Question -
Answer -
(i) Indecreasing order of the pKbvalues:
C2H5NH2, C6H5NHCH3, (C2H5)2NHand C6H5NH2
(ii) Inincreasing order of basic strength:
C6H5NH2, C6H5N(CH3)2, (C2H5)2NHand CH3NH2
(iii) In increasing order of basic strength:
(a) Aniline, p-nitroanilineand p-toluidine
(b) C6H5NH2, C6H5NHCH3, C6H5CH2NH2.
(iv) Indecreasing order of basic strength in gas phase:
C2H5NH2, (C2H5)2NH,(C2H5)3N and NH3
(v) Inincreasing order of boiling point:
C2H5OH,(CH3)2NH, C2H5NH2
(vi) Inincreasing order of solubility in water:
C6H5NH2,(C2H5)2NH,C2H5NH2
Answer
(i) In C2H5NH2,only one −C2H5 group is present while in (C2H5)2NH,two −C2H5 groups are present. Thus, the +I effect is more in(C2H5)2NH than in C2H5NH2.Therefore, the electron density over the N-atom is more in (C2H5)2NHthan in C2H5NH2. Hence, (C2H5)2NHis more basic than C2H5NH2.
Also, both C6H5NHCH3 andC6H5NH2 are less basic than (C2H5)2NHand C2H5NH2 due to the delocalization of the lone pair in theformer two. Further, among C6H5NHCH3 andC6H5NH2, the former will be more basic due to the +T effect of−CH3 group. Hence, the order of increasing basicity ofthe given compounds is as follows:
C6H5NH2 6H5NHCH3 < C2H5NH2 <(C2H5)2NH
We know that the higher the basic strength,the lower is the pKb values.
C6H5NH2 >C6H5NHCH3 > C2H5NH2 >(C2H5)2NH
(ii) C6H5N(CH3)2 ismore basic than C6H5NH2 due to the presence of the +I effect of two −CH3 groupsin C6H5N(CH3)2. Further, CH3NH2 containsone −CH3 group while (C2H5)2NHcontains two −C2H5 groups. Thus, (C2H5)2 NHis more basic than C2H5NH2.
Now, C6H5N(CH3)2 isless basic than CH3NH2 because of the−R effect of −C6H5 group.
Hence, the increasing order of the basic strengths of thegiven compounds is as follows:
C6H5NH2 6H5N(CH3)2 < CH3NH2 < (C2H5)2NH
(iii) (a)
In p-toluidine, the presence ofelectron-donating −CH3 group increases the electron density on the N-atom.
Thus, p-toluidine is more basicthan aniline.
On the other hand, the presence of electron-withdrawing
−NO2 group decreasesthe electron density over the N−atom in p-nitroaniline. Thus, p-nitroanilineis less basic than aniline.
Hence, the increasing order of the basic strengths of thegiven compounds is as follows:
p-Nitroaniline< Aniline < p-Toluidine
(b) C6H5NHCH3 ismore basic than C6H5NH2 due to the presence of electron-donating −CH3 groupin C6H5NHCH3.
Again, in C6H5NHCH3, −C6H5 groupis directly attached to the N-atom. However, it is not so in C6H5CH2NH2.Thus, in C6H5NHCH3, the −R effect of −C6H5 groupdecreases the electron density over the N-atom. Therefore, C6H5CH2NH2 ismore basic than C6H5NHCH3.
Hence, the increasing order of the basic strengths of thegiven compounds is as follows:
C6H5NH2 6H5NHCH3 < C6H5CH2NH2.
(iv) Inthe gas phase, there is no solvation effect. As a result, the basic strengthmainly depends upon the +I effect. The higher the +I effect, the stronger isthe base. Also, the greater the number of alkyl groups, the higher is the +Ieffect. Therefore, the given compounds can be arranged in the decreasing orderof their basic strengths in the gas phase as follows:
(C2H5)3N> (C2H5)2NH > C2H5NH2 >NH3
(v) Theboiling points of compounds depend on the extent of H-bonding present in thatcompound. The more extensive the H-bonding in the compound, the higher is theboiling point. (CH3)2NH contains only one H−atom whereas C2H5NH2 containstwo H-atoms. Then, C2H5NH2 undergoes more extensive H-bonding than (CH3)2NH.Hence, the boiling point of C2H5NH2 ishigher than that of (CH3)2NH.
Further, O is more electronegative than N.Thus, C2H5OH forms stronger H−bonds than C2H5NH2. Asa result, the boiling point of C2H5OHis higher than that of C2H5NH2 and (CH3)2NH.
Now, the given compounds can be arranged in the increasingorder of their boiling points as follows:
(CH3)2NH< C2H5NH2 < C2H5OH
(vi) Themore extensive the H−bonding, the higher is the solubility. C2H5NH2 containstwo H-atoms whereas (C2H5)2NH contains only one H-atom. Thus, C2H5NH2 undergoesmore extensive H−bonding than (C2H5)2NH.Hence, the solubility in water of C2H5NH2 ismore than that of (C2H5)2NH.
Further, the solubility of amines decreaseswith increase in the molecular mass. This is because the molecular mass ofamines increases with an increase in the size of the hydrophobic part. The molecularmass of C6H5NH2 is greater than that of C2H5NH2 and(C2H5)2NH.
Hence, the increasing order of their solubility in wateris as follows:
C6H5NH2 <(C2H5)2NH < C2H5NH2