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: Recall the relation between basic strength and \(pK_b\).
\[ \text{Higher basic strength} \Rightarrow \text{Lower } pK_b \] and
\[ \text{Lower basic strength} \Rightarrow \text{Higher } pK_b \] Thus, decreasing order of \(pK_b\) means decreasing order of weakness of bases.

Step 2: Analyze aromatic amines.
In compounds containing benzyl group, the phenyl ring exerts electron withdrawing effect.
This decreases electron density on nitrogen and reduces basicity.
Between \((c)\) and \((d)\), compound \((d)\) contains an additional methyl group attached to nitrogen, which introduces steric hindrance and reduces solvation of conjugate acid.
Therefore, \((d)\) becomes less basic than \((c)\).
Hence,
\[ d\gt c \] in \(pK_b\).

Step 3: Compare aliphatic amines.
For aliphatic amines in aqueous medium, methylamine is more basic than trimethylamine because steric hindrance in trimethylamine decreases effective solvation.
Thus,
\[ (CH_3)_3N \] is less basic than
\[ CH_3NH_2 \] Therefore,
\[ b\gt a \] in \(pK_b\).

Step 4: Combine the order.
Since weakest base has highest \(pK_b\), the decreasing order of \(pK_b\) becomes
\[ d\gt c\gt b\gt a \]

Step 5: Final conclusion.
Hence, the correct decreasing order of \(pK_b\) values is
\[ \boxed{d\gt c\gt b\gt a} \]