NCERT Solutions for Class 11 Chemistry Part 2 Chapter 2 - Organic Chemistry

Carbon dioxide is an acidic substance, whereas potassium hydroxide is a strong basic. As a result, when carbon dioxide combines with potassium hydroxide, potassium carbonate and water are formed.

2KOH + CO2→K2CO3 + H2O

As a result, the mass of the KOH containing U-tube increases. This rise represents the amount of CO2 created. The proportion of carbon in the organic compound can be determined based on its mass.

 

 The following are the first 5 members of each homologous series, starting with the given compounds: 

(a) H -- COOH: Methanoic acid

CH3 -- COOH : Ethanoic acid

CH3 -- CH2 -- COOH : Propanoic acid 

CH3 -- CH2 -- CH2 -- COOH : Butanoic acid

CH3 -- CH2 -- CH2 -- CH2 -- COOH : Pentanoic acid

(b) CH3COCH3 : Propanone 

CH3COCH2CH3 : Butanone

CH3COCH2CH2CH3 : Pentan - 2 - one 

CH3COCH2CH2CH2CH3 : Hexan - 2 - one

CH3COCH2CH2CH2CH2CH3 : Heptan - 2 - one

(c) H -- CH  =  CH2 : Ethene

CH3 -- CH  =  CH2 : Propene 

CH3 -- CH2 -- CH  =  CH2 : 1 - Butene

CH3 -- CH2 -- CH2 -- CH  =  CH2 : 1 - Pentene

CH3 -- CH2 -- CH2 -- CH2 -- CH  =  CH2 : 1 - Hexene

 

a. C^1⁡H2 = C^2⁡ =  O

In the given compound, carbon-1 is bonded to two hydrogen atoms and one carbon atom. So, according to the VSEPR theory, the steric number is 3 which corresponds to sp2 hybridization. Carbon-2 is bonded to one carbon atom and one oxygen atom. So, the steric number is 2 which corresponds to sp hybridization.

b. C1⁡H3C2⁡H  = C3⁡H2

In the given compound, carbon-1 is bonded to three hydrogen atoms and one carbon atom. So, according to the VSEPR theory, the steric number is 4 which corresponds to sp3 hybridization. Carbon-2 is bonded to two carbon atoms and one hydrogen atom. So, the steric number is 3 which corresponds to sp2 hybridization.

c. C1⁡H3C2⁡OC3⁡H3

In the given compound, carbon-1 and carbon-3 are bonded to three hydrogen atoms and one carbon atom. So, according to the VSEPR theory, the steric number is 4 which corresponds to sp3 hybridization. Carbon-2 is bonded to two carbon atoms and one oxygen atom. So, the steric number is 3 which corresponds to sp2 hybridization. 

d. C1⁡H2 = C2⁡HC3⁡N

In the given compound, carbon-1 is bonded to two hydrogen atoms and one carbon atom. So, according to the VSEPR theory, the steric number is 3 which corresponds to sp2 hybridization. Carbon-2 is bonded to two carbon atoms and one hydrogen atom. So, the steric number is 3 which corresponds to sp2 hybridization. Carbon-3 is bonded to one carbon atom and one nitrogen atom. So, the steric number is 2 which corresponds tosp hybridization.

e. C6H6

In the given compound, all carbon atoms are bonded to two carbon atoms and one hydrogen atom. So, according to the VSEPR theory, the steric number is 3 which corresponds to sp2 hybridization.

 

The structure of C6H6 is shown below.

 

Single bond contains only one sigma bond and double bond contains one sigma and one pi bond. In this structure, nine single bonds and three double bonds are present. So, there are 12 σ and 3 π  bonds present.

The structure of C6H12 is shown below.

 

In this structure, eighteen single bonds are present. So, there are only 18 σ  bonds present. The structure of CH2Cl2 is shown below.

 

In this structure, four single bonds are present. So, there are only 4 σ  bonds present.

The structure of CH2 =  C  =  CH2 is shown below.

 

In this structure, four single bonds and two double bonds are present. So, there are 6 σ and 2 π bonds present.

The structure of CH3NO2 is shown below.

 

In this structure, five single bonds and one double bond are present. So, there are 6 σ and 1 π bonds present.

The structure of HCONHCH3 is shown below.

 

In this structure, seven single bonds and one double bond are present. So, there are 8 σ and 1 π bonds present.

 

The bond line formulas for the given compounds are shown below.

1. Isopropyl Alcohol: 

2. 2,3 − Dimethylbutanal

3. Heptan − 4 − one

(a)

 

In this structure, the longest chain of carbon contains 3 carbon atoms and one phenyl group is present. So, the IUPAC name is 1-phenyl propane. 

(b)

 

In this structure, the longest chain of carbon contains four carbon atoms. Also, one methyl and one cyanide groups are attached with carbon-2 and carbon-1 respectively. So, the IUPAC name is 2-methyl-1-cyanobutane.

(c)

 

In this structure, the longest chain of carbon contains 7 carbon atoms and 2 methyl are positioned at carbon-2 and carbon-5. So, the IUPAC name of the compound is 2, 5-dimethyl heptane.

(d)

 

In this structure, the longest chain of carbon contains 7 carbon atoms. Also, 1 chloro and 1 bromo groups are present at carbon-3. So, the IUPAC name of the compound is 3-bromo-3-chloroheptane.

(e)

 

In this structure, the longest chain of carbon contains 3 carbon atoms. Also, 1 aldehyde and 1 chloro functional groups are present at carbon-1 and carbon-3 respectively. So, the IUPAC name of the compound is 3-chloropropanal.

(f)

 

In this structure, the longest chain of carbon contains 2 carbon atoms. Also, 1 alcohol and 2 chloro functional groups are present at carbon-1 and carbon-2 respectively. So, the IUPAC name of the compound is 2,2-dichloro-1-ethanol.

 

1. In the IUPAC nomenclature, the prefix Di- denotes that the parent chain has two identical substituent groups. As there are two methyl groups in the C–2 of the parent chain of the given chemical compound, the correct IPUAC name is 2,2 − Dimethylpentane.

2. The locant number 2, 4, 7 is less than the locant number 2, 5, 7. As a result, the given compound's IUPAC nomenclature is 2,4,7 − Trimethyloctane

3. If the functional groups are present at the parent chain's equivalent position, the lower number is assigned to the one that appears first in the name in alphabetical order. As a result, the proper IUPAC nomenclature for the given compound is 2 − Chloro − 4 − methylpentane.

4. The given molecule contains two functional groups: alcoholic and alkyne. The alcoholic group is the principal functional group. As a result, the parent chain will be terminated with ol. The alkyne group can be found in the parent chain's C–3. As a result, But − 3 − yn − 1 − ol is the correct IUPAC nomenclature for the given compound.

 

a. 2,2,4 − Trimethylpentane

Condensed formula:(CH3)2CHCH2C (CH3)3

Bond Line Formula:

b. 2 − Hydroxy − 1,2,3 − propanetricarboxylic acid

Condensed Formula: 

(COOH)CH2C(OH)(COOH)CH2(COOH)

Bond Line Formula:

In the given compound, the functional groups present are carboxylic acid

(−COOH) and alcoholic group (−OH).

c. Hexanedial 

Condensed Formula:

(CHO)(CH2)4(CHO)

Bond Line Formula:

In the given compound, the functional group present is aldehyde

(−CHO).

 

a. In the given compound, aldehyde (−−CHO), hydroxyl(−−OH) and methoxy(−−OMe) functional groups are present.

b. In the given compound, amino (−−NH2), ketone(C = O), diethylamine

(N(C2H5)2) functional groups are present.

c. In the given compound, nitro group (−NO2) and C = C double bond are present.

 

 The nitro group is an electron-withdrawing group. As a result, it exhibits the -I effect. This group reduces the compound's negative charge by attracting electrons towards it and stabilizing it. The ethyl group, on the other hand, is an electron-releasing group. As a result, the ethyl group exhibits the +I effect. This raises the compound's negative charge, making it unstable. As a result,

O2NCH2CH2O -- is projected to be more stable than CH3CH2O --.

 

a. The bond cleavage can be illustrated using curved arrows to show the electron flow of the given reaction as

As one of the shared pairs in a covalent bond travels with the connected atom, it is an example of homolytic cleavage. A free radical is produced as a reaction intermediate.

b. The bond cleavage can be illustrated using curved arrows to show the electron flow of the given reaction as

It is an example of heterolytic cleavage because the bond breaks in such a way that the shared pair of electrons stay with the carbon of propanone. The reactive intermediate created is carbanion.

c. The bond cleavage can be illustrated using curved arrows to show the electron flow of the given reaction as

It is an example of heterolytic cleavage because the bond breaks in such a way that the bromine ion retains the shared pair of electrons. A carbocation is produced as a reaction intermediate.

d. The bond cleavage can be illustrated using curved arrows to show the electron flow of the given reaction as

The bond breaks in such a way that the shared pair of electrons stays with one of the fragments, resulting in a heterolytic cleavage. A carbocation is created as an intermediate.

 

When an alkyl group is linked to a π system, the process of hyperconjugation causes it to act as an electron-donor group. Let us use propene as an example to better comprehend this topic.

 

The sigma electrons of an alkyl group's C-H bond are delocalised during hyperconjugation. This group is directly connected to an unsaturated system's atom. The delocalisation happens as a result of a partial overlap of a sp3 sigma bond orbital with an empty p orbital of the pi bond of a nearby carbon atom's bond.

1. The resonating structures for C6H5OH are shown below.

 

2. The resonating structures for C6H5NO2 are shown below.

 

3. The resonating structures for CH3CH  =  CHCHO are shown below.

 

4. The resonating structures of C6H5 -- CHO are shown below.

 

5. The resonating structures of C6H5 -- C+⁡H2 are shown below.

 

6. The resonating structures of CH3CH  =  CHC+⁡H2 are shown below.

 

An electrophile is a reagent that removes a pair of electrons. In other words, an electrophile (E+) is a reagent that seeks electrons. Electrophiles are electron-deficient and receiver of an electron pair.

Electrophiles include carbocations and (CH3CH2+) neutral compounds with functional groups such as the carbonyl group.

A reagent that produces an electron pair is known as a nucleophile. A nucleus-seeking reagent is referred to as a nucleophile (Nu:).

For instance, OH−, NC−, carbanions, (R3C−), and so on.

Because of the presence of a lone pair, neutral molecules such as water and ammonia also serve as nucleophiles.

 

Electrophiles are entities that are electron-deficient and can accept an electron pair. Nucleophiles, on the other hand, are electron-rich entities that can transfer their electrons.

a. In this equation, OH− serves as a nucleophile because it is an electron-rich species.

b. In this equation, CN−serves as a nucleophile because it is an electron-rich species.

c. In this equation,CH3C + ⁡O serves as an electrophile because it is an electron-deficient species.

 

1. The given reaction is a type of a substitution reaction because the bromine group in bromoethane is substituted by thionyl group.

2. The given reaction is a type of an addition reaction because two reactant molecules react to produce a single product.

3. The given reaction is a type of an elimination reaction because hydrogen and bromine are eliminated by bromoethane in order to produce ethene.

4. Substitution takes place in this reaction, followed by atom and atom group rearrangement.

 

a. Compounds with the same chemical formula but different structures are referred to as structural isomers. The mentioned compounds have the same molecular formula, but the position of the functional group differs (ketone group).

The ketone group is at C-3 of the parent chain (hexane chain) in structure I, and at C-2 of the parent chain in structure II (hexane chain). As a result, the given pair of compounds are structural isomers.

b. Geometrical isomers are compounds that have the same chemical formula, composition, and series of covalent bonds but have different relative positions of their atoms in space.

The relative positions of Deuterium (D) and hydrogen (H) in space differ in structures I and II. As a result, the given pairs are geometrical isomers.

c. The structures shown are either canonical or contributing structures. They are fictitious and do not represent any actual molecule. As a result, the given pair represents resonance structures known as resonance isomers.

 

Inductive effect: It refers to the permanent shift of sigma electrons along a saturated chain whenever an electron withdrawing or electron donating group is present. The inductive effect could be either a + I effect or − I effect. When an atom or group attracts electrons more strongly than hydrogen, it is said to have the − I effect. Example, 

F−←CH2−←CH2−←CH2−←CH3

When an atom or set of atoms attracts electrons less strongly than hydrogen, this is referred to as the + I effect. For instance,

CH3−→CH2−→Cl

Electromeric effect: In the presence of an attacking agent, it involves the complete transfer of the shared pair of pi electrons to either of the two atoms joined by multiple bonds. For example,

 

Electrometric effects might be + E or − E. The + E effect occurs when electrons are transferred to the assaulting reagent. Whereas when electrons are moved away from the attacking reagent, this is referred to as the − E effect.

(a) Cl3CCOOH  >  Cl2CHCOOH  >  ClCH2COOH 

On the basis of the inductive effect(− I), the order of acidity can be explained. The − I effect increases as the number of chlorine atoms increases. The acid strength increases in proportion to the increase in − I effect.

 

(b) CH3CH2COOH  > (CH3)2CHCOOH  > (CH3)3C.COOH

The inductive effect (+ I effect) can be used to explain the order of acidity. The

+ I effect increases in proportion to the quantity of alkyl groups. The acid strength increases in proportion to the increase in + I effect.

a. Crystallisation: It is one of the most widely used methods for purifying solid chemical molecules.

Principle: The differential in the solubilities of the compound and the impurities in a solvent. The impure substance dissolves in the solvent, which is only sparingly soluble at ambient temperature but completely soluble at higher temperatures. The solution is concentrated until it is virtually saturated. When the solution is cooled, the pure compound crystallises and is extracted by filtration. 

Pure aspirin, for example, is created by recrystallising crude aspirin. 2 - 4 g of crude aspirin is dissolved in approximately 20 mL of ethyl alcohol. To ensure complete decomposition, the solution is heated.

b. Distillation: This approach is used to separate volatile liquids from nonvolatile contaminants or a mixture of those liquids with a large enough difference in boiling points.

Principle: It works on the principle that liquids with different boiling points vapourize at different temperatures. The vapours are then cooled, and the liquids that form are separated. 

For example: Distillation can be used to separate a combination of chloroform and aniline. The mixture is placed in a condenser-equipped round bottom flask. After that, it is heated. Because chloroform is more volatile, it vaporizes first and enters the condenser. The vapours condense in the condenser, and chloroform trickles down. Aniline is left in the round bottom flask.

c. Chromatography: It is one of the most effective technologies for organic chemical separation and purification.

Principle: The differential in movement of separate components of a mixture through the stationary phase under the influence of the mobile phase is the basis for the principle.

Example: Chromatography, can separate a combination of red and blue ink. On the chromatogram, a drop of the combination is inserted. The component of the ink that is less adsorbed on the chromatogram travels with the mobile phase, whereas the component that is more adsorbed remains stationary.

 

The process of fractional crystallisation is used to separate two compounds with varying solubilities in a solvent S. The fractional crystallisation process is divided into four phases.

1. Solution’s Preparation: The powdered combination is placed in a flask, and the solvent is slowly and simultaneously added and agitated. The solvent is added until the solute is completely dissolved in it. This saturated solution is then boiled.

2. Solution’s Filtration: In a China dish, the hot saturated solution is filtered through filter paper.

3. Fractional Crystallization: Allow the solution in a China dish to cool. To begin with, the less soluble molecule crystallizes, while the more soluble compound remains in solution. During the separation of these crystals from the liquor, the latter is concentrated once more. The hot solution is allowed to cool, resulting in the formation of crystals of the more soluble chemical.

4. Isolation and Drying: Filtration is used to separate the crystals from the mother liquor. The crystals are finally dried.

 

The differences between distillation, reduced pressure distillation, and steam distillation are shown below.

Distillation: It is used to purify chemicals that are related with non-volatile impurities or liquids that do not disintegrate when heated. In other words, distillation is used to separate volatile liquids from nonvolatile contaminants or a mixture of such liquids with significant boiling point difference. This process separates a mixture of fuel and kerosene.

Reduced Pressure Distillation: This procedure is used to purify a liquid that decomposes when heated. Under reduced pressure, the liquid will boil at a lower temperature than its boiling point and so will not decompose.

This Process is Used to Purify Glycerol. At a temperature of 593 K, it boils with decomposition. It boils at 453 K without breakdown at a lower pressure.

Steam Distillation: It is used to purify an organic molecule that is steam volatile and water immiscible. The compound is heated by passing steam, and the steam is condensed to water. After some time, the water-liquid mixture begins to boil and flows through the condenser. A separating funnel is then used to separate this condensed mixture of water and liquid.

A solution of water and aniline can be separated by this method.

 

Lassaigne's Test: This test detects the presence of nitrogen, sulfur, halogens, and phosphorus in an organic compound. An organic compound has these elements in covalent form. By fusing the chemical with sodium metal, these are transformed into the ionic form.

Na  +  C  +  N → NaCN

Na  +  S  +  C  +  N → NaSCN

2Na  +  S → Na2S

Na  +  X → NaX

The sodium cyanide, sulphide, and halide are removed from the fused material by boiling it in distilled water. Lassaigne's extract is the name given to the extract obtained in this manner. The presence of nitrogen, sulphur, halogens, and phosphorus is then determined in this Lassaigne's extract.

1. Chemistry of Nitrogen Test: The sodium fusion extract is heated with iron (II) sulphate and then acidified with sulphuric acid in the Lassaigne's test for nitrogen in an organic molecule. Sodium cyanide first interacts with iron (II) sulphate to generate sodium hexacyanoferrate in this method (II). The iron (II) is then oxidized by sulphuric acid to generate iron (III) hexacyanoferrate (II), which is Prussian blue. The reactions involved are shown below.

6CN -  +  Fe2 + → [Fe(CN)6]4 - 

3[Fe(CN)6]4 -  +  4Fe3 + → Fe4[Fe(CN)6]3

2. Test for Sulphur: 

(i) Lassaignes's extract  +  Lead acetate→ Black precipitate

Chemistry of the test: The sodium fusion extract is acidified with acetic acid and then lead acetate is added to it in the Lassaigne's test for sulphur in an organic compound. The presence of sulphur in the combination is shown by the precipitation of lead sulphide, which is black in color.

S2 -  +  Pb2 +  → PbS

(ii) Lassaignes's extract  +  Sodium nitroprusside → Violet colour

Chemistry of the Test: Sodium nitroprusside is used to treat the sodium fusion extract. The presence of sulphur in the compound is also indicated by the presence of violet color.

S2 -  +  [Fe(CN)5NO]2 - → [Fe(CN)5NOS] - 4 

If both nitrogen and sulphur are present in an organic molecule, NaSCN is formed instead of NaCN.

Na  +  C  +  N  +  S → NaSCN

This NaSCN (sodium thiocyanate) color is blood red. The absence of free cyanide ions prevents the formation of prussian color.

3. Test for Halogen: The sodium fusion extract is acidified with nitric acid and then treated with silver nitrate in the Lassaigne's test for halogens in an organic molecule. If the organic compound contains both nitrogen and sulphur, the Lassaigne's extract is heated to remove the nitrogen and sulphur, which would otherwise interfere with the halogen test.

 

In the Dumas method, a specified amount of a nitrogen-containing organic molecule is burned strongly with an excess of copper oxide in a carbon dioxide atmosphere to yield free nitrogen as well as carbon dioxide and water.

The process can also yield traces of nitrogen oxides, which can be converted to dinitrogen by passing the gaseous mixture over a heated copper gauge. The dinitrogen generated is collected over a potassium hydroxide aqueous solution. At room temperature and atmospheric pressure, the volume of nitrogen generated is then measured.

Kjeldahl's method, on the other hand, involves heating a known quantity of a nitrogen-containing organic molecule with concentrated sulphuric acid. The compound's nitrogen is quantitatively transformed into ammonium sulphate. It is then distilled with a high concentration of sodium hydroxide. The ammonia produced by this method is injected into a known volume of H2SO4.The chemical equations involved are shown below.

Volumetric analysis (titration against a standard alkali) is used to measure the quantity of acid that is left unused, and the amount of ammonia produced can be calculated. As a result, the percentage of nitrogen in the compound can be calculated. This approach is inapplicable to compounds with nitrogen in a ring structure, and it is also inapplicable to compounds with nitro and azo groups.

 

Estimation of Halogens: The Carius method is used to calculate halogens. In this approach, a known amount of organic chemical is heated with fuming nitric acid in the presence of silver nitrate in a Carius tube, which is placed in a furnace. The carbon and hydrogen in the compound are oxidized to generate CO2 and H2O, respectively, while the halogen in the compound is transformed into the form of AgX . After that, the AgX is filtered, washed, dried, and weighed.

Estimation of Sulphur: In this approach, a known amount of organic substance is heated in a hard glass tube called the Carius tube with either fuming nitric acid or sodium peroxide. The compound's sulphur is oxidized to generate sulphuric acid. The precipitation of barium sulphate occurs when an excess of barium chloride is added to it. After that, the precipitate is filtered, washed, dried, and weighed.

Estimation of Phosphorus: A known amount of organic substance is heated with fuming nitric acid in this procedure. The compound's phosphorus is oxidized to generate phosphoric acid. Phosphorus can be precipitated as ammonium phosphomolybdate by adding ammonia and ammonium molybdate to the solution. Phosphorus can also be measured by precipitating it as

MgNH4PO4 and then on igniting it yeild Mg2P2O7.

 

Paper chromatography employs the use of chromatography paper. This paper includes trapped water, which behaves as the stationary phase. The solution of the mixture is spotted on the base of this chromatography paper. The paper strip is then hung in a suitable solvent, which serves as the mobile phase. By capillary action, this solvent moves up the chromatography paper and flows over the spot during the procedure. Different component spots travel to different heights with mobile phase. The resulting paper is known as a chromatogram.

 

The Lassaigne's extract is first heated with weak nitric acid before being tested for the presence of halogens. This is done to convert NaCN to HCN and Na2S to H2S and then expel these gases. If there is any nitrogen or sulphur present in the form of NaCN or Na2S, it is eliminated. The chemical equations involved are shown below.

NaCN  +  HNO3  → HCN  +  NaNO3

Na2S  +  2HNO3 → H2S  +  2NaNO3

 

Organic chemicals are covalently bonded to nitrogen, sulfur, and halogens. They must first be changed to ionic form in order to be detected. This is accomplished by combining the organic chemical with sodium metal. This is known as the "Lassaigne's test." The chemical equations used in the test are as follows:

Na  +  C  +  N → NaCN

Na  +  S  +  C  +  N → NaSCN

2Na  +  S → Na2S

Na  +  X → NaX

Here, X is halogen like chlorine, bromine, iodine.

Organic compounds contain carbon, nitrogen, sulfur, and halogen.

A combination of camphor and calcium sulphate is separated by sublimation. The sublimable chemical converts from solid to vapour without going through the liquid state throughout this procedure. Camphor is a sublimable molecule, whereas calcium sulphate is a non-sublimable solid. As a result, when heated, camphor will sublime while calcium sulphate will remain.

 

In the process of steam distillation, the organic liquid begins to boil when the total of the organic liquid’s vapour pressure (p1) and the water’s vapour 

(p2) equals atmospheric pressure

(p), i.e. p = 

(p), i.e. p = p1 + p2. Therefore, p1 < p2

 Hence, organic liquid will vapourize at a lower temperature than its boiling point.

 

When CCl4 heated with silver nitrate, it does not produce the white precipitate of AgCl. This is because the chlorine atoms in CCl4 are covalently connected to carbon. It must be present in ionic form in order to obtain the precipitate, which requires preparing the Lassaigne's extract of CCl4.

 

Due to the common ion effect, the addition of sulphuric acid would precipitate lead sulphate, but the addition of acetic acid would assure complete precipitation of sulphur in the form of lead sulphate. As a result, acetic acid must be used to acidify sodium extract before testing for sulphur using the lead acetate test.

 

The percentage of carbon in organic compound is 69 %.

It means 100 g of organic compound contains 69 g of carbon. So, the mass of carbon contained in 0.20 g of organic compound is calculated as

Mass of C  = 69 × 0.20100 

= 0.138 g of C

The molecular mass of carbon dioxide is 44 g/mol. Therefore,

12g of carbon is contained in 44 g of CO2.

Therefore, the mass of CO2 contained in 0.138 g of carbon is calculated as 

Mass of CO2 = 44 × 0.13812 

 = 0.506 g

Thus, 0.506 g of CO2 will be produced on complete combustion of 0.2 g of organic compound.

The percentage of hydrogen in organic compound is 4.8 % . It means 100 g of organic compound contains 4.8 g of hydrogen.

Therefore, the mass of hydrogen contained in 0.20 g of organic compound is calculated as

Mass of H  = 4.8 × 0.20100 

= 0.0096 g of H

It is known that the molecular mass of water is 18 g/mol.

Thus, 2 g of hydrogen is contained in 18 g of water. Therefore, the mass of water contained in 0.0096 g of hydrogen is calculated as

Mass of H2O  = 18 × 0.00962  

= 0.0864 g

 

Given: The total mass of organic compound is 0.50 g. 60 mL of 0.5 M solution of NaOH was required for neutralisation by residual acid.

60 mL of 0.5 M NaOH solution  = 602 mL of 0.5M H2SO4

602 mL of 0.5M H2SO4 =  30 mL of 0.5 M H2SO4

Therefore, acid consumed in absorption of evolved ammonia is calculated as 

 = (50 - 30) mL 

 =  20 mL

Again,

20 mL of 0.5 M H2SO4 =  40 mL of 0.5 M NH3

Also, since 1000 mL of 1 M NH3 contains 14 g of nitrogen, the mass of nitrogen in 40 mL of 0.5 M NH3 is calculated as

Mass of N  = 14 × 40 × 0.5 / 1000 

= 0.28 g

Therefore, the percentage of nitrogen in 0.50 g of organic compound is calculated as

Percentage of nitrogen  = 0.280.5 × 100% 

=  56 % 

 

Given: The mass of organic compound is 0.3780 g and the mass of silver chloride formed is 0.5740 g. The molar mass of AgCl is 143.32 g/mol and the molar mass of chlorine is 35.5 g/mol.

One mole of silver chloride contains one mole of chlorine. So, the mass of chlorine in 0.5740 g of silver chloride is calculated as

Mass of AgCl  = 35.5 × 0.5740143.32  

=  0.1421 g

Thus, the percentage of chlorine is calculated as

Percentage of chlorine  = 0.14210.3780 × 100% 

 =  37.59 % 

Hence, the percentage of chlorine is 37.59 % .

 

Given: The total mass of organic compound is 0.468 g and the mass of barium sulphate formed is 0.668 g.

The molar mass of BaSO4 is 233 g/mol and the molar mass of sulphur is 32 g/mol.

So, 1 mol of BaSO4 =  233 g of BaSO4 

 =  32 g of sulphur

Thus, 0.668 g of BaSO4 = 32 × 0.668233 g of sulphur

=  0.0917 g of sulphur

Therefore, the percentage of sulphur is calculated as

Percentage of sulphur  = 0.09170.468 × 100% 

=  19.59 % 

Hence, the percentage of sulphur is 19.59 % .

 

The given organic compound is shown below.

C1⁡H2 = C2⁡H -- C3⁡H2 -- C4⁡H2 -- C5⁡ ≡ C6⁡H

The hybridization of carbon atoms numbered as 1, 2, 3, 4, 5 and 6 are

sp, sp, sp3 , sp3 , sp2 , and sp2 respectively. Therefore, the pair of hybridized orbitals involved in the formation of C2 -- C3 bond is sp −− sp3.

 

The sodium fusion extract is heated with iron (II) sulphate and then acidified with sulphuric acid in the Lassaigne's test for nitrogen in an organic molecule. Sodium cyanide initially interacts with iron (II) sulphate to generate sodium hexacyanoferrate (II). The iron (II) is then oxidized by sulphuric acid to generate iron (III) hexacyanoferrate (II), which is Prussian blue. The chemical reactions involved are shown below. 

6CN− + Fe2+ →[Fe(CN)6]4−

3[Fe(CN)6]4− + 4Fe3+ →x.H2O Fe4[Fe(CN)6]3.xH2O Prussian Blue 

Therefore, the Prussian blue color is formed due to the formation of complex

Fe4[Fe(CN)6]3.

 

(CH3)3C +  is a tertiary carbocation. Because of the electron-releasing effect of three methyl groups, a tertiary carbocation is the most stable carbocation. Three methyl groups increase the + I effect, which stabilizes the positive charge on the carbocation. The stability of tertiary carbocations is also due to hyperconjugation and resonance energy.

 

Chromatography is the most effective and up-to-date method of separating and purifying organic molecules. Initially, it was employed to separate a combination of colored substances.

 

The reaction is given as:

CH3CH2I  +  KOH(aq) → CH3CH2OH  +  KI

It shows a nucleophilic substitution reaction. The hydroxyl group of potassium hydroxide with a lone pair serves as a nucleophile, substituting iodide ion in CH3CH2I to produce ethanol.