$A$ halide $C_5H_{11}Br$ on treatment with alc. $KOH$ gives $2$-pentene only. The halide will be
- A$CH_3-CH_2-CH_2-CH_2-CH_2-Br$
- B$CH_3-CH_2-CH_2-CH(Br)-CH_3$
- C$CH_3-CH_2-CH(Br)-CH_2-CH_3$
- D$CH_3-CH(CH_3)-C(Br)(CH_3)-CH_3$
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$(i) \, (CH_3)_2CH-CH_2Br \xrightarrow{C_2H_5OH} (CH_3)_2CH-CH_2OC_2H_5 + HBr$
$(ii) \, (CH_3)_2CH-CH_2Br \xrightarrow{C_2H_5O^-} (CH_3)_2CH-CH_2OC_2H_5 + Br^-$
The reaction mechanisms for $(i)$ and $(ii)$ are respectively:
$(ii) \, (CH_3)_2CH-CH_2Br \xrightarrow{C_2H_5O^-} (CH_3)_2CH-CH_2OC_2H_5 + Br^-$
The reaction mechanisms for $(i)$ and $(ii)$ are respectively:
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