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Q. When 2-bromobutane reacts with alcoholic $ KOH, $ the reaction is

Rajasthan PMTRajasthan PMT 2003Hydrocarbons

Solution:

$CH_3 - \underset{\overset{|}{B}r}{CH} - CH_2 -CH_3 \xrightarrow[(-HBr)]{alc. KOH} $
$\underset{\text{Butene - 1}}{CH_2 = CH - CH_2 -CH_3}$
$2$-bromobutane on reaction with alc. $KOH$ eliminates $HBr$. Therefore, this reaction is known as dehydrohalogenation.