hich of the following haloalkanes is optically active?(a) 1-Butanol (b) 1-Propanol (c) 2-Chlorobutane (d) 4-Hydroxybutanal

The optically active compound among the given options is (c) 2-Chlorobutane.

Optical activity is a form of isomerism that arises due to the presence of a chiral center in a molecule. A chiral center is a carbon atom that is attached to four different groups.

Let's analyze each option:

(a) 1-Butanol: This is an alcohol, not a haloalkane. It does not have a chiral center as the carbon atom is attached to two hydrogen atoms.

(b) 1-Propanol: This is also an alcohol, not a haloalkane. It also does not have a chiral center as the carbon atom is attached to two hydrogen atoms.

(c) 2-Chlorobutane: This is a haloalkane. The second carbon atom is attached to four different groups: a methyl group (CH3), a hydrogen atom (H), a chlorine atom (Cl), and an ethyl group (C2H5). Therefore, it is a chiral center and the compound is optically active.

(d) 4-Hydroxybutanal: This is an aldehyde, not a haloalkane. It does not have a chiral center as the carbon atom is attached to two hydrogen atoms.

So, the correct answer is (c) 2-Chlorobutane.