A) $ a=Br_2/CCl_4, $ $ b=aq.KOH $
B) $ a=Br_2/{H^{+}}, $ $ b=aq.KOH $
C) $ a=Br_2/{H^{+}}, $ $ b=alc.KOH $
D) $ a=Br_2/H{O^{-}}, $ $ b=aq.KOH $
Correct Answer: C
[c]
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