NCERT Solutions for Class 11 Chemistry Chapter 12: Organic Chemistry – Some Basic Principles and Techniques

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12.1. What are hybridisation states of each carbon atom in the following compounds?
CH₂=C=O, CH₃CH=CH₂, (CH₃)₂CO, CH₂=CHCN, C₆H₆

In this compound, carbon in –CH₂ i.e., C–1 is sp² hybridised and the next carbon wich is bonded to oxygen (C=O) i.e., C–2 is sp hybridised.

Carbons in CH₃– group i.e., C-1 and C-3 are sp³ hybridised and carbon in

i.e., C-2 is sp² hybridised.

Carbon in –CH₂ i.e., C-1 is sp² hybridised.
Carbon in =CH i.e., C-2 is sp² hybridised.
Carboin in –CN i.e., C-3 is sp hybridised.

All the carbons in this structure are sp² hybrised.

12.2. Indicate the σ and π bonds in the following molecules:
C₆H₆, C₆H₁₂, CH₂Cl₂, CH₂=C=CH₂, CH₃NO₂, HCONHCH₃

$$In\space C_6H_6, three\space C=C pi (π_{C=C}) bonds,\space six\space C–C sigma\space (\sigma _{C–C}) bonds\space and\space six\space C–H\space sigma bonds\space (\sigma _{C–H}) bonds\space are\space present.$$

$$In\space C_6H_{12}, twelve\space C–H sigma\space (\sigma _{C–H}) bond\space and\space six\space C–C sigma\space (\sigma _{C–C}) bonds\space are\space present.$$

$$In\space CH_2Cl_2, two\space C–H sigma\space (\sigma _{C–H}) bond\space and\space two\space C–Cl sigma\space (\sigma _{C–Cl}) bonds\space are\space present.$$

$$In\space CH_2=C=CH_2, four\space C–H sigma\space (\sigma _{C–H}) bonds,\space two\space C–C\space sigma\space (σ _{C–C}) bonds\space and\space two\space C=C pi (π_{C=C}) bonds\space are\space present.$$

$$In\space CH_3NO_2, three\space C–H\space sigma\space (σ_{C–H}) bond,\space one\space C–N\space sigma\space (σ_{C–N}) bond,\space one\space N–O\space sigma\space (σ_{N–O})\space bond\space and\space one\space N=O pi\space (π_{N=O}) bond\space is\space present.$$

$$In\space HCONHCH_3, four\space C–H sigma\space (σ_{C–H}) bond,\space two\space C–N sigma\space (σ_{C–N}) bonds,\space one N–H\space sigma\space (\sigma _{N–H})\space bond\space one\space C–O\space sigma\space (\sigma _{C=O})\space bond,\space and\space one\space C=O pi (π_{C=O})\space bond\space is\space present.$$

12.3. Write bond line formulas for :Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one.

12.4. Give the IUPAC names of the following compounds :

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.

Ans. (a) 2,2-Dimethylpentane or 2-Dimethylpentane 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–H2 of the parent chain of the given compound. Therefore, the correct IUPAC name of the given compound is 2,2-Dimethylpentane.
(b) 2,4,7-Trimethyloctane or 2,5,7-Trimethyloctane According to IUPAC nomenclature rules, the numbering is done in such a way that the branched carbon atoms get the lowest possible numbers. This means locant number 2,4,7 is lower than 2,5,7. Therefore, the correct IUPAC name of the given compound is 2,4,7- Trimethyloctane.

(c) 2-Chloro-4-methylpentane or 4-Chloro-2- methylpentane 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. Therefore, the correct IUPAC name of the given compound is 2-Chloro-4-methylpentane.

(d) But-3-yn-1-ol or But-4-ol-1-yne According to IUPAC nomenclature rules, in the case of polyfunctional compounds, one of the functional groups is chosen as the principal functional group and the compound is then named on that basis. The principal functional group is the alcoholic group. Hence, the parent chain will be suffixed with ol. Hence, the correct IUPAC name of the given compound is But-3-yn-1-ol.

12.6. Draw formulas for the first five members of each homologous series beginning with the following compounds.
(a) H–COOH (b) CH₃COCH₃ (c) H–CH=CH₂
Ans. The first five members of each homologous series beginning with the given compounds are as follows:
(a) H–COOH: Methanoic acid
CH₃–COOH: Ethanoic acid
C₂H₅–COOH: Propanoic acid
C₃H₇–COOH: Butanoic acid
C₄H₉–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

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.
Ans. (a) 2,2,4-Trimethylpentane Condensed formula: (CH₃)₂CHCH₂C(CH₃)₃ Bond line formula:

This compound is alkane and no other functional group is present.

(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.

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

12.8. Identify the functional groups in the following compounds:

The functional groups present in the above
structure are:
1. Aldehyde (–CHO)
2. Hydroxyl (–OH)
3. Methoxy (–OMe)
4. C=C double bond

The functional groups present in the above
structure are :
1. Amino i.e., primary amine (–NH₂)
2. Ester (–O–CO–)
3. Tertiary amine (–NR₂)
4. C=C double bond

1. The functional groups present in the above structure are : Nitro group (–NO₂)
2. C=C doube bond

12.9. Which of the two: O₂NCH₂CH₂O^–or CH₃CH₂O^–is expected to be more stable and why?

Ans. –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 (C₂H₅–) group is an electron-donating group. Hence, the C₂H₅– 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–
.

12.10. Explain why alkyl groups act as electron donors when attached to a π system.
Ans. When an alkyl group is attached to a π system, it acts as an electron-donor group by the process of hyperconjugation. Let us take the example of propene to understand this process.

In hyperconjugation, the sigma electrons ofthe 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 n bond of an adjacent carbon atom.

The process of hyperconjugation in propene is shown as follows.

This type of overlap leads to a delocalisation (also known as no-bond resonance) of the π electrons, making the molecule more stable.

12.11. Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.
(a) C6H5OH
(b) C6H5NO2
(c) CH3CH=CHCHO
(d) C6H5–CHO

+
(e) C H6 5CH2
+
(f) CH3 2 CH = CHC

Ans. (a) C₆H₅OH
The structure of C₆H₅OH is as follows:

The resonating structures of phenol are represented as:

(b) C₆H₅NO₂
The structure of C₆H₅NO₂ is as follows:

(d) C6H5–CHO
The structure of C6H5CHO is:
The resonating structures of the given compounds are represented as:

(e) C6H5CH2⁺
The resonating structures of the given compound are represented as:

(f) CH₃CH=CHCH₂⁺
The resonating structures of the given compound are represented as:

12.12. What are electrophiles and nucleophiles ? Explain with examples.
Ans. A reagent that takes away an electron pair from reactive site is called as an electrophile (E+). In other words, an electrophile is electron deficient or electron seeking reagent. . The reactions in which electrophiles participate are called as electrophilic reactions. Examples of electrophiles include electron deficient centres such as carbocations (CH₃⁺) and neutral molecules having functional groups like carbonyl group (>C=O) or alkyl halides (R₃C–X, where X is a halogen atom).

A reagent that brings an electron pair to the reactive site is called as a nucleophile (Nu^–) . In other words, nucleophile is a nucleus seeking reagent. The reactions in which nucleophile participate are called as nucleophilic reaction. Some examples of nucleophiles are the negatively charged ions with lone pair of electrons such as hydroxide (HO^–), cyanide (NC^–) ions and carbanions (R₃C^–).

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

$$\text{(a) CH}_3\text{COOH} + \text{HO} →\text {CH}_3\text{COO}^–+ \text{H}_2\text{O}\\\text{ (b) CH}_3\text{COCH}_3+\={C}N \rightarrow(\text{CH}_3)_2(\text{CN})+(\text{OH})\\\text{(c) C}_6\text{H}_5+\text{CH}_3 \text{C}^+\text{O}\space \rightarrow \text{C}_6\text{H}_5\text{COCH}_3$$

Ans. $$\text{(a) CH}_3\text{COOH} + \text{HO}\rightarrow \text{CH}_3\text{COO}^–+\text{H}_2\text{O}$$

Here, HO^– acts as a nucleophile as it is an electron-rich species, i.e., it is a nucleus seeking species.

$$\text{(b) CH}_3\text{COCH}_3+\text{CN}^-\rightarrow(\text{CH}_3)_2\text{C(CN)}+(\text{OH})$$

Here, CN^–
acts as a nucleophile as it is an electron-rich species, i.e., it is a nucleus seeking species.

$$\text{C}_6\text{H}_5+\text{CH}_3\text{CO}^+\rightarrow \text{C}_6\text{H}_5\text{CHOCH}_3$$

Here, CH₃CO⁺ acts as a electrophile as it is an electron-deficient species.

12.14. Classify the following reactions in one of the reaction type studied in this unit.

$$\text{(a)CH}_3\text{CH}_2\text{Br}+\text{HS}^-\rightarrow \text{CH}_3\text{CH}_2\text{SH}+\text{Br}^-\\\text{(b)}(\text{CH}_3)_2\text{C}=\text{CH}_2+\text{HCI}\rightarrow (\text{CH}_3)_2\text{CIC}-\text{CH}_3\\\text{(C)CH}_3\text{CH}_2\text{Br}+\text{HO}^-\rightarrow \text{CH}_2=\text{CH}_2+\text{H}_2\text{O}+\text{Br}^-\\\text{(d)}(\text{CH}_3)_3 \text{C}-\text{CH}_2\text{OH}+\text{HBr}\rightarrow(\text{CH}_3)_2 \text{CBrCH}_2\text{CH}_3+\text{H}_2\text{O}$$

Ans. $$\text{(a) CH}_3\text{CH}_2\text{Br} + \text{HS}^– \rightarrow \text{CH}_3\text{CH}_2\text{SH} +\text{ Br}^–$$

Br^– and HS^– are electron rich species. Hence, it is nucleophile. A nucleophile (Br^–) is replaced by other nucleophile (HS^–). Hence, it is nucleophilic substitution reaction.

$$\text{(b)} (\text{CH}_3)_2\text {C} = \text{CH}_2 + \text{HCl} \rightarrow (\text{CH}_3)_2 \text{ClC}–\text{CH}_3$$

HCl is added to the double bond.
HCl is electron - loving species. Hence, it is
electrophilic addition reaction.

$$\text{(c) CH}_3\text{CH}_2\text{Br} + \text{HO}^– \rightarrow \text{CH}_2 = \text{CH}_2 + \text{H}_2\text{O}+ \text{Br}–$$

H and Br are eliminated from successive carbon atoms to form ethene. Hence, it is β - elimination reaction.

$$\text{(d)} (\text{CH}_3)_3 \text{C} – \text{CH}_2 \text{OH} + \text{HBr} \rightarrow (\text{CH}_3)_2 \text{CBrCH}_2\text{CH}_3 + \text{H}_2\text{O}$$

OH^– and Br^– are electron rich species. Hence, it is nucleophile and we can see that nucleophile (OH^–) is replaced by nucleophile (Br^–) and after that reaction products are rearranged. Hence, it is nucleophilic substitution reaction with rearrangement.

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

Ans. (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 shows functional group isomerism which is sub-class of structural isomerism.

(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.

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.

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

It is an example of homolytic cleavage as one of the shared pair in a covalent bond goes with the bonded atom. The reaction intermediate formed is a free radical.

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

It is an example of heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with
the carbon of propanone. The reaction intermediate formed is carbanion.
(c) The bond cleavage using curved-arrows to show the electron flow of the given reaction can be represented as

It is an example of heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with the bromine ion. The reaction intermediate formed is a carbocation.

(d) 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.

12.17. Explain the terms inductive and electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids?

$$\text{(a)Cl}_3\text{CCOOH}>\text{Cl}_2\text{CHCOOH}>\text{ClCH}_2\text{COOH}\\\text{(b)CH}_3\text{CH}_2\text{COOH} >(\text{CH}_3)_2 \text{CHCOOH} >(\text{CH}_3)_3 \text{C.COOH}$$

Ans. Inductive effect : The permanent displacement of sigma (σ) electrons along a saturated chain, whenever an electron-with drawing or electron donating group is present, is called the 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,

When an atom or group attracts electrons towards itself less strongly than hydrogen, it is said to possess +I effect. For example :

Electrometric effect : It involves the complete transfer of the shared pair of π electrons to either of the two atoms linked by multiple bond in the presence of an attacking agent. For example,

Electrometric effect could be +E effect of –E effect.
(i) +E effect : When the electrons are transferred towards the attacking reagent.
(ii) –E effect : When the electrons are transferred away from the attacking reagent.

(a) Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH Halides are electron withdrawing groups. Hence, the order of acidity can be explained on the basis of Inductive effect (–I effect). As the number of chlorine atoms increases, the - I effect increases. With the increase in –I effect, the acid strength also increases accordingly.

(b) CH₃CH₂COOH > (CH₃)₂ CHCOOH > (CH₃)₃ C.COOH
Alkyl groups are electron donating groups. Hence, the order of acidity can be explained on the basis of inductive effect (+ I effect). As the number of alkyl groups increases, the + I effect also increases. With the increase in + I effect, the acid strength also increases accordingly.

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.

Ans. (a) Crystallisation : Crystallisation techniques is commonly used for the purification of solid organic compounds. Principle: It is based on the difference in the solubilities of the compound and the impurities in a suitable solvent. The impure compound is dissolved in the solvent in which it is sparingly soluble at room temperature, but 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 as 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: Liquids having different boiling points vaporise at different temperatures. The vapours are cooled and the liquids so formed are collected separately.
For example, a mixture of chloroform (b.p. = 334 K) and aniline (b.p. = 457K) 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, vapourises first and passes into the condenser. In the condenser, the vapour condense chloroform trickles down. In the round bottom flask, aniline is left behind.
(c) Chromatography : Chromatography is an important technique extensively used to separate mixtures into their components, purify compounds and also to test the purity of compounds. Principle: In this technique, the mixture of substances is applied onto a stationary phase, which may be a solid or a liquid. A pure solvent, a mixture of solvents, or a gas is allowed to move slowly over the stationary phase. The components of the mixture get gradually separated from one
another. The moving phase is called the 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.

12.19. Describe the method, which can be used to separate two compounds with different solubilities in a solvent S.

Ans. Fractional crystallisation is used to separate 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.

12.20. What is the difference between distillation, distillation under reduced pressure and steam distillation?

	Distillation	Distillation under reduced pressure	steam distilation
1.	This important method is used to separate (i) volatile liquids from non-volatile impurities and (ii) the liquids having sufficient difference in their boiling points. Liquids having different boiling points vaporise at different temperatures. The vapours are cooled and the liquids so formed are collected separately.	This method is used to purify liquids having very high boiling points and those, which decompose at or below their boiling points. Such liquids are made to boil at a temperature lower than their normal boiling points by reducing the pressure on their surface. Under the condition of reduced pressure, the liquid will boil at lower temperature than its boiling point and therefore, will not decompose.	This technique is applied to separate substances which are steam volatile and are immiscible with water. In steam distillation, steam from a steam generator is passed through a heated flask containing the liquid to be distilled. The mixture of steam and the volatile organic compound is condensed and collected. The compound is later separated from water using a separating funnel.
2.	Chloroform (b.p. 334 K) and aniline(b.p. 457 K) are easily separated by the technique of distillation	Glycerol can be separated from spent-lye in soap industry by using this technique. Glycerol boils with decomposition at a temperature of 593K. At a reduced pressure, it boils at 453K without decomposition.	Aniline is separated by this technique from aniline–water mixture

12.21. Discuss the chemistry of Lassaigne’s test.

Ans. Nitrogen, sulphur, halogens and phosphorus present in an organic compound are detected by “Lassaigne’s test”.
These 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:

$$\text{Na} + \text{C} + \text{N} \xrightarrow{\Delta} \text{NaCN}\\2\text{Na} + \text{S} \xrightarrow{\Delta} \text{Na}_2\text{S}\\\text{Na} + \text{X}\xrightarrow{\Delta} \text{Na X}\\(\text{X} = \text{Cl, Br or I})$$

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 or Lassaigne’s extract which is tested for the presence of nitrogen, sulphur, phosphorous and halogens.

(a) Chemistry of Lassaigne’s test for nitrogen: In the Lassaigne’s test for nitrogen, the sodium fusion extract is boiled with iron(II) sulphate and then acidified with concentrated 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 formation of Prussian blue colour confirms the presence of nitrogen.

$$\underset{\downarrow {\text{Conc. Sulphuric\space acid}}} {\text{Lassaigne's\space extract}+\text{ Ferrous\space sulphat}}\\\text{Prussian blue colour}$$

The chemical equation involved in the reaction can be represented as:

$$6\text{CN}^– + \text{Fe}^2+ \rightarrow [\text{Fe(CN)}_6]^{4–}\\3[\text{Fe(CN)}_6]^{4–} + 4\text{Fe}^{3+}\underset{\underset{\text{Prussian\space blue}}{\text{Fe}_4[\text{Fe(CN)}_6]_3.\text{xH}_2\text{O}}} {\underrightarrow{\text{xH}_2\text{O}}}$$

(b) Chemistry of Lassaigne’s test for sulphurIn the Lassaigne’s test for sulphur, 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.

(i) Lassaignes’s extract + Lead acetate

$$\underrightarrow{\text{acetic\space acid}}\space \text{Black\space precipitate}\\\text{S}^{2–} + \text{Pb}^{2+}\text{PbS}\\\text{Black}$$

(ii) On treating sodium fusion extract with sodium nitroprusside, appearance of a violet colour further indicates the presence of sulphur in the compound.

Lassaignes’s extract + Sodium nitroprusside

$$\rightarrow \text{Violet\space colour}\\\text{S}^{2–} + [\text{Fe(CN)}_5\text{NO}]^{2–} \rightarrow \underset{\text{Violet}}{[\text{Fe(CN)}_5 \text{NOS}]^{4–}}$$

In case, nitrogen and sulphur both are present in an organic compound, sodium thiocyanate (NaSCN) is formed which gives blood red colour. Since there are no free cyanide ions, therefore, prussian blue colour is not obtained. The chemical equation involved in the reaction can be represented as:

$$\text{Na} +\text{C} + \text{N} + \text{S} \rightarrow \text{NaSCN}\\\text{Fe}^{3+} + \text{SCN}^– \rightarrow\underset{\text{Blood red}} {[\text{Fe(SCN)}]^{2+}}$$

(c) Chemistry of Lassaigne’s test for halogens : In the Lassaigne’s test for halogens in an organic compound, 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.

$$\text{X}^+ +\text{Ag}^+ \rightarrow \text{AgX}$$

X represents a halogen – Cl, Br or I

L.E + Dilute nitric acid + Silver nitrate

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 first for halogens.

12.22. Differentiate between the principle of estimation of nitrogen in an organic compound by (i) Dumas method and (ii) Kjeldahl’smethod.

Ans. (i) Dumas Method : 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 :

$$\text{CxHyNz} + (2\text{x} + \text{y}/2)\text{CuO} \rightarrow \text{xCO}_2\\+ \text{y}/2\text{H}_2\text{O} + \text{z}/2\text{N}_2 + (2\text{x} + \text{y}/2)\text{Cu}$$

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.

The organic compound yields nitrogen gas on heating it with copper (II) oxide in the presence of CO₂ gas. The mixture of gases is collected over potassium hydroxide solution in which CO₂ is absorbed and volume of nitrogen gas is determined by the formula:

$$\text{Volume\space of\space Nitrogen\space at\space STP}= \frac{\text{P}_1\text{V}_1×273}{760×\text{T}_1}$$

% of N

$$=\frac{28 × \text{volume\space of\space N}_2\space\text{at\space STP}\space 100}{22400 × mass\space of\space organic\space compound}$$

(ii) Kjeldahl’s Method: The compound containing nitrogen is heated with concentrated sulphuric acid. The nitrogen present in the compound gets quantitatively converted into ammonium sulphate. 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. The chemical reaction involved in the process can be represented as:

$$\text{Organic\space compound}\space\underrightarrow {\text{Conc.H}_2\text{SO}_4}(\text{NH}_4)_2\text{SO}_4\\(\text{NH}_4)_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{NH}_3+ 2\text{H}_2\text{O}\\2\text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4$$

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.The difference between the initial amount of acid taken and that left after the reaction gives the amount of acid reacted with ammonia.

% of N

$$= \frac{1.4×M×\lgroup{V-\frac{V_1}{2}}\rgroup}{m}$$

Where, M = Molarity of H₂SO₄ = Molarity of NH₃V = Volume of H₂SO₄ taken initially V_1 = Volume of NaOH m = Mass of organic compound Hence, 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.

12.23. Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.

Ans. 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 oxidised to form CO₂ and H₂O respectively and the halogen present in the compound is converted to form 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, Mass of halogen in m₁ g of AgX

$$=\frac{(\text{Atomic\space mass\space of\space X}×\text{m}_1\text{g})}{\text{Molecular\space mass\space of\space AgX}}$$

Thus, % of halogen will be

$$=\frac{(\text{Atomic\space mass\space of\space X}×\text{m}_1×100)}{\text{Molecular\space mass\space of\space AgX}×\text{m}}$$

Estimation of Sulphur : A known mass of an organic compound is heated in a hard glass tube called as the Carius tube with either 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 be m g.

Mass of BaSO₄ formed = m₁ g

1 mol of BaSO₄ = 233 g BaSO₄ = 32 g of Sulphur
Therefore, m ₁ g of BaSO ₄ contains

$$\frac{32×\text{m}_1}{233}\text{g}\space \text{sulphur\\Thus,\space percentage\space of\space sulphur\space} = \frac{32×\text{m}_1×100}{233×\text{m}}$$

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
Thus, percentage of phosphorus

$$=\frac{(31×\text{m}_1×100)}{(1877×\text{m})}%$$

$$\text{If\space P\space is\space estimated\space as\space Mg}_2P_2O_7,\\ \text{Then,\space Percentage\space of\space phosphorus}\space=\frac{(62×\text{m}_1×100)}{222×\text{m}}%$$

12.24. Explain the principle of paper chromatography.

Ans. Paper chromatography is a type of partition chromatography. In paper chromatography, a special quality paper known as chromatography paper is used. Chromatography paper contains water trapped in it, which acts as the stationary phase.

A strip of chromatography paper spotted at the base with the solution of the mixture is suspended in a suitable solvent or a mixture of solvents. This solvent acts as the mobile phase. The solvent rises up the paper by capillary action and it also flows over the spot.

The paper selectively retains different components on the paper according to their differing partition in the two phases.

The paper strip so developed is known as a chromatogram. The spots of the separated coloured compounds are visible at different heights from the position of initial spot on the chromatogram.

12.25. Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?

Ans. If the sodium fusion extract is first boiled with concentrated nitric acid then it decompose cyanide or sulphide of sodium formed during Lassaigne’s test due to the presence of nitrogen and sulphur ions in compound. This removes all the nitrogen and sulphur present in the form of NaCN and Na₂S. Hence, only halogen is available for testing and these ions do not interfere with silver nitrate test for halogens.

Therefore, nitric acid is added to sodium extract before adding silver nitrate for testing halogens.

12.26. Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.
Ans. In organic compounds, nitrogen, sulphur and halogens are bonded by covalent bond. Detection of these elements are possible if they are in ionic form. Therefore, for their detection, they have to be first converted to ionic form. Thus, by fusing the compound with sodium metal the elements present in the compound are converted from covalent form into the ionic form. Hence, an organic compound is fused with sodium metal for testing nitrogen, sulphur and halogens. This is called Lassaigne’s test and the chemical equations involved in the test are:

$$\text{Na} + \text{C} + \text{N} \rightarrow \text{NaCN}\\\text{Na} + \text{S} + \text{C} + \text{N} \rightarrow \text{NaSCN}\\2\text{Na} + \text{S} \rightarrow \text{Na}_2\text{S}\\\text{Na} + \text{X} \rightarrow \text{NaX}$$
(X = Cl, Br, I)

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

12.27. Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.
Ans. Camphor is a solid substance which changes from solid to vapour state without passing through liquid state on heating and this phenomenon is known as sublimation.

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

12.28. Explain, why an organic liquid vaporises at a temperature below its boiling point in its steam distillation?
Ans. In steam distillation, the organic liquid starts to boil when the sum of vapour pressures due to the organic liquid (p1) and that due to water (p2) becomes equal to the atmospheric pressure (p), i.e. p = p₁+ p₂.

Since p₁ is lower than p₂, the organic liquid vaporises at lower temperature than its boiling point i.e.vapour pressure of water is higher than organic liquid. Therefore, organic liquid vaporises at a temperature below its boiling point in steam distillation.

12.29. Will CCl4 give white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.

Ans. CCl₄ will not give white precipitate of AgCl on heating it with silver nitrate. This is because in CCl₄, the chlorine atoms are covalently bonded to carbon. Thus, CCl₄ is a covalent compound and the covalent compound does not break into ions. To form AgCl, chloride ions should react with silver ions. Therefore, the CCl₄ does not give a white precipitate of AgCl on heating with silver nitrate i.e., no reaction. Hence, to obtain the precipitate, it should be present in ionic form and therefore, it is necessary to prepare the Lassaigne’s extract of CCl₄.

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?

Ans. Carbon dioxide is acidic in nature which reacts with basic potassium hydroxide to form potassium carbonate and water. The chemical reaction can be represented as:

$$2\text{KOH} + \text{CO}_2 \rightarrow \text{K}_2\text{CO}_3 + \text{H}_2\text{O}$$

The increase in the mass of U-tube due to formation of potassium carbonate corresponds to the mass of carbon dioxide produced and from this, the percentage of carbon in the organic compound is estimated. Hence, to estimate percentage of carbon dioxide in an organic compound a solution of potassium hydroxide used to absorb carbon dioxide.

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?
Ans. When sulphuric acid is added the complete precipitate formation of lead sulphate does not take place. While the addition of acetic acid will ensure a complete precipitation 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.

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.

Ans. Given, Percentage of carbon in organic compound
= 69%
So, 100 g of organic compound contains 69 g of
carbon.
∴ 0.2 g of organic compound will contain
= 69 × 0.2 /100 = 0.138 g of carbon
Molecular mass of carbon dioxide, CO₂ = 44 g
So, 12 g of carbon is contained in 44 g of CO₂.
∴ 0.138 g of carbon will be contained in

$$=×\frac{0.138}{12}$$

= 0.506 g of carbon
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.

∴ 0.2 g of organic compound will contain

$$=4.8×\frac{0.2}{100}\\=0.0096\space g\space of\space H$$

Molecular mass of water (H₂O) is 18 g.
In 18 gm of water, mass of hydrogen is 2 gm. i.e.,
2 gm of hydrogen contains 18 gm water

∴ 0.0096 g of hydrogen will be contained in

$$=18×\frac{0.0096}{2} = 0.0864\space \\text{g\space of\space water}$$

Hence, 0.0864 g of water will be produced on complete combustion of 0.2 g of the organic compound.

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.

Ans. Given :
Total mass of organic compound = 0.50 g
60 mL of 0.5 M solution of NaOH required for
neutralisation by residual acid.
60 mL of 0.5 M NaOH solution

$$=\frac{60}{2}\text{mL\space of\space 0.5\space M H}_2\text{SO}_4\\=30\space \text{mL\space of}\space 0.5\space \text{M H}_2\text{SO}_4$$

Therefore,
Acid consumed in absorption of evolved
ammonia is (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, So, 40 mL of 0.5 M NH₃ will contain

$$=\frac{14×14}{1000}×0.5\\
=0.28\space \text{g\space of\space N}$$

Therefore, percentage of nitrogen in 0.50 g of organic compound.

$$=\frac{0.28}{0.50}×100$$

=56%

Hence, the percentage composition of nitrogen in the compound is 56%.

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.
Ans. Given :
Mass of organic compound is 0.3780 g
Mass of AgCl formed = 0.5740 g
1 mol of AgCl contains 1 mol of Cl.
0.3780 g of organic compound gives 0.5740 g of
AgCl.
143.32g of AgCl contains 35.5 g of Cl.
Thus,

$$0.5740\space \text{g\space of\space AgCl\space contains}\space \frac{35.5×0.5740}{143.32}= 0.1421g$$

of chlorine.
The percentage of chlorine in the organic chloro

$$\text{compound\space is}\frac{0.1421×100}{0.3780}=37.6\%$$

Hence, the percentage of chlorine present in the given organic chloro compound is 37.6%.

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.
Ans. Given: Total mass of organic compound = 0.468 g
Mass of barium sulphate formed = 0.668 g
Molar mass of BaSO₄ is 233 g/mol.
Thus,
233 g of BaSO₄ corresponds to 32 g of sulphur.

$$0.668\space \text{g\space of\space BaSO}_4 \text{will contain} \frac{32×0.668}{233}$$

= 0.0917g of sulphur.
Therefore, the percentage of sulphur in the given
compound is:

$$\frac{0.0917×100}{0.468}=19.6\%$$

Hence, the percentage of sulphur in the given compound is 19.6%.

12.36. In the organic compound CH₂ = CH – CH₂ – CH₂ – C ≡ CH, the pair of hybridized orbitals involved in the formation of: C₂ – C₃ bond is:
(a) sp – sp₂

(b) sp – sp₃
(c) sp₂ – sp₃

(d) sp₃ – sp₃

Ans. (c) sp2 – sp3

Explanation:

$$\text{C}^6\text{H}_2 =\text{C}^5\text{H}- \text{C}^4\text{H}_2- \text{C}^3\text{H}_2- \text{C}^2 \text{C}^1\text{H}$$

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 hybridised respectively. Thus, the pair of hybridized orbitals involved in the formation of C²–C³ bond is sp–sp³

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
Ans. (b) Fe4[Fe(CN)6]3

Explanation:
In the Lassaigne’s test for nitrogen in an organic compound, the sodium fusion extracted 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 ferrous hexacyanoferrate(II), which is Prussian blue in colour. The chemical equations involved in the reaction can be represented as :

$$6\text{CN}^– + \text{Fe}^{2+} \rightarrow [\text{Fe(CN)}_6]^{4–}\\3[\text{Fe(CN)}_6]^{4–} + 4\text{Fe}^{3+}\underrightarrow{\text{xH}_2O}\underset{\text{Prussian blue}}{\text{Fe}_4[\text{Fe(CN)}_6]_3.\text{xH}_2\text{O}}$$

Hence, the Prussian blue colour is due to the
formation of Fe₄[Fe(CN)₆]₃.

12.38. Which of the following carbocation is most stable ?

$$\text{(a) (CH}_3)_3\text{C. C}^+\text{H}_2\\
\text{(b)} (\text{CH}_3)_3 \text{C}^+\\
\text{(c) CH}_3\text{CH}_2 \text{C}^+\text{H}_2\\
\text{(d) CH}_3\text{C}^+\text{HCH}_2\text{CH}_3$$

Ans. (b) (CH₃)₃ C⁺

Explanation:
(CH₃)₃ C⁺ is a tertiary carboncation. 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 stabilises the positive charge on the carbocation.

12.39. The best and latest technique for isolation, purification and separation of organic compounds is:
(a) Crystallisation

(b) Distillation
(c) Sublimation

(d) Chromatography

Ans. (d) Chromatography

Explanation:
The best and latest technique for isolation, purification and separation of organic compounds is chromatography. It is the most useful and latest technique of separation and purification of organic compound. It is used to separate a mixture of coloured substances.

12.40. The reaction: CH₃CH₂ I + KOH (aq) → CH₃CH₂OH + KI is classified as :
(a) electrophilic substitution
(b) nucleophilic substitution
(c) elimination
(d) addition
Ans. (b) Nucleophilic substitution

Explanation:
The reaction:

$$\underset{Ethyl iodide}{CH_3CH_2I + KOH (aq)} → \underset{Ethanol}CH_3CH_2OH + KI$$

is classified as nucleophilic substitution reaction. This is because the hydroxyl group of KOH (OH–) with a lone pair of electron itself acts as a nucleophile and substitutes iodide ion in CH₃CH₂I to form ethanol.

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