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(B) The reaction is a dehydrohalogenation of $2$-bromobutane using alcoholic $KOH$.When an alkyl halide can undergo dehydrohalogenation to form more t

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Saytzeff's rule

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(B) The reaction is a dehydrohalogenation of $2$-bromobutane using alcoholic $KOH$.When an alkyl halide can undergo dehydrohalogenation to form more than one alkene,the major product is the most highly substituted alkene (the one with the most alkyl groups attached to the double-bonded carbons).According to Saytzeff's rule,$CH_3CH=CHCH_3$ ($2$-butene) is more substituted than $CH_2=CHCH_2CH_3$ ($1$-butene),making it the major product.Therefore,the correct option is $(b)$.

Which of the following rules applies in the reaction:
$CH_3CHBrCH_2CH_3 \xrightarrow{alc. KOH}$
$(i)$ $CH_3CH = CHCH_3$ (major product)
$(ii)$ $CH_2 = CHCH_2CH_3$ (minor product)

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Choose the $most\ appropriate$ answer from the options given below:

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Which of the following rules applies in the reaction:$CH_3CHBrCH_2CH_3 \xrightarrow{alc. KOH}$$(i)$ $CH_3CH = CHCH_3$ (major product)$(ii)$ $CH_2 = CHCH_2CH_3$ (minor product)