What will be the increasing order of basic strength of the following compounds: \(\text{C}_2\text{H}_5\text{NH}_2, (\text{C}_2\text{H}_5)_2\text{NH},(\text{C}_2\text{H}_5)_3\text{N},\text{C}_6\text{H}_5\text{NH}_2\)
- \(\text{C}_2\text{H}_5\text{NH}_2 < (\text{C}_2\text{H}_5)_2\text{NH} < (\text{C}_2\text{H}_5)_3\text{N} < \text{C}_6\text{H}_5\text{NH}_2\)
- \(\text{C}_6\text{H}_5\text{NH}_2 < \text{C}_2\text{H}_5\text{NH}_2 < (\text{C}_2\text{H}_5)_3\text{N} < (\text{C}_2\text{H}_5)_2\text{NH}\)
- \((\text{C}_2\text{H}_5)_3\text{N} < (\text{C}_2\text{H}_5)_2\text{NH} < \text{C}_6\text{H}_5\text{NH}_2 < \text{C}_2\text{H}_5\text{NH}_2\)
- \((\text{C}_2\text{H}_5)_2\text{NH} < (\text{C}_2\text{H}_5)_3\text{N} < \text{C}_2\text{H}_5\text{NH}_2 < \text{C}_6\text{H}_5\text{NH}_2\)
The Correct Option is B
Solution and Explanation
To determine the increasing order of basic strength, we need to consider the factors that affect the basicity of amines:
- Aliphatic vs. Aromatic Amines: Aliphatic amines (like C2H5NH2, (C2H5)2NH, (C2H5)3N) are generally more basic than aromatic amines (like C6H5NH2). This is because in aromatic amines, the lone pair of electrons on the nitrogen is delocalized into the benzene ring, making it less available for protonation.
- Alkyl Groups: Alkyl groups are electron-donating, which increases the electron density on the nitrogen atom, making it more basic.
- Steric Hindrance: In tertiary amines, the bulky alkyl groups can hinder the approach of a proton, reducing basicity.
- Solvation Effects: In aqueous solutions, solvation of the protonated amine plays a significant role. Secondary amines are generally more solvated than tertiary amines.
Let's analyze the given compounds:
C2H5NH2 (Ethylamine): A primary aliphatic amine.
(C2H5)2NH (Diethylamine): A secondary aliphatic amine.
(C2H5)3N (Triethylamine): A tertiary aliphatic amine.
C6H5NH2 (Aniline): An aromatic amine.
Based on these factors:
Aniline (C6H5NH2) is the least basic due to resonance delocalization.
Among the aliphatic amines, secondary amines are generally the most basic due to a combination of inductive and solvation effects.
Primary amines are less basic than secondary amines but more basic than tertiary amines due to steric hindrance in tertiary amines.
Therefore, the increasing order of basic strength is:
C6H5NH2 < C2H5NH2 < (C2H5)3N < (C2H5)2NH
The correct answer is:
Option 2: C6H5NH2 < C2H5NH2 < (C2H5)3N < (C2H5)2NH
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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