NCERT Solutions Class 11 Chemistry Chapter 12 Organic Chemistry

Question 12.1: What are hybridisation states of each carbon atom in the following compounds?
CH₂ = C = O, CH₃ CH = CH₂, (CH₃)₂CO, CH₂ = CHCN, C₆H₆
Answer:

C–1 is sp² hybridised.
C–2 is sp² hybridised.
C–3 is sp hybridised.

(v) C₆H₆
All the 6 carbon atoms in benzene are sp2 hybridised.

2. Indicate the σ and π bonds in the following molecules:
C₆H₆, C₆H₁₂, CH₂Cl₂, CH₂ = C = CH₂, CH₃NO₂, HCO NHCH₃
Answer:
(i) C₆H₆

Question 12.3: Write bond line formulas for: Isopropyl alcohol, 2,3-Dimethyl butanal, Heptan-4-one.
Answer The bond line formulae of the given compounds are:
(a) Isopropyl alcohol

Question 12.4: Give the IUPAC names of the following compounds:

(f) Cl₂CHCH₂OH
Answer

Question 12.5: Which of the following represents the correct IUPAC name for the compounds concerned?
(a) 2,2-Dimethylpentane or 2-Dimethylpentane
(b) 2,4,7-Trimethyloctane or 2,5,7-Trimethyloctane
(c) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane
(d) But-3-yn-1-ol or But-4-ol-1-yne
Answer (a) The prefix di in the IUPAC name indicates that two identical substituent groups are present in the parent chain. Since two methyl groups are present in the C–2 of the parent chain of the given compound, the correct IPUAC name of the given compound is 2, 2–dimethylpentane.
(b) Locant number 2, 4, 7 is lower than 2, 5, 7. Hence, the IUPAC name of the given compound is 2, 4, 7–trimethyloctane.
(c) If the substituents are present in the equivalent position of the parent chain, then the lower number is given to the one that comes first in the name according to the alphabetical order. Hence, the correct IUPAC name of the given compound is 2–chloro– 4–methylpentane.
(d) Two functional groups – alcoholic and alkyne – are present in the given compound.
The principal functional group is the alcoholic group. Hence, the parent chain will be suffixed with ol. The alkyne group is present in the C–3 of the parent chain. Hence, the correct IUPAC name of the given compound is But–3–yn–1–ol.

Question 12.6: Draw formulas for the first five members of each homologous series beginning with thefollowing compounds.
(a) H – COOH (b) CH₃COCH₃ (c) H – CH = CH₂
Answer The first five members of each homologous series beginning with the given compounds are shown as follows:
(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)
CH₃ COCH₃ :Propanone
CH₃ COCH₂ CH₃: Butanone
CH₃ COCH₂ CH₂ CH₃ : Pentan-2-one
CH₃ COCH₂ CH₂CH₂ CH₃ : Hexan-2-one
CH₃ COCH₂ CH₂ CH₂ CH₂ CH₃ : 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

Question 12.7: Give condensed and bond line structural formulas and identify the functional group (s) present, if any, for :
(a) 2,2,4-Trimethylpentane
(b) 2-Hydroxy-1,2,3-propanetricarboxylic acid
(c) Hexanedial
Answer (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:
The functional groups present in the given compound are carboxylic acid (–COOH) and
alcoholic (–OH) groups.
(c) Hexanedial
Condensed Formula:
(CHO) (CH₂)₄ (CHO)
Bond line Formula:

The functional group present in the given compound is aldehyde (–CHO).

Question 12.8: Identify the functional groups in the following compounds
(a)

Answer
The functional groups present in the given compounds are:
(a) Aldehyde (–CHO),
Hydroxyl (–OH),
Methoxy (–OMe),

Question 12.9: Which of the two: O₂NCH₂CH₂O– or CH₃CH₂O– is expected to be more stable and why?
Answer NO₂ group is an electron-withdrawing group. Hence, it shows –I effect. By withdrawing the electrons toward it, the NO₂
group decreases the negative charge on the compound, thereby stabilising it. On the other hand, ethyl group is an electron-releasing group.
Hence, the ethyl group shows +I effect. This increases the negative charge on the compound, thereby destabilising it. Hence, O₂NCH₂CH₂O^– is expected to be more stable than CH₃CH₂O^–
.
Question 12.10: Explain why alkyl groups act as electron donors when attached to a π system.
Answer When an alkyl group is attached to a π system, it acts as an electron-donor group by the process of hyperconjugation. To understand this concept better, let us take the example of propene.

In hyperconjugation, the sigma electrons of the C–H bond of an alkyl group are delocalised. This group is directly attached to an atom of an unsaturated system. The delocalisation occurs because of a partial overlap of a sp³ –s sigma bond orbital with an empty p orbital of the π bond of an adjacent carbon atom.
The process of hyperconjugation in propene is shown as follows:

Question 12.11: Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.
(a) C₆H₅ OH
(b) C₆H₅ NO₂
(c) CH₃CH = CH – CHO
(d) C₆H₅ CHO
(e) C₆H₅ - CH₂
(f) CH₃ CH = CHCH₂
Answer
(a) The structure of C₆H₅OH is:

Question 12.12: What are electrophiles and nucleophiles? Explain with examples.
Answer An electrophile is a reagent that takes away an electron pair. In other words, an electron-seeking reagent is called an electrophile (E+). Electrophiles are electrondeficient and can receive an electron pair.

A nulceophile is a reagent that brings an electron pair. In other words, a nucleus-seeking
reagent is called a nulceophile (Nu:).
For example: OH^–, NC^–, carbanions (R₃C^–), etc.
Neutral molecules such as H₂Ö and ammonia also act as nulceophiles because of the
presence of a lone pair.

Question 12.13: Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles:

Answer Electrophiles are electron-deficient species and can receive an electron pair. On the other hand, nucleophiles are electron-rich species and can donate their electrons.

Here, CH₃ CO acts as an electrophile as it is an electron-deficient species

Question 12.14: Classify the following reactions in one of the reaction type studied in this unit.
(a) CH₃ CH₂Br + HS^– → CH₃ CH₂ SH + Br^–
(b) (CH₃)₂ C=CH2 + HCl→ (CH₃)₂ ClC – CH³
(c) CH₃ CH₂Br + HO^– → CH₂ =CH₂ + H2O + Br^–
(d) (CH₃)₃C–CH2 OH + HBr → (CH₃)₂ CBr CH₂CH₃ + H₂O
Answer: (a) Substitution reaction since bromine group gets substituted by –SH group.
(b) Addition reaction since two reactant molecules combines to form a single product.
(c) Elimination reaction since reaction hydrogen and bromine are removed to form ethene.
(d) Substitution reaction since rearrangement of atoms takes place.

Question 12.15: What is the relationship between the members of following pairs of structures? Are they structural or geometrical isomers or resonance contributors?

Answer:  (a) Compounds having the same molecular formula but with different structures are called structural isomers. The given compounds have the same molecular formula but they differ in the position of the functional group (ketone group).

In structure I, ketone group is at the C-3 of the parent chain (hexane chain) and in structure II, ketone group is at the C-2 of the parent chain (hexane chain). Hence, the given pair represents structural isomers.
(b) Compounds having the same molecular formula, the same constitution, and the sequence of covalent bonds, but with different relative position of their atoms in space are called geometrical isomers.

In structures I and II, the relative position of Deuterium (D) and hydrogen (H) in space are different. Hence, the given pairs represent geometrical isomers.
(c) The given structures are canonical structures or contributing structures. They are hypothetical and individually do not represent any real molecule. Hence, the given pair represents resonance structures, called resonance isomers.

Question 12.16: For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion.

Answer
(a) The bond cleavage using curved-arrows to show the electron flow of the given reaction can be represented as

It is a heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with one of the fragments. The intermediate formed is a carbocation.

Question 12.17: Explain the terms Inductive and Electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids?
(a) Cl₃C COOH > Cl₂CH COOH > ClCH₂ COOH
(b) CH₃CH₂ COOH > (CH₃)₂ CH COOH > (CH₃)₃C. COOH
Answer Inductive effect
The permanent displacement of sigma (σ) electrons along a saturated chain, whenever an electron withdrawing or electron donating group is present, is called inductive effect. Inductive effect could be + I effect or – I effect. When an atom or group attracts electrons towards itself more strongly than hydrogen, it is said to possess – I effect. For example,

Question 12.18: Give a brief description of the principles of the following techniques taking an example in each case.
(a) Crystallisation (b) Distillation (c) Chromatography
Answer (a) Crystallisation
Crystallisation is one of the most commonly used techniques for the purification of solid organic compounds.
Principle: It is based on the difference in the solubilites of the compound and the impurities in a given solvent. The impure compound gets dissolved in the solvent in which it is sparingly soluble at room temperature, but appreciably soluble at higher temperature. The solution is concentrated to obtain a nearly saturated solution. On cooling the solution, the pure compound crystallises out and is removed by filtration.
For example, pure aspirin is obtained by recrystallising crude aspirin. Approximately 2 – 4 g of crude aspirin is dissolved in about 20 mL of ethyl alcohol. The solution is heated (if necessary) to ensure complete dissolution. The solution is then left undisturbed until some crystals start to separate out. The crystals are then filtered and dried.
(b) Distillation
This method is used to separate volatile liquids from non-volatile impurities or a mixture of those liquids that have a sufficient difference in their boiling points.
Principle: It is based on the fact that liquids having different boiling points vapourise at different temperatures. The vapours are then cooled and the liquids so formed are collected separately.
For example, a mixture of chloroform (b.p = 334 K) and aniline (b.p = 457 K) can be separated by the method of distillation. The mixture is taken in a round bottom flask fitted with a condenser. It is then heated. Chloroform, being more volatile, vaporizes first and passes into the condenser. In the condenser, the vapours condense and chloroform trickles down. In the round bottom flask, aniline is left behind.
(c) Chromatography
It is one of the most useful methods for the separation and purification of organic compounds.
Principle: It is based on the difference in movement of individual components of a mixture through the stationary phase under the influence of mobile phase.
For example, a mixture of red and blue ink can be separated by chromatography. A drop of the mixture is placed on the chromatogram. The component of the ink, which is less adsorbed on the chromatogram, moves with the mobile phase while the less adsorbed component remains almost stationary.

Question 12.19: Describe the method, which can be used to separate two compounds with different solubilities in a solvent S.
Answer Fractional crystallisation is the method used for separating two compounds with different solubilities in a solvent S. The process of fractional crystallisation is carried out in four steps.
(a) Preparation of the solution: The powdered mixture is taken in a flask and the solvent is added to it slowly and stirred simultaneously. The solvent is added till the solute is just dissolved in the solvent. This saturated solution is then heated.
(b) Filtration of the solution: The hot saturated solution is then filtered through a filter paper in a China dish.
(c) Fractional crystallisation: The solution in the China dish is now allowed to cool.
The less soluble compound crystallises first, while the more soluble compound remains in the solution. After separating these crystals from the mother liquor, the latter is concentrated once again. The hot solution is allowed to cool and consequently, the crystals of the more soluble compound are obtained.
(d) Isolation and drying: These crystals are separated from the mother liquor by filtration. Finally, the crystals are dried.'

Question 12.20: What is the difference between distillation, distillation under reduced pressure and steam distillation ?
Answer The differences among distillation, distillation under reduced pressure, and steam distillation are given in the following table.

	Distillation	Distillation under reduced pressure	Steam distillation
1	It is used for the purification of compounds that are associated with non-volatile impurities or those liquids, which do not decompose on boiling. In other words, distillation is used to separate volatile liquids from non volatile impurities or a mixture of those liquids that have sufficient difference in boiling points.	This method is used to purify a liquid that tends to decompose on boiling. Under the conditions of reduced pressure, the liquid will boil at a low temperature than its boiling point and will, therefore, not decompose.	It is used to purify an organic compound, which is steam volatile and immiscible in water. On passing steam, the compound gets heated up and the steam gets condensed to water. After some time, the mixture of water and liquid starts to boil and passes through the condenser. This condensed mixture of water and liquid is then separated by using a separating funnel.
2	Mixture of petrol and kerosene is separated by this method.	Glycerol is purified by this method. It boils with decomposition at a temperature of 593 K. At a reduced pressure, it boils at 453 K without decomposition.	A mixture of water and aniline is separated by steam distillation.

Question 12.21: Discuss the chemistry of Lassaigne’s test.
Answer Lassaigne’s test
This test is employed to detect the presence of nitrogen, sulphur, halogens, and phosphorous in an organic compound. These elements are present in the covalent form in an organic compound. These are converted into the ionic form by fusing the compound with sodium metal.

The cyanide, sulphide, and halide of sodium formed are extracted from the fused mass by boiling it in distilled water. The extract so obtained is called Lassaigne’s extract. This
Lassaigne’s extract is then tested for the presence of nitrogen, sulphur, halogens, and phosphorous.
(a) Test for nitrogen

Chemistry of the test
In the Lassaigne’s test for nitrogen in an organic compound, the sodium fusion extract is boiled with iron (II) sulphate and then acidified with sulphuric acid. In the process, sodium cyanide first reacts with iron (II) sulphate and forms sodium hexacyanoferrate (II). Then, on heating with sulphuric acid, some iron (II) gets oxidised to form iron (III) hexacyanoferrate (II), which is Prussian blue in colour. The chemical equations involved in the reaction can be represented as

The sodium fusion extract is treated with sodium nitroprusside. Appearance of violet colour also indicates the presence of sulphur in the compound.

If nitrogen and sulphur both are present in the organic compound, then the Lassaigne’s extract is boiled to expel nitrogen and sulphur, which would otherwise interfere in the test for halogens.

Question 12.22: Differentiate between the principle of estimation of nitrogen in an organic compound by
(i) Dumas method and (ii) Kjeldahl’s method.
Answer In Dumas method, a known quantity of nitrogen containing organic compound is heated strongly with excess of copper oxide in an atmosphere of carbon dioxide to produce free nitrogen in addition to carbon dioxide and water. The chemical equation involved in the process can be represented as

The traces of nitrogen oxides can also be produced in the reaction, which can be reduced
to dinitrogen by passing the gaseous mixture over a heated copper gauge. The
dinitrogen produced is collected over an aqueous solution of potassium hydroxide. The
volume of nitrogen produced is then measured at room temperature and atmospheric
pressure.
On the other hand, in Kjeldahl’s method, a known quantity of nitrogen containing organic
compound is heated with concentrated sulphuric acid. The nitrogen present in the
compound is quantitatively converted into ammonium sulphate. It is then distilled with
excess of sodium hydroxide. The ammonia evolved during this process is passed into a
known volume of H2SO4. The chemical equations involved in the process are

The acid that is left unused is estimated by volumetric analysis (titrating it against a standard alkali) and the amount of ammonia produced can be determined. Thus, the percentage of nitrogen in the compound can be estimated. This method cannot be applied to the compounds, in which nitrogen is present in a ring structure, and also not applicable to compounds containing nitro and azo groups.

Question 12.23: Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.
Answer Estimation of halogens
Halogens are estimated by the Carius method. In this method, a known quantity of organic compound is heated with fuming nitric acid in the presence of silver nitrate, contained in a hard glass tube called the Carius tube, taken in a furnace. Carbon and hydrogen that are present in the compound are oxidized to form CO₂ and H₂O respectively and the halogen present in the compound is converted to the form of AgX.
This AgX is then filtered, washed, dried, and weighed.
Let the mass of organic compound be m g.
Mass of AgX formed = m1 g
1 mol of Agx contains 1 mol of X.
Therefore,

Estimation of Sulphur
In this method, a known quantity of organic compound is heated with either fuming nitric acid or sodium peroxide in a hard glass tube called the Carius tube. Sulphur, present in the compound, is oxidized to form sulphuric acid. On addition of excess of barium chloride to it, the precipitation of barium sulphate takes place. This precipitate is then filtered, washed, dried, and weighed.
Let the mass of organic compound be m g.
Mass of BaSO₄ formed = m1 g
1 mol of BaSO₄ = 233 g BaSO₄ = 32 g of Sulphur

Estimation of phosphorus
In this method, a known quantity of organic compound is heated with fuming nitric acid. Phosphorus, present in the compound, is oxidized to form phosphoric acid. By adding ammonia and ammonium molybdate to the solution, phosphorus can be precipitated as ammonium phosphomolybdate.
Phosphorus can also be estimated by precipitating it as MgNH4PO4 by adding magnesia mixture, which on ignition yields Mg2P2O7.
Let the mass of organic compound be m g.
Mass of ammonium phosphomolybdate formed = m1 g
Molar mass of ammonium phosphomolybdate = 1877 g

Question 12.24: Explain the principle of paper chromatography.
Answer In paper chromatography, chromatography paper is used. This paper contains water trapped in it, which acts as the stationary phase. On the base of this chromatography paper, the solution of the mixture is spotted. The paper strip is then suspended in a suitable solvent, which acts as the mobile phase. This solvent rises up the chromatography paper by capillary action and in the procedure, it flows over the spot.
The components are selectively retained on the paper (according to their differing partition in these two phases). The spots of different components travel with the mobile phase to different heights. The paper so obtained (shown in the given figure) is known as a chromatogram.

Question 12.25: Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?
Answer While testing the Lassaigne’s extract for the presence of halogens, it is first boiled with dilute nitric acid. This is done to decompose NaCN to HCN and Na₂S to H2S and to expel these gases. That is, if any nitrogen and sulphur are present in the form of NaCN and Na₂S, then they are removed. The chemical equations involved in the reaction are represented as'

Question 12.26: Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.
Answer Nitrogen, sulphur, and halogens are covalently bonded in organic compounds. For their detection, they have to be first converted to ionic form. This is done by fusing the organic compound with sodium metal. This is called “Lassaigne’s test”. The chemical equations involved in the test are

Carbon, nitrogen, sulphur, and halogen come from organic compounds.

Question 12.27: Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.
Answer The process of sublimation is used to separate a mixture of camphor and calcium sulphate. In this process, the sublimable compound changes from solid to vapour state without passing through the liquid state. Camphor is a sublimable compound and calcium sulphate is a non-sublimable solid. Hence, on heating, camphor will sublime
while calcium sulphate will be left behind.

Question 12.28: Explain, why an organic liquid vaporises at a temperature below its boiling point in its steam distillation?
Answer In steam distillation, the organic liquid starts to boil when the sum of vapour pressure due to the organic liquid (p₁) and the vapour pressure due to water (p₂) becomes equal to atmospheric pressure (p), that is, p = p₁ + p₂ Since p₁ < p₂, organic liquid will vapourise at a lower temperature than its boiling point.

Question 12.29: Will CCl4 give white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.
Answer CCl₄ will not give the white precipitate of AgCl on heating it with silver nitrate. This is because the chlorine atoms are covalently bonded to carbon in CCl₄. To obtain the precipitate, it should be present in ionic form and for this, it is necessary to prepare the Lassaigne’s extract of CCl₄.

Question 12.30:  Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?
Answer Carbon dioxide is acidic in nature and potassium hydroxide is a strong base. Hence, carbon dioxide reacts with potassium hydroxide to form potassium carbonate and water as
2KOH + CO₂  →  K₂CO₂ + H₂O
Thus, the mass of the U-tube containing KOH increases. This increase in the mass of Utube gives the mass of CO₂ produced. From its mass, the percentage of carbon in the organic compound can be estimated.

Question 12.31: Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?
Answer Although the addition of sulphuric acid will precipitate lead sulphate, the addition of acetic acid will ensure a complete precipitation of sulphur in the form of lead sulphate due to common ion effect. Hence, it is necessary to use acetic acid for acidification of sodium extract for testing sulphur by lead acetate test.

Question 12.32: An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion.
Answer Percentage of carbon in organic compound = 69 %
That is, 100 g of organic compound contains 69 g of carbon.
∴0.2 g of organic compound will contain = 69 x 0.2 / 100 = 0.138g of C
Molecular mass of carbon dioxide, CO₂. = 44 g
That is, 12 g of carbon is contained in 44 g of CO₂.
Therefore, 0.138 g of carbon will be contained in 44 x 0.138/12 = 0.506 g of CO₂.
Thus, 0.506 g of CO₂. will be produced on complete combustion of 0.2 g of organic compound.
Percentage of hydrogen in organic compound is 4.8.
i.e., 100 g of organic compound contains 4.8 g of hydrogen.
Therefore, 0.2 g of organic compound will contain 4.8 x 0.2/100 = 0.0096g of H
It is known that molecular mass of water (H₂O) is 18 g.
Thus, 2 g of hydrogen is contained in 18 g of water.
∴0.0096 g of hydrogen will be contained in 18 x 0.0096 / 2  = 0.0864g of water
Thus, 0.0864 g of water will be produced on complete combustion of 0.2 g of the organic compound.

Question 12.33: A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method.
The ammonia evolved was absorbed in 50 mL of 0.5 M H₂SO₄. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound.
Answer Given that, total mass of organic compound = 0.50 g
60 mL of 0.5 M solution of NaOH was required by residual acid for neutralisation.
60 mL of 0.5 M NaOH solution  = 60/2 mL of 0.5M H₂SO₄ = 30 mL of 0.5 M H₂SO₄
∴Acid consumed in absorption of evolved ammonia is (50–30) mL = 20 mL
Again, 20 mL of 0.5 MH₂SO₄ = 40 mL of 0.5 MNH₃
Also, since 1000 mL of 1 MNH₃ contains 14 g of nitrogen,
∴ 40 mL of 0.5 M NH₃ will contain 14 x 40 / 1000 x 0.5 = 0.28 g of N
Therefore, percentage of nitrogen in 0.50 g of organic compound 0.28 x .050 / 100 = 56%

Question 12.34: 0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.
Answer Given that,
Mass of organic compound is 0.3780 g.
Mass of AgCl formed = 0.5740 g
1 mol of AgCl contains 1 mol of Cl.
Thus, mass of chlorine in 0.5740 g of AgCl

Question 12.35: In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound.
Answer Total mass of organic compound = 0.468 g [Given]
Mass of barium sulphate formed = 0.668 g [Given]
1 mol of BaSO₄ = 233 g of BaSO₄ = 32 g of sulphur

Question 12.36: In the organic compound CH₂ = CH – CH₂ – CH₂ – C ≡ CH, the pair of hydridised orbitals involved in the formation of: C₂ – C₃ bond is:
(a) sp – sp² (b) sp – sp³ (c) sp² – sp³ (d) sp³– sp³
Answer

In the given organic compound, the carbon atoms numbered as 1, 2, 3, 4, 5, and 6 are sp, sp, sp³, sp³, sp², and sp² hybridized respectively. Thus, the pair of hybridized orbitals involved in the formation of C₂-C₃ bond is sp – sp³.

Question 12.37: In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of:
(a) Na₄[Fe(CN)₆] (b) Fe₄[Fe(CN)₆]₃ (c) Fe₂[Fe(CN)₆] (d) Fe₃[Fe(CN)₆]₄
Answer In the Lassaigne’s test for nitrogen in an organic compound, the sodium fusion extract is boiled with iron (II) sulphate and then acidified with sulphuric acid. In the process, sodium cyanide first reacts with iron (II) sulphate and forms sodium hexacyanoferrate
(II). Then, on heating with sulphuric acid, some iron (II) gets oxidised to form iron (III) hexacyanoferrate (II), which is Prussian blue in colour. The chemical equations involved in the reaction can be represented as

Question 12.38: Which of the following carbocation is most stable?

Answer (CH₃)₃C⁺ is a tertiary carbocation. A tertiary carbocation is the most stable carbocation due to the electron releasing effect of three methyl groups. An increased + I effect by three methyl groups stabilizes the positive charge on the carbocation.

Question 12.39: The best and latest technique for isolation, purification and separation of organic compounds is:
(a) Crystallisation (b) Distillation (c) Sublimation (d) Chromatography
Answer Chromatography is the most useful and the latest technique of separation and purification of organic compounds. It was first used to separate a mixture of coloured substances.

Question 12.40: The reaction:

is classified as :
(a) electrophilic substitution (b) nucleophilic substitution
(c) elimination (d) addition
Answer

It is an example of nucleophilic substitution reaction. The hydroxyl group of KOH (OH–) with a lone pair of itself acts as a nucleophile and substitutes iodide ion in CH₃CH₂I to form ethanol.