Total number of isomeric compounds (including stereoisomers) formed by monochlorination of 2-methylbutane is ______.

Let us first draw the structure of 2-methylbutane. It can be written as:

$$\text{CH}_3\text{-CH(CH}_3\text{)-CH}_2\text{-CH}_3$$

Next, we identify all distinct types of hydrogen atoms that can be replaced by chlorine.

1. Primary hydrogens at C1 (three equivalent H)
2. Tertiary hydrogen at C2 (one H)
3. Secondary hydrogens at C3 (two equivalent H)
4. Primary hydrogens at C4 (three equivalent H)
5. Primary hydrogens on the methyl substituent at C2 (three equivalent H)

Each distinct type of H gives a different monochlorinated product. Thus we have five structural isomers:

(i) 1-chloro-2-methylbutane
(ii) 2-chloro-2-methylbutane
(iii) 3-chloro-2-methylbutane
(iv) 4-chloro-2-methylbutane
(v) 2-(chloromethyl)butane

Now we check for chirality. In 3-chloro-2-methylbutane the carbon bearing Cl (C3) has four different groups, so it is a chiral centre. Hence 3-chloro-2-methylbutane exists as two enantiomers (R and S).

Therefore, total isomers including stereoisomers =

$$1\;(\text{from C1}) \;+\;1\;(\text{from C2}) \;+\;2\;(\text{from C3, R \& S}) \;+\;1\;(\text{from C4}) \;+\;1\;(\text{from methyl substituent}) \;=\;6.$$

Final Answer: 6