2-Methylbutane undergoes monochlorination in sunlight. The total number of monochloro products formed, counting stereoisomers separately, is

Four different types of hydrogen exist in $\mathrm{(CH_3)_2CH-CH_2-CH_3}$. Replacing H on the two equivalent methyls of the isopropyl end gives 1-chloro-2-methylbutane, with a chiral centre at C-2: 2 enantiomers. Replacing the tertiary H gives 2-chloro-2-methylbutane: 1 product. Replacing H on the $\mathrm{CH_2}$ gives 2-chloro-3-methylbutane, chiral at C-2: 2 enantiomers. Replacing H on the terminal $\mathrm{CH_3}$ gives 1-chloro-3-methylbutane: 1 product. Total $=2+1+2+1=6$.