The correct order of decreasing basicities of \( C_{2}H_{5}NH_{2} \), \( (C_{2}H_{5})_{2}NH \) and \( (C_{2}H_{5})_{3}N \) in aqueous solution is :
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Methyl amines: \( 2^\circ \gt 1^\circ \gt 3^\circ \) (213 rule)
Ethyl amines: \( 2^\circ \gt 3^\circ \gt 1^\circ \) (231 rule)
Secondary is always the strongest!
- \( (C_{2}H_{5})_{3}N \gt (C_{2}H_{5})_{2}NH \gt C_{2}H_{5}NH_{2} \)
- \( (C_{2}H_{5})_{2}NH \gt C_{2}H_{5}NH_{2} \gt (C_{2}H_{5})_{3}N \)
- \( C_{2}H_{5}NH_{2} \gt (C_{2}H_{5})_{2}NH \gt (C_{2}H_{5})_{3}N \)
- \( (C_{2}H_{5})_{2}NH \gt (C_{2}H_{5})_{3}N \gt C_{2}H_{5}NH_{2} \)
The Correct Option is D
Solution and Explanation
• The basicity of amines in aqueous solution is determined by the stability of the substituted ammonium cation formed after accepting a proton.
• Three major factors influence this stability: The inductive effect (+I effect of alkyl groups), the solvation effect (hydration through hydrogen bonding), and steric hindrance.
• In the gas phase, basicity follows the simple order: \( 3^\circ \gt 2^\circ \gt 1^\circ \) due to the +I effect.
• In aqueous solution, the interplay between these three factors leads to an anomalous order depending on the size of the alkyl group.
Step 1: Evaluate the effect of alkyl groups (Inductive Effect)
Alkyl groups like ethyl (\( -C_{2}H_{5} \)) are electron-releasing groups.
They increase electron density on the Nitrogen atom, making the lone pair more available for donation.
Based solely on this, the tertiary amine \( (C_{2}H_{5})_{3}N \) should be the strongest base.
Step 2: Analyze the Solvation Effect
When an amine accepts a proton, it forms a conjugate acid (cation).
This cation is stabilized by water molecules via hydrogen bonding.
The more Hydrogens attached to the Nitrogen in the cation, the better it is solvated.
Order of stability via solvation: \( 1^\circ \gt 2^\circ \gt 3^\circ \).
Step 3: Consider Steric Hindrance
Bulky ethyl groups hinder the approach of water molecules for solvation and the approach of a proton (\( H^+ \)).
This is particularly high in the tertiary amine \( (C_{2}H_{5})_{3}N \), which reduces its stability significantly compared to the secondary amine.
Step 4: Determine the combined final order
For Ethyl-substituted amines, the secondary amine is the most stable because it balances +I effect and solvation perfectly.
The tertiary amine follows, as the +I effect of three ethyl groups compensates for the steric hindrance more effectively than in methyl amines.
Final Order: Secondary (\( 2^\circ \)) \( \gt \) Tertiary (\( 3^\circ \)) \( \gt \) Primary (\( 1^\circ \)).
\[ (C_{2}H_{5})_{2}NH \gt (C_{2}H_{5})_{3}N \gt C_{2}H_{5}NH_{2} \]
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(i) (CH3 )2CHNH2 (ii) CH3 (CH2 )2NH2 (iii) CH3NHCH(CH3 )2
(iv) (CH3 )3CNH2 (v) C6H5NHCH3 (vi) (CH3CH2 )2NCH3 (vii) m–BrC6H4NH2
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Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
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(v) Aniline does not undergo Friedel-Crafts reaction.
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Arrange the following:
(i) In decreasing order of the pKb values: C2H5NH2 , C6H5NHCH3 , (C2H5 )2NH and C6H5NH2
(ii) In increasing order of basic strength: C6H5NH2 , C6H5N(CH3 )2 , (C2H5 )2NH and CH3NH2
(iii) In increasing order of basic strength: (a) Aniline, p-nitroaniline and p-toluidine Exercises Rationalised 2023-24 279 Amines (b) C6H5NH2 , C6H5NHCH3 , C6H5CH2NH2 .
(iv) In decreasing order of basic strength in gas phase: C2H5NH2 , (C2H5 )2NH, (C2H5 )3N and NH3
(v) In increasing order of boiling point: C2H5OH, (CH3 )2NH, C2H5NH2
(vi) In increasing order of solubility in water: C6H5NH2 , (C2H5 )2NH, C2H5NH2 .
How will you convert:
(i) Ethanoic acid into methanamine
(ii) Hexanenitrile into 1-aminopentane
(iii) Methanol to ethanoic acid
(iv) Ethanamine into methanamine
(v) Ethanoic acid into propanoic acid
(vi) Methanamine into ethanamine
(vii) Nitromethane into dimethylamine
(viii) Propanoic acid into ethanoic acid?
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