Ch 12 Chemistry Class 11 Ncert Solutions

Organic chemistry as a chapter name might sound difficult to some students who are not very inclined towards Chemistry. NCERT understands the preference and requirements of its students. Hence, NCERT Exemplar Class 11 Chemistry solutions Chapter 12 is designed to ease the pressure and burden of the students. What do you understand by the term organic chemistry? Unlike the intense name, this branch of chemistry is quite engaging as it deals with carbon- containing compounds along with its structure, compositions, reactions and preparations. The chapter would be very interesting for students who are curious about compounds in general and Carbon in particular.

NCERT exemplar class 11 Chemistry solutions chapter 12: MCQ (Type 1)

Question:1

Which of the following is the correct IUPAC name?
(i) 3-Ethyl-4, 4-dimethylheptane
(ii) 4,4-Dimethyl-3-ethylheptane
(iii) 5-Ethyl-4, 4-dimethylheptane
(iv) 4,4-Bis(methyl)-3-ethylheptane
Answer:

The answer is the option (i) 3-Ethyl-4, 4-dimethylheptane
Explanation: According to the IUPAC naming, in case of presence of more than one different types of alkyl groups, they are written in alphabetical order and hence ethyl is wrritten first and then methyl.

Question:2

The IUPAC name for is _______:

(i) 1-hydroxypentane-1,4-dione
(ii) 1,4-dioxopentanol
(iii) 1-carboxybutan-3-one
(iv) 4-oxopentanoic acid
Answer:

The answer is the option (iv) 4-oxopentanoic acid
Explanation: According to the IUPAC naming, in case of presence of more than one different types of functional groups, one functional group will be taken as the main functional group on a priority basis and is mentioned as a suffix, while the other functional group is written as a prefix.

Question:3

The IUPAC name for

(i) 1-Chloro-2-nitro-4-methylbenzene
(ii) 1-Chloro-4-methyl-2-nitrobenzene
(iii) 2-Chloro-1-nitro-5-methylbenzene
(iv) m-Nitro-p-chlorotoluene
Answer:

The answer is the option (ii) 1-Chloro-4-methyl-2-nitrobenzene
Explanation: The substituent of the base compound is considered to be number 1, and then the direction of numbering is chosen such that the next substituent group is getting the lowest number. The substituents appear in the name, arranged in alphabetical order.

Question:5

In which of the following, functional group isomerism is not possible?
(i) Alcohols
(ii) Aldehydes
(iii) Alkyl halides
(iv) Cyanides
Answer:

The answer is the option (iii) Alkyl halides
Explanation: Isomerism is a phenomenon in which two compounds have the exact same molecular formula, but the different structural formula and such compounds are called isomers, and Functional isomers have same molecular formula but a different functional group. Alcohols are functional isomers of ethers. Aldehydes are functional isomers of ketones. Cyanides are functional isomers of isocyanides. Only alkyl halides do not show functional isomerism.

Question:8

The principle involved in paper chromatography is
(i) Adsorption
(ii) Partition
(iii) Solubility
(iv) Volatility
Answer:

The answer is the option (ii) Partition
Explanation: Partition chromatography is based on the concept of continuous differential partitioning of components of a mixture between stationary and mobile phases.

Question:9

What is the correct order of decreasing stability of the following cations?

(i) II > I > III
(ii) II > III > I
(iii) III > I > II
(iv) I > II > III
Answer:

(i) II > I > III
Explanation: In the case of (I) +vely charged C is attached to two alkyl groups and so +I effect stabilises the carbocation. In the case of (II) +R effect of $OCH_3$ group stabilises the carbocation. In the case of (III), - I effect of $-OCH_3$ gp, destabilises the carbocation; hence, the order of stability will be such: II > I >III.

Question:10

Correct IUPAC name for is _____________ .
(i) 2- ethyl-3-methylpentane
(ii) 3,4- dimethyl hexane
(iii) 2-sec-butylbutane
(iv) 2, 3-dimethylbutane
Answer:

(ii) 3,4- dimethyl hexane
Explanation: In the longest continuous chain, both ethyl groups are included, and 3rd and 4th carbon have methyl groups present as the substituent.

Question:19

Which of the following pairs are position isomers?

(i) I and II
(ii) II and III
(iii) II and IV
(iv) III and IV
Answer:

The answer is the option (ii) II and III
Explanation: When two or more compounds have functional groups or substituent atoms attached at different positions on the carbon skeleton, they are termed as position isomers, and this phenomenon is termed as position isomerism.

Question:20

Which of the following pairs are not functional group isomers?

(i) II and III
(ii) II and IV
(iii) I and IV
(iv) I and II
Answer:

The answer is the option (i) and (iii)
Explanation: Isomerism is a phenomenon in which two compounds have the exact same molecular formula, but the different structural formula and such compounds are called isomers, and Functional isomers have same molecular formula but a different functional group.

Question:21

Nucleophile is a species that should have
(i) a pair of electrons to donate
(ii) positive charge
(iii) negative charge
(iv) electron deficient species
Answer:

The answer is the option (i) and (iii)
Explanation: Nucleophiles are -vely charged or electron rich (lone pair of electrons) species.

Question:22

Hyperconjugation involves delocalisation of ______.
(i) electrons of carbon-hydrogen $\sigma$ bond of an alkyl group directly attached to an atom of unsaturated system.
(ii) electrons of carbon-hydrogen $\sigma$ bond of alkyl group directly attached to the positively charged carbon atom.
(iii) $\pi$ -electrons of carbon-carbon bond
(iv) lone pair of electrons
Answer:

The answer is the option (i) and (ii)
Explanation: Hyperconjugation means the delocalisation of electrons of the C-H bond of an alkyl group directly attached to an atom of unsaturated system or to an atom with an unshared p orbital. Electrons of C-H bond of the alkyl group enter into the partial conjugation with the attached unsaturated system or with the unshared p orbital. Thus, Hyperconjugation is a permanent effect.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 12: Short answer type

Question:23

Which of the above compounds form pairs of metameres?

Answer:

Compounds with same molecular formula but with different alkyl groups on either side of the functional groups are called metameres. In compounds above, structures V and VI or VI and VII or V and VII form a pair of metameres.

Question:24

Identify the pairs of compounds which are functional group isomers.

Answer:

Isomerism is a phenomenon in which two compounds have the exact same molecular formula, but the different structural formula and such compounds are called isomers, and Functional isomers have same molecular formula but a different functional group. Hence, I and V, I and VI, I and VII; II and V, II and VI, II and VII; III and V, III and VI; III and VII; IV and V, IV and VI, IV and VI are functional group isomers.

Question:25

Identify the pairs of compounds that represents position isomerism.

Answer:

When two or more compounds have functional groups or substituent atoms attached at different positions on the carbon skeleton, they are termed as position isomers, and this phenomenon is termed as position isomerism. In the given structures, I and II, III and IV, and VI and VII are position isomers.

Question:26

Identify the pairs of compounds that represents chain isomerism.

Answer:

Two or more compounds having similar molecular formula but different carbon skeletons, these are referred to as chain isomers, and the phenomenon is termed as chain isomerism. So, the answer will be I and III, I and IV, II and III, II and IV.

Question:28

What is the hybridisation of each carbon in $H_{2}C = C = CH_{2}$ .
Answer:

All three carbon atoms are linked to each other by double bonds only. Carbon at 1 and 3 are $sp^{2}$ hybridised because it has $3\sigma$ and $1\pi$ bonds, whereas carbon at 2 has 2 $\sigma$ bonds and $2\pi$ bonds, so it is sp hybridised.

Question:40

Name the compounds whose line formulae are given below:

Answer:

(i) 3-Ethyl-4-methylheptan-5-en-2-one (because the longest chain of carbon atoms is selected in such a way that the functional group > C = O gets lowest possible locant number)
(ii) 3-Nitrocyclohex-1-ene. (Carbon atoms of the ring are numbered in such a way that double bonded carbon gets the lowest number followed by the nitro group – $NO_{2}$ )

Question:43

Identify the most stable species in the following set of ions giving reasons:

Answer:

(i) $CH_{3}^{+}$ will be more stable. This is because bromine atom destabilises the positive charge on a carbon atom. Bromine atom has a lone pair of electrons and is from the electron-withdrawing group.
(ii) $^-CCl_{3}$ will be most stable. This is because chlorine is more electron-withdrawing atom. The negative charge on carbon will be stabilised by the chlorine atom. As the number of chlorine atoms that are attached to carbocation increase, its stability also enhances.

Question:44

Give three points of differences between inductive effect and resonance effect.
Answer:

S.No	Inductive effect	Resonance effect
1	This is due to displacement of $\sigma$ electrons in saturated compounds.	This is due to s displacement of the $\pi$ electrons or the lone pair of electrons in unsaturated and conjugated compounds.
2.	Partial +ve or –ve charge is developed.	Complete +ve or –ve charge is developed.
3	Inductive effect is effective only upto 3 to 4 carbons.	Movement of electrons takes place along the length of the conjugated system.

Question:46

Why does $SO_{3}$ act as an electrophile?
Answer:

$SO_{3}$ has total three highly electronegative oxygen atoms, that are attracted to the sulphur atom in $SO_{3}$ thus making sulphur electron deficient, that is it gets +ve and thus acts as an electrophile.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 12: Long Answer Type

Question:61

What is meant by hybridisation? Compound $CH_{2} = C = CH_{2}$ contains sp or $sp^{2}$ hybridized carbon atoms. Will it be a planar molecule?

Answer:

Atomic orbitals combine together to form a new set of orbitals which are termed as the hybrid orbitals and different from pure orbitals, these hybrid orbitals are used in the bond formation and this particular phenomenon is known as hybridisation, which can be defined as the process of intermixing of orbitals of different energy levels so as to redistribute their energies, which results in the formation of a new set of orbitals of equivalent energies and shape.
As the two-terminal Carbon atoms are forming three sigma bonds, therefore the hybridisation will be $sp^{2}$ . Whereas the central carbon atom is forming only two sigma bonds so, therefore the hybridisation will be sp.
It is known to us that the hybridisation of atoms is predicted by the total numbers of the sigma bonds formed by that particular atom and the lone pair of electrons present at the atom.
The p-orbitals in one plane overlap with one of the p-orbital of left terminal carbon atom and the p-orbital in other plane overlaps with p-orbital of right side terminal carbon atom. This fixes the position of two terminal carbon atoms and the hydrogen atoms attached to them in planes perpendicular to each other. Due to this the pair of hydrogen atoms attached to terminal carbon atoms are present in different planes.

Hence, the molecule $CH_{2} = C = CH_{2}$ is an not a planar molecule.

Question:62

Benzoic acid is an organic compound. Its crude sample can be purified by crystallisation from hot water. What characteristic differences in the properties of benzoic acid and the impurity make this process of purification suitable?
Answer:

As per the principle of crystallisation, it is known to us that the solute must be more soluble in hot water and less soluble in the cold water or any other similar solvent.
Now, the solubility of Benzoic acid in water is lesser as it is an organic compound. Moreover, it is less polar, whereas, in comparison, the water molecule is highly polar. But when the temperature of the water is increased, we observed that the solubility of benzoic acid is better in hot water in comparison to the cold water. The other reason being that the impurities are not insoluble in the water. Therefore, they can be filtered from the benzoic acid solution (if not soluble) or they would remain dissolved even after cooling the solution(if completely soluble in cold water). They will not be interfering with recrystallisation of the benzoic acid, as the extra solution can be discarded once the process of crystallisation is over.
The properties of benzoic acid and the impurity which makes this process of purification suitable are: -

Impurities which are present in benzoic acid are either insoluble in water or soluble in water to such an extent that they remain in solution.
The solubility of benzoic acid is slightly higher in hot water as compared to cold water.

Question:66

A liquid with high boiling point decomposes on simple distillation but it can be steam distilled for its purification. Explain how is it possible?
Answer:

Steam distillation is a different type of separation process for temperature sensitive materials like natural organic compounds. What happens with some organic compounds is that they decompose at higher temperatures and thus normal distillation does not suit the purpose. So, steam or water is added to the apparatus and the temperature of the compounds are depressed as a reason of that, evaporation happens at lower temperatures. Steam distillation is useful for separation if substances that are volatile, insoluble in water and have high vapour pressure at the boiling point of water i.e. $100^{\circ}C$ . Then after distillation is over vapours are condensed and hence constituents separate at ease.

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

NCERT solutions for Exemplar Class 11 Chemistry for Chapter 12 includes shapes, structural representations, classification, Nomenclature of organic compounds which makes it easier for the students to comprehend as it is given an ideal pattern. The concept of Isomerism that is presence of isomers is also covered in the NCERT Exemplar Class 11 Chemistry solutions Chapter 12. Isomers are polyatomic ions having the same molecular formulas with different arrangements/composition. These concepts will be dealt separately in detail further in the chapter.

Students looking forward to scoring well not just in 11th grade but also in Higher secondary examination must acquire Class 11 Chemistry NCERT Exemplar solutions chapter 12. You can make use of NCERT Exemplar class 11 Chemistry solutions chapter 12 PDF download from the official website's solution page.

NCERT Exemplar solutions for Class 11 Chemistry chapter 12 cover the following topics-

General Introduction
Tetravalence Of Carbon: Shapes Of Organic Compounds
Structural Representations Of Organic Compounds
1. Complete, Condensed And Bond-line Structural Formulas
2. Three-dimensional representation Of Organic Molecules
Classification Of Organic Compounds
Nomenclature Of Organic Compounds
1. The IUPAC System Of Nomenclature
2. Iupac Nomenclature Of Alkanes
3. Nomenclature Of Organic Compounds Having Functional Group(S)
4. Nomenclature Of Substituted Benzene Compounds
Isomerism
1. Structural Isomerism
2. Stereoisomerism
Fundamental Concepts In Organic Reaction Mechanism
1. Fission Of A Covalent Bond
2. Nucleophiles And Electrophiles
3. Electron Movement In Organic Reactions
4. Electron Displacement Effects In Covalent Bonds
5. Inductive Effect
6. Resonance Structure
7. Resonance Effect
8. Electromeric Effect (E Effect)
9. Hyperconjugation
10. Types Of Organic Reactions And Mechanisms
Methods Of Purification Of Organic Compounds
1. Sublimation
2. Crystallization
3. Distillation
4. Differential Extraction
5. Chromatography
Qualitative Analysis Of Organic Compounds
1. Detection Of Carbon And Hydrogen
2. Detection Of Other Elements
Quantitative Analysis
1. Carbon And Hydrogen
2. Nitrogen
3. Halogens
4. Sulphur
5. Phosphorus
6. Oxygen

What will students learn from NCERT Exemplar Class 11 Chemistry Solutions Chapter 12?

Organic chemistry along with its basic principles and techniques will be understood by the students after studying from NCERT Exemplar solutions Chapter 12 Class 11 Chemistry. These solutions are designed keeping in mind students having different IQs. The answers are explained in the simplest yet unique manner. Electron Movement in Organic Reactions, Electron Displacement Effects In Covalent Bonds, Inductive Effect, Resonance Structure, Resonance Effect etc which are complex concepts are explained in the solution.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 12

Important topics to cover for exams from NCERT Exemplar Class 11 Chemistry Solutions Chapter 12 Organic Chemistry - Some Basic Principles and Techniques

Despite the chapter being technical it is important from the academic point of view. Continuation of these concepts in 12th grade is a possibility. Hence it is pivotal to be familiar with organic compounds with its basic principles and techniques. Following are the pointers outlining important concepts covered in NCERT Exemplar Class 11 Chemistry Solutions Chapter 12.

o The Shapes of organic molecules and tetravalence of carbon and its causes will be clearly understood by the students after studying from the Class 11 Chemistry NCERT Exemplar solutions chapter 12.

o The aim of the chapter is to prepare students in such a manner that they can name the compounds as per system of Nomenclature provided by IUPAC. In addition to this, they will also be able to draw structures simply by looking at the names.

o Organic reaction mechanism, purification of organic compounds are some other topics taken care of in NCERT Exemplar Class 11 Chemistry Solutions Chapter 12.

o Qualitative and Quantitative analysis of carbon compounds is the ultimate explanation to comprehend previously learnt concepts.

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Frequently Asked Question (FAQs) - NCERT Exemplar Class 11 Chemistry Solutions Chapter 12 Organic Chemistry Some Basic Principles and Technique

Question: 1. Why is it important to study this chapter?

Answer:

A: The chapter is extremely important as it is a bit technical and can be tested thoroughly in the examination. However accustoming to organic chemistry will help you save time without constantly going back and forth in the chapter.

Question: 2. Will these solutions be useful for competitive exams?

Answer:

A: Yes, these solutions in the NCERT book will be useful not just while preparing for boards but also while studying for competitive exams like JEE etc.

Question: 3. Is this chapter important from the point of view of boards?

Answer:

A: Definitely, this chapter is important from a board point of view as these concepts are important to understand and are required in further studies. So, the authorities might test this chapter at length.

Question: 4. How can I download a solution for this particular chapter?

Answer:

A: You can download NCERT Exemplar Class 11 Chemistry Solutions Chapter 12 by simply going to the official website and downloading the PDF available on the solution page.

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NCERT Exemplar Class 11 Chemistry Solutions Chapter 12 Organic Chemistry Some Basic Principles and Technique

Ch 12 Chemistry Class 11 Ncert Solutions

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Ch 12 Chemistry Class 11 Ncert Solutions

NCERT exemplar class 11 Chemistry solutions chapter 12: MCQ (Type 1)

Question:1

NCERT Exemplar Class 11 Chemistry Solutions Chapter 12: Short answer type

NCERT Exemplar Class 11 Chemistry Solutions Chapter 12: Long Answer Type

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

NCERT Exemplar solutions for Class 11 Chemistry chapter 12 cover the following topics-

What will students learn from NCERT Exemplar Class 11 Chemistry Solutions Chapter 12?

NCERT Exemplar Class 11 Chemistry Solutions Chapter 12

Important topics to cover for exams from NCERT Exemplar Class 11 Chemistry Solutions Chapter 12 Organic Chemistry - Some Basic Principles and Techniques

NCERT Exemplar Class 11 Solutions

Also, check Chapter wise NCERT Solutions for Class 11 Chemistry

Frequently Asked Question (FAQs) - NCERT Exemplar Class 11 Chemistry Solutions Chapter 12 Organic Chemistry Some Basic Principles and Technique

Question: 1. Why is it important to study this chapter?

Question: 2. Will these solutions be useful for competitive exams?

Question: 3. Is this chapter important from the point of view of boards?

Question: 4. How can I download a solution for this particular chapter?

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