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

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

Edited By Sumit Saini | Updated on Sep 10, 2022 06:00 PM IST

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 _______:

1660629392817(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
capture-24

(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:4

Electronegativity of carbon atoms depends upon their state of hybridisation. In which of the following compounds, the carbon marked with asterisk is most electronegative?
(i) CH_3 - CH_{2}-^{\ast }CH_{2} -CH_{3}
(ii) CH_{}3 - ^{\ast}CH = CH - CH_{3}
(iii) CH_{3} - CH_{2} - C \equiv ^{\ast } CH
(iv) CH_{3} - CH_{2} - CH = ^{\ast}CH_{2}
Answer:

The answer is the option (iii) CH_{3} - CH_{2} - C \equiv ^{\ast }CH
Explanation: The electronegativity of the carbon atom depends on its hybridisation state and the percentage of ‘s’ character. It is directly proportional to the ‘s’ character. Here we have sp hybridised carbon having 50 percent s character.

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:6

The fragrance of flowers is due to the presence of some steam volatile organic compounds called essential oils. These are generally insoluble in water at room temperature but are miscible with water vapour in vapour phase. A suitable method for the extraction of these oils from the flowers is:
(i) Distillation
(ii) Crystallisation
(iii) Distillation under reduced pressure
(iv) Steam distillation
Answer:

The answer is the option (iv) Steam distillation
Explanation: This is a technique which is applied to separate substances which are steam volatile and are immiscible with water, differently. In case of 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 then condensed and collected together. The compound is later separated from water using a separating funnel.

Question:7

During hearing of a court case, the judge suspected that some changes in the documents had been carried out. He asked the forensic department to check the ink used at two different places. According to you which technique can give the best results?
(i) Column chromatography
(ii) Solvent extraction
(iii) Distillation
(iv) Thin layer chromatography
Answer:

The answer is the option (iv) Thin layer chromatography
Explanation: Thin-layer chromatography (TLC) is a method for analysing mixtures by separating the components of the mixture. This method can be used to determine the number of components in the mixture, to identity the compounds and the purity of compounds by observing the appearance of a product or the disappearance of a reactant, whichever is the case.

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?
capture-25

(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 capture-26is _____________ .
(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:11

In which of the following compounds the carbon marked with asterisk is expected to have greatest positive charge?
(i) ^{\ast}CH_{3} -CH_{2}-Cl
(ii) ^{\ast}CH_{3} -CH_{2}-Mg^{+}Cl^{-}
(iii) ^{\ast}CH_{3} -CH_{2}-Br
(iv) ^{\ast}CH_{3} -CH_{2}-CH_{3}
Answer:

The answer is the option (i) ^{\ast}CH_{3} -CH_{2}-Cl
Explanation: Order of electronegativity is as follows: Cl > Br > C > Mg. The more electronegative group attached to the C will give a more positive charge. Therefore, in (i) case, asterisk C will have the greatest positive charge.

Question:12

Ionic species are stabilised by the dispersal of charge. Which of the following carboxylate ion is the most stable?
capture-27
Answer:

The answer is the option capture-28
Explanation: The dispersal of the negative charge depends on the stabilisation of the carboxylate ion. The negative charge which is shown is dispersed by two factors i.e +R effect of carboxylate ion and -I effect of halogens. In the above cases, +R effect is common, but halogen atoms are different and hence dispersal of the negative charge depends upon halogen atoms, where F is most electronegative.

Question:13

Electrophilic addition reactions proceed in two steps. The first step involves the addition of an electrophile. Name the type of intermediate formed in the first step of the following addition reaction.
H_{3}C-HC = CH_{2} + H^{+} \rightarrow ?

(i) 2^{\circ} Carbanion
(ii) 1^{\circ} Carbocation
(iii) 2^{\circ} Carbocation
(iv) 1^{\circ} Carbanion
Answer:

The answer is the option (iii) 2^{\circ} Carbocation
Explanation: When H+ attacks on propene delocalisation of electrons can take place in two possible ways:
capture-29

Question:14

Covalent bond can undergo fission in two different ways. The correct representation involving a heterolytic fission of CH_{3}-Br is
capture-30
Answer:

The answer is the option (ii).
Explanation: Br is more electronegative than C, hence heterolytic fission takes place in that case and so electrons displace from carbon to Br. Therefore,CH_{3} gets the positive charge and Br gets the negative charge.

Question:15

The addition of HCl to an alkene proceeds in two steps. The first step is the attack of H+ ion to> C= C< portion which can be shown as
capture-31
(iv) All these are possible
Answer:

The answer is the option (ii) because the double bond is an electron source. The charge flows from higher electron density; thus, the proton is attacked by n electrons of the double bond.
capture-32
It can be shown as
capture-33

NCERT Exemplar Class 11 Chemistry Solutions Chapter 12: MCQ (Type 2)

Question:16

Which of the following compounds contain all the carbon atoms in the same hybridisation state?
(i) H-C \equiv C-C\equiv C-H
(ii) CH_{3}-C \equiv C-CH_{3}
(iii) CH_{2} = C = CH_{2}
(iv) CH_{2} = CH-CH = CH_{2}
Answer:

The answer is the option (i) H-C \equiv C-C\equiv C-H and (iv) CH_{2} = CH-CH = CH_{2}
Explanation: In these two compounds only all carbon atoms are in the same hybridisation state, which is sp and sp2 hybridised, respectively.

Question:17

In which of the following representations given below spatial arrangement of group/ atom different from that given in structure ‘A’?
capture-34

capture-35
Answer:

The answer is the option (i), (iii) and (iv)
Explanation: The spatial arrangement of groups of atoms can be checked by interchanging and bringing the H below the plane of the paper and then finding out the sequence of the remaining groups in a particular order whether clockwise or anticlockwise where it is started from the atom with the highest atomic number to the atom with the lowest atomic numbers.

Question:18

Electrophiles are electron seeking species. Which of the following groups contain only electrophiles?
(i)BF_{3}, NH_{3},H_{2}O
(ii)AlCl_{3}, SO_{3},NO_{2}^{+}
(iii)NO_{2}^{+},CH_{3}^{+},CH_{3}-C^{+}=O
(iv)C_{2}H_{5}^{-},C^{\cdot }_{2}H_{5},C_{2}H_{5}^{+}
Answer:

The answer is the option (ii) and (iii)
Explanation: Electrophiles are the +vely charged species or the electron-deficient species. They are Lewis acids. AlCl_{3}, SO_{3} and are Lewis acids, (iii)NO_{2}^{+},CH_{3}^{+},CH_{3}-C^{+}=O are +vely charged species

Question:19

Which of the following pairs are position isomers?
capture-36
(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?
capture-36
(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.
capture-37

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

Question:23

Which of the above compounds form pairs of metameres?
capture-38
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.
capture-38
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.
capture-38
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.
capture-38
Answer:

compounds I and III, I and IV, II and III and II and IV represent chain isomerism.

Question:26

Identify the pairs of compounds that represents chain isomerism.
capture-38
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:27

For testing halogens in an organic compound with AgNO_{3} solution, sodium extract (Lassaigne’s test) is acidified with dilute HNO_{3}. What will happen if a student acidifies the extract with dilute H_{2}SO_{4} in place of dilute HNO_{3} ?
Answer:

Elements like nitrogen, halogens, sulphur and phosphorous present in organic compounds converted into ions by fusing with sodium metal which is then plucked in distilled water getting sodium extract. On adding diluted H_{2}SO_{4} in place of diluted HNO3 for testing halogens by AgNO_{3}, a white precipitate of Ag_2SO_{4} is formed. This will lead to the wrong result of chloride. Hence, only HNO_{3} is used instead of diluted H_{2}SO_{4}.

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:29

Explain, how is the electronegativity of carbon atoms related to their state of hybridisation in an organic compound?
Answer:

Electronegativity of carbon is directly proportional to the ‘s’ character. If C is sp^{3}hybridised then V character is 25%, sp^{2} hybridised then V character is 33% and if sp hybridised then V character is 50%. Hence, sp hybridised carbon has strong hybridisation, s electrons are more strongly attracted by nucleus than p-electrons thus electronegativity of carbon increases with increase in ‘s’ character.

Question:30

Show the polarisation of carbon-magnesium bond in the following structure.
CH_{3}-CH_{2}-CH_{2}-CH_{2}-Mg-X

Answer:

Carbon is more electronegative than magnesium; hence Mg has a partial positive charge, and C has a partial negative charge because the bonded pair of electrons are attracted towards carbon.
CH_{3}-CH_{2}-CH_{2} -CH_{2}^{-\delta} -Mg^{+\delta}- X

Question:31

Compounds with same molecular formula but differing in their structures are said to be structural isomers. What type of structural isomerism is shown by
capture-39

Answer:

These are Metamers. When two or more compounds have differnet arrangement of atoms attached at different positions on the carbon skeleton, they are termed as metamers, and this phenomenon is termed as Metamarism.

Question:32

Which of the following selected chains is correct to name the given compound according to IUPAC system.
capture-40
Answer:

IUPAC says that the selected longest carbon chain must have maximum functional groups present in the compound. Hence 4 carbon chain is correct to name the given compound according to IUPAC system

Question:33

In DNA and RNA, nitrogen atom is present in the ring system. Can Kjeldahl method be used for the estimation of nitrogen present in these? Give reasons.
Answer:

In the case of DNA and RNA, nitrogen is present in the heterocyclic base and in the ring not as a substituent. Therefore, nitrogen present in the ring cannot be converted into(NH_{4} )_{2}SO_{4}. Hence, it cannot be estimated by the Kjeldahl method.

Question:34

If a liquid compound decomposes at its boiling point, which method (s) can you choose for its purification. It is known that the compound is stable at low pressure, steam volatile and insoluble in water.
Answer:

Steam distillation can be used for its purification for a compound when the compound decomposes at its boiling point but is steam volatile and insoluble in water and is also stable at low pressures

Question:35

Draw the possible resonance structures for
capture-41
and predict which of the structures is more stable. Give a reason for your answer.

Answer:

capture-42
C has a more stable structure than A because. This is because in C, the octet of all the atoms is complete, but, in A, C-atom with a positive charge does not have 8 electrons in the valence shell.

Question:36

Which of the following ions is more stable? Use resonance to explain your answer.
capture-43

Answer:

Structure A is more stable. Carbocation A is more Planar, and \pi electrons from the ring shift to side group and are stabilised by resonance. Moreover, the double bond is more stable within the ring as compared to the side chain.
capture-44

Question:37

The structure of triphenylmethyl cation is given below. This is very stable, and some of its salts can be stored for months. Explain the cause of high stability of this cation.
capture-45

Answer:

Triphenylmethyl cation is very stable because the +ve charge of methyl carbon is delocalised in three phenyl rings. In each phenyl ring, +ve charge is developed on 2 ortho position and para position, i.e. three resonating structures. Thus, the total resonating structures given by triphenylmethyl cation are nine. Hence, it is very stable.

Question:38

Write structures of various carbocation’s that can be obtained from 2- methylbutane. Arrange these carbocation’s in order of increasing stability.
Answer:

capture-46
The stability order according to increasing stability is (III) > (II) > (IV) > (I)
because (I) and (IV) are primary, (II) is secondary carbocation, and (III) is tertiary carbocation.

Question:40

Name the compounds whose line formulae are given below:
capture-47

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:
capture-52

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:45

Which of the following compounds will not exist as resonance hybrid. Give reason for your answer:
(i) CH_{3} OH
(ii) R-CONH_{2}
(iii) CH_{3}CH = CHCH_{2}NH_{2}

Answer:

(i) CH_{3} OH: It does not show resonance because there are no electrons.
(ii) R-CONH_{2} : It shows resonance due to the presence of lone pair of electrons on N atom and electrons on C = O bond. Hence it can be represented by three resonating structures.
(iii) CH_{3}CH = CHCH_{2}NH_{2} : Lone pair of electrons on N atom, is not conjugated with the \pi electrons of double bond, so no resonance.

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.

Question:47

Resonance structures of propenal are given below. Which of these resonating structures is more stable? Give reason for your answer.
capture-53

Answer:

Structure I, propenal, is more stable than structure II because structure I has more covalent bonds in its resonating structures and moreover in structure II has only six electrons on its terminal carbon making it less stable.

Question:48

By mistake, an alcohol (boiling point 97^{\circ}C) was mixed with a hydrocarbon (boiling point 68^{\circ}C). Suggest a suitable method to separate the two compounds. Explain the reason for your choice.
Answer:

Because both the components above have a large difference in their boiling points, so simple distillation method is used to separate. The temperature when it rises to the low boiling point the hydrocarbon vapours are only formed without any contamination of alcohol.

Question:49

Which of the two structures (A) and (B) given below is more stabilised by resonance? Explain.
(A)
CH_{3}COOH and
capture-54

Answer:

Structure ‘B’ is more stabilised as it does not involve charge separation.. Hence CH_{3}COO^{-} is more stable.
(A)
CH_{3}COOH

1660630083602

capture-54

1660630180860

NCERT Exemplar Class 11 Chemistry Solutions Chapter 12: Matching Type

Question:50

Match the type of mixture of compounds in Column I with the technique of separation/purification given in Column II.

Column I

Column II

(i) Two solids which have different solubilities in a solvent, and which do not undergo reaction when dissolved in it.

(a) Steam distillation

(ii) Liquid that decomposes at its boiling point

(b) Fractional distillation

(iii) Steam volatile liquid

(c) Simple distillation

(iv) Two liquids which have boiling points close to each other

(d) Distillation under reduced pressure

(v) Two liquids with a large difference in boiling points.

(e) Crystallisation

Answer:

(i) → (e); (ii) → (d); (iii) → (a);(iv) → (b); (v) → (c)

Question:51

Match the terms mentioned in Column I with the terms in Column II.

Column I

Column II

(i) Carbocation

(a) Cyclohexane and 1-hexene

(ii) Nucleophile

(b) Conjugation of electrons of C – H \sigma bond with empty p-orbital present at adjacent positively charged carbon.

(iii) Hyperconjugation

(c) sp2 hybridised carbon with empty p-orbital

(iv) Isomers

(d) Ethyne

(v) sp hybridisation

(e) Species that can receive a pair of electrons

(vi) Electrophile

(f) Species that can supply a pair of electrons

Answer:

(i) →(c);(ii) → (f); (iii) → (b); (iv) → (a); (v) → (d); (vi) → (e)

Question:52

Match Column I with Column II.

Column I

Column II

(i) Dumas method

(a) AgNO_{3}

(ii) Kjeldahl method

(b) Silica gel

(iii) Carius method

(c) Nitrogen gel

(iv) Chromatography

(d) Free radicals

(v) Homolysis

(e) Ammonium sulphate

Answer:

(i) → (c); (ii) → (e); (iii) → (a); (iv) → (b);(v) → (d)

Question:53

Match the intermediates given in Column I with their probable structure in Column II.

Column I

Column II

(i) Free radical

(a) Trigonal planar

(ii) Carbocation

(b) Pyramidal

(iii) Carbanion

(c) Linear

Answer:

(i) → (a); (ii) → (a); (iii) → (b)

Question:54

Match the ions given in Column I with their nature-given in Column II.

Column I

Column II

(i) capture-55

(a) Stable due to resonance

(ii) capture-56

(b) Destabilised due to inductive effect

(iii) capture-57

(c) Stabilised due to hyperconjugation

(iv) capture-58

(d) A secondary carbocation

Answer:

(i) → (a), (b), (d) (ii) → (b) (iii) → (b) (iv) → (c), (d)

NCERT Exemplar Class 11 Chemistry Solutions Chapter 12: Assertion and Reason Type

Question:55

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Simple distillation can help in separating a mixture of propan-1-ol (boiling point 97{^\circ}C) and propanone (boiling point 56{^\circ}C).
Reason (R): Liquids with a difference of more than 20{^\circ}C in their boiling points can be separated by simple distillation.

(i) Both A and R are correct, and R is the correct explanation of A.
(ii) Both A and R are correct, but R is not the correct explanation of A.
(iii) Both A and R are not correct.
(iv) A is not correct but R is correct.
Answer:

The answer is the option (i) Both A and R are correct, and R is the correct explanation of A.
Explanation: The liquids given above have quite a difference in their boiling points and liquids which have different boiling points also vaporise at different temperatures. Vapours gets condensed and is then collected separately.

Question:56

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Energy of resonance hybrid is equal to the average of energies of all canonical forms.
Reason (R): Resonance hybrid cannot be presented by a single structure.

(i) Both A and R are correct, and R is the correct explanation of A.
(ii) Both A and R are correct, but R is not the correct explanation of A.
(iii) Both A and R are not correct.
(iv) A is not correct but R is correct.
Answer:

The answer is the option (iv) A is not correct, but R is correct.
Explanation: Canonical structures always have more energy as compared to the resonance hybrid, and so the Resonance hybrids are always more stable than any of the canonical structures. The delocalisation of electrons lowers the orbitals energy in case of resonance hybrid and gives stability.

Question:57

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Pent- 1- ene and pent- 2- ene are position isomers.
Reason (R): Position isomers differ in the position of functional group or a substituent.

(i) Both A and R are correct, and R is the correct explanation of A.
(ii) Both A and R are correct, but R is not the correct explanation of A
(iii) Both A and R are not correct.
(iv) A is not correct but R is correct
Answer:

The answer is the option (i) Both A and R are correct, and R is the correct explanation of A.
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:58

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): All the carbon atoms in H_{2}C=C=CH_{2} are sp^{2} hybridised
Reason (R): In this molecule, all the carbon atoms are attached to each other by double bonds.

(i) Both A and R are correct, and R is the correct explanation of A.
(ii) Both A and R are correct, but R is not the correct explanation of A.
(iii) Both A and R are not correct.
(iv) A is not correct but R is correct.
Answer:

The answer is the option (iv) A is not correct, but R is correct.
Explanation: Hybridisation of C can be found out by counting the number of \sigma bonds and \pi bonds present on the C atom. If C has 3\sigma bonds, it is sp^2 hybridised. If C has 2\sigma bonds, it is sp hybridised.

Question:59

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Sulphur present in an organic compound can be estimated quantitatively by Carious method.
Reason (R): Sulphur is separated easily from other atoms in the molecule and gets precipitated as light yellow solid.

(i) Both A and R are correct, and R is the correct explanation of A.
(ii) Both A and R are correct, but R is not the correct explanation of A.
(iii) Both A and R are not correct.
(iv) A is not correct but R is correct.
Answer:

The answer is the option (iii) Both A and R are not correct.
Explanation: Sulphur is estimated by the Carius method in the form of white precipitate of BaSO_{4} on heating with fuming and BaCl_{2}. If light yellow solid is obtained it means that impurities are present. It is then filtered, washed, and then dried to get pure BaSO_{4}.

Question:60

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Components of a mixture of red and blue inks can be separated by distributing the components between stationary and mobile phases in paper chromatography.
Reason (R): The colored components of inks migrate at different rates because paper selectively retains different components according to the difference in their partition between the two phases.

Answer:

The answer is the option (i) Both A and R are correct, and R is the correct explanation of A.
Explanation: In paper chromatography what is used, is a special quality paper known as chromatography paper, which contains water trapped in it, and which acts as the stationary phase. A strip of chromatography paper which is spotted at its base is suspended with the solution of the mixture in a suitable solvent or a mixture of solvents. This solvent acts as the mobile phase. The solvent rises up the paper by the capillary action and flows all over to the spot. The paper now selectively retains different components according to their differing partition in the two phases. The paper strip so developed is now known as a chromatogram.

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:63

Two liquids (A) and (B) can be separated by the method of fractional distillation. The boiling point of liquid (A) is less than boiling point of liquid (B). Which of the liquids do you expect to come out first in the distillate? Explain.
Answer:

The two liquids which we are considering above does not have much difference in their boiling points, and so simple distillation cannot just be used for the separation process. So the technique of fractional distillation is used in very such cases, which is a technique where vapours of a liquid mixture are passed through a fractionating column before condensation and the column used for fractionating is fitted over the mouth of the round bottom flask and the liquid with low boiling point distils first.

When vapours of a liquid mixture are passed through a fractionating column, the capours of the low boiling liquid (A) will move up while those of the high boiling liquid will condense and fall back into the flask. Therefore, liquid (A) with low boiling point will distill first.

capture-59

Question:64

You have a mixture of three liquids A, B and C. There is a large difference in the boiling points of A and rest of the two liquids i.e., B and C. Boiling point of liquids B and C are quite close. Liquid A boils at a higher temperature than B and C and boiling point of B is lower than C. How will you separate the components of the mixture. Draw a diagram showing set up of the apparatus for the process.
Answer:

As per the information available in the question, liquid A boils at a higher temperature than B and C and boiling point of B is lower than C.
Therefore, the correct order of boiling point for all three compounds is:
A > C > B
As it has been provided in the question that there is a significant difference in the boiling points of A and B and C. So, it is possible to separate liquid A from the mixture through the process of simple distillation. The setup for simple distillation is: -

capture-60

As the remaining two liquids B and C have close boiling points; thus the process of fractional distillation will be used for the separation of the two. The setup for fractional distillation is
capture-61

Question:65

Draw a diagram of bubble plate type fractionating column. When do we require such type of a column for separating two liquids? Explain the principle involved in the separation of components of a mixture of liquids by using fractionating column. What industrial applications does this process have?

Answer:

The two liquids which we are considering above does not have much difference in their boiling points and so dimple distillation cannot just be used for the separation process. So, the technique of fractional distillation is used in such very cases, which is a technique where vapours of a liquid mixture are passed through a fractionating column before condensation and the column used for fractionating is fitted over the mouth of the round bottom flask and the liquid with low boiling point distills first.
capture-62

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-

  1. General Introduction

  2. Tetravalence Of Carbon: Shapes Of Organic Compounds

  3. Structural Representations Of Organic Compounds

    1. Complete, Condensed And Bond-line Structural Formulas

    2. Three-dimensional representation Of Organic Molecules

  4. Classification Of Organic Compounds

  5. 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

  6. Isomerism

    1. Structural Isomerism

    2. Stereoisomerism

  7. 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

  8. Methods Of Purification Of Organic Compounds

    1. Sublimation

    2. Crystallization

    3. Distillation

    4. Differential Extraction

    5. Chromatography

  9. Qualitative Analysis Of Organic Compounds

    1. Detection Of Carbon And Hydrogen

    2. Detection Of Other Elements

  10. 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.

Also, Check Chapter Wise NCERT Solutions for Class 11 Chemistry

Chapter-1 - Some Basic Concepts of Chemistry

Chapter-2 - Structure of Atom

Chapter-3 - Classification of Elements and Periodicity in Properties

Chapter-4 - Chemical Bonding and Molecular Structure

Chapter-5 - States of Matter

Chapter-6 - Thermodynamics

Chapter-7 - Equilibrium

Chapter-8 - Redox Reaction

Chapter-9 - Hydrogen

Chapter-10 - The S-Block Elements

Chapter-11 - The P-Block Elements

Chapter-12 - Organic chemistry- some basic principles and techniques

Chapter-14 - Hydrocarbons

Chapter-15 - Environmental Chemistry

NCERT Exemplar Class 11 Solutions

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Frequently Asked Question (FAQs)

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

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.


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

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 Main, etc.

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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.

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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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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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