Arrange the following:
(i) in decreasing order of pKb values C2H5NH2, C6H5NHCH3,(C2H5)2 and C6H5NH2.
(ii) in increasing order of basic strength C6H5NH2, C6H5N(CH3)2, (C2H5)2
(iii) increasing order of basic strength
(a) Aniline, p-nitroaniline and p-toluidine (b) C6H5NH2, C6H5NHCH3, C6H5CH2NH2
(iv)Decreasing order of basic strength in the gas phase C2H5NH2, (C2H5)2 NH, (C2H5)3 N and NH3 and NH3 .
(v) Increasing order of boiling point C2H5OH, (CH3)2, NH C2H5 NH2.
(vi) Increasing order of solubility in water C6H5NH2, (C2H5)2 NH, C2H5NH2.
(i) Due to delocalization of lone pair of electrons of the N-atom over the benzene ring, C6H5NH2 and C6H5NHCH 3 are far less basic than C2H5NH2 and (C2H5)2NH. Further, due to +I-effect of the CH3 group, C6H5NHCH3 is little more basic than C6H5NH2. Among C2H5NH2 and C2H5)2NH,(C2H5)2NH, is more basic than C2H5NH2 due to greater +I-effect of the two C2H5 groups and stabilization of its conjugate acid by H-bonding. Combining all these facts, the relative basic strength of these four amines decreases in the order: (C2H5)2NH> C2H5NH2>C6H5NHCH3> C6H5NH2
Since a stronger base has a lower pKb value, therefore, pKb values decrease in the reverse order:
C6H5NH2> C6H5NHCH3>C2H5NH2> (C2H5)2NH
(ii) We have already explained in ans. (i) above that the relative basic strength of the amines, C6H5NH2, C6H5NHCH3 and (C2H5)2NH decreases in the order: (C2H5)2 NH > C6H5NHCH3 > C6H5NH2.
Conversely, C6H5NH2 < C6H5NHCH3 < (C2H5)2 NH the basic strength of these amines increases in the order:
Among CH3NH2 and (C2H5)2 NH, primarily due to the greater +I-effect of the two C2H5 groups over one CH3 group, (C2H5)2NH is more basic than CH3NH2. Thus, the basic strength of the four amines increases in the order: C6H5NH2 < C6H5NHCH3 < CH3NH2 < (C2H5)2NH.
(iii) (a) The electron-donating groups increase while the electron-withdrawing groups decrease the basic strength of amines. Therefore, p-nitroaniline is the weakest base followed by aniline while p-toluidine is the strongest base. In other words, basicity increases in order : p-nitroaniline < aniline < p-toluidine.
(b) In C6H5NH2 and C6H5NHCH3, N is directly attached to the benzene ring. As a result, the lone pair of electrons on the N-atom is delocalized over the benzene ring. Therefore, both C6H5NH2 and C6H5NHCH3 are weaker bases than C6H5CH2NH2. Further, due to +I-effect of the CH3 group, C6H5NHCH3 is a stronger base that C6H5NH2.
In other words, basic strength increases in the order: C6H5NH2 < C6H5NHCH3 < C6H5CH2 NH2
(iv) In the gas phase, solvent effects, i.e., stabilization of the conjugate acids due to H-bonding, are absent. Therefore, in the gas phase, basic strength mainly depends upon the +I-effect of the alkyl groups. Since the +I-effect increases with the number of alkyl groups, therefore, the basic strength of the amines decreases as the number of ethyl groups decreases from 3 in (C2H5)3N to 2 in (C2H5)2NH to 1 in C2H5NH2 and zero in NH3. In other words, basic strength in the gas phase decreases in the order:
(C2H5)3 N >(C2H5)2 NH > C2H5NH2 > NH3
(v) Since the electronegativity of O is higher than that of N, therefore, alcohols form stronger H-bonds than amines. In other words, the boiling points of alcohols are higher than those of amines of comparable molecular masses. Therefore, the b.p. of C2H5OH (molecular mass = 46) is higher than those of (CH3)2 NH and C2H5NH2 (each having a molecular mass of 45). Further, since, the extent of H-bonding depends upon the number of H-atoms on the N-atom. Therefore, 1° amines with two H-atoms on the N-atom have higher b.ps than 2° amines (of comparable molecular mass) having only one H-atom. In other words, the b.p. of C2H5NH2 is higher than that of (CH3)2NH. Thus, the b.ps. of the given three compound increases in the order: (CH3)2 NH < C2H5NH2 < C2H5OH
(vi) Solubility decreases with increase in molecular mass of amines due to increase in the size of the hydrophobic hydrocarbon part and with decrease in the number of H-atoms on the N-atom which undergo H-bonding. Now among the given compounds, C6H5NH2 has the highest molecular mass of 93 followed by (C2H5)2NH with molecular mass of 73 while C2H5NH2 has the lowest molecular mass of 45. Thus, the solubility increases in the order in which molecular mass decreases, i.e., C6H5NH2 <(C2H5)2 NH < C2H5NH2