Selectivity In Radical Halogenation

Monochlorination replaces one hydrogen with a chlorine atom, so the number of distinct monochloro products equals the number of structurally different types of hydrogen in the molecule. The structure of 2-methylbutane is $(C{H}_3{)}_{2}CH-C{H}_2-C{H}_3$. Examining each carbon for unique hydrogen environments: the two equivalent methyl groups on $C_2$ form one type, the single tertiary hydrogen on $C_2$ forms a second type, the methylene $-C{H}_2-$ hydrogens form a third type, and the terminal methyl on $C_4$ forms a fourth type. These four distinct hydrogen environments give four different monochloro structural isomers. A count of $3$ is wrong because it overlooks one of the inequivalent positions, likely merging two distinct environments. A count of $5$ is incorrect because it treats the two equivalent branch methyl groups as different, which they are not by symmetry. A count of $2$ greatly underestimates the variety of hydrogen environments and is not consistent with the structure. This is a classic JEE application of identifying equivalent hydrogens, related to the NCERT discussion of substitution. A consistency check: drawing each replacement and removing duplicates by symmetry confirms exactly four unique products.