Which of the following is an optically active compound?

Question

What is the preliminary test for nanoparticles?

Answer 1

To determine which of the given compounds is optically active, we need to understand the concept of optical activity. A compound is optically active if it can rotate the plane of polarized light. This typically occurs when a compound has a chiral center, which is a carbon atom bonded to four different groups.

Let's analyze each of the given options:

1-Butanol: The structural formula for 1-butanol is CH₃CH₂CH₂CH₂OH. Here, the hydroxyl group (-OH) is attached to the terminal carbon (the first carbon), which is connected to three hydrogen atoms and one butyl group (CH₃CH₂CH₂-). All these groups are not different, so 1-butanol does not have a chiral center and is not optically active.
1-Propanol: The structural formula for 1-propanol is CH₃CH₂CH₂OH. Similar to 1-butanol, the terminal carbon has two hydrogen atoms, one hydroxyl group, and a propyl group (CH₃CH₂-). Again, the groups are not all different, so 1-propanol is not optically active.
2-Chlorobutane: The structural formula for 2-chlorobutane is CH₃CHClCH₂CH₃. In this compound, the second carbon (CHCl) is bonded to four different groups: a chlorine atom (Cl), a hydrogen atom (H), a methyl group (CH₃), and an ethyl group (CH₂CH₃). This makes the second carbon a chiral center, hence 2-chlorobutane is optically active.
4-Hydroxyheptane: The structural formula is longer, but similar to the other alcohols listed, we would expect the hydroxyl group to be bonded to the fourth carbon in a continuous chain, which usually does not grant chirality unless there are different groups around the central carbon. As provided in the simple hydrocarbon chain example (heptane), this would usually not tend toward having a chiral center without additional functional groups or branches.

Hence, among the given options, 2-chlorobutane is the only compound with a chiral center, making it optically active.

Answer 2

To determine which of the given compounds is optically active, we need to understand the concept of optical activity. A compound is optically active if it can rotate the plane of polarized light. This typically occurs when a compound has a chiral center, which is a carbon atom bonded to four different groups.

Let's analyze each of the given options:

1-Butanol: The structural formula for 1-butanol is CH₃CH₂CH₂CH₂OH. Here, the hydroxyl group (-OH) is attached to the terminal carbon (the first carbon), which is connected to three hydrogen atoms and one butyl group (CH₃CH₂CH₂-). All these groups are not different, so 1-butanol does not have a chiral center and is not optically active.
1-Propanol: The structural formula for 1-propanol is CH₃CH₂CH₂OH. Similar to 1-butanol, the terminal carbon has two hydrogen atoms, one hydroxyl group, and a propyl group (CH₃CH₂-). Again, the groups are not all different, so 1-propanol is not optically active.
2-Chlorobutane: The structural formula for 2-chlorobutane is CH₃CHClCH₂CH₃. In this compound, the second carbon (CHCl) is bonded to four different groups: a chlorine atom (Cl), a hydrogen atom (H), a methyl group (CH₃), and an ethyl group (CH₂CH₃). This makes the second carbon a chiral center, hence 2-chlorobutane is optically active.
4-Hydroxyheptane: The structural formula is longer, but similar to the other alcohols listed, we would expect the hydroxyl group to be bonded to the fourth carbon in a continuous chain, which usually does not grant chirality unless there are different groups around the central carbon. As provided in the simple hydrocarbon chain example (heptane), this would usually not tend toward having a chiral center without additional functional groups or branches.

Hence, among the given options, 2-chlorobutane is the only compound with a chiral center, making it optically active.

Which of the following is an optically active compound?

The Correct Option is C

Solution and Explanation

Top Questions on Haloalkanes and Haloarenes

Questions Asked in NEET (UG) exam