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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