
Arrange the following in decreasing order of their basic character:
$C{{H}_{3}}N{{H}_{2}}$, ${{(C{{H}_{3}})}_{2}}NH$, ${{(C{{H}_{3}})}_{3}}N$, ${{C}_{6}}{{H}_{5}}N{{H}_{2}}$, ${{C}_{6}}{{H}_{5}}C{{H}_{2}}N{{H}_{2}}$
Answer
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Hint: Amines are the functional groups containing nitrogen atoms with a lone pair. Its structure resembles ammonia ($N{{H}_{3}}$ ) but in amines, one or more hydrogen can be replaced by alkyl or aryl groups. As the nature of alkyl group (R) is electron releasing, so it shifts the electron density towards nitrogen and thus, makes it available for an unshared electron pair for protons of acid. Thus, amines are more basic than ammonia. Also, with increase in alkyl groups, the basicity of compound increases.
Complete Step by Step Solution:
We should know that aromatic amines are less basic than ammonia, which is less basic than aliphatic amines.
In aqueous solution, with an increase in the number of alkyl groups in the amines, the basicity of the compound increases. That is why secondary amines are more basic than primary amines. But in the case of methyl amines, the tertiary amines are less basic than primary amines. (${{2}^{o}}>{{1}^{o}}>{{3}^{o}}$ ).
Now, the aniline (${{C}_{6}}{{H}_{5}}N{{H}_{2}}$) contains ${{C}_{6}}{{H}_{5}}$group, which is an electron withdrawing group. As a result, the lone pairs on nitrogen atoms are less available for protons of acid, and its basic character is very small, even less than that of ammonia.
In ${{C}_{6}}{{H}_{5}}C{{H}_{2}}N{{H}_{2}}$ , the ${{C}_{6}}{{H}_{5}}C{{H}_{2}}$ group is electron releasing, which makes it more basic than aniline.
The correct decreasing order of the basic character of given compounds is:
${{(C{{H}_{3}})}_{2}}NH$> $C{{H}_{3}}N{{H}_{2}}$ > ${{(C{{H}_{3}})}_{3}}N$ > ${{C}_{6}}{{H}_{5}}C{{H}_{2}}N{{H}_{2}}$ > ${{C}_{6}}{{H}_{5}}N{{H}_{2}}$
Note: It should be kept in mind that in the case of ethyl amines, tertiary amines are more basic than primary amines. So, the order of basicity will be ${{2}^{o}}$ amines >${{3}^{o}}$ amines >${{1}^{o}}$ amines.
Complete Step by Step Solution:
We should know that aromatic amines are less basic than ammonia, which is less basic than aliphatic amines.
In aqueous solution, with an increase in the number of alkyl groups in the amines, the basicity of the compound increases. That is why secondary amines are more basic than primary amines. But in the case of methyl amines, the tertiary amines are less basic than primary amines. (${{2}^{o}}>{{1}^{o}}>{{3}^{o}}$ ).
Now, the aniline (${{C}_{6}}{{H}_{5}}N{{H}_{2}}$) contains ${{C}_{6}}{{H}_{5}}$group, which is an electron withdrawing group. As a result, the lone pairs on nitrogen atoms are less available for protons of acid, and its basic character is very small, even less than that of ammonia.
In ${{C}_{6}}{{H}_{5}}C{{H}_{2}}N{{H}_{2}}$ , the ${{C}_{6}}{{H}_{5}}C{{H}_{2}}$ group is electron releasing, which makes it more basic than aniline.
The correct decreasing order of the basic character of given compounds is:
${{(C{{H}_{3}})}_{2}}NH$> $C{{H}_{3}}N{{H}_{2}}$ > ${{(C{{H}_{3}})}_{3}}N$ > ${{C}_{6}}{{H}_{5}}C{{H}_{2}}N{{H}_{2}}$ > ${{C}_{6}}{{H}_{5}}N{{H}_{2}}$
Note: It should be kept in mind that in the case of ethyl amines, tertiary amines are more basic than primary amines. So, the order of basicity will be ${{2}^{o}}$ amines >${{3}^{o}}$ amines >${{1}^{o}}$ amines.
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