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Question
How to find the decreasing order of basic strength of a compound?
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Solution
Two of the factors which influence the strength of bases are:
The ease with which the lone pair picks up a hydrogen ion,
The stability of the ions being formed.
For example:
(i) The compounds are given in decreasing order of the pKbvalues:
C2H5NH2,C6H5NHCH3,(C2H5)2NHandC6H5NH2
(ii)The compounds are given in increasing order of basic strength:
C6H5NH2,C6H5N(CH3)2,(C2H5)2NH and CH3NH2
Solutions: (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)2NH than in C2H5NH2. Hence, (C2H5)2NH is more basic than C2H5NH2.
Also, both C6H5NHCH3 and C6H5NH2 are less basic than (C2H5)2NH and C2H5NH2 due to the delocalization of the lone pair in the former two. Further, among C6H5NHCH3 and C6H5NH2, the former will be more basic due to the +T effect of −CH3 group. Hence, the order of increasing basicity of the given compounds is as follows:
C6H5NH2<C6H5NHCH3<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 is more basic than C6H5NH2 due to the presence of the +I effect of two −CH3 groups in C6H5N(CH3)2. Further, CH3NH2 contains one −CH3 group while (C2H5)2NH contains two −C2H5 groups. Thus, (C2H5)2NH is more basic than C2H5NH2.
Now, C6H5N(CH3)2 is less basic than CH3NH2 because of the -R effect of −C6H5 group.
Hence, the increasing order of the basic strengths of the given compounds is as follows:
C6H5NH2<C6H5N(CH3)2<C2H5NH2<(C2H5)2NH
The ease with which the lone pair picks up a hydrogen ion,
The stability of the ions being formed.
For example:
(i) The compounds are given in decreasing order of the pKbvalues:
C2H5NH2,C6H5NHCH3,(C2H5)2NHandC6H5NH2
(ii)The compounds are given in increasing order of basic strength:
C6H5NH2,C6H5N(CH3)2,(C2H5)2NH and CH3NH2
Solutions: (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)2NH than in C2H5NH2. Hence, (C2H5)2NH is more basic than C2H5NH2.
Also, both C6H5NHCH3 and C6H5NH2 are less basic than (C2H5)2NH and C2H5NH2 due to the delocalization of the lone pair in the former two. Further, among C6H5NHCH3 and C6H5NH2, the former will be more basic due to the +T effect of −CH3 group. Hence, the order of increasing basicity of the given compounds is as follows:
C6H5NH2<C6H5NHCH3<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 is more basic than C6H5NH2 due to the presence of the +I effect of two −CH3 groups in C6H5N(CH3)2. Further, CH3NH2 contains one −CH3 group while (C2H5)2NH contains two −C2H5 groups. Thus, (C2H5)2NH is more basic than C2H5NH2.
Now, C6H5N(CH3)2 is less basic than CH3NH2 because of the -R effect of −C6H5 group.
Hence, the increasing order of the basic strengths of the given compounds is as follows:
C6H5NH2<C6H5N(CH3)2<C2H5NH2<(C2H5)2NH
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