Question

\( CH _{3} CH _{2} CH _{2} OH \xrightarrow[623 K ]{ Al _{2} O _{3}}(A) \xrightarrow{ HBr }(B) \xrightarrow{ KOH (\text { aq. })}(C) \)

Answer 1

Al2O3 is dehydrating agent

A. is propene

B is 2-Bromo propane by markovnikov addition

C  is propan-2-ol

Answer 2

Since Al2O3 is a dehydrating agent an aqueous koh is further used in the reaction rather than alc. koh so the products will be

(A)  CH3CH2=CH2  → propene

(B) CH3CH2(Br)—CH3   (intermediate formed ⇒CH3CH2^ð+—CH2^ð-)  →2-bromo propane

(C) CH3CH2(OH)—CH3 → propane-2-ol

if alc. koh would have been used then product C would be propene