1. Write IUPAC names of the following compounds and classify them into primary, secondary, and tertiary amines.
(i) (CH3)2 CH NH 2
(ii) CH3 (CH2)2 NH 2
(iii) CH3 NH CH (CH3)2
(iv) (CH3)3 CNH 2
(v) C6H5 NH CH3
(vi) (CH3 CH2)2 N CH3
(vii) m – Br C6 H4 NH2
(i) 1 – Methylethanamine (10 amines)
(ii) Propan – 1 – amine (10 amines)
(iii) N – Methyl – 2 – methyl ethanolamine (20 amines)
(iv) 2 – Methylpropan – 2 – amine (10 amines)
(v) N – Methyl benzamine or N – methylamine (20 amines)
(vi) N – Ethyl – N – methyl ethanolamine (30 amines)
(vii) 3 – Bromobenzenamine or 3 – bromoaniline (10 amines)
Arrange the following:
(i) In decreasing order of basic strength in the gas phase: C2H5NH2, (C2H5)2NH, (C2H5)3N and NH3
(ii) In increasing order of boiling point: C2H5OH, (CH3)2NH, C2H5NH2
(iii) In increasing order of solubility in water: C6H5NH2, (C2H5)2NH, C2H5NH2.
(iv) In decreasing order of the pKb values:
C2H5NH2, C6H5NHCH3, (C2H5)2NH and C6H5NH2
(v) In increasing order of basic strength:
C6H5NH2, C6H5N(CH3)2, (C2H5)2NH and CH3NH2
(i) In the gas phase, basicity follows the order: (C2H5)3N > (C2H5)2NH > C2H5NH2 > NH3. Because solvation does not stabilise basic strength in the gas phase, basic strength follows the predicted order based on the +I action of alkyl groups.
(ii) (CH3)2NH < C2H5NH2 < C2H5OH
The extent of H-bonding in a chemical determines the boiling point of that component. The boiling point of a compound increases as the H-bonding in the compound becomes more extensive.
(iii) C6H5NH2 < (C2H5)2NH < C2H5NH2
Amines are soluble in water because they may form hydrogen bonds with it. However, when the bulk of the hydrocarbon portion grows, the solubility drops.
(iv) Only one C2H5 group is present in C2H5NH2, but two C2H5 groups are present in (C2H5)2NH. As a result, the +I effect in (C2H5)2NH is greater than in C2H5NH2. (C2H5)2NH has a higher electron density over the N-atom than C2H5NH2. Thus, (C2H5)2NH has a higher basicity than C2H5NH2.
Furthermore, due to the delocalization of the lone pair in the former two, both C6H5NHCH3 and C6H5NH2 are less basic than (C2H5)2NH and C2H5NH2. In addition, because of the +T effect of the CH3 group, C6H5NHCH3 will be more basic than C6H5NH2. As a result, the following is the order in which the compounds' basicity increases:
C6H5NH2 < C6H5NHCH3 < C2H5NH2 < (C2H5)2NH
We know that the higher the basic strength, the lower is the pKb value.
Therefore, the correct order is:
C6H5NH2 > C6H5NHCH3 > C2H5NH2 > (C2H5)2NH
(v) The presence of the +I effect of two CH3 groups in C6H5N(CH3)2 makes it more basic than C6H5NH2. Furthermore, CH3NH2 has one CH3 group, whereas (C2H5)2NH has two C2H5 groups. C2H5NH2 is thus more basic than (C2H5)2 NH.
Due to the R effect of the C6H5 group, C6H5N(CH3)2 is now less basic than CH3NH2.
As a result, the basic strengths of the given compounds are listed in ascending order:
C6H5NH2 < C6H5N(CH3)2 < CH3NH2 < (C2H5)2NH