chlorine atoms whne it is heated with silver nitrate solution then the number of moles of AgCl formed is 1) 4 2) 2 3) 0 4) 3 0.2 g of an organic compound on complete combustion produces 0.18 g of water, then the percentage of hydrogen in its (Aq. tension at 288K = 12.7mm) 1) 5 2) 10 3) 20 4) 15 208

0.303 g of an organic compound was analysed for nitrogen by Kjeldhahl’s method. The ammonia gas evolved was absorbed in 50ml of 0.05M H2SO4. The excess acid required 25ml of 0.1M NaOH for neutralization. The percentage of nitrogen in the given compound is 1) 11.5 2) 23 30 12.5 4) 14.5 0.50g of an organic compound was Kjeldahlised and the NH 3 evolved was absorbed in 50ml of 0.5M H2SO4. The residual acid required 60cm3 of 0.5M NaOH. The percentage of nitrogen in the organic compound is 1) 14 2) 28 3) 56 4) 42 0.99g of organic compound containing halogen when heated with furning HNO3 is presence of AgNO3 in a carius tube gave 0.287g of white precipitate. The percentage of halogen is 1) 29.6 2) 71.7 3) 35.4 4) 64.2 Insulin contains 3.4% sulphur, the minimum molecular weight of insulin is 1) 350 2) 470 3) 560 4) 940 0.12g of an organci compound gave 0.22g Mg 2 P 2 O 7 by the usual analysis. The percentage of phosphorus in the compound is 1) 15.23 2) 38.75 3) 51.20 4) 60.92 In Dumas method 0.5 g of an organic compound containing nitrogen gave 112 ml of nitrogen at S.T.P. The percentage of nitrogen in the given compound is 1) 28 2) 38 3) 18 4) 48 Compound A contains 20% C, 46.66% n and 6.66% H. It gave NH3 on heating with NaOH. A can be 1) CH3CONH2 2) NH2CONH2 3) C6H5CONH2 4) CH3NHCONH2

01) 4 07) 1 13) 2

02) 1 08) 3

03) 1 09) 2

04) 3 10) 4

05) 3 06) 2 11) 3 12) 3

GENERAL ORGANIC CHEMISTRY