2-chlorobutane + $$Cl_2 \rightarrow C_4H_8Cl_2$$ (isomers)
Total number of optically active isomers shown by $$C_4H_8Cl_2$$, obtained in the above reaction is ________.

2-chlorobutane: CH$$_3$$CHClCH$$_2$$CH$$_3$$. Monochlorination gives C$$_4$$H$$_8$$Cl$$_2$$ isomers.

Possible positions for the second Cl:

1. C1: CH$$_2$$ClCHClCH$$_2$$CH$$_3$$ — C2 is chiral, optically active (2 enantiomers)

2. C2: CH$$_3$$CCl$$_2$$CH$$_2$$CH$$_3$$ — no chiral center, not optically active

3. C3: CH$$_3$$CHClCHClCH$$_3$$ — two chiral centers, can have (R,R), (S,S), and meso. The (R,R) and (S,S) are optically active = 2 enantiomers. Also (R,S) which is meso.

4. C4: CH$$_3$$CHClCH$$_2$$CH$$_2$$Cl — C2 is chiral, 2 enantiomers

Total optically active isomers: 2 + 0 + 2 + 2 = 6.

Therefore, the answer is $$\boxed{6}$$.