KCET · Chemistry · s Block Elements
-propyl bromide on treating with alcoholic \(\mathrm{KOH}\) produces
- A propane
- B propene
- C propyne
- D propanol
Answer & Solution
Correct Answer
(B) propene
Step-by-step Solution
Detailed explanation
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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