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NCERT Solutions Class 11 Chemistry Chapter 8 – Organic Chemistry: Basic Principles
NCERT Solutions for Class 11 Chemistry Chapter 8 (Organic Chemistry – Some Basic Principles and Techniques) help students understand the basic rules of organic chemistry and how carbon forms a wide range of compounds. This chapter covers important topics such as catenation, hybridisation, types of organic compounds, and their structures. These concepts help students understand the structure and properties of compounds found in living things, medicines, and everyday products.
NCERT Solutions for Class 11 Chemistry Chapter 8 by ALLEN provide answers to NCERT textbook questions with simple explanations. Students can learn IUPAC naming, electronic effects, reaction intermediates, and methods of purifying organic compounds. These solutions help students understand the chapter, solve textbook questions, and prepare for CBSE exams, JEE, and NEET.
1.0Class 11 Chemistry Chapter 8 Organic Chemistry: Key Concepts
This chapter explains how organic compounds are named, classified, purified, and studied. The key concepts include:
- Tetravalence of Carbon and Hybridisation: Understanding how carbon forms four bonds and how sp, sp², and sp³ hybridisation affect the shape of molecules.
- Structural Representations: Learning to write and read bond-line, condensed, and wedge-dash structures of organic compounds.
- Classification of Organic Compounds: Grouping organic compounds into open-chain, alicyclic, aromatic, and heterocyclic compounds.
- IUPAC Nomenclature: Learning the rules for naming alkanes, alkenes, alkynes, and compounds with different functional groups.
- Isomerism: Understanding structural isomerism, including chain, position, functional, and metamerism, along with the basics of stereoisomerism.
- Breaking of Covalent Bonds: Learning the difference between homolytic and heterolytic bond fission.
- Nucleophiles and Electrophiles: Identifying electron-rich species that donate an electron pair and electron-deficient species that accept an electron pair.
- Electronic Effects: Understanding the inductive effect, electromeric effect, resonance effect, and hyperconjugation, and how they affect organic compounds.
- Purification of Organic Compounds: Learning methods such as sublimation, crystallisation, simple and fractional distillation, vacuum distillation, steam distillation, and chromatography.
- Qualitative and Quantitative Analysis: Learning how to detect elements in organic compounds using Lassaigne’s test and calculate the percentage of carbon, hydrogen, nitrogen, and halogens using methods such as Dumas and Kjeldahl.
2.0Detailed NCERT Textbook Solutions : Class 11 Chemistry Chapter 8
- What are hybridisation states of each carbon atom in the following compounds?
Ans.
(i) is hybridised. C-2 is sp hybridised.
(ii)
is hybridised.
C-2 is hybridised.
C-3 is hybridised.
(iii)
and are hybridised.
C-2 is hybridised.
(iv) is hybridised. C-2 is hybridised. C-3 is sp hybridised.
(v) All the 6 carbon atoms in benzene are hybridised.
- Indicate the and bonds in the following molecules:
Ans.
- Write bond line formulas for: Isopropyl alcohol, 2,3-Dimethyl butanal, Heptan-4-one. Ans. The bond line formulae of the given compounds are:
4.Give the IUPAC names of the following compounds:
Ans.
- 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 2,5,7Trimethyloctane (c) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane (d) But-3-yn1-ol or But-4-ol-1-yne
Ans.
(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, 2dimethylpentane.
(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-4methylpentane.
(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.
- Draw formulas for the first five members of each homologous series beginning with the following compounds. (a) H-COOH (b) (c)
Ans. The first five members of each homologous series beginning with the given compounds are shown as follows:
(a) H-COOH : Methanoic acid : Ethanoic acid : Propanoic acid : Butanoic acid : Pentanoic acid
(b) : Propanone : Butanone : Pentan-2-one : Hexan-2-one : Heptan-2-one
(c) : Ethene : Propene -Butene -Pentene -Hexene
- 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:
(b) 2-hydroxy-1, 2, 3-propane tricarboxylic acid Condensed Formula:
The functional groups present in the given compound are carboxylic acid (-COOH) and alcoholic (-OH) groups.
(c) Hexanedial Condensed Formula:
The functional group present in the given compound is aldehyde (-CHO).
- dentify the functional groups in the following compounds
Ans. The functional groups present in the given compounds are:
(a) Aldehyde (-CHO), Hydroxyl(-OH), Methoxy - OMe), double bond
(b) Amino ; primary amine, Ester (-O-CO-), Triethylamine ; tertiary amine
(c) Nitro , double bond
- Which of the two: - or is expected to be more stable and why? Ans. group is an electron-withdrawing group. Hence, it shows -I effect. By withdrawing the electrons toward it, the 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, is expected to be more stable than .
- Explain why alkyl groups act as electron donors when attached to a system. Ans. When an alkyl group is attached to a system, it acts as an electron-donor group by the process of hyper conjugation. To understand this concept better, let us take the example of propene.
In hyper conjugation, 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 sigma bond orbital with an empty p orbital of the bond of an adjacent carbon atom.
The process of hyper conjugation in propene is shown as follows:
This type of overlap leads to a delocalisation (also known as no-bond resonance) of the electrons, making the molecule more stable.
- Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation. (a) (b) (c) (d) (e) (f)
Ans.
- What are electrophiles and nucleophiles? Explain with examples. Ans. An electrophile is a reagent that takes away an electron pair. In other words, an electron-seeking reagent is called an electrophile ( ). Electrophiles are electron-deficient and can receive an electron pair. Carbocation's and neutral molecules having functional groups such as carbonyl group are examples of electrophiles. A nucleophile is a reagent that brings an electron pair. In other words, a nucleus-seeking reagent is called a nucleophile (Nu:). For example: , carbanions , etc. Neutral molecules such as and ammonia also act as nulceophiles because of the presence of a lone pair.
- Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles: (a) (b) (c)
Ans. 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.
(a) Here, 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) Here, acts as an electrophile as it is an electron-deficient species.
- Classify the following reactions in one of the reaction type studied in this unit. (a) (b) (c) (d)
Ans.
(a) It is an example of substitution reaction as in this reaction the bromine group in bromoethane is substituted by the -SH group.
(b) It is an example of addition reaction as in this reaction two reactant molecules combine to form a single product.
(c) It is an example of elimination reaction as in this reaction hydrogen and bromine are removed from bromoethane to give ethane.
(d) In this reaction, substitution takes place, followed by a rearrangement of atoms and groups of atoms.
- 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 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.
- 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.
- Explain the terms Inductive and Electrometric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids? (a) (b)
Ans. Inductive effect The permanent displacement of sigma ( ) 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,
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 electrons to either of the two atoms linked by multiple bonds in the presence of an attacking agent. For example,
Electrometric effect could be + E effect or - E effect.
- +E effect: When the electrons are transferred towards the attacking reagent
- -E effect: When the electrons are transferred away from the attacking reagent
(a) 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) 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.
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 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.
- Describe the method, which can be used to separate two compounds with different solubilities in a solvent S.
Ans. 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 lesssoluble 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.
- What is the difference between distillation, distillation under reduced pressure and steam distillation ? Ans. The differences among distillation, distillation under reduced pressure, and steam distillation are given in the following table.
#Please refer attached PDF for the next set of questions and solutions.
3.0NCERT Class 11 Biology – Chapter-Wise Solutions & Explanations
Explore NCERT Solutions for Class 11 Chemistry with answers to textbook questions from every chapter. Revise key topics, practise questions, and prepare for school exams.
4.0Key Features of NCERT Solutions for Class 11 Chemistry Chapter 8
- IUPAC Naming: Practice naming organic compounds by following the rules and priority order of functional groups.
- Stability of Reaction Intermediates: Step-by-step explanations of carbocations, carbanions, and free radicals, including why tertiary carbocations are more stable than primary carbocations.
- Purification Methods: Clear explanations of purification techniques and when to use them, such as vacuum distillation for liquids that may decompose at high temperatures.
- Numerical Problems: Step-by-step solutions for calculating the percentage of carbon, hydrogen, and nitrogen in organic compounds.
- Prepared by ALLEN Faculty: Solutions prepared by ALLEN subject experts to help students understand key concepts and prepare for CBSE exams, JEE, and NEET.