An halide C5H11Br on treatments with Alc. KOH given 2–Pentene only the halide will be

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CH3-CH2-CH2-CH2-CH2-Br

CH3–CH2–CH2–CHBr–CH3

CH3–CH2–CHBr−CH2–CH3

CH3-CHCH3-CH(Br)-CH3

answer is C.

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Detailed Solution

1-Bromo pentane (option 1) on dehydro halogenation gives 1-pentene.2-Bromo pentane (option 2) on dehydro halogenation gives 1-pentene and 2-pentenene2-Bromo-3-methyl pentane (option 4) on dehydro halogenation gives 2-methyl-2-butene .Therefore option 3 is correct.since it on treatments with Alc. KOH gives 2–Pentene only.

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