What is the product formed when n-propyl brom | vedclass

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(B) The reaction of $n$-propyl bromide $(CH_3-CH_2-CH_2-Br)$ with ethanolic $KOH$ is a dehydrohalogenation reaction (an elimination reaction). In this

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Propene

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(B) The reaction of $n$-propyl bromide $(CH_3-CH_2-CH_2-Br)$ with ethanolic $KOH$ is a dehydrohalogenation reaction (an elimination reaction). In this reaction,the base $(OH^-)$ abstracts a proton from the $\beta$-carbon,leading to the formation of a double bond between the $\alpha$ and $\beta$ carbons. The reaction is: $CH_3-CH_2-CH_2-Br + KOH \xrightarrow{C_2H_5OH} CH_3-CH=CH_2 + KBr + H_2O$. Thus,the product formed is propene.

What is the product formed when $n$-propyl bromide is reacted with ethanolic potassium hydroxide?

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