Some Basic Principles and Techniques

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

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

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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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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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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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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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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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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-