NCERT Solutions for Class 11 Chemistry Chapter 12 Organic Chemistry Some Basic Principles and Techniques- In this chapter, students will get to know about some basic principles and techniques of analysis which are required for understanding the properties and formation of organic compounds. The NCERT solutions for Class 11 Chemistry Chapter 12 Organic Chemistry Some Basic Principles and Techniques are prepared by our subject experts to help students in understanding the basic concepts of organic chemistry in a holistic manner. Chapter 12 Organic Chemistry Some Basic Principles and Techniques is an important chapter for a chemistry student from the perspective of getting hold of basic concepts of organic chemistry which will eventually help in subsequent chapters. Since this is a theoretical chapter and many concepts and facts will be learnt from exercise questions directly, hence It is must to go through NCERT solutions for Class 11 Chemistry Chapter 12 Organic Chemistry Some Basic Principles and Techniques. In this chapter, there are a total of 40 questions in the exercise. You can also refer to NCERT solutions with the detailed explanation of each and every question of NCERT textbook which will help you in preparation of CBSE class 11 final examination as well as in various competitive exams like NEET, JEE Main, BITSAT, VITEEE, etc.
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This chapter along with NCERT solutions for Class 11 Chemistry Chapter 12 Organic Chemistry Some Basic Principles and Techniques discusses some basic concepts in the reactivity and structure of organic compounds which are formed due to the covalent bonding, IUPAC nomenclature and classification of organic compounds, electromeric effect, inductive effect, hyperconjugation, and resonance. All related information about the heterolytic fission and homolytic fission of a covalent bond- carbocations, carbanions, free radicals, nucleophiles, and electrophiles is discussed in the later part of the chapter. By referring to the NCERT solutions for class 11 , students can understand all the important concepts and practice questions well enough before their examination
Organic compounds are essential for sustaining life's on the earth and include complex molecules like protein and DNA that constitutes important compounds of our blood, skin, and muscles. Some important areas of application of organic compounds are medicines, dyes, fuels, polymers, and clothing. These compounds can be classified on the basis of their functional group or the structure.
What is Functional Group?
A functional group is defined as an atom or group of atoms bonded together in a specific manner, which gives the chemical and physical properties of the organic compounds and they are the centres of the chemical reactivity.
Some important points of Chapter 12 Organic Chemistry Some Basic Principles and Techniques of Class 11 Chemistry-
1. Organic compounds are formed due to covalent bonding.
2. According to orbital hybridization concept carbon can have and hybridized orbitals.
3. The 3D representation of organic compounds on paper can be drawn by dash and wedge formula.
4. Compounds having the same molecular formula but differ in their physical and chemical properties are known as isomers and the phenomenon is called isomeris m.
5. Addition, substitution, elimination and rearrangement reactions are the types of organic reaction.
Topics of NCERT Grade 11 Chemistry Chapter 12 Organic Chemistry Some Basic Principles and Techniques
12.1 General Introduction
12.2 Tetravalence of Carbon: Shapes of Organic Compounds
12.3 Structural Representations of Organic Compounds
12.4 Classification of Organic Compounds
12.5 Nomenclature of Organic Compounds
12.7 Fundamental Concepts in Organic Reaction Mechanism
12.8 Methods of Purification of Organic Compounds
12.9 Qualitative Analysis of Organic Compounds
12.10 Quantitative Analysis
NCERT Solutions for Class 11 Chemistry Chapter 12 Organic Chemistry Some Basic Principles and Techniques
Q 12.1(iii) What are hybridisation states of each carbon atom in the following compounds ?
Here is the structure
Here and carbon atom is in hybridistaion and the carbon is
Q 12.2(i) Indicate the and bonds in the following molecule :
There are six C-C sigma bonds, six C-H sigma bonds and three -bonds between the carbon atom in the benzene.
Q 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 .
(a) The prefix in the IUPAC nomenclature is indicating two identical substituent group in the parent chain. Since two methyl group present at of the parent chain. So, the correct IUPAC name is 2, 2-dimethylpentane.
(b) In IUPAC nomenclature the sum of locant number should be minimum. Here the sum of 2,4,7 (=13)is less than the sum of 2, 5, 7 (=14). Thus the correct IUPAC name is 2,4,7- Trimethyloctane
(c) In IUPAC nomenclature, if the substituent group acquires the equivalent position of the parent chain then the lower number is given to one that comes first in alphabetical order. Hence 2-Chloro-4-methyl pentane is the correct IUPAC name of the compound.
(d) If the two functional groups are present in the parent chain then the suffix of the IUPAC name depends on the principal functional group. Here alcohol is the principal function group so the suffix should be and alkyne group considered as a substituent group. Therefore the correct IUPAC name of the compound is But-3-yn-1-ol.
Q 12.9 Which of the two: or is expected to be more stable and why ?
Since Nitro group is an electron withdrawing group. So, it shows effect, By reducing the electron negative charge of the compound, it stabilises the compound. On the other hand, the methyl group is an electron donor group so it shows effect. This increases the negative charge on the compound and destabilises the compound.
Hence is more stable than the .
Q 12.12 What are electrophiles and nucleophiles ? Explain with examples.
Electrophile- It is an electron deficient species, which seeking for an electron pair. This reagent takes away an electron pair. It is denoted as . For example, carbocationised neutral molecule having functional groups such as carbonyl group are an example of the electrophile.
A nucleophile is a reagent that brings an electron pair. In other words, it is nucleus- seeking reagent called nucleophile.
For examples- and carbanions ( ) etc. Ammonia and water also act as a nucleophile due to the presence of lone pair of electron.
Q 12.13 Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles:
Electrophiles are electron deficient species and can be seeking for an electron pair. On the other hand, the nucleophile is electron rich reagents and they can electron donor(nucleus seeking).
(a) have one lone pair of an electron, electron rich species. So, it is a nucleophile
(b) acts as a nucleus seeking reagent and act as a nucleophile
(c) , it is an electron deficient species, So, it is electrophile.
Q 12.14.(a) Classify the following reactions in one of the reaction type studied in this unit.
It is a substitution reaction as in this above equation the bromine group is replaced by the group.
Q 12.14.(b) Classify the following reactions in one of the reaction type studied in this unit.
The given reaction is an example of an addition reaction because in this reaction the two reactant molecule combined to form a single product. Also, we can say that the Hydrogen and chlorine of is added in the two different carbon atom of the same compound.
Q 12.15 What is the relationship between the members of following pairs of structures ? Are they structural or geometrical isomers or resonance contributors ?
These are pairs are structural isomers.
Compounds having the same molecular formula but different structures are called structural isomers. The above compounds have the same molecular formula but the structures are different due to the difference in the position of the carbonyl group.
In structure one -CO group is present at position and in second the -CO group is present at position.
Q 12.17. Explain the terms Inductive and Electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids?
Inductive effect- The permanent displacement of sigma ( ) electrons along the saturated chain, whenever an electron donating or withdrawing group is present, is known as the inductive effect. It could be effect.
Electromeric effect- The complete transfer of the shared pair of -electrons to one of the atoms linked by multiple bonds on the demand of attacking reagent. It is a temporary effect. It may be +E(electrons transferred towards attacking reagent) or -E effect(electron transferring away from the attacking reagent).
(a) The order of acidity can be explained by the negative inductive effect ( ). As the no. of chlorine atom increases, the effect also increases and so, the acidic strength also increases.
(b) This can be explained by the effect of the alkyl group. As the number of electron donor group increases, effect will also increase. With the increases in effect, the acidic character decreases accordingly.
Q 12.18 (a) Give a brief description of the principles of the following techniques taking an example in each case.
It is one of the most commonly used techniques for the purification of solid organic compounds. Its principle is based on the difference in the solubilities of the compounds and the solvent's impurities. The impure compounds are dissolved in solvent but they are sparingly soluble at room temperature but soluble at the higher temperature. On cooling the compound, the pure compounds get crystallise and removed by filtration.
For example - pure aspirin is obtained by recrystallising crude aspirin. Around 2 - 4 g of crude aspirin is dissolved in 20 mL of ethyl alcohol and the solution is heated for complete dissolution. Then after crystal formation, they can filter out and dried.
Q 12.18 (b) Give a brief description of the principles of the following techniques taking an example in each case.
This method is used for the purification of liquids from non-volatile impurities. It is based on the fact that fluids of having different boiling points vaporise at different temperatures. The impure liquid is boiled in a flask, and initially, the vapours of lower boiling points component are formed. The vapours are condensed by using a condenser, and the liquid is collected in a receiver.
For example- Organic liquids such as benzene, toluene, xylene etc can be purified by this method.
Q 12.19 Describe the method, which can be used to separate two compounds with different solubilities in a solvent S.
Fractional distillation can be used to separate the two compounds with different solubilities in a solvent S. The following steps are carried out in this process-
- The powdered mixture is taken in a flask, and the solvent is added into it and stirred simultaneously. We have to prepare a saturated solution and then heat it.
- After heating, this hot saturated solution can be filtered with filter paper in a china dish.
- Now the solution is allowed to cool. The less soluble compounds crystalise first, and more soluble compounds remain in the solution. After removing these crystals, the latter is concentrated once again. The hot solution is allowed to cool, and then the crystals of the more soluble compound are obtained.
- The last step is isolation and drying of crystals from the mother liquor.
Q 12.20 What is the difference between distillation, distillation under reduced pressure and steam distillation ?
The difference between distillation, distillation under reduced pressure and steam distillation are-
| Distillation || Distillation under reduced pressure || Steam Distillation |
| 1. Used for the purification of the compounds that are non-volatile impurities or liquids, which don't decompose on boiling. || 1. Used to purify liquids, which tends to decompose on boiling. Under the condition of reduced pressure. || 1.Used to purify the organic compounds, which is steam volatile and immiscible with water. |
| SImply, to separate the volatile liquids from non-volatile impurities. Or a mixture of liquids having sufficient Boiling point difference. || The liquid will boil at low temperature than its boiling points, therefore, it does not decompose. || The mixture of water and aniline can be separated by this method. |
| The mixture of petrol and kerosene is separated by this method. || Glycerol is purified by this method. |
Q 12.21. Discuss the chemistry of Lassaigne’s test.
This test is used to detect the presence of nitrogen, sulphur, halogens and sulphur in organic compounds. These elements are present in the covalent form in an organic compound. So, they are converted into the ionic form by fusing the compound with the sodium metal.
(X = Cl, Br, I)
Cyanide, sulphide or halide of sodium are extracted from the fused mass by boiling it with distilled water. This Is known as sodium extract method or Lassaigne's extract.
Q 12.22(i) Differentiate between the principle of estimation of nitrogen in an organic compound by
In the Dumas method, Nitrogen containing organic compound is heated with the copper oxide in carbon dioxide atmosphere, yields free nitrogen in addition to and water.
Traces of nitrogen oxides can also be formed in the reaction, which can be reduced to nitrogen by passing the gaseous mixture over the heated copper gauge. The produced mixture of gases is collected over by an aqueous solution of , it absorbs . Nitrogen is collected in the upper part of the graduated tube.
Q 12.22(ii) Differentiate between the principle of estimation of nitrogen in an organic compound by
In Kjeldahl’s method, the nitrogen-containing organic compound is heated with concentrated sulphuric acid. Nitrogen is converted into ammonium sulphate. It is then passed into the known volume of sulphuric acid. The amount of ammonia produced can be estimated by the amount of consumed in the reaction.
It is done by estimating the amount of unreacted left after the absorption of ammonia by titrating it with a standard alkali solution. This method is not applicable for the compound contain nitrogen in -nitro form or nitrogen present in the ring structure.
Q 12.23 Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.
Estimation of halogen is done by Carius method. In this method, a known quantity of organic compound is heated with fuming (nitric acid) with the presence of silver nitrate, contained in a hard glass tube, known as carius tube. C and H present in the compound are oxidised to carbon dioxide and water. And halogens are into to form and then it is filtered, dried, and weighed.
Let the mass of the organic compound be gram.
Mass of formed = gram
1 mol of contains 1 mol of X.
Therefore, Mass of halogen inm1 g of AgX = (Atomic mass of ) (Molecular mass of )
Thus, % of halogen will be = (Atomic mass of mo l. wt of
Estimation of sulphur- In this method, Organic compound is heated either fuming nitric acid or sodium peroxide in a hard glass tube called carius tube. Sulphur present in the compound is oxidised to form sulphuric acid. It is precipitated by as barium sulphate by adding barium chloride solution in water. Then ppt is filtered, washed and weighed.
Let the mass of organic compound taken = g
and the mass of barium sulphate formed = g
1 mol of = 233 g = 32 g sulphur
g contains = g sulphur
Percentage (%)of sulphur =
Estimation of phosphorus- In this process, a known mass of an organic compound is heated with fuming nitric acid and the phosphorus gets oxidised to phosphoric acid. By adding ammonia and ammonium molybdate, phosphorus can be precipitated as ammonium phosphomolybdate, . It can be also estimated as by precipitating it as by adding magnesia mixture which on ignition yields .
Let the mass of organic compound taken = g and mass of ammonium phosphomolybdate = g
Molar mass of g
Percentage(%) of phosphorus =
If phosphorus is estimated as ,
Percentage(%) of phosphorus = %
Q 12.24 Explain the principle of paper chromatography.
In paper chromatography, chromatography paper is used. It contains water trapped in it, which acts as the stationary phase. The solution of the mixture is spotted on this base of chromatography paper. The strip of paper is suspended in a suitable solvent, which is the mobile phase. Due to capillary action, the solvent rises up in the paper and it flows over the spot. The spots of the different component travel with the mobile phase to a different level of heights. The obtained paper is called chromatogram.
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