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(B) Dehydrohalogenation of haloalkanes with alcoholic $KOH$ follows Zaitsev's rule,where the more substituted alkene is the major product. However,the

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$(i), (iv)$

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(B) Dehydrohalogenation of haloalkanes with alcoholic $KOH$ follows Zaitsev's rule,where the more substituted alkene is the major product. However,the target product is $pent-1-ene$ $(CH_3-CH_2-CH_2-CH=CH_2)$.$(i) \ CH_3-CH(Br)-CH_2-CH_2-CH_3$ on dehydrohalogenation gives $pent-1-ene$ and $pent-2-ene$.$(ii) \ CH_3-CH_2-CH(Br)-CH_2-CH_3$ on dehydrohalogenation gives $pent-2-ene$.$(iii) \ CH_3-CH_2-CH_2-CH_2-CH_2Br$ on dehydrohalogenation gives $pent-1-ene$.$(iv) \ CH_3-CH_2-CH_2-CH(Br)-CH_3$ on dehydrohalogenation gives $pent-1-ene$ and $pent-2-ene$.Thus,reactants $(i)$,$(iii)$,and $(iv)$ can produce $pent-1-ene$. Given the options,$(i)$ and $(iv)$ are the most appropriate choices for producing the terminal alkene.

Find the reactants which on heating with alcoholic $KOH$ produce the compound $CH_3-CH_2-CH_2-CH=CH_2$.
$(i) \ CH_3-CH(Br)-CH_2-CH_2-CH_3$
$(ii) \ CH_3-CH_2-CH(Br)-CH_2-CH_3$
$(iii) \ CH_3-CH_2-CH_2-CH_2-CH_2Br$
$(iv) \ CH_3-CH_2-CH_2-CH(Br)-CH_3$

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Find the reactants which on heating with alcoholic $KOH$ produce the compound $CH_3-CH_2-CH_2-CH=CH_2$.$(i) \ CH_3-CH(Br)-CH_2-CH_2-CH_3$$(ii) \ CH_3-CH_2-CH(Br)-CH_2-CH_3$$(iii) \ CH_3-CH_2-CH_2-CH_2-CH_2Br$$(iv) \ CH_3-CH_2-CH_2-CH(Br)-CH_3$