Question

An optically active compound among the following is

• A. 1-Chlorobutane
• B. neo-Pentyl chloride
• C. Isobutyl chloride
• D. tert-Butyl chloride
• E. 2-Chlorobutane

Hint:

A carbon with four different substituents is a chiral center. Look for carbons attached to four different atoms/groups.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
Optical activity is shown by chiral molecules, which typically have a carbon atom bonded to four different substituents (a chiral center or stereocenter).

Step 2: Detailed Explanation:
Examine each compound for a chiral carbon:
• 1-Chlorobutane (CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)Cl): C1 is bonded to Cl, H, and two identical groups (CH\(_2\)CH\(_2\)CH\(_3\) and H? No, C1 has Cl, H, and a CH\(_2\)CH\(_2\)CH\(_3\) group. Wait, C1 is -CH\(_2\)Cl, so it has two H atoms. Not chiral.
• neo-Pentyl chloride ((CH\(_3\))\(_3\)C-CH\(_2\)Cl): The carbon with the Cl is a -CH\(_2\)Cl group, not chiral. The central quaternary carbon has four identical methyl groups? It has three methyls and one CH\(_2\)Cl. All four groups are not different. Wait, (CH\(_3\))\(_3\)C-CH\(_2\)Cl: The central carbon (C) is bonded to three CH\(_3\) groups and one CH\(_2\)Cl group. Three of the groups are identical (CH\(_3\)), so it's not a chiral center. The CH\(_2\)Cl carbon has two H atoms.
• Isobutyl chloride ((CH\(_3\))\(_2\)CH-CH\(_2\)Cl): The CH\(_2\)Cl carbon has two H atoms. The CH carbon is bonded to two methyl groups, an H, and a CH\(_2\)Cl group. Two of the groups are identical (CH\(_3\)), so not chiral.
• tert-Butyl chloride ((CH\(_3\))\(_3\)C-Cl): The central carbon is bonded to three identical methyl groups. Not chiral.
• 2-Chlorobutane (CH\(_3\)-CHCl-CH\(_2\)CH\(_3\)): The carbon bearing the chlorine (C2) is bonded to four different groups: a chlorine atom, a hydrogen atom, a methyl group (CH\(_3\)-), and an ethyl group (-CH\(_2\)CH\(_3\)). This is a chiral center, and the compound is optically active.

Step 3: Final Answer:
2-Chlorobutane is optically active.