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 Optical Activity:
A compound is optically active if it is chiral. Chirality is the property of a molecule that is non-superimposable on its mirror image. The most common cause of chirality in organic molecules is the presence of a chiral center (or asymmetric carbon atom). A chiral center is a carbon atom that is bonded to four different atoms or groups.
Step 2: Analysis of Structures:
Let's draw the structure of each compound and look for a chiral center.
(A) 1-Chlorobutane: CH\(_{2}\)(Cl)-CH\(_{2}\)-CH\(_{2}\)-CH\(_{3}\). The C1 carbon is bonded to two H atoms, so it's not chiral. No other carbon is bonded to four different groups. It is achiral.
(B) neo-Pentyl chloride: (CH\(_{3}\))\(_{3}\)C-CH\(_{2}\)Cl. The C1 carbon has two H atoms. The C2 carbon has three identical methyl groups. It is achiral.
(C) Isobutyl chloride: (CH\(_{3}\))\(_{2}\)CH-CH\(_{2}\)Cl. The C1 carbon has two H atoms. The C2 carbon has two identical methyl groups. It is achiral.
(D) tert-Butyl chloride: (CH\(_{3}\))\(_{3}\)C-Cl. The central carbon is bonded to three identical methyl groups and a chlorine atom. It is achiral.
(E) 2-Chlorobutane: CH\(_{3}\)-CH(Cl)-CH\(_{2}\)-CH\(_{3}\). Let's examine the C2 carbon. It is bonded to:
1. A hydrogen atom (-H)
2. A chlorine atom (-Cl)
3. A methyl group (-CH\(_{3}\))
4. An ethyl group (-CH\(_{2}\)CH\(_{3}\))
Since the C2 carbon is attached to four different groups, it is a chiral center. The molecule is chiral and therefore optically active.
Step 3: Final Answer:
2-Chlorobutane is the only compound among the options that has a chiral center, making it optically active. This corresponds to option (E).