NEET (UG) 2025 — Question 70

The substrate is 2-methylbutane, $\ce{CH3-CH(CH3)-CH2-CH3}$, with four distinct types of hydrogens: the two equivalent methyl groups $\ce{C1/C5}$, the tertiary $\ce{C2}$, the methylene $\ce{C3}$, and the terminal methyl $\ce{C4}$. Chlorination at $\ce{C1}$ (or equivalent branch methyl) gives $\ce{ClCH2-CH(CH3)-CH2-CH3}$, which has a chiral center at $\ce{C2}$, giving 2 enantiomers. Chlorination at $\ce{C2}$ gives $\ce{(CH3)2C(Cl)-CH2-CH3}$, which is achiral, giving 1 product. Chlorination at $\ce{C3}$ gives $\ce{CH3-CH(CH3)-CH(Cl)-CH3}$, with the $\ce{Cl}$-bearing carbon chiral, giving 2 enantiomers. Chlorination at $\ce{C4}$ gives $\ce{CH3-CH(CH3)-CH2-CH2Cl}$, which is achiral, giving 1 product. Total products including stereoisomers = $2 + 1 + 2 + 1 = 6$. Answer: D.