Question:medium
Which from following compounds has lowest \(\text{pK}_\text{b}\) value?
Which from following compounds has lowest \(\text{pK}_\text{b}\) value?
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In aqueous solution: \(2^\circ>1^\circ>NH_3>\text{aromatic amine}\) due to +I effect and solvation.
Updated On: May 14, 2026
- N -Ethylethanamine
- Propan-2-amine
- \(\text{NH}_3\)
- Benzenamine
Show Solution
The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
The base dissociation constant, \(\text{K}_\text{b}\), measures the strength of a base. A stronger base has a higher \(\text{K}_\text{b}\) value. The \(\text{pK}_\text{b}\) is the negative logarithm of \(\text{K}_\text{b}\) (\(\text{pK}_\text{b} = -\log_{10} \text{K}_\text{b}\)). Therefore, a stronger base will have a lower \(\text{pK}_\text{b}\) value. We need to identify the strongest base among the options.
Step 2: Key Formula or Approach:
Evaluate the basicity of each compound based on inductive (+I) effects, steric hindrance, and resonance effects. The strongest base will correspond to the lowest \(\text{pK}_\text{b}\).
Step 3: Detailed Explanation:
Let's evaluate the basicity of the given compounds:
- (A) N-Ethylethanamine: This is a secondary aliphatic amine (\(\text{CH}_3\text{CH}_2\text{-NH-CH}_2\text{CH}_3\), also known as diethylamine). The two ethyl groups provide a strong +I (inductive) effect, increasing electron density on nitrogen and making it very basic.
- (B) Propan-2-amine: This is a primary aliphatic amine (isopropylamine). It has one alkyl group providing a +I effect, making it more basic than ammonia but generally less basic than a secondary amine in aqueous solution due to a combination of inductive effects and solvation.
- (C) \(\text{NH}_3\) (Ammonia): It has no alkyl groups, so no +I effect. It is less basic than aliphatic amines.
- (D) Benzenamine (Aniline): This is an aromatic amine. The lone pair on the nitrogen atom is delocalized into the benzene ring due to resonance. This makes the lone pair less available for protonation, making aniline the weakest base among the choices.
In general, for aliphatic amines in aqueous solution, secondary amines are stronger bases than primary amines. Thus, N-ethylethanamine is the strongest base here and will have the lowest \(\text{pK}_\text{b}\) value.
Step 4: Final Answer:
N-Ethylethanamine is the correct answer.
The base dissociation constant, \(\text{K}_\text{b}\), measures the strength of a base. A stronger base has a higher \(\text{K}_\text{b}\) value. The \(\text{pK}_\text{b}\) is the negative logarithm of \(\text{K}_\text{b}\) (\(\text{pK}_\text{b} = -\log_{10} \text{K}_\text{b}\)). Therefore, a stronger base will have a lower \(\text{pK}_\text{b}\) value. We need to identify the strongest base among the options.
Step 2: Key Formula or Approach:
Evaluate the basicity of each compound based on inductive (+I) effects, steric hindrance, and resonance effects. The strongest base will correspond to the lowest \(\text{pK}_\text{b}\).
Step 3: Detailed Explanation:
Let's evaluate the basicity of the given compounds:
- (A) N-Ethylethanamine: This is a secondary aliphatic amine (\(\text{CH}_3\text{CH}_2\text{-NH-CH}_2\text{CH}_3\), also known as diethylamine). The two ethyl groups provide a strong +I (inductive) effect, increasing electron density on nitrogen and making it very basic.
- (B) Propan-2-amine: This is a primary aliphatic amine (isopropylamine). It has one alkyl group providing a +I effect, making it more basic than ammonia but generally less basic than a secondary amine in aqueous solution due to a combination of inductive effects and solvation.
- (C) \(\text{NH}_3\) (Ammonia): It has no alkyl groups, so no +I effect. It is less basic than aliphatic amines.
- (D) Benzenamine (Aniline): This is an aromatic amine. The lone pair on the nitrogen atom is delocalized into the benzene ring due to resonance. This makes the lone pair less available for protonation, making aniline the weakest base among the choices.
In general, for aliphatic amines in aqueous solution, secondary amines are stronger bases than primary amines. Thus, N-ethylethanamine is the strongest base here and will have the lowest \(\text{pK}_\text{b}\) value.
Step 4: Final Answer:
N-Ethylethanamine is the correct answer.
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