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A: n - propanal
B: 2 - chlorobutane
C: n - butanal
D: 3 - pentanol
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Explanation:
The molecule should have a chiral carbon for optical activity. Chiral carbon has four valencies, each of which is satisfied by a different group.
One chiral carbon exists in 2-chlorobutane. As a result, it's optically active.
Final Answer: Hence, the correct answer is B i.e. 2 - chlorobutane
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