Free-radical monochlorination of 2-methylbutane gives a mixture of monochloro products. The total number of distinct monochloro compounds formed…

2-Methylbutane, $\mathrm{(CH_3)_2CH\!-\!CH_2\!-\!CH_3}$, has four sets of non-equivalent hydrogens, so four constitutional monochloro isomers are formed:
(i) 1-chloro-2-methylbutane (from the $\mathrm{CH_3}$ on $\mathrm{C2}$ side; $\mathrm{C2}$ is a stereocentre: 2 enantiomers);
(ii) 2-chloro-2-methylbutane (tertiary; no stereocentre: 1);
(iii) 2-chloro-3-methylbutane (from the $\mathrm{CH_2}$; $\mathrm{C2}$ and $\mathrm{C3}$ - only $\mathrm{C2}$ is a stereocentre: 2 enantiomers);
(iv) 1-chloro-3-methylbutane (from the terminal $\mathrm{CH_3}$ of the ethyl end; no stereocentre: 1). Total $=2+1+2+1=6$. Radical chlorination is not very selective but all positions react, so all six compounds are formed.