a) \(\ce{C=C(C)C=C} + HBr \xrightarrow{0^\circ C} \) b) \(\ce{C=C(C)C=C} + HCl \xrightarrow{100^\circ C} \)
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If we have an alkene as the starting material, the HBr will add across the double bond to form a bromoalkane. Show more…
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