The optically inactive compound from the following is:

To determine the optically inactive compound, we need to recall that optical inactivity occurs when a compound does not rotate the plane of polarized light. This typically happens if the molecule is achiral, meaning it either lacks a chiral center or possesses a plane of symmetry. A chiral center is a carbon atom bonded to four different groups. Let's analyze each option step by step.

Starting with Option A: 2-chloropropanal. The structure is $$CH_{3}-CHCl-CHO$$. The carbon at position 2 is bonded to four groups: hydrogen (H), chlorine (Cl), methyl group ($$CH_{3}$$), and aldehyde group ($$CHO$$). Since $$CH_{3}$$ and $$CHO$$ are different, and all four groups are distinct (H, Cl, $$CH_{3}$$, $$CHO$$), this carbon is a chiral center. The molecule has no plane of symmetry, so it is chiral and optically active.

Next, Option B: 2-chloro-2-methylbutane. The structure is $$CH_{3}-C(Cl)(CH_{3})-CH_{2}-CH_{3}$$, which can be written as:

$$CH_{3}$$
 | 
$$CH_{3}-C-CH_{2}-CH_{3}$$
 | 
$$Cl$$
The carbon at position 2 is bonded to four groups: chlorine (Cl), two methyl groups (both $$CH_{3}$$), and an ethyl group ($$CH_{2}CH_{3}$$). Here, the two methyl groups are identical. Therefore, the four groups are not all different (Cl, $$CH_{3}$$, $$CH_{3}$$, $$CH_{2}CH_{3}$$). Since two groups are the same, this carbon is not a chiral center. Moreover, the molecule has a plane of symmetry that passes through the chlorine atom, the central carbon, and the ethyl group, bisecting the angle between the two methyl groups. This plane of symmetry makes the molecule achiral, so it is optically inactive.

Moving to Option C: 2-chlorobutane. The structure is $$CH_{3}-CHCl-CH_{2}-CH_{3}$$. The carbon at position 2 is bonded to H, Cl, $$CH_{3}$$, and $$CH_{2}CH_{3}$$. These groups are all different: methyl ($$CH_{3}$$) and ethyl ($$CH_{2}CH_{3}$$) have different sizes and compositions. Thus, carbon-2 is a chiral center, and the molecule lacks a plane of symmetry, making it chiral and optically active.

Finally, Option D: 2-chloropentane. The structure is $$CH_{3}-CHCl-CH_{2}-CH_{2}-CH_{3}$$. The carbon at position 2 is bonded to H, Cl, $$CH_{3}$$, and propyl group ($$CH_{2}CH_{2}CH_{3}$$). All four groups are distinct (H, Cl, $$CH_{3}$$, $$CH_{2}CH_{2}CH_{3}$$), so carbon-2 is a chiral center. The molecule has no plane of symmetry, rendering it chiral and optically active.

Therefore, the only optically inactive compound is Option B, 2-chloro-2-methylbutane, due to its plane of symmetry and absence of a chiral center.

Hence, the correct answer is Option B.