Selectivity In Radical Halogenation

Question

When 2-methylbutane undergoes monochlorination, several constitutional isomers form, and the total number of distinct monochloro structural isomers possible must be counted.

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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 $(CH_3)_2CH-CH_2-CH_3$. 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 $-CH_2-$ 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.

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Selectivity In Radical Halogenation

Select the correct option:

🔒 Solution Hidden from View

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About This Question

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A hydrocarbon with the molecular formula $C_{6} H_{10}$ is being analysed, and the total number of degr...