Write the structure of 2-methylbutane

2-methylbutane has the structure: $\mathrm{CH_3-CH(CH_3)-CH_2-CH_3}$

Label the carbons: $\mathrm{C_1H_3-C_2H(C_5H_3)-C_3H_2-C_4H_3}$

Here, $C_5$ is the methyl branch attached to $C_2$.

Identify distinct types of hydrogens

On monochlorination, one H is replaced by Cl. We check all non-equivalent H positions.

In 2-methylbutane, there are 4 different hydrogen environments:

• Type A: Hydrogens on the two equivalent methyl groups attached to $C_2$ ($C_1$ and $C_5$)
• Type B: Hydrogen on $C_2$ (tertiary H)
• Type C: Hydrogens on $C_3$ ($CH_2$)
• Type D: Hydrogens on terminal $C_4$ methyl group

So constitutionally, monochlorination can give 4 different structural isomers.

Now include stereoisomers

We examine whether each product is chiral.

(i) Chlorination at Type A: one of the equivalent methyl groups attached to $C_2$

Product: $\mathrm{CH_2Cl-CH(CH_3)-CH_2-CH_3}$

Now $C_2$ is attached to four different groups:

• $H$
• $CH_3$
• $CH_2Cl$
• $CH_2CH_3$

So $C_2$ becomes a chiral center.

Hence this gives 2 enantiomers.

(ii) Chlorination at Type B: tertiary hydrogen on $C_2$

Product: $\mathrm{CH_3-C(Cl)(CH_3)-CH_2-CH_3}$

At $C_2$, two substituents are identical methyl groups, so it is not chiral.

Hence this gives 1 isomer.

(iii) Chlorination at Type C: on $C_3$ ($CH_2$)

Product: $\mathrm{CH_3-CH(CH_3)-CH(Cl)-CH_3}$

Now $C_3$ is attached to:

• $H$
• $Cl$
• $CH_3$
• $CH(CH_3)CH_3$

These are four different groups, so $C_3$ is chiral.

Hence this gives 2 enantiomers.

(iv) Chlorination at Type D: terminal methyl group $C_4$

Product: $\mathrm{CH_3-CH(CH_3)-CH_2-CH_2Cl}$

Check chirality: no carbon has four different substituents.

Hence this gives 1 isomer.

Total number of isomeric compounds

Adding all structural isomers and stereoisomers: $2+1+2+1=6$

Final answer

$\boxed{6}$

Comparison with stored correct answer

Stored correct answer = $6$

Our derived answer also = $6$, so they agree.