CH_3CH(Br)CH_2CH_3 + text{alc. } KOH to ? | vedclass

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(B) The reaction of $2-bromobutane$ with alcoholic $KOH$ is a dehydrohalogenation reaction (elimination reaction).According to $Saytzeff$ rule,the maj

Vedclass · Accepted Answer

$But-2-ene$

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(B) The reaction of $2-bromobutane$ with alcoholic $KOH$ is a dehydrohalogenation reaction (elimination reaction).According to $Saytzeff$ rule,the major product is the more substituted alkene.$CH_3-CH(Br)-CH_2-CH_3 + 	ext{alc. } KOH 	o CH_3-CH=CH-CH_3$ ($But-2-ene$,major) $+ CH_3-CH_2-CH=CH_2$ ($But-1-ene$,minor).Since $But-2-ene$ is the major product,it is the expected answer.

$CH_3CH(Br)CH_2CH_3 + \text{alc. } KOH \to$ ?

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