Question 28

Halogens attached to benzene rings exert – I and +R effect. +R effect dominates – I effect and increases the electron density at ortho and para positions of the benzene ring with respect to halogens. 33. 2-Methyl butane is . Possible compounds are A, B and C given below : 171 Hydrocarbons Nine possibilities for compound ‘A’ because nine methyl hydrogens are present in 2-methylbutane. Two possibilities for ‘B’ compound because two CH hydrogens are present in 2-methylbutane. Only one possibility for ‘C’ compound because one CH hydrogen is present in 2- methylbutane. Relative amounts of A, B = number of hydrogen × relative reactivity and C compounds A (1°) B (2°) C (3°) Relative amount 9×1= 9 2×3.8 = 7.6 1×5=5 Total Amount of monohaloginated compounds = 9 + 7.6 + 5 = 21.6 Percentage of A = × 100 = 41.7% 21.6 7.6 Percentage of B = × 100 = 35.2% 21.6 Percentage of C = × 100 = 23.1% 21.6 35. Radical I is tertiary where as radical II is primary. Radical I is more stable due to hyperconjugation. 36. 37. A = Planar ring, all atoms of the ring sp2 hybridised, has six delocalised π electrons, follows Huckel rule. It is aromatic. B = Has six π electrons, but the delocalisation stops at sp3 hybridised CH2- carbon. Hence, not aromatic. C = Six delocalised π-electrons (4 π electrons + 2 unshared electrons on negatively charged carbon) in a planar ring, follows Huckel’s rule. It is aromatic. D = Has only four delocalised π-electrons. It is non aromatic. E = Six delocalised π-electrons follows Huckel’s rule. π electrons are in sp2 hybridised orbitals, conjugation all over the ring because of positively charged carbon. The ring is planar hence is aromatic. F = Follows Huckel’s rule, has 2 π electrons i.e. (4n+2) π-electrons where (n=0), delocalised π-electrons. It is aromatic. G = 8π electrons, does not follow Huckel’s rule i.e., (4n+2) π-electrons rule. It is not aromatic. 38. A = Has 8π electrons, does not follow Huckel rule. Orbitals of one carbon atom are not in conjugation. It is not aromatic. B = Has 6π delocalised electrons. Hence, is aromatic. C = Has 6π electrons in conjugation but not in the ring. Non aromatic. D = 10π electrons in planar rings, aromatic. E = Out of 8π electrons it has delocalised 6π electrons in one six membered planar ring, which follows Huckel’s rule due to which it will be aromatic. F = 14 π electrons are in conjugation and are present in a ring. Huckel’s rule is being followed. Compound will be aromatic if ring is planar. IV. Matching Type 40. (i) → (d) (ii) → (a) (iii) → (e) (iv) → (c) (v) → (b) 41. (i) → (b) (ii) → (c) (iii) → (a) 42. (i) → (d) (ii) → (c) (iii) → (b) (iv) → (a) 43. (i) → (d) (ii) → (a) (iii) → (b) (iv) → (c) 173 Hydrocarbons V. Assertion and Reason Type 44. (i) 45. (i) 46. (i) 47. (iii) VI. Long Answer Type Br in CS alc.KOH 48. C5H11Br ⎯⎯⎯⎯⎯ → Alkene (C5H10) ⎯⎯⎯⎯⎯⎯ 2 2 → C5H10Br2 (A) (B) (C) Alc.KOH Na–liq.NH 1 –2HBr → C5H8 ⎯⎯⎯⎯⎯⎯ ⎯⎯⎯⎯⎯ → C5H7–Na + H 2 2 D (Alkyne) Sodium alkylide The reactions suggest that (D) is a terminal alkyne. This means triple bond is at the end of the chain. It could be either (I) or (II). CH3— CH2—CH2—CH2 ≡ CH I II Since alkyne ‘D’ on hydrogenation yields straight chain alkane, therefore structure I is the structure of alkyne (D). Hence, the structures of A, B and C are as follows : (A) CH3—CH2—CH2—CH2—CH2Br (B) CH3—CH2—CH2—CH CH2 (C) CH3—CH2—CH2—CH (Br)—CH2Br 49. Step I 896 mL vapour of CxHy (A) weighs 3.28 g 3.28 × 22700 –1 22700 mL vapour of CxHy (A) weighs g mol = 83.1 g mol–1 Step II Element (%) Atomic Relative ratio Relative no. Simplest mass of atoms ratio C 87.8 12 7.31 1 3 H 12.19 1 12.19 1.66 4.985 Empirical formula of ‘A’ C3H5 Empirical formula mass = 35 + 5 = 41 u Molecular mass 83.1 n= = = 2.02 ≈ 2 Empirical formula mass 41 Molecular mass is double of the empirical formula mass. ∴ Molecular Formula is C6H10 Step III Structure of 2-methylpentane is Hence, the molecule has a five carbon chain with a methyl group at the second carbon atom. ‘A’ adds a molecule of H2O in the presence of Hg2+ and H+, it should be an alkyne. Two possible structures for ‘A’ are : I II Since the ketone (B) gives a positive iodoform test, it should contain a —COCH3 group. Hence the structure of ketone is as follows : Therefore structure of alkyne is II. 50. Two molecules of hydrogen add on ‘A’ this shows that ‘A’ is either an alkadiene or an alkyne. On reductive ozonolysis ‘A’ gives three fragments, one of which is dialdehyde. Hence, the molecule has broken down at two sites. Therefore, ‘A’ has two double bonds. It gives the following three fragments : OHC—CH2—CH2—CHO, CH3CHO and CH3—CO—CH3 Hence, its structure as deduced from the three fragments must be (A) Reactions (A) ⎯⎯⎯⎯ Ozone → Zn /H2O ⎯⎯⎯⎯⎯ → CH3—CHO + OHC—CH2—CH2—CHO + 175 Hydrocarbons