Question

Complete the following chemical reaction:
\( CH_3CH_2CH_2OH \xrightarrow{PBr_3} [A] \xrightarrow[\Delta]{Alc.\ KOH} [B] \)

Hint:

\( PBr_3 \) replaces \( -OH \) with \( -Br \); alcoholic KOH then eliminates \( HBr \) to give an alkene.

Answer 1

Step 1: Identify the starting material and first reagent.
The starting compound is propan-1-ol, \( CH_3CH_2CH_2OH \), a primary alcohol. \( PBr_3 \) (phosphorus tribromide) converts alcohols into alkyl bromides by replacing the \( -OH \) group with a \( -Br \) group.

Step 2: Form product [A].
\( 3\,CH_3CH_2CH_2OH + PBr_3 \rightarrow 3\,CH_3CH_2CH_2Br + H_3PO_3 \)
So [A] is 1-bromopropane, \( CH_3CH_2CH_2Br \).

Step 3: React [A] with alcoholic KOH and heat.
Alcoholic KOH is a strong base and promotes elimination (dehydrohalogenation). It removes one \( H \) from the carbon next to the \( C-Br \) bond and the \( Br \) atom, forming a double bond (Saytzeff/beta elimination).

Step 4: Form product [B].
\( CH_3CH_2CH_2Br \xrightarrow[\Delta]{Alc.\ KOH} CH_3CH=CH_2 + KBr + H_2O \)
So [B] is propene, \( CH_3CH=CH_2 \).

Final answer: \[\boxed{[A] = CH_3CH_2CH_2Br,\quad [B] = CH_3CH=CH_2}\]