Amines have a lone pair of electrons on the nitrogen atom, due to which they behave as Lewis bases. Greater the value of \( K_b \) or smaller the value of \( pK_b \), stronger is the base. Amines are more basic than alcohols, ethers, esters, etc. The basic character of aliphatic amines should increase with the increase of alkyl substitution. However, it does not occur in a regular manner, as a secondary aliphatic amine is unexpectedly more basic than a tertiary amine in aqueous solutions. Aromatic amines are weaker bases than ammonia and aliphatic amines. Electron-releasing groups such as \( -CH_3 \), \( -NH_2 \), etc., increase the basicity, while electron-withdrawing substituents such as \( -NO_2 \), \( -CN \), halogens, etc., decrease the basicity of amines. The effect of these substituents is more pronounced at the para-position than at the meta-position.
Arrange the following in increasing order of their basic character. Give reason:
Amines have a lone pair of electrons on the nitrogen atom, due to which they behave as Lewis bases. Greater the value of \( K_b \) or smaller the value of \( pK_b \), stronger is the base. Amines are more basic than alcohols, ethers, esters, etc. The basic character of aliphatic amines should increase with the increase of alkyl substitution. However, it does not occur in a regular manner, as a secondary aliphatic amine is unexpectedly more basic than a tertiary amine in aqueous solutions. Aromatic amines are weaker bases than ammonia and aliphatic amines. Electron-releasing groups such as \( -CH_3 \), \( -NH_2 \), etc., increase the basicity, while electron-withdrawing substituents such as \( -NO_2 \), \( -CN \), halogens, etc., decrease the basicity of amines. The effect of these substituents is more pronounced at the para-position than at the meta-position.
Arrange the following in increasing order of their basic character. Give reason:
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Solution and Explanation
To arrange the given amines in increasing order of their basic character, we need to consider the electronic effects of the substituents on the aromatic ring and their positions relative to the amino group (\(-NH_2\)).
Given Compounds:
1. Aniline (\(C_6H_5NH_2\)) 2. p-Nitroaniline (\(NO_2-C_6H_4-NH_2\)) 3. p-Toluidine (\(CH_3-C_6H_4-NH_2\))
Key Concepts:
- Basicity of Amines: The lone pair on nitrogen determines basicity. Electron-donating groups (\(-CH_3\)) increase electron density on nitrogen, enhancing basicity, while electron-withdrawing groups (\(-NO_2\)) decrease it.
- Positional Effects: The effect of substituents is stronger at the para position than at the meta position due to resonance and inductive effects.
Analysis of Given Amines:
- p-Nitroaniline (\(NO_2-C_6H_4-NH_2\)): The \(-NO_2\) group is strongly electron-withdrawing through both resonance and inductive effects, significantly reducing the electron density on nitrogen. This makes it the weakest base among the three.
- Aniline (\(C_6H_5NH_2\)): The \(-NH_2\) group in aniline is less basic than aliphatic amines due to resonance delocalization of the lone pair into the benzene ring. It has no additional substituents to alter its basicity further.
- p-Toluidine (\(CH_3-C_6H_4-NH_2\)): The \(-CH_3\) group is electron-donating (+I effect), increasing electron density on nitrogen. This makes it the strongest base among the three.
Increasing Order of Basicity:
\[ \text{p-Nitroaniline} < \text{Aniline} < \text{p-Toluidine} \]
Reason:
The \(-NO_2\) group destabilizes the protonated form (ammonium ion) by withdrawing electron density, while the \(-CH_3\) group stabilizes it by donating electron density. Aniline, with no substituents, has intermediate basicity.
Final Answer:
The increasing order of basic character is: \[ \boxed{NO_2-C_6H_4-NH_2 \ < \ C_6H_5NH_2 \ < \ CH_3-C_6H_4-NH_2} \]
Top Questions on Amines
- Arrange the following compounds as asked:
(a) In decreasing order of pKb, values: C$_6$H$_5$NH$_2$, (C$_2$H$_5$)$_2$NH, C$_6$H$_5$NHCH$_2$CH$_3$, C$_6$H$_5$NH$_3$
(b) Increasing order of boiling point: C$_6$H$_5$OH, C$_2$H$_5$NH$_2$, (C$_2$H$_5$)$_2$NH, C$_6$H$_5$NH$_2$
(c) Increasing order of solubility in water: C$_6$H$_5$NH$_2$, (C$_2$H$_5$)$_2$NH, C$_6$H$_5$NHCH$_2$CH$_3$ - Give plausible explanation for each of the following:
Amides are less basic than amines. Amines are usually formed from amides, imides, halides, nitro compounds, etc. They exhibit hydrogen bonding which influences their physical properties. In alkyl amines, a combination of electron releasing, steric and H-bonding factors influence the stability of the substituted ammonium cations in protic polar solvents and thus affect the basic nature of amines. Alkyl amines are found to be stronger bases than ammonia. Amines being basic in nature, react with acids to form salts. Aryldiazonium salts, undergo replacement of the diazonium group with a variety of nucleophiles to produce aryl halides, cyanides, phenols and arenes.
How can you convert the following:
(i) Ethanoic acid to methanamine- Give the structures of A, B, and C in the following reaction sequences: (i)\; \(\mathrm{CH_3CH_2I} \xrightarrow[\ ]{\mathrm{NaCN}} A \xrightarrow[\ ]{\;\;HO^-\;/\;H_2O\;\;} B \xrightarrow[\;Br_2/NaOH\;]{} C\)
(ii)\; \(\mathrm{C_6H_5N_2^+Cl^-} \xrightarrow[\ ]{\mathrm{CuCN}} A \xrightarrow[\ ]{\;H_3O^+\;} B \xrightarrow[\ \Delta\ ]{\;NH_3\;} C\)
(iii)\; \(\mathrm{CH_3CH_2Br} \xrightarrow[\ ]{\mathrm{KCN}} A \xrightarrow[\ ]{\mathrm{LiAlH_4}} B \xrightarrow[\;273\,K\;]{\;HNO_2\;} C\) - How can you obtain the following?
(a) Aniline from ammonium benzoate
(b) Benzene diazonium chloride from nitrobenzene
(c) 2,4,6-Tribromoaniline from aniline
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