How would you arrange the following amines in order of decreasing base strength?
$\begin{align}
NH_{3},CH_{3}NH_{2},(CH_{3})_{2}NH,NH_{2}Br
\end{align}$
Answer
601.2k+ views
Hint: Basicity of amines can be determined with the help of factors like steric hindrance, hydrogen bonding and electronegativity. Now considering this, we can easily solve the given question.
Complete step by step answer:
- It is known to us that amines are basic in nature as they have a lone pair of electrons on nitrogen. Therefore, they have a strong tendency to donate the lone pair of electrons to electron acceptors.
- Now, the basicity of amines depends on the electronic properties of each substituent and the degree of salvation of protonated amine, which includes steric hindrance by the group on nitrogen.
- To solve the given question, we need to remember that as electron density around nitrogen increases, the availability to donate electrons also increases. This results in the compound becoming more Lewis basic.
- $C{H_{3}}N{H_{2}}$ consists of an electron donating group, which is a methyl group. So, it donates electrons activating the Lewis base behaviour of nitrogen. Therefore, it is more basic than ammonia.
- Since $(C{H_{3}})_{2}NH$ consists of a second methyl group, it experiences an even greater electron donating effect and as a result is more basic than methylamine.
- In $N{H_{2}}Br$, nitrogen is more electronegative than bromine. But bromine is more inductively electron-withdrawing than nitrogen group. The N-H bonding electron pairs get polarized towards Br. This results in bromine pulling electron density away from nitrogen. This leaves less electron density to nitrogen. Therefore, it is less basic than ammonia.
- The order of decreasing base strength will be
$\begin{align}
(CH_{3})_{2}NH>CH_{3}NH_{2}>NH_{3}>NH_{2}Br
\end{align}$
Note: It should be noted that aliphatic amines are more basic compared to aromatic amines. This happens because the lone pair present on the nitrogen atom is delocalized. So, the availability of the lone pair of electrons will be less and hence will be less basic than aliphatic amines.
Complete step by step answer:
- It is known to us that amines are basic in nature as they have a lone pair of electrons on nitrogen. Therefore, they have a strong tendency to donate the lone pair of electrons to electron acceptors.
- Now, the basicity of amines depends on the electronic properties of each substituent and the degree of salvation of protonated amine, which includes steric hindrance by the group on nitrogen.
- To solve the given question, we need to remember that as electron density around nitrogen increases, the availability to donate electrons also increases. This results in the compound becoming more Lewis basic.
- $C{H_{3}}N{H_{2}}$ consists of an electron donating group, which is a methyl group. So, it donates electrons activating the Lewis base behaviour of nitrogen. Therefore, it is more basic than ammonia.
- Since $(C{H_{3}})_{2}NH$ consists of a second methyl group, it experiences an even greater electron donating effect and as a result is more basic than methylamine.
- In $N{H_{2}}Br$, nitrogen is more electronegative than bromine. But bromine is more inductively electron-withdrawing than nitrogen group. The N-H bonding electron pairs get polarized towards Br. This results in bromine pulling electron density away from nitrogen. This leaves less electron density to nitrogen. Therefore, it is less basic than ammonia.
- The order of decreasing base strength will be
$\begin{align}
(CH_{3})_{2}NH>CH_{3}NH_{2}>NH_{3}>NH_{2}Br
\end{align}$
Note: It should be noted that aliphatic amines are more basic compared to aromatic amines. This happens because the lone pair present on the nitrogen atom is delocalized. So, the availability of the lone pair of electrons will be less and hence will be less basic than aliphatic amines.
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