Question

Which among following amines has lowest \( pK_b \) values?

• A. \( \text{CH}_3\text{CH}_2\text{NH}_2 \)
• B. \( (\text{CH}_3\text{CH}_2)_2\text{NH} \)
• C. \( (\text{CH}_3\text{CH}_2)_3\text{N} \)
• D. \( \text{C}_6\text{H}_5\text{NH}_2 \)

Hint:

In aliphatic amines, the basicity order in water: secondary > primary > tertiary. For aniline, the lone pair is involved in resonance with the ring, making it much less basic.

Answer 1

The Correct Option is B

Step 1: Understand the question.
We must find the amine with the lowest $pK_b$. A lower $pK_b$ means a stronger base. So we are really hunting for the strongest base.
Step 2: Recall the order for aliphatic amines in water.
In water the strength usually goes: secondary > primary > tertiary. The tertiary one is weaker because its crowded shape makes it hard for water to support the charged form.
Step 3: Place aniline.
Aniline ($C_6H_5NH_2$) is an aromatic amine. Its lone pair spreads into the ring, so it is a very weak base with a high $pK_b$.
Step 4: List the four amines.
Ethylamine is primary, diethylamine is secondary, triethylamine is tertiary, and aniline is aromatic.
Step 5: Pick the strongest.
By the order in Step 2, the secondary amine diethylamine is the strongest base here, so it has the lowest $pK_b$.
Step 6: Choose the answer.
The lowest $pK_b$ belongs to diethylamine, which is option 2. \[ \boxed{(CH_3CH_2)_2NH} \]