Step 1: Understanding the Question:
We need to determine the structural class (primary, secondary, or tertiary) of the given IUPAC-named amines and select the one that is a primary ($1^\circ$) amine.
Step 2: Detailed Explanation:
Amines are classified strictly based on the number of carbon groups attached directly to the central nitrogen atom:
- Primary ($1^\circ$): One carbon group ($\text{R-NH}_2$).
- Secondary ($2^\circ$): Two carbon groups ($\text{R}_2\text{NH}$). In IUPAC naming, this is indicated by a single 'N-' prefix.
- Tertiary ($3^\circ$): Three carbon groups ($\text{R}_3\text{N}$). In IUPAC naming, this is indicated by 'N,N-' prefixes.
Let's deduce the structures from the given names:
(a) N-methylmethanamine: $\text{CH}_3\text{-NH-CH}_3$. The nitrogen is bonded to two methyl groups. This is a secondary ($2^\circ$) amine (Dimethylamine).
(c) N-Phenylbenzenamine: $\text{C}_6\text{H}_5\text{-NH-C}_6\text{H}_5$. The nitrogen is bonded to two phenyl rings. This is a secondary ($2^\circ$) amine (Diphenylamine).
(d) N-Ethyl-N-methylpropan-2-amine: $\text{CH}_3\text{-CH}(\text{CH}_3)\text{-N}(\text{CH}_3)(\text{C}_2\text{H}_5)$. The nitrogen is bonded to three different carbon groups (an ethyl, a methyl, and an isopropyl). This is a tertiary ($3^\circ$) amine.
(b) 4-Bromobenzenamine: $p\text{-Br-C}_6\text{H}_4\text{-NH}_2$. This is a benzene ring substituted with a bromine atom at position 4 and an amine group at position 1. The nitrogen is bonded to only one carbon group (the aromatic ring). This features the $-\text{NH}_2$ group, defining it as a primary ($1^\circ$) amine (p-bromoaniline).
Step 3: Final Answer:
4-Bromobenzenamine is a primary amine, matching option (b).