A) \[FeBr_{4}^{-}\]
B) \[Br_{{}}^{+}\]
C) \[Br_{{}}^{-}\]
D) \[B{{r}_{2}}\]molecule
Correct Answer: B
Solution :
The metallic halide (Lewi acid) polarises the halogen molecule to yield the electrophite \[{{X}^{+}}\] (halonium ion) \[FeB{{r}_{3}}+B{{r}_{2}}\xrightarrow[{}]{{}}FeB{{r}_{4}}^{-}+B{{r}^{+}}\]You need to login to perform this action.
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