How many chiral compounds are possible on monochlorination of $$2-$$methyl butane?

• Substitution at Position (1): 1-Chloro-2-methylbutane

  $$\text{ClCH}_2-\text{C}^* \text{H}(\text{CH}_3)-\text{CH}_2-\text{CH}_3$$

  Carbon-2 ($$\text{C}^*$$) is bonded to four different groups: $$-\text{CH}_2\text{Cl}$$, $$-\text{H}$$, $$-\text{CH}_3$$, and $$-\text{CH}_2\text{CH}_3$$. Therefore, it is a chiral center, generating a pair of enantiomers:

  $$\rightarrow \mathbf{2 \text{ chiral compounds }} (d \text{ and } l)$$




• Substitution at Position (2): 2-Chloro-2-methylbutane

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

  Carbon-2 is bonded to two identical methyl ($-\text{CH}_3$) groups, making the molecule achiral.

  $$\rightarrow 0 \text{ chiral compounds}$$




• Substitution at Position (3): 2-Chloro-3-methylbutane

  $$\text{CH}_3-\text{CH}(\text{CH}_3)-\text{C}^* \text{H}(\text{Cl})-\text{CH}_3$$

  Carbon-3 ($$\text{C}^*$$) is bonded to four different groups: $$-\text{CH}(\text{CH}_3)_2$$, $$-\text{H}$$, $$-\text{Cl}$$, and $$-\text{CH}_3$$. This creates a chiral center, giving rise to another pair of enantiomers:

  $$\rightarrow \mathbf{2 \text{ chiral compounds }} (d \text{ and } l)$$




• Substitution at Position (4): 1-Chloro-3-methylbutane

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

  None of the carbon atoms in this molecule are bonded to four unique groups, making it entirely achiral.

  $$\rightarrow 0 \text{ chiral compounds}$$

Conclusion

Summing up all the optically active stereo-isomeric forms produced:

$$\text{Total Chiral Compounds} = 2 \text{ (from 1-chloro-2-methylbutane)} + 2 \text{ (from 2-chloro-3-methylbutane)} = 4$$

Answer: Option C — 4