CH_3 - CH_2 - CH_2Br + KOH (alc.) to ProductT | vedclass

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(A) The reaction of $1$-bromopropane $(CH_3 - CH_2 - CH_2Br)$ with alcoholic $KOH$ is a dehydrohalogenation reaction.This is an elimination reaction (

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$CH_3 - CH = CH_2$

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(A) The reaction of $1$-bromopropane $(CH_3 - CH_2 - CH_2Br)$ with alcoholic $KOH$ is a dehydrohalogenation reaction.This is an elimination reaction ($E2$ mechanism) where a hydrogen atom is removed from the $\beta$-carbon and the bromine atom is removed from the $\alpha$-carbon.This results in the formation of a double bond between the $\alpha$ and $\beta$ carbons.The reaction is: $CH_3 - CH_2 - CH_2Br + KOH 	ext{ (alc.)} 	o CH_3 - CH = CH_2 + KBr + H_2O$.The product formed is propene $(CH_3 - CH = CH_2)$.

$CH_3 - CH_2 - CH_2Br + KOH$ (alc.) $\to$ Product
The product in the above reaction is:

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$CH_3 - CH_2 - CH_2Br + KOH$ (alc.) $\to$ ProductThe product in the above reaction is: