Chemistry NCERT Exemplar Solutions Class 11th Chapter Twelve: Overview, Questions, Preparation

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39. Option (iv)

Steam distillation is correct since this technique is applicable to separate the substances which are steam volatile and are immiscible with water. In steam distillation, steam from a 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 is collected. The compound is later separated from water using the separating funnel.

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40. Option (iv)

Thin layer chromatography is correct since thin layer chromatography is a method for analysing the mixtures by separating the components of the mixture. This technique can be used to determine the number of components in a mixture, the identity of compounds and the purity of compounds by observing the appearance of a product or the disappearance of a reactant. It can also be used to analyse the progress of a reaction. 

5. Whenever the difference in the boiling points of two liquids is not much than simple distillation technique cannot be used to separate them. The vapours of such liquids are formed within the same temperature range and are condensed simultaneously and hence the technique of fractional distillation is used in such cases. In this technique, vapours of a liquid mixture are passed through a fractionating column before condensation. The fractionating column is fitted over the mouth of the round bottomed flask. The vapours of the liquid with higher boiling point condense before the vapours of the liquid with lower boiling point. The vapours rising up in the fractionating column become richer in more volatile components and by the time the vapours reach to the top of the fractionating column, these become rich in more volatile components.

6. Steam distillation is a special type of separation technique for temperature sensitive materials such as natural organic compounds. Few organic compounds tend to decompose at higher temperatures and normal distillation does not fit for this purpose. So, water is added to the apparatus and the temperature of the compounds is depressed, evaporating them at lower temperature. Once the distillation is accomplished, the vapours are condensed and  there occurs,   the separation of the constituents at ease.

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34. Option (i) 3-Ethyl-4, 4-dimethylheptane is correct since in IUPAC naming the different alkyl groups present are written in alphabetical order. Di, tri, tetra are not included in the alphabetical order. So, “e” of the ethyl group comes before the “m” of the dimethyl group. The number 3 is given to the ethyl group and is written first. 

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35. Option (iv) 4-oxopentanoic acid is correct  since more than one functional group is present in the chain so, out of the two functional groups one will be given more priority as a result numbering is done from that side and mentioned as a sec. suffix, while the other one is written as prefix and mentioned as oxo along with the locant.

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41. Option (ii) Partition is correct since partition chromatography is based on the principle of continuous differential partitioning of components of a mixture between a stationary and a mobile phase. Paper chromatography is a type of partition chromatography only.

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42. Option (i) II > I > III is correct since in case (I), positively charged C is attached to two alkyl groups and is stabilized through  +I effect. In case (II), +R effect of - OCH3group stabilizes this carbocation. In case (III), - I effect of - OCH3 group destabilizes the carbocation.

Hence, the order of stability is given as: 

II > I >III.

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44. Option (i)

is found to be the correct one since the order of electronegativity of the attached group is found to be: 

Cl > Br > C > Mg. 

Greater the electronegativity of the group attached to the C  greater will be the positive charge. Therefore, in case (i) asterisk C will have the greatest positive charge.

4. Liquids which have different boiling points vaporize at different temperatures. The vapours are cooled and then liquids so formed are collected separately. Liquid A can be separated from B and C due to the large difference in boiling points. Liquid B and Liquid C have boiling points very close to each other and cannot be separated by simple distillation and are separated by fractional distillation. Liquid B distilled first since the order of boiling points of A, B and C is as follows: B < C < A.

7. If two or more compounds have the same molecular formula but different alkyl groups on either side of the functional groups then the compounds are said to be metamers. 

In the given structures- 

V and VI and VI and VII form a pair of metamers because alkyl groups are different on either side of the functional groups. 

8. The compounds with the same molecular formula but different functional groups are said to be functional group isomers. 

Alcohols are found to be the functional isomers of ether. In the given structures, I, II, III, IV  alcohol functional group is present and V, VI, VII contains ether 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 etc. are functional group isomers.

16. In accordance with the IUPAC nomenclature, the longest carbon chain is selected such that it must have maximum functional groups present in the compound. Therefore, only in one selected chain of 4 carbon atoms including both the functional group is corrected one. 

While in the other three, carbon atoms are in the selected chain, but both the functional groups are not included. 

17. In DNA and RNA, nitrogen is present in heterocyclic base and also present in the ring but not as a substituent. Therefore, nitrogen present in the ring cannot be converted into (NH₄ )2SO₄. Hence, it cannot be estimated by Kjeldahl method. 

18. If a liquid compound decomposes at its boiling point but is steam volatile and insoluble in water and stable at low pressure. Then steam distillation method can be used for its purification. 

Note : Answer the questions 35 to 38 on the basis of information given below:

“Stability of carbocations depends upon the electron releasing inductive effect of groups adjacent to positively charged carbon atom involvement of neighbouring groups in hyperconjugation and resonance.”

19. The resonating structures can be drawn as-

In structure I, CH₂ group has the positive charge which means octet is not completed, but in structure II, both the carbon atoms and oxygen atom have complete octet hence, more stable.

21. Triphenylmethyl cation is found to be very much stable as this positive charge on methyl carbon is delocalized in three phenyl rings. In each phenyl ring, positive charge is developed on 2 ortho position and para position, i.e., three resonating structures. Total resonating structures given by triphenylmethyl cation are nine. Hence, it is very stable. These structures can be shown as.

22. Four possible carbocations can be obtained from 2- methyl butane  are 

Generally, order of stability follows the order-

1° carbocation < 2° carbocation < 3° carbocation 

Order of increasing stability of the given structures is-

I < IV < II < III.

23. Whenever an organic compound contains both N and S, then on fusion with Na metal compound gives NaSCN or NaCN and Na₂S depending on the quantity of Na metal. If  Na metal is less, then only NaSCN is formed. In that case L.E on treating with FeSO₄ and H₂SO₄ gives red colour due to the formation of ferric thiocyanide Fe (SCN)₃. In case of NaCN, L.E on treating with FeSO₄ and H₂SO₄gives Prussian blue colour. 

Chemical reactions: Fe₂+ →Fe₃+ (Oxidation) 

Fe₃+ + 3NaSCN →Fe (SCN)₃ + 3Na⁺ 

                                 Red colour

If Na is in excess then : 

NaSCN + 2Na   NaCN + Na₂S 

6CN- + Fe₂+ →  [Fe (CN) ₆] ^4- 

3 [Fe (CN)₆ ]^4- + 4Fe₃+  Fe₄  [Fe (CN)₆ ]3 .xH₂O

Prussian blue   

24. In this structure i.e., 3-ethyl, 4-methylhept-5-en-2-one. (the longest carbon  chain is selected in such a way that the functional group > C = O gets the lowest possible locant.

28.

29. (i) It does not show resonance due to the absence of π electrons. 

(ii) It shows resonance due to the presence of lone pairs on the Nitrogen atom and π electrons on the C = O bond. Hence, amide can be represented by three resonating structures.

(iii) In this structure, the lone pair of electrons present on Nitrogen atom, is not conjugated with the π electrons of double bond. Hence, it does not show resonance.

30. SO₃ acts as an electrophile as in this, three highly electronegative oxygen atoms are attached to the sulphur atom which makes it electron deficient. It can be shown by resonance that it gets positive charge and acts as an electrophile.

31. The structure-I of propenal is found to be more stable because structure having more covalent bonds in a resonating structure is found to have more stability, in structure II, the terminal carbon has only six electrons which makes it less stable. 

32. Mixture of alcohol and hydrocarbon can be separated by simple distillation because both the components have large differences in their boiling points. When temperature is raised to the low boiling liquid i.e., hydrocarbon, the vapours consist only of hydrocarbons without any contamination of alcohol.

33. Resonating structure of A

In this case structure II is less stable than I because in structure II separation of positive and negative charges is there. Therefore, contribution of structure (II) is less than that of (I), these hybrids of compound (A) i.e., CH₃COOH.

Resonating structure of B

While structure (III) and (IV) are of equal energy and also contribute equally towards, resonance hybrid of compound (B). Therefore, CH3COO- is more stable.

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36. Option (ii) 1-Chloro-4-methyl-2-nitrobenzene is correct since the substituent of the base compound is given number 1 and then the direction of numbering is chosen in such a manner that the next substituent gets the lowest number. The substituents that appear in the IUPAC name are in alphabetical order.  Hence in this case numbering can be shown as-

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37. Option

(iii) CH₃ - CH₂ - C ≡ *CH is correct since electronegativity of carbon depends on its hybridized state and percentage of s-character. Electronegativity ∝ percentage of  s-character.

So, the order of percentage ‘s’ character is given as-

 sp³ (25%) < sp² (33%) < sp (50%)

In option (iii) Carbon linked through triple bond and is sp hybridized hence it is highly electronegative.

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38. Option (iii)

Alkyl halides is correct since functional isomers are the ones which have the same molecular formula but different functional groups being present in them. 

Alcohols are functional isomers of ethers while aldehydes are the functional isomers of ketones whereas cyanides are functional isomers of isocyanides.  Only alkyl halides are the ones which do not show functional isomerism.

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43. Option (ii) 3,4- dimethyl hexane is correct since in selecting the longest continuous carbon chain both the ethyl groups are included while the third and fourth carbons have methyl groups as a substituent.

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45. Option (iv) is found to be the correct one since the stabilisation of carboxylate ion depends upon the dispersal of the negative charge. And the dispersal depends upon two factors, i.e., +R effect of the carboxylate ion and Inductive effect of the halogens. In all the above structures, +R effect is common, but halogen atoms are uncommon. Therefore, dispersal of negative charge depends upon the halogen atoms. Fluorine is most electronegative, in structure (iv) two F atoms are present and hence more dispersal of negative charge is there.

1. The atomic orbitals combine to give rise to a new set of equivalent orbitals called hybrid orbitals and are used in bond formation. This phenomenon is known as hybridization which can be defined as the process of intermixing of the orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of new sets of orbitals having equivalent energies and shape. In H₂C=C=CH₂ (allene) carbon atom l and 3 are sp² hybridized as each one has 3σ bonds while carbon atom 2 has 2σ bonds so it is sp hybridized while the allene molecule as a whole is found to be non-planar. 

2.  Benzoic acid can be purified by hot water because of following characteristics.
(i) Benzoic acid is more soluble in hot water and less soluble in cold water.
(ii) Impurities present in benzoic acid are either insoluble in water or more soluble in water to such an extent that they remain in solution as the mother liquor upon crystallisation.

3. When the difference in boiling points of two liquids is not much then simple distillation cannot be used to separate them. Vapours of such liquids are formed within the same temperature range and are condensed simultaneously. The technique of fractional distillation is used in such cases. In this technique, vapours of a liquid mixture are passed through a fractionating column before condensation. The fractionating column is fitted over the mouth of the round bottom flask. The liquid (A) with low boiling point will distill first.

9. If two or more compounds differ in the position of substituent, functional group, or multiple bonds but molecular formula is same then these are said to be position isomers as they exhibit position isomerism. 

In the given structures, I and II, III and IV, VI and VII are found to be the position isomers.

10. If two or more compounds have the same molecular formula but different carbon skeletons then these compounds are referred to as chain isomers and the phenomenon is termed as chain isomerism. The pairs of compounds that represent chain isomerism are- I, II, III, IV.

11. Elements such as nitrogen, halogens, Sulphur, and phosphorus present in an organic compound converted into ions by fusing with sodium metal followed by plucking in distilled water getting sodium extract (Lassaigne's test). On adding dil. H₂SO₄ in place of dil. HNO₃for testing halogens by AgNO₃, white precipitate of AgSO₄ is formed. This will lead wrong result of chloride. Hence, only HNO₃ is used instead of dil. H₂SO₄

12. The given structure is of C₃H₄ in which all three carbon atoms are linked to each other by double bonds. Carbons are marked as Cl,  C₂ and C₃. Cl and C₃ are sp₂ hybridized as containing 3σ bonds and 1π bond. C₂ has 2 σ bonds and 2 π bonds so, it is sp hybridized.

13. Electronegativity of an element is directly proportional to its  s-character. If C is sp3hybridized then s- character is 25%, if it is sp2 hybridized then s-character is 33% and if sp hybridized then s-character is 50%. Hence, sp hybridized carbon has strong hybridization, s electrons are more strongly attracted by nucleus than p-electrons thus electronegativity of carbon increases with increase in s-character.

14. Carbon is known to be more electronegative than Magnesium and hence Mg acquires partial positive charge while C acquires partial negative charge because the bonded pair of electrons is attracted towards it.

15. These are position isomers as if two or more compounds differ in the position of substituents, functional groups, or multiple bonds but if the molecular formula is same then these are said to be position isomers. Here the functional groups are attached to different carbon atoms in the carbon chain, but the molecular formula is the same. They are not metamers because the number of C atoms on either side of-S- are the same.

20. The structure A is more stable than structure B. This is because carbocation A is more Planar and? electrons from the ring shift to the side group and are stabilized by resonance while structure B is non-planar and does not undergo resonance. Also, double bond is more stable within the ring as compared to the side chain.

(i) In this case CH3+ is the most stable species because replacement of H atom by Br atom (-Inductive effect) increases positive (ii

(ii)  is most stable because negative charge on carbon atom is dispersed due to -I effect of Cl. More Cl atoms, more dispersal of negative charge and more stabilized.

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Carbocation is correct since H⁺ attacks on propene and the delocalization of electrons can take place in two possible ways which can be shown as-

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47. Bromine is found to be more electronegative than Carbon which leads to heterolytic fission. Due to the higher electronegativity the electrons displace from carbon to Br. Therefore, CH₃ gets positive charge and Br gets negative charge and hence option (ii) is found to be the correct one.

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48. Option (ii) is correct since the arrow shows the direction of electron transfer. In the given reaction, the electron flows from π bond towards the electron-deficient H⁺ group . This can be represented as-

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49. Options (i) and (iv) are correct since only in these two compounds all the carbon atoms are having same hybridization i.e., sp hybridized in option (i) and sp2 hybridized in option (iv).

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50. Options (i), (iii) and (iv) are correct since the spatial arrangement of groups or atoms can be checked by doing two interchanges and bringing the H atom below the plane of the paper. Then find out the sequence of the left-out groups in a particular order whether clockwise or anticlockwise starting from the atom with the highest atomic number to the atom with lower atomic numbers. So, out of the given options option (ii) has the same spatial arrangement as (A) while in rest three spatial arrangements are different.

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51. Options (ii) and (iii) are correct since the electrophiles are positively charged or electron deficient species. They act as Lewis acids. AlCl₃, SO₃are Lewis acids while  NO₂⁺, CH₃⁺, CH₃-C⁺ = O are positively charged species. Therefore, the species involved in the above-mentioned options act as electrophiles.

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52. Option (ii) II and III since in position isomerism, two or more compounds differ in the position of substituents, functional group, or multiple bonds but their molecular formula is same. In pentanone-2 and pentanone-3, position of ketonic group is different. CH₃-CH₂-CH₂-CO-CH₃ and CH₃ - CH₂-CO-CH₂- CH₃

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53. Options (i) and (ii) are correct since two or more compounds with same molecular formula but different functional groups are known as functional isomers. In the given compounds- 

I. Aldehydic group. II. Ketonic group. III. Ketonic group. IV. Aldehydic group. So, II and III, I and IV are not functional isomers.

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54. Options (i) and (iii) are correct since nucleophiles are negatively charged or electron rich (as contains lone pair of electrons) species. 

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55. Options (i) and (ii) are correct since hyperconjugation involves the delocalization of electrons of C—H bond of an alkyl group directly attached to an atom of unsaturated system or to an atom with an unshared p-orbital. The electrons of C—H bond of the alkyl group enter into partial conjugation with the attached unsaturated system or with the unshared p-orbital. This is a permanent effect. Hyperconjugation can be shown as-

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

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

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58. (i) →   (a); (ii) →   (a); (iii) →   (b)

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59.  (i) → (a); (ii) → (b); (iii) → (b) : (iv) → (c)

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60. Option (i)

i.e.,   Both A and R are correct, and R is the correct explanation of A is the answer. Since the liquids are having sufficient difference in their boiling points. Liquids having different boiling points vaporize at different temperatures. The vapours are cooled, and the liquids so formed are collected separately. 

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61. Option (iv)

i.e.,   A is not correct, but R is correct is the answer. This is because canonical structures always have more energy than the resonance hybrid. Resonance hybrids are always found to be more stable than any of the canonical structures since the delocalization of electrons lowers the orbitals energy and gives stability.

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62. Option (i) i.e.,  

Both A and R are correct, and R is the correct explanation of A is the answer as when two or more compounds differ in the position of substituent atom or functional group on the carbon skeleton, they are known as position isomers and this phenomenon is termed as position isomerism. Pent-2-ene and pent-l-ene are position isomers because they differ in the position of the double bond. 

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63. Option (iv) i.e., A is not correct, but R is correct is the answer since hybridization of C can be found out by counting σ bonds and π bonds present on C atom. This can be shown as-

If C has 3σ bonds then it is sp² hybridized. If C has 2σ bonds then it is sp hybridized.

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64. Option (iii) i.e., Both A and R are not correct is the answer since sulphur is estimated by Carius method in the form of white precipitate of BaSO₄ on heating with fuming and BaCl₂. If light yellow solid is obtained means impurities are present. It is filtered, washed and then dried to get pure BaSO₄ . 

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65. Option (i)

i.e.,   Both A and R are correct, and R is the correct explanation of A is the answer since in paper chromatography, a special quality of paper known as chromatography paper is used. Chromatography paper contains water molecules trapped in it which acts as the stationary phase. A strip of chromatography paper spotted at the base with the solution of the mixture is suspended in a suitable solvent or a mixture of solvents which acts as the mobile phase. The solvent rises up the paper by capillary action and flows over the spot. The paper selectively retains different components according to their differing partition in the two phases. During this process the paper strip so developed is known as a chromatogram and the spots of the separated coloured compounds are visible at different heights from the position of the initial spot on the chromatogram.