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NCERT/CBSE CHEMISTRY CLASS 12 textbook http://www.TutorBreeze.com Answers to NCERT/CBSE Chemistry (Class XII)textbook Chapter 13 AMINES 13.4 Arrange the following: (i) In decreasing order of the pKb values: C2H5NH2, C6H5NHCH3, (C2H5)2NH and C6H5NH2 In aromatic ammine there is delocalization of electrons of the N-atom over the benzene ring.In aliphatic amines the +I effect of two C2H5 groups in (C2H5)2NH is more than that of C2H5NH2. Therefore (C2H5)2NH is more basic than C2H5NH2.In C6H5NHCH3 and C6H5NH2 , C6H5NH2 is more basic than C6H5NHCH3 due to delocalization of electrons on the N-atom.Therfore the overall decreasing order of the pKb values are C6H5NH2 > C6H5NHCH3 > C2H5NH2 >(C2H5)2NH (ii) In increasing order of basic strength: C6H5NH2, C6H5N(CH3)2, (C2H5)2NH and CH3NH2 . In aromatic ammine there is delocalization of electrons of the N-atom over the benzene ring.In aliphatic amines the +I effect of two C2H5 groups in (C2H5)2NH is more than that of CH3NH2. Therefore (C2H5)2NH is more basic than CH3NH2.In C6H5NHCH3 and C6H5NH2 , C6H5NH2 is more basic than C6H5NHCH3 due to delocalization of electrons on the N-atom.So the increasing order of basic strength of the amines are C6H5NH2 > C6H5NHCH3 > CH3NH2 > C2H5NH2. (iii) In increasing order of basic strength: (a) Aniline, p-nitroaniline and p-toluidine p-nitroaniline < Aniline < p-toluidine (b) C6H5NH2, C6H5NHCH3, C6H5CH2NH2 In C6H5NH2, C6H5NHCH3 the N is directly attached to the benzene ring therefore the lone pair of electrons of the N atom is the delocalized over the bezene ring.There is a +I effect on C6H5CH2NH2 due to the presence of CH3 group of C6H5NHCH3 , Therefore C6H5NH2, C6H5NHCH3 are weaker bases than C6H5NH2.From the above explanations the increasing order of basic strength can be given as below C6H5NH2 < C6H5NHCH3 < C6H5CH2NH2 Please do not copy the answer given here Write to us for help NCERT/CBSE CHEMISTRY CLASS 12 textbook http://www.TutorBreeze.com (iv) In decreasing order of basic strength in gas phase: C2H5NH2, (C2H5)2NH, (C2H5)3N and NH3 In gas phase there is no hydrogen bonding , therefore stabalisation due to hydrogen bonding is not there. Therfore the only effect to determine the strength is the inductive effect.The +I effect increases with increase in the alkyl group.Therefore the basic strength will be the highest in (C2H5)3N and least in NH3. Therefore the decreasing order of basic strength in gas phase will be (C2H5)3N > (C2H5)2NH > C2H5NH2 > NH3. (v) In increasing order of boiling point: C2H5OH, (CH3)2NH, C2H5NH2 The elcetronegativity of O is more than N. Therefore C2H5OH forms hydrogen bonding and because of the hydrogen bonding the alcohols will have higher boiling point than the amines with comparable molecular mass.Now between (CH3)2NH and C2H5NH2 , C2H5NH2 has more hydrogen atom attached the th Nitrogen.Therfore the extent of hydrogen bonding will be more in primary amines i.e C2H5NH2 as compared to sec or ter amines.Therfore C2H5NH2 will have higher boiling point as compared to (CH3)2NH .Therefore the increasing order of boiling point will be as given below. (CH3)2NH < (CH3)2NH < C2H5OH (vi) In increasing order of solubility in water: C6H5NH2, (C2H5)2NH, C2H5NH2. Solubility of the amines will decrease with increase in the molecular mass and with decrease in the number of hydrogen atom s attached to the nitrogen. C2H5NH2 has the least molar mass and also has 2 hydrogen atoms attached to the nitrogen. Therefore this is most soluble. C6H5NH2 has the highest molar mass . Therefore the order in the increasing order of solubility in water is C6H5NH2 < (C2H5)2NH < C2H5NH2 Please do not copy the answer given here Write to us for help