Class 11 Chemistry Chapter 12 NCERT Solutions: Organic Chemistry- Some Basic Principles and Techniques NCERT PDF

Single bond contains only one sigma bond and double bond contains one sigma and one pi bond.

1. In this structure, nine single bonds and three double bonds are present. So, there are 12σ and 3π bonds present.
2. In this structure, eighteen single bonds are present. So, there are only 18σ bonds present.
3. In this structure, four single bonds are present. So, there are only 4σ bonds present.
4. In this structure, four single bonds and two double bonds are present. So, there are 6σ and 2π bonds present.
5. In this structure, five single bonds and one double bond are present. So, there are 6σ and 1π bonds present.
6. In this structure, seven single bonds and one double bond are present. So, there are 8σ and 1π bonds present.

(a) Propylbenzene (b) 3-Methylpentanenitrite (c) 2, 5-Dimethylheptane
(d) 3-Bromo- 3-chloroheptane (e) 3-Chloropropanal (f) 2, 2-Dichloroethanol

(a) 2, 2-Demethylpentane

(b) 2, 4, 7-Trimethyloctane. For two alkyl groups on the same carbon its locant is repeated twice, 2, 4, 7-locant set is lower than 2, 5, 7.
(c) 2- Chloro-4-methylpentane. Alphabetical order of substituents, (d) But-3-yn-1-ol. Lower locant for the principal functional group, i.e., alcohol.

(a) H-COOH

CH₃—COOH
CH₃CH₂—COOH

CH₃CH₂CH₂—COOH
CH₃CH₂CH₂CH₂—COOH

(b) CH₃COCH₃
CH₃COCH₂CH₃
CH₃COCH₂CH₂CH₃
CH₃COCH₂CH₂CH₂CH₃
CH₃CO (CH₃)₄CH₃

(c) H—CH=CH₂
CH₃CH=CH₂

CH₃CH₂CH=CH₂
CH₃CH₂CH₂CH=CH₂
CH₃CH₂CH₂CH₂CH=CH₂

Nitroethoxide ion (O₂NCH₂CH₂O^–) is more stable than ethoxide ion (CH₃CH₂O^–) due to -I effect of nitro group which decreases the negative charge of oxide ion of nitroethoxide ion leading to stability. In contrast, CH₃CH₂ has +I-effect. It, therefore, tends to intensify the -ve charge and hence destabilizes it.

Due to hyperconjugation, alkyl groups act as electron donors when attached to a π - system as shown below:

Electrophiles: The name electrophiles mean electron loving. Electrophiles are electron deficient. They may be positive ions or neutral molecules.
Ex: H⁺, Cl⁺, Br⁺, NO₂⁺, R₃C⁺, RN₂⁺, AlCl₃, BF₃
Nucleophiles: The name nucleophiles means 'nucleus loving' and indicates that it attacks the region of low electron density (positive centres) in a substrate molecule. They are electron rich they may be negative ions or neutral molecules.
Ex: Cl^– Br^–, CN^–, OH^–, RCR₂^–, NH₃, RNH₂, H₂O, ROH etc.

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

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

(c) CH₃C⁺O acts as an electrophile as it is an electron deficient species.

(a) Nucleophilic substitution (b) Electrophilic addition
(c) Bimolecular elimination (d) Nucleophilic substitution with rearrangement.

  (a) They are structural isomers. The given compounds have the same molecular formula but they differ in the position of the functional group (here ketone group: first one at C-3 and second one at C-2 positions).

(b) They are geometrical isomers. Compounds having the same molecular formula, the same constitution, and the sequence of covalent bonds, but with the different relative position of their atoms in space are called geometrical isomers.

(c) They are resonance contributors because they differ in the position of electrons but not atoms.

Inductive effect: The inductive effect refers to the polarity produced in a molecule as a result of higher electronegativity of one atom compared to another. Atoms or groups which lose electron towards a carbon atom are said to have +I Effect. Examples of +I effect are (Electron releasing)
(CH₃)₂C—, (CH₃)₂CH—, CH₃CH₂— CH₃— etc.
Those atoms or groups which draw electron away from a carbon atom are said to have -I Effect. Examples of -I effect are:
NO₂, F, Cl, Br, I, OH etc.

Electromeric effect: The electromeric effect refers to the polarity produced in a multiple bonded compound as it is approached by a reagent.

The atom A has lost its share of electron pair and B has gained this share.
As a result A acquires a positive charge and B a negative charge. It is a temporary effect and takes place only in the presence of a reagent.
(a) –I effect as shown below:
As the number of halogen atoms decreases, the overall -I effect decreases and the acid strength decreases accordingly.

(b) +I effect as shown below:
As the number of alkyl groups increases, the +I effect increases and the acid strength decreases accordingly.

(a) Crystallisation: In this process the impure solid is dissolved in the minimum volume of a suitable solvent. The soluble impurities pass into the solution while the insoluble ones left behind. The hot solution is then filtered and allowed to cool undisturbed till crystallisation is complete. The crystals are then separated from the mother liquor by filtration and dried.
Example: crystallisation of sugar.

(b) Distillation: The operation of distillation is employed for the purification of liquids from non-volatile impurities. The impure liquid is boiled in a flask and the vapours so formed are collected and condensed to give back pure liquid in another vessel. Simple organic liquids such as benzene toluene, xylene etc. can be purified.

(c) Chromatography: Chromatography is based on the principle of selective distribution of the components of a mixture between two phases, a stationary phase and a moving phase. The stationary phase can be a solid or liquid, while the moving phase is a liquid or a gas. When the stationary phase is solid the basis is adsorption and when it is a liquid the basis is partition. Chromatography is generally used for the separation of coloured substances such as plant pigments or dyestuffs.

Fractional crystallisation is the method that can be used to separate two compounds with different solubilities in a solvent S. A hot saturated solution of these two compounds is allowed to cool, the less soluble compound crystallises out while the more soluble remains in the solution. The crystals are separated from the mother liquor and the mother liquor is again concentrated and the hot solution again allowed cooling when the crystals of the second compound are obtained. These are again filtered and dried.

Distillation is used in case of volatile liquid mixed with non-volatile impurities.
Distillation under reduced pressure: This method is used to purify such liquids which have very high boiling points and which decompose at or below their boiling points.
Steam distillation is used to purify steam volatile liquids associated with water immiscible impurities.

Lassaigne's test: In organic compounds, nitrogen, sulphur and halogens are covalently bonded. Their detection in 'Lassaigne's test' is possible if they are in the ionic form. This can be achieved by fusing the organic compound with sodium metal.

Chemistry for test for nitrogen: 

Sodium fusion extract is boiled with ferrous sulphate and acidified with sulphuric acid. Sodium cyanide reacts with ferrous sulphate and forms sodium hexacyanoferrate (II).  On heating with sulphuric acid, some ferrous is oxidized to ferric hexacyanoferate (II) Fe₄ [Fe (CN)₆]₃ which is prussian blue in colour.

Chemistry of the test for sulphur:

Acetic acid is added to sodium fusion extract. Complete precipitation of sulphur in the form of lead sulphate occurs which is black in colour. Also, the sodium fusion extract is treated with sodium nitroprusside to obtain violet color. However if N and S are present, then instead of NaCN, NaSCN is obtained. This gives blood red colour on reaction with Fe³⁺ ions.

Chemistry of test for halogens:

Sodium fusion extract is acidified with nitric acid and then treated with silver nitrate. However, if nitrogen and sulphur both are present, then Lassaigne's extract is boiled to expel nitrogen and sulphur which would otherwise interfere in the test for halogens.

(i) Dumas method: The nitrogen containing organic compound, when heated with copper oxide in an atmosphere of carbon dioxide, yields free nitrogen in addition to carbon dioxide and water

(ii)Kjeldahl's method: A known mass of the organic compound is heated strongly with conc. H₂SO₄,  a little potassium sulphate and a little mercury (a catalyst). As a result, the nitrogen present in the organic compound is converted to ammonium sulphate.

Estimation of halogens: It involves oxidising the organic substance with fuming nitric acid in the presence of silver nitrate. The halogen of the substance is thus converted to silver halide which is separated and weighed:
Weight of organic compound = W gm
weight of silver halide = x g.

% of halogen =

Estimation of sulphur: The organic substance is heated with fuming nitric acid but no silver nitrate is added. The sulphur of the substance is oxidised to sulphuric acid which is then precipitated as barium sulphate by adding excess of barium chloride solution. From the weight of BaSO₄ so obtained the percentage of sulphur can be calculated.

% of sulphur =  x

Estimation of phosphorous: The organic substance is heated with fuming nitric acid whereupon phosphorous is oxidised to phosphoric acid. The phosphoric acid is precipitated as ammonium phosphomolybdate, (NH₄)₃PO₄.12MoO₃, by the addition of ammonia and ammonium molybdate solution which is then separated, dried and weighed.

% of phosphorus =,

Where, molar mass of (NH₄)₃PO₄.12MoO₃ = 1877 g.

If phosphorus is estimated as Mg₂P₂O₇

% of P =   

This is the simplest form of chromatography. Here a strip of paper acts as an adsorbent. It is based on the principle which is partly adsorption. The paper is made of cellulose fibres with molecules of water adsorbed on them. This acts as stationary phase. The mobile phase is the mixture of the components to be identified prepared in a suitable solvent.

Nitric acid is added to sodium extract so as to decompose NaCN to HCN and Na₂S to H₂S and to expel these gases.

NaCN + HNO₃ ——-> NaNO₃ + HCN
Na₂S + 2HNO₃ ——> 2NaNO₃ + H₂S

Organic compound is fused with sodium metal so as to convert organic compounds into NaCN,  Na₂S, NaX and Na₃PO₄. Since these are ionic compounds and become more reactive and thus can be easily tested by suitable reagents.

Sublimation can be used for the separation of the two compounds because camphor can sublime whereas CaSO₄ does not.

It is because in steam distillation the sum of vapour pressure of organic compound and steam should be equal to atmospheric pressure.

No. CCl₄ is a completely non-polar covalent compound whereas AgNO₃ is ionic in nature. Therefore, they are not expected to react and thus a white precipitate of silver chloride will not be formed.

CO₂ is acidic in nature and therefore, it reacts with the strong base KOH to form K₂CO₃.

2KOH + CO₂ ? K₂CO₃+ H₂O.

It necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test because sulphuric acid will react with lead acetate to form a white precipitate of lead sulphate which will interfere in the test of sulphur.

Pb (OCOCH₃)₂ + H₂SO₄ →PbSO₄  + 2CH₃COOH

Step 1: Calculation of mass of CO₂ produced

Mass of compound = 0.20 g

% of carbon = 69%

i.e. 12/44 x  = Mass of Carbondioxide formed / Mass of Compound = 69/100

Therefore, mass of CO₂formed = (69 x 44 x 0.20) / (12 x 100) = 0.506 g

Step 2: Calculation of mass of H₂O produced

Mass of compound = 0.20 g

% of hydrogen = 4.8%

i.e. 2/18 x Mass of Water Formed/Mass of Compound = 4.8/100

Therefore, mass of H₂O formed = 4.8*18*0.20/2*100 = 0.0864 g

Step I: Calculation of volume of unused acid i.e. V2 =?

V1 = Volume of NAOH solution required = 60 cm³ 

N1 = Normality of NaOH solution = ½ N

N2 = Normality of H₂SO₄ = 1N

Applying N1V1 = N2V2

½ N x 60 cm³ = 1N x V2

Or V2 = 30 cm³

Step II: calculation of volume of acid used

Volume of acid added = 50 cm³

Volume of unused acid = 30 cm³

Volume of acid used = 50 – 30 = 20 cm³

Step III: Calculation of % of nitrogen

Mass of compound = 0.50 g

Volume of acid used = 20 cm³

Normality of acid used = 1 N

% of nitrogen = (1.4 x 20 x 1) / 0.50 = 56%

Mass of the compound = 0.3780 g
Mass of silver chloride = 0.5740 g

% of chlorine =

Mass of the compound = 0.468 g
Mass of barium sulphate= 0.668 g

% of sulphur = 

                     

(b) Iron (III) hexacyanidoferrate (II) (or ferriferrocyanide) Fe₄ [Fe (CN)₆]₃  is the correct answer.

(b) Is the most stable since it is a tertiary carbocation.

(d) Is the correct answer.

(b) It is a nucleophilic substitution reaction. KOH (aq) provides OH- ion for the nucleophile attack.

(c) In homolytic cleavage, one of the electrons of the shared pair in a covalent bond goes with each of the bonded atoms. Thus, in homolytic cleavage, the movement of a single electron takes place instead of an electron pair.

A heterolytic cleavage yields carbocations or carbanions, while a homolytic cleavage gives free radicals as reactive intermediate.

(b) The difference in energy between the actual structure and the lowest energy resonance structure is called the resonance stabilisation energy or simply the resonance energy. The more the number of important contributing structures, the more is the resonance energy.

(d) Electromeric effect is a temporary effect. The organic compounds having a multiple bond (a doubleor triple bond) show this effect in the presence of an attacking reagent only. It is defined as the complete transfer of a shared pair of? -electrons to one of the atoms joined by multiple bond on the demand of an attacking reagent.

(a) Permanent effect

(d) All of the above

(d) Both b and c

(b) Carbon and hydrogen

(a) Brings an electron pair to the reactive site

When an organic compound is present in anaqueous medium, it is separated by shaking it with an organic solvent in which it is more soluble than in water. The organic solvent and the aqueous solution should be immiscible with each other so that they form two distinct layers which can be separated by separatory funnel. The organic solvent is later removed by distillation or by evaporation to get backthe compound. Differential extraction is carried out in a separatory funnel.

If the organic compound is less soluble in the organic solvent, a very large quantity of solvent would be required to extract even a very small quantity of the compound. The technique of continuous extraction is employed in such cases. In this technique same solvent is repeatedly used for extraction of the compound.

Buckminsterfullerene is a common name given to the newly discovered C60 cluster (a form of carbon) noting its structural similarity to the geodesic domes popularised by the famous architect R. Buckminster Fuller.

The resonance effect is defined as 'the polarity produced in the molecule by the interaction of two π-bonds or between a π -bond and lone pair of electrons present on an adjacent atom'. The effect is transmitted through the chain. There are two types of resonance or mesomeric effect designated as R or M effect.

A heterolytic cleavage yields carbocations or carbanions, while a homolytic cleavage gives free radicals as reactive intermediate.

(i) By fractional crystallisation

(ii) Simple distillation

(iii Distillation under reduced pressure.

Steam distillation 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. In steam distillation, the liquid boils 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 p, the organic liquid vaporises at lower temperature than its boiling point.

Hyperconjugation is a general stabilising interaction. It involves delocalisation of σ electrons of C—H bond of an alkyl group directly attached to an atom of unsaturated system or to an atom with an unshared p orbital. The σ electrons of C—H bond of the alkyl group enter into partial conjugation with the attached unsaturated system or with the unshared p orbital. Greater the hyperconjugation, greater will be the stability of alkenes.

No, hyperconjugation is not a temporary effect. It is a permanent effect.

The presence of alternate single and double bonds in an open chain or cyclic system is termed as a conjugated system. These systems often show abnormal behaviour. The examples are 1, 3- butadiene, aniline and nitrobenzene etc. In such systems, the π -electrons are delocalised and the system develops polarity.

The resonance effect is defined as 'the polarity produced in the molecule by the interaction of two π -bonds or between a π -bond and lone pair of electrons present on an adjacent atom'. The effect is transmitted through the chain. There are two types of resonance or mesomeric effect designated as +R and-R effect.

The atoms or substituent groups, whichrepresent +R or –R electron displacementeffects are as follows:

+R effect: – halogen, –OH, –OR, –OCOR, –NH2, –NHR, –NR2, –NHCOR,

– R effect: – COOH, –CHO, >C=O, – CN, –NO2

The resonance structures have

(i) The same positions of nuclei, and

(ii) The same number ofunpaired electrons.

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As per NCERT the purification of an organic compounds are done in three ways - sublimation, crystallization, and distillation. The selection of the technique is based on the properties of compound and present of impurities.

No, students must solve some extra NCERT Class 11 Chemistry Chapter 12 to understand what types of questions are prepared in the exam. Moroever, solving more questions will help to obtain a good marks.