Question

Which one of the following compounds can exhibit both optical isomerism and geometrical isomerism?

• A. \({2-chloropent-2-ene} \)
• B. \({5-chloropent-2-ene} \)
• C. \({4-chloropent-2-ene} \)
• D. \({3-chloropent-1-ene} \)
• E. \({3-chloropent-2-ene} \)

Hint:

For a compound to exhibit both geometrical and optical isomerism: It must have a double bond with different groups on each carbon for geometrical isomerism. It must have a chiral center for optical isomerism.

Answer 1

The Correct Option is C

Step 1: Understanding geometrical isomerism
Geometrical isomerism (cis-trans or E-Z isomerism) arises due to restricted rotation around a double bond. The presence of two different groups on each carbon of the double bond is required. 
Step 2: Understanding optical isomerism Optical isomerism occurs when a compound has a chiral center (a carbon attached to four different groups). The presence of a chiral center leads to non-superimposable mirror images (enantiomers). 
Step 3: Analyzing each option \({2-chloropent-2-ene}\): Lacks a chiral center. 
\({5-chloropent-2-ene}\): No chiral center. 
\({4-chloropent-2-ene}\): - Double bond at C2-C3 ensures geometrical isomerism. 
- The chiral center at C4 (\(-{Cl}, -{H}, -{CH}_3, -{CH}_2CH_3\)) leads to optical isomerism. 
\({3-chloropent-1-ene}\): No geometrical isomerism due to terminal double bond. 
\({3-chloropent-2-ene}\): No chiral center. 
Step 4: Conclusion Only \({4-chloropent-2-ene}\) satisfies both conditions. Thus, it exhibits both geometrical and optical isomerism.