$CH_3-CH_2-CH_2-CH_3 \xrightarrow[HBr]{AlCl_3}$ Product in the above reaction is:
- A$CH_3-CH(Br)-CH_2-CH_3$
- B$CH_3-CH(CH_3)-CH_3$
- C$Br-CH_2-CH_2-CH_2-CH_3$
- DAll of these
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