The elements present in organic compounds are carbon and hydrogen. In
addition to these, they may also contain oxygen, nitrogen, sulphur,
halogens and phosphorus.
12.9.1 Detection of Carbon and Hydrogen
Carbon and hydrogen are detected by heating the compound with copper(II) oxide. Carbon present in the compound is oxidised to carbon dioxide (tested with lime-water, which develops turbidity) and hydrogen to water (tested with anhydrous copper sulphate, which turns blue).
12.9.2 Detection of Other Elements
Nitrogen, sulphur, halogens and phosphorus present in an organic compound are detected by “Lassaigne’s test”. The elements present in the compound are converted from covalent form into the ionic form by fusing the compound with sodium metal. Following reactions take place:
C, N, S and X come from organic compound.
Cyanide, sulphide and halide of sodium so formed on sodium fusion are extracted from the fused mass by boiling it with distilled water. This extract is known as sodium fusion extract.
(A) Test for Nitrogen
The sodium fusion extract is boiled with iron(II) sulphate and then acidified with concentrated sulphuric acid. The formation of Prussian blue colour confirms the presence of nitrogen. Sodium cyanide first reacts with iron(II) sulphate and forms sodium hexacyanoferrate(II). On heating with concentrated sulphuric acid some iron(II) ions are oxidised to iron(III) ions which react with sodium hexacyanoferrate(II) to produce iron(III) hexacyanoferrate(II) (ferriferrocyanide) which is Prussian blue in colour.
(B) Test for Sulphur
(a) The sodium fusion extract is acidified with acetic acid and lead acetate is added to it. A black precipitate of lead sulphide indicates the presence of sulphur.
S2-+Pb2+→PbS
Black
(b) On treating sodium fusion extract with sodium nitroprusside, appearance of a violet colour further indicates the presence of sulphur.
S2-+[Fe(CN)5NO]2-→[Fe(CN)5NOS]4-
Violet
In case, nitrogen and sulphur both are present in an organic compound, sodium thiocyanate is formed. It gives blood red colour and no Prussian blue since there are no free cyanide ions.
Na+C+N+S→NaSCN
Fe3++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 give their usual tests.
NaSCN + 2Na → NaCN+Na2S
(C) Test for Halogens
The sodium fusion extract is acidified with nitric acid and then treated with silver nitrate. A white precipitate, soluble in ammonium hydroxide shows the presence of chlorine, a yellowish precipitate, sparingly soluble in ammonium hydroxide shows the presence of bromine and a yellow precipitate, insoluble in ammonium hydroxide shows the presence of iodine.
X-+Ag+→AgX
X represents a halogen -Cl, Br or I
If nitrogen or sulphur is also present in the compound, the sodium fusion extract is first boiled with concentrated nitric acid to decompose cyanide or sulphide of sodium formed during Lassaigne’s test. These ions would otherwise interfere with silver nitrate test for halogens.
(D) Test for Phosphorus
The compound is heated with an oxidising agent (sodium peroxide). The phosphorus present in the compound is oxidised to to phosphate. The solution is boiled with nitric acid and then treated with ammonium molybdate. A yellow colouration or precipitate indicates the presence of phosphorus.
Na3PO4+3HNO3→H3PO4+3NaNO3
H3PO4+12(NH4)
Ammonium molybdate
2MoO4+21HNO3→(NH4)
33PO4.12MoO3+21NH4NO3+12H2O
Ammonium phosphomolybdate
12.10 QUANTITATIVE ANALYSIS
The percentage composition of elements present in an organic compound is determined by the methods based on the following principles:
12.10.1 Carbon and Hydrogen
Both carbon and hydrogen are estimated in one experiment. A known mass of an organic compound is burnt in the presence of excess of oxygen and copper(II) oxide. Carbon and hydrogen in the compound are oxidised to carbon dioxide and water respectively.
CxHy + (x + y/4) O2 → x CO2 + (y/2) H2O
The mass of water produced is determined by passing the mixture through a weighed U-tube containing anhydrous calcium chloride. Carbon dioxide is absorbed in another U-tube containing concentrated solution of potassium hydroxide. These tubes are connected in series (Fig.12.14). The increase in masses of calcium chloride and potassium hydroxide gives the amounts of water and carbon dioxide from which the percentages of carbon and hydrogen are calculated.
Let the mass of organic compound be m g, mass of water and carbon dioxide produced be m1 and m2 g respectively;
Percentage of carbon=12×m2×100/44×m
Percentage of hydrogen=2×m1×100/18×m
Problem 12.20
On complete combustion, 0.246 g 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.
Solution
Percentage of carbon = 12×0.198×100/44×0.246
= 21.95%
Percentage of hydrogen = 2×0.1014× 100/ 18× 0.246
= 4.58%
12.10.2 Nitrogen
There are two methods for estimation of nitrogen: (i) Dumas method and (ii) Kjeldahl’s method.
(i) Dumas method: The nitrogen containing organic compound, when heated with copper oxide in an atmosphere of carbon dioxide, yields free nitrogen in addition to carbon dioxide and water.
CxHyNz + (2x + y/2) CuO → x CO2 + y/2 H2O + z/2 N2 + (2x + y/2) Cu
Traces of nitrogen oxides formed, if any, are reduced to nitrogen by passing the gaseous mixture over a heated copper gauze. The mixture of gases so produced is collected over an aqueous solution of potassium hydroxide which absorbs carbon dioxide. Nitrogen is collected in the upper part of the graduated tube (Fig.12.15).
Let the mass of organic compound = m g
Volume of nitrogen collected = V1 mL
Room temperature = T1K
Volume of nitrogen at STP=p1V1×273/760×T1
(Let it be V mL)
Where p1 and V1are the pressure and volume of nitrogen, p1 is different from the atmospheric pressure at which nitrogen gas is collected. The value of p1is obtained by the relation:
p1= Atmospheric pressure – Aqueous tension 22400 mL N2 at STP weighs 28 g.
V mL N2 at STP weighs =(28×V/22400)g.
Percentage of nitrogen =
28×V× 100/22400×m
Problem 12.21
In Dumas’ method for estimation of nitrogen, 0.3g of an organic compound gave 50mL of nitrogen collected at 300K temperature and 715mm pressure.Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300K=15 mm)
Solution
Volume of nitrogen collected at 300K and 715mm pressure is 50 mL
Actual pressure = 715-15 =700 mm
Volume of nitrogen at STP =273×700×50/300×760
=41.9mL
22,400 mL of N2 at STP weighs = 28 g
28 41.9
41.9 mL of nitrogen weighs=(28×41.9/22400)g
Percentage of nitrogen = (28× 41.9× 100/22400× 0.3)
=17.46%
(ii) Kjeldahl’s method: The compound containing nitrogen is heated with concentrated sulphuric acid. Nitrogen in the compound gets converted to ammonium sulphate (Fig. 12.16). The resulting acid mixture is then heated with excess of sodium hydroxide. The liberated ammonia gas is absorbed in an excess of standard solution of sulphuric acid. The amount of ammonia produced is determined by estimating the amount of sulphuric acid consumed in the reaction. It is done by estimating unreacted sulphuric acid left after the absorption of ammonia by titrating it with standard alkali solution. The difference between the initial amount of acid taken and that left after the reaction gives the amount of acid reacted with ammonia.
2NH3+H2SO4 → (NH4)2SO4
Let the mass of organic compound taken = m g
Volume of H2SO4 of molarity, M,taken = V mL
Volume of NaOH of molarity, M, used for titration of excess of H2SO4 = V1 mL V1mL of NaOH of molarity M = V1/2 mL of H2SO4 of molarity M
Volume of H2SO4 of molarity M unused =(V-V1/2) mL
(V- V1/2) mL of H2SO4 of molarity M
= 2(V-V1/2) mL of NH3 solution of molarity M.
1000 mL of 1 M NH3 solution contains 17g
NH3 or 14 g of N
2(V-V1/2) mL of NH3 solution of molarity M contains:
(14×M×2(V-V1/2)/1000)gN
Percentage of N=(14×M×2(V-V1/2)/1000)×100/m
=1.4×M×2(V-V1/2)/m
Kjeldahl method is not applicable to compounds containing nitrogen in nitro and azo groups and nitrogen present in the ring (e.g. pyridine) as nitrogen of these compounds does not change to ammonium sulphate under these conditions.
Problem 12.22
During estimation of nitrogen present in an organic compound by Kjeldahl’s method, the ammonia evolved from 0.5 g of the compound in Kjeldahl’s estimation of nitrogen, neutralized 10 mL of 1 M H2SO4. Find out the percentage of nitrogen in the compound.
Solution
1 M of 10 mL H2SO4=1M of 20 mL NH3
1000 mL of 1M ammonia contains 14 g nitrogen
20 mL of 1M ammonia contains (14×20/1000) g nitrogen
Percentage of nitrogen = (14×20×100/1000×0.5)=56.0%
12.10.3 Halogens
Carius method: A known mass 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, (Fig.12.17) in a furnace. Carbon and hydrogen present in the compound are oxidised to carbon dioxide and water. The halogen present forms the corresponding silver halide (AgX). It is filtered, washed, dried and weighed.
Let the mass of organic
compound taken = m g
Mass of AgX formed = m1 g
1 mol of AgX contains 1 mol of X
Mass of halogen in m1g of AgX
=atomic mass of X ×m1g/molecular mass of AgX
Percentage of halogen = atomic mass of X ×m1×100/molecular mass of AGX× m
Problem 12.23
In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12 g of AgBr. Find out the percentage of bromine in the compound.
Solution
Molar mass of AgBr = 108 + 80
= 188 g mol-1
188g AgBr contains 80 g bromine
0.12g AgBr contains= 80×012/188 g bromine
Percentage of bromine=80× 0.12× 100/188× 0.15
= 34.04%
12.10.4 Sulphur
A known mass of an organic compound is heated in a Carius tube with sodium peroxide or fuming nitric acid. Sulphur present in the compound is oxidised to sulphuric acid. It is precipitated as barium sulphate by adding excess of barium chloride solution in water. The precipitate is filtered, washed, dried and weighed. The percentage of sulphur can be calculated from the mass of barium sulphate.
Let the mass of organic compound taken = m g
and the mass of barium sulphate formed = m1g
1 mol of BaSO4 = 233 g BaSO4 = 32 g sulphur m1 g BaSO4 contains 32×m1/233g sulphur
percentage of sulphur= 32×m1×100/233×m
Problem 12.24
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?
Solution
Molecular mass of BaSO4 = 137+32+64
= 233 g
233 g BaSO4 contains 32 g sulphur
0.4813 g BaSO4 contains 32× 0.4813/233g
sulphur
Percentage of sulphur= 32×0.4813×100/233× 0.157
= 42.10%
12.10.5 Phosphorus
A known mass of an organic compound is heated with fuming nitric acid whereupon phosphorus present in the compound is oxidised to phosphoric acid. It is precipitated as ammonium phosphomolybdate, (NH4)3 PO4.12MoO3, by adding ammonia and ammonium molybdate. Alternatively, phosphoric acid may be precipitated as MgNH4PO4 by adding magnesia mixture which on ignition yields Mg2P2O7.
Let the mass of organic compound taken = m g and mass of ammonium phosphomolydate = m1g
Molar mass of (NH4)3PO4.12MoO3 = 1877 g
Percentage of phosphorus =(31×100/1877×m)%
If phosphorus is estimated as Mg2P2O7,
Percentage of phosphorus = (62×m1 100/222×m)%
where, 222 u is the molar mass of Mg2P2O7, m, the mass of organic compound taken, m1, the mass of Mg2P2O7 formed and 62, the mass of two phosphorus atoms present in the compound Mg2P2O7.
12.10.6 Oxygen
The percentage of oxygen in an organic compound is usually found by difference between the total percentage composition (100) and the sum of the percentages of all other elements. However, oxygen can also be estimated directly as follows:
A definite mass of an organic compound is decomposed by heating in a stream of nitrogen gas. The mixture of gaseous products containing oxygen is passed over red-hot coke when all the oxygen is converted to carbon monoxide. This mixture is passed through warm iodine pentoxide (I2O5) when carbon monoxide is oxidised to carbon dioxide producing iodine.
On making the amount of CO produced in equation (A) equal to the amount of CO used in equation (B) by multiplying the equations (A) and (B) by 5 and 2 respectively; we find that each mole of oxygen liberated from the compound will produce two moles of carbondioxide.
Thus 88 g carbon dioxide is obtained if 32 g oxygen is liberated.
Let the mass of organic compound taken be m g Mass of carbon dioxide produced be m1 g
m1g carbon dioxide is obtained from
(32 m1/88) g O2
∴Percentage of oxygen = (32 m1 100/ 88 m)%
The percentage of oxygen can be derived from the amount of iodine produced also.
Presently, the estimation of elements in an organic compound is carried out by using microquantities of substances and automatic experimental techniques. The elements, carbon, hydrogen and nitrogen present in a compound are determined by an apparatus known as CHN elemental analyser. The analyser requires only a very small amount of the substance (1-3 mg) and displays the values on a screen within a short time. A detailed discussion of such methods is beyond the scope of this book.
SUMMARY
In this unit, we have learnt some basic concepts in structure and reactivity of organic compounds, which are formed due to covalent bonding. The nature of the covalent bonding in organic compounds can be described in terms of orbitals Hybridisation concept, according to which carbon can have sp3, sp2 and sp hybridised orbitals. The sp3, sp2 and sp hybridised carbons are found in compounds like methane, ethene and ethyne respectively. The tetrahedral shape of methane, planar shape of ethene and linear shape of ethyne can be understood on the basis of this concept. A sp3 hybrid orbital can overlap with 1s orbital of hydrogen to give a carbon – hydrogen (C-H) single bond (sigma, σ bond). Overlap of a sp2 orbital of one carbon with sp2 orbital of another results in the formation of a carbon-carbon σ bond. The unhybridised p orbitals on two adjacent carbons can undergo lateral (side-byside) overlap to give a pi (π) bond. Organic compounds can be represented by various structural formulas. The three dimensional representation of organic compounds on paper can be drawn by wedge and dash formula.Organic compounds can be classified on the basis of their structure or the functional groups they contain. A functional group is an atom or group of atoms bonded together in a unique fashion and which determines the physical and chemical properties of the compounds. The naming of the organic compounds is carried out by following a set of rules laid down by the International Union of Pure and Applied Chemistry (IUPAC). In IUPAC nomenclature, the names are correlated with the structure in such a way that the reader can deduce the structure from the name.
Organic reaction mechanism concepts are based on the structure of the substrate molecule, fission of a covalent bond, the attacking reagents, the electron displacement effects and the conditions of the reaction. These organic reactions involve breaking and making of covalent bonds. A covalent bond may be cleaved in heterolytic or homolytic fashion. A heterolytic cleavage yields carbocations or carbanions, while a homolytic cleavage gives free radicals as reactive intermediate. Reactions proceeding through heterolytic cleavage involve the complimentary pairs of reactive species. These are electron pair donor known as nucleophile and an electron pair acceptor known as electrophile. The inductive, resonance, electromeric and hyperconjugation effects may help in the polarisation of a bond making certain carbon atom or other atom positions as places of low or high electron densities.
Organic reactions can be broadly classified into following types; substitution, addition, elimination and rearrangement reactions.
Purification, qualitative and quantitative analysis of organic compounds are carried out for determining their structures. The methods of purification namely : sublimation, distillation and differential extraction are based on the difference in one or more physical properties.Chromatography is a useful technique of separation, identification and purification of compounds. It is classified into two categories : adsorption and partition chromatography.Adsorption chromatography is based on differential adsorption of various components of a mixture on an adsorbent. Partition chromatography involves continuous partitioning of the components of a mixture between stationary and mobile phases. After getting the compound in a pure form, its qualitative analysis is carried out for detection of elements present in it. Nitrogen, sulphur, halogens and phosphorus are detected by Lassaigne’s test. Carbon and hydrogen are estimated by determining the amounts of carbon dioxide and water produced. Nitrogen is estimated by Dumas or Kjeldahl’s method and halogens by Carius method. Sulphur and phosphorus are estimated by oxidising them to sulphuric and phosphoric acids respectively. The percentage of oxygen is usually determined by difference between the total percentage (100) and the sum of percentages of all other elements present
12.9.1 Detection of Carbon and Hydrogen
Carbon and hydrogen are detected by heating the compound with copper(II) oxide. Carbon present in the compound is oxidised to carbon dioxide (tested with lime-water, which develops turbidity) and hydrogen to water (tested with anhydrous copper sulphate, which turns blue).
12.9.2 Detection of Other Elements
Nitrogen, sulphur, halogens and phosphorus present in an organic compound are detected by “Lassaigne’s test”. The elements present in the compound are converted from covalent form into the ionic form by fusing the compound with sodium metal. Following reactions take place:
C, N, S and X come from organic compound.
Cyanide, sulphide and halide of sodium so formed on sodium fusion are extracted from the fused mass by boiling it with distilled water. This extract is known as sodium fusion extract.
(A) Test for Nitrogen
The sodium fusion extract is boiled with iron(II) sulphate and then acidified with concentrated sulphuric acid. The formation of Prussian blue colour confirms the presence of nitrogen. Sodium cyanide first reacts with iron(II) sulphate and forms sodium hexacyanoferrate(II). On heating with concentrated sulphuric acid some iron(II) ions are oxidised to iron(III) ions which react with sodium hexacyanoferrate(II) to produce iron(III) hexacyanoferrate(II) (ferriferrocyanide) which is Prussian blue in colour.
(B) Test for Sulphur
(a) The sodium fusion extract is acidified with acetic acid and lead acetate is added to it. A black precipitate of lead sulphide indicates the presence of sulphur.
S2-+Pb2+→PbS
Black
(b) On treating sodium fusion extract with sodium nitroprusside, appearance of a violet colour further indicates the presence of sulphur.
S2-+[Fe(CN)5NO]2-→[Fe(CN)5NOS]4-
Violet
In case, nitrogen and sulphur both are present in an organic compound, sodium thiocyanate is formed. It gives blood red colour and no Prussian blue since there are no free cyanide ions.
Na+C+N+S→NaSCN
Fe3++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 give their usual tests.
NaSCN + 2Na → NaCN+Na2S
(C) Test for Halogens
The sodium fusion extract is acidified with nitric acid and then treated with silver nitrate. A white precipitate, soluble in ammonium hydroxide shows the presence of chlorine, a yellowish precipitate, sparingly soluble in ammonium hydroxide shows the presence of bromine and a yellow precipitate, insoluble in ammonium hydroxide shows the presence of iodine.
X-+Ag+→AgX
X represents a halogen -Cl, Br or I
If nitrogen or sulphur is also present in the compound, the sodium fusion extract is first boiled with concentrated nitric acid to decompose cyanide or sulphide of sodium formed during Lassaigne’s test. These ions would otherwise interfere with silver nitrate test for halogens.
(D) Test for Phosphorus
The compound is heated with an oxidising agent (sodium peroxide). The phosphorus present in the compound is oxidised to to phosphate. The solution is boiled with nitric acid and then treated with ammonium molybdate. A yellow colouration or precipitate indicates the presence of phosphorus.
Na3PO4+3HNO3→H3PO4+3NaNO3
H3PO4+12(NH4)
Ammonium molybdate
2MoO4+21HNO3→(NH4)
33PO4.12MoO3+21NH4NO3+12H2O
Ammonium phosphomolybdate
12.10 QUANTITATIVE ANALYSIS
The percentage composition of elements present in an organic compound is determined by the methods based on the following principles:
12.10.1 Carbon and Hydrogen
Both carbon and hydrogen are estimated in one experiment. A known mass of an organic compound is burnt in the presence of excess of oxygen and copper(II) oxide. Carbon and hydrogen in the compound are oxidised to carbon dioxide and water respectively.
CxHy + (x + y/4) O2 → x CO2 + (y/2) H2O
The mass of water produced is determined by passing the mixture through a weighed U-tube containing anhydrous calcium chloride. Carbon dioxide is absorbed in another U-tube containing concentrated solution of potassium hydroxide. These tubes are connected in series (Fig.12.14). The increase in masses of calcium chloride and potassium hydroxide gives the amounts of water and carbon dioxide from which the percentages of carbon and hydrogen are calculated.
Let the mass of organic compound be m g, mass of water and carbon dioxide produced be m1 and m2 g respectively;
Percentage of carbon=12×m2×100/44×m
Percentage of hydrogen=2×m1×100/18×m
Problem 12.20
On complete combustion, 0.246 g 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.
Solution
Percentage of carbon = 12×0.198×100/44×0.246
= 21.95%
Percentage of hydrogen = 2×0.1014× 100/ 18× 0.246
= 4.58%
12.10.2 Nitrogen
There are two methods for estimation of nitrogen: (i) Dumas method and (ii) Kjeldahl’s method.
(i) Dumas method: The nitrogen containing organic compound, when heated with copper oxide in an atmosphere of carbon dioxide, yields free nitrogen in addition to carbon dioxide and water.
CxHyNz + (2x + y/2) CuO → x CO2 + y/2 H2O + z/2 N2 + (2x + y/2) Cu
Traces of nitrogen oxides formed, if any, are reduced to nitrogen by passing the gaseous mixture over a heated copper gauze. The mixture of gases so produced is collected over an aqueous solution of potassium hydroxide which absorbs carbon dioxide. Nitrogen is collected in the upper part of the graduated tube (Fig.12.15).
Let the mass of organic compound = m g
Volume of nitrogen collected = V1 mL
Room temperature = T1K
Volume of nitrogen at STP=p1V1×273/760×T1
(Let it be V mL)
Where p1 and V1are the pressure and volume of nitrogen, p1 is different from the atmospheric pressure at which nitrogen gas is collected. The value of p1is obtained by the relation:
p1= Atmospheric pressure – Aqueous tension 22400 mL N2 at STP weighs 28 g.
V mL N2 at STP weighs =(28×V/22400)g.
Percentage of nitrogen =
28×V× 100/22400×m
Problem 12.21
In Dumas’ method for estimation of nitrogen, 0.3g of an organic compound gave 50mL of nitrogen collected at 300K temperature and 715mm pressure.Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300K=15 mm)
Solution
Volume of nitrogen collected at 300K and 715mm pressure is 50 mL
Actual pressure = 715-15 =700 mm
Volume of nitrogen at STP =273×700×50/300×760
=41.9mL
22,400 mL of N2 at STP weighs = 28 g
28 41.9
41.9 mL of nitrogen weighs=(28×41.9/22400)g
Percentage of nitrogen = (28× 41.9× 100/22400× 0.3)
=17.46%
(ii) Kjeldahl’s method: The compound containing nitrogen is heated with concentrated sulphuric acid. Nitrogen in the compound gets converted to ammonium sulphate (Fig. 12.16). The resulting acid mixture is then heated with excess of sodium hydroxide. The liberated ammonia gas is absorbed in an excess of standard solution of sulphuric acid. The amount of ammonia produced is determined by estimating the amount of sulphuric acid consumed in the reaction. It is done by estimating unreacted sulphuric acid left after the absorption of ammonia by titrating it with standard alkali solution. The difference between the initial amount of acid taken and that left after the reaction gives the amount of acid reacted with ammonia.
2NH3+H2SO4 → (NH4)2SO4
Let the mass of organic compound taken = m g
Volume of H2SO4 of molarity, M,taken = V mL
Volume of NaOH of molarity, M, used for titration of excess of H2SO4 = V1 mL V1mL of NaOH of molarity M = V1/2 mL of H2SO4 of molarity M
Volume of H2SO4 of molarity M unused =(V-V1/2) mL
(V- V1/2) mL of H2SO4 of molarity M
= 2(V-V1/2) mL of NH3 solution of molarity M.
1000 mL of 1 M NH3 solution contains 17g
NH3 or 14 g of N
2(V-V1/2) mL of NH3 solution of molarity M contains:
(14×M×2(V-V1/2)/1000)gN
Percentage of N=(14×M×2(V-V1/2)/1000)×100/m
=1.4×M×2(V-V1/2)/m
Kjeldahl method is not applicable to compounds containing nitrogen in nitro and azo groups and nitrogen present in the ring (e.g. pyridine) as nitrogen of these compounds does not change to ammonium sulphate under these conditions.
Problem 12.22
During estimation of nitrogen present in an organic compound by Kjeldahl’s method, the ammonia evolved from 0.5 g of the compound in Kjeldahl’s estimation of nitrogen, neutralized 10 mL of 1 M H2SO4. Find out the percentage of nitrogen in the compound.
Solution
1 M of 10 mL H2SO4=1M of 20 mL NH3
1000 mL of 1M ammonia contains 14 g nitrogen
20 mL of 1M ammonia contains (14×20/1000) g nitrogen
Percentage of nitrogen = (14×20×100/1000×0.5)=56.0%
12.10.3 Halogens
Carius method: A known mass 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, (Fig.12.17) in a furnace. Carbon and hydrogen present in the compound are oxidised to carbon dioxide and water. The halogen present forms the corresponding silver halide (AgX). It is filtered, washed, dried and weighed.
Let the mass of organic
compound taken = m g
Mass of AgX formed = m1 g
1 mol of AgX contains 1 mol of X
Mass of halogen in m1g of AgX
=atomic mass of X ×m1g/molecular mass of AgX
Percentage of halogen = atomic mass of X ×m1×100/molecular mass of AGX× m
Problem 12.23
In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12 g of AgBr. Find out the percentage of bromine in the compound.
Solution
Molar mass of AgBr = 108 + 80
= 188 g mol-1
188g AgBr contains 80 g bromine
0.12g AgBr contains= 80×012/188 g bromine
Percentage of bromine=80× 0.12× 100/188× 0.15
= 34.04%
12.10.4 Sulphur
A known mass of an organic compound is heated in a Carius tube with sodium peroxide or fuming nitric acid. Sulphur present in the compound is oxidised to sulphuric acid. It is precipitated as barium sulphate by adding excess of barium chloride solution in water. The precipitate is filtered, washed, dried and weighed. The percentage of sulphur can be calculated from the mass of barium sulphate.
Let the mass of organic compound taken = m g
and the mass of barium sulphate formed = m1g
1 mol of BaSO4 = 233 g BaSO4 = 32 g sulphur m1 g BaSO4 contains 32×m1/233g sulphur
percentage of sulphur= 32×m1×100/233×m
Problem 12.24
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?
Solution
Molecular mass of BaSO4 = 137+32+64
= 233 g
233 g BaSO4 contains 32 g sulphur
0.4813 g BaSO4 contains 32× 0.4813/233g
sulphur
Percentage of sulphur= 32×0.4813×100/233× 0.157
= 42.10%
12.10.5 Phosphorus
A known mass of an organic compound is heated with fuming nitric acid whereupon phosphorus present in the compound is oxidised to phosphoric acid. It is precipitated as ammonium phosphomolybdate, (NH4)3 PO4.12MoO3, by adding ammonia and ammonium molybdate. Alternatively, phosphoric acid may be precipitated as MgNH4PO4 by adding magnesia mixture which on ignition yields Mg2P2O7.
Let the mass of organic compound taken = m g and mass of ammonium phosphomolydate = m1g
Molar mass of (NH4)3PO4.12MoO3 = 1877 g
Percentage of phosphorus =(31×100/1877×m)%
If phosphorus is estimated as Mg2P2O7,
Percentage of phosphorus = (62×m1 100/222×m)%
where, 222 u is the molar mass of Mg2P2O7, m, the mass of organic compound taken, m1, the mass of Mg2P2O7 formed and 62, the mass of two phosphorus atoms present in the compound Mg2P2O7.
12.10.6 Oxygen
The percentage of oxygen in an organic compound is usually found by difference between the total percentage composition (100) and the sum of the percentages of all other elements. However, oxygen can also be estimated directly as follows:
A definite mass of an organic compound is decomposed by heating in a stream of nitrogen gas. The mixture of gaseous products containing oxygen is passed over red-hot coke when all the oxygen is converted to carbon monoxide. This mixture is passed through warm iodine pentoxide (I2O5) when carbon monoxide is oxidised to carbon dioxide producing iodine.
On making the amount of CO produced in equation (A) equal to the amount of CO used in equation (B) by multiplying the equations (A) and (B) by 5 and 2 respectively; we find that each mole of oxygen liberated from the compound will produce two moles of carbondioxide.
Thus 88 g carbon dioxide is obtained if 32 g oxygen is liberated.
Let the mass of organic compound taken be m g Mass of carbon dioxide produced be m1 g
m1g carbon dioxide is obtained from
(32 m1/88) g O2
∴Percentage of oxygen = (32 m1 100/ 88 m)%
The percentage of oxygen can be derived from the amount of iodine produced also.
Presently, the estimation of elements in an organic compound is carried out by using microquantities of substances and automatic experimental techniques. The elements, carbon, hydrogen and nitrogen present in a compound are determined by an apparatus known as CHN elemental analyser. The analyser requires only a very small amount of the substance (1-3 mg) and displays the values on a screen within a short time. A detailed discussion of such methods is beyond the scope of this book.
SUMMARY
In this unit, we have learnt some basic concepts in structure and reactivity of organic compounds, which are formed due to covalent bonding. The nature of the covalent bonding in organic compounds can be described in terms of orbitals Hybridisation concept, according to which carbon can have sp3, sp2 and sp hybridised orbitals. The sp3, sp2 and sp hybridised carbons are found in compounds like methane, ethene and ethyne respectively. The tetrahedral shape of methane, planar shape of ethene and linear shape of ethyne can be understood on the basis of this concept. A sp3 hybrid orbital can overlap with 1s orbital of hydrogen to give a carbon – hydrogen (C-H) single bond (sigma, σ bond). Overlap of a sp2 orbital of one carbon with sp2 orbital of another results in the formation of a carbon-carbon σ bond. The unhybridised p orbitals on two adjacent carbons can undergo lateral (side-byside) overlap to give a pi (π) bond. Organic compounds can be represented by various structural formulas. The three dimensional representation of organic compounds on paper can be drawn by wedge and dash formula.Organic compounds can be classified on the basis of their structure or the functional groups they contain. A functional group is an atom or group of atoms bonded together in a unique fashion and which determines the physical and chemical properties of the compounds. The naming of the organic compounds is carried out by following a set of rules laid down by the International Union of Pure and Applied Chemistry (IUPAC). In IUPAC nomenclature, the names are correlated with the structure in such a way that the reader can deduce the structure from the name.
Organic reaction mechanism concepts are based on the structure of the substrate molecule, fission of a covalent bond, the attacking reagents, the electron displacement effects and the conditions of the reaction. These organic reactions involve breaking and making of covalent bonds. A covalent bond may be cleaved in heterolytic or homolytic fashion. A heterolytic cleavage yields carbocations or carbanions, while a homolytic cleavage gives free radicals as reactive intermediate. Reactions proceeding through heterolytic cleavage involve the complimentary pairs of reactive species. These are electron pair donor known as nucleophile and an electron pair acceptor known as electrophile. The inductive, resonance, electromeric and hyperconjugation effects may help in the polarisation of a bond making certain carbon atom or other atom positions as places of low or high electron densities.
Organic reactions can be broadly classified into following types; substitution, addition, elimination and rearrangement reactions.
Purification, qualitative and quantitative analysis of organic compounds are carried out for determining their structures. The methods of purification namely : sublimation, distillation and differential extraction are based on the difference in one or more physical properties.Chromatography is a useful technique of separation, identification and purification of compounds. It is classified into two categories : adsorption and partition chromatography.Adsorption chromatography is based on differential adsorption of various components of a mixture on an adsorbent. Partition chromatography involves continuous partitioning of the components of a mixture between stationary and mobile phases. After getting the compound in a pure form, its qualitative analysis is carried out for detection of elements present in it. Nitrogen, sulphur, halogens and phosphorus are detected by Lassaigne’s test. Carbon and hydrogen are estimated by determining the amounts of carbon dioxide and water produced. Nitrogen is estimated by Dumas or Kjeldahl’s method and halogens by Carius method. Sulphur and phosphorus are estimated by oxidising them to sulphuric and phosphoric acids respectively. The percentage of oxygen is usually determined by difference between the total percentage (100) and the sum of percentages of all other elements present
Great. Precise. Will benifit anyone who reads it. Thanks.
ReplyDeleteI like this. it's good as many examples have been targeted
ReplyDeleteA large part of Alfa Chemistry's customers are pharmaceutical and biotechnology companies, including Pfizer, Novartis, Merck & Co., Johnson & Johnson, AstraZeneca, and Bayer. Alfa Chemistry is also a preferred partner for many universities and non-profit institutes. yariv reagent
ReplyDelete