Question

Arrange the following in decreasing order of their \(pK_b\) values

• A. \(d\gt a\gt c\gt b\)
• B. \(a\gt b\gt d\gt c\)
• C. \(d\gt c\gt b\gt a\)
• D. \(a\gt c\gt d\gt b\)

Hint:

Remember: \[ \text{Strong base} \Rightarrow \text{Small } pK_b \] Steric hindrance and electron withdrawing groups generally decrease the basic strength of amines.

Answer 1

The Correct Option is C

Step 1: Link basic strength to $pK_b$.
A stronger base has a smaller $pK_b$, while a weaker base has a larger $pK_b$. So arranging in decreasing $pK_b$ means arranging from weakest base to strongest base.
Step 2: Look at the benzyl-type amines (c) and (d).
The phenyl ring withdraws electron density and lowers the basicity of these amines. Between them, (d) carries an extra methyl on nitrogen that hinders solvation of its conjugate acid, so (d) ends up less basic than (c).
Step 3: Order the benzyl amines by $pK_b$.
Since (d) is the weaker base, it has the higher $pK_b$, so $d > c$ in $pK_b$.
Step 4: Compare the aliphatic amines (a) and (b).
In water, methylamine $CH_3NH_2$ is more basic than trimethylamine $(CH_3)_3N$, because the crowded trimethylamine is poorly solvated when protonated.
Step 5: Order the aliphatic amines by $pK_b$.
Here (b) is the weaker base of this pair, so it has the higher $pK_b$, giving $b > a$ in $pK_b$.
Step 6: Combine into one chain.
Joining both comparisons, the decreasing order of $pK_b$ is $d > c > b > a$, which is option 3.
\[ \boxed{d > c > b > a} \]