1. Understand the reaction

We start with 2-chlorobutane:

$\mathrm{CH_3-CH(Cl)-CH_2-CH_3}$

On chlorination with $\mathrm{Cl_2}$, one more hydrogen is replaced by chlorine, giving dichlorobutanes of formula:

$\mathrm{C_4H_8Cl_2}$

We must count the optically active isomers among the products obtainable from 2-chlorobutane.

2. Possible positions for introducing the second chlorine

In 2-chlorobutane, the available hydrogens are on:

• C1: terminal $\mathrm{CH_3}$
• C2: the carbon already bearing Cl
• C3: $\mathrm{CH_2}$
• C4: terminal $\mathrm{CH_3}$

So the possible constitutional isomers formed are:

1. 1,2-dichlorobutane
2. 2,2-dichlorobutane
3. 2,3-dichlorobutane
4. 1,3-dichlorobutane

(Chlorination at C1 and C4 give different structures because the starting compound is already substituted at C2.)

3. Now examine optical activity of each product

(i) 1,2-dichlorobutane

Structure:

$\mathrm{ClCH_2-CH(Cl)-CH_2-CH_3}$

At C2, the four groups are:

• H
• Cl
• $\mathrm{CH_2Cl}$
• $\mathrm{CH_2CH_3}$

All four are different, so C2 is a chiral center.

Hence 1,2-dichlorobutane exists as 2 enantiomers, both optically active.

Count = 2

(ii) 2,2-dichlorobutane

Structure:

$\mathrm{CH_3-CCl_2-CH_2-CH_3}$

C2 has two identical substituents (two Cl), so it is not chiral. No other stereocenter exists.

Count = 0

(iii) 2,3-dichlorobutane

Structure:

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

There are two stereocenters (C2 and C3). Total stereoisomers possible:

• $(R,R)$
• $(S,S)$
• $(R,S)$ / meso form

Because the molecule is symmetrical, $(R,S)$ is meso and optically inactive.

So optically active isomers are only:

• $(R,R)$
• $(S,S)$

Count = 2

(iv) 1,3-dichlorobutane

Structure:

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

At C3, the four groups are:

• H
• Cl
• $\mathrm{CH_3}$
• $\mathrm{CH_2CH_2Cl}$

All four are different, so C3 is a chiral center.

Hence 1,3-dichlorobutane exists as 2 enantiomers, both optically active.

Count = 2

4. Total optically active isomers

Adding all optically active isomers:

$2 + 0 + 2 + 2 = 6$

5. Final answer

The total number of optically active isomers is:

$\boxed{6}$

6. Comparison with stored answer

Stored correct answer = 6

Our derived answer = 6

So they agree.