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-propyl bromide on treating with alcoholic $\mathrm{KOH}$ produces
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The correct answer is:
propene
Alcoholic $\mathrm{KOH}$ is a dehydrohalogenating reagent, so when -propyl bromide is treated with alcoholic $\mathrm{KOH}$, propene is obtained.
$$
\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}+\text { alc } \mathrm{KOH} \longrightarrow
$$
n-propyl bromide
$\mathrm{CH}_{3} \mathrm{CH}=\underset{\text { Propene }}{=} \mathrm{CH}_{2}+\mathrm{HBr}$
$$
\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}+\text { alc } \mathrm{KOH} \longrightarrow
$$
n-propyl bromide
$\mathrm{CH}_{3} \mathrm{CH}=\underset{\text { Propene }}{=} \mathrm{CH}_{2}+\mathrm{HBr}$
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