Step 1: Identify the longest carbon chain. The compound $CH_3CH(CH_3)CH(Br)CH_3$ has a maximum of 4 carbons in the continuous chain. The parent chain is butane. Step 2: Number the chain to give lowest locants. Numbering from the end nearest to bromine gives Br at C2 and methyl at C3 (locant sum = 5). Numbering from the other end gives methyl at C2 and Br at C3 (locant sum = 5). Applying alphabetical priority, 'bromo' (b) is cited before 'methyl' (m), so bromo should get the lower number. Numbering from the bromo end: Br at C2, methyl at C3. Step 3: Name the substituents. At C2: bromo. At C3: methyl. Parent chain: butane. Step 4: Assemble the name. IUPAC name: 2-bromo-3-methylbutane.
