Which of the following compounds possess the highest \(pK_b\) value?

The correct answer is 4.

The \(pK_b\) value of an amine is a measure of its basicity. The higher the \(pK_b\) value, the less basic the amine. This is because a higher \(pK_b\) value means that the amine is less likely to lose a proton \((H^+)\) and form a conjugate base.

Aniline (1), phenylmethanamine (2), and N-methylbenzanamine (3) are all primary amines. This means that they have one nitrogen atom that is bonded to one alkyl group. Primary amines are relatively basic, with pKb values in the range of 3-4.

N,N-dimethylbenzanamine (4) is a secondary amine. This means that it has two nitrogen atoms, each of which is bonded to one alkyl group. Secondary amines are less basic than primary amines, with pKb values in the range of 5-6.

Therefore, N,N-dimethylbenzanamine has the highest \(pK_b\) value of the given compounds.