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: Nature of each reagent.
\( PBr_3 \) is a brominating agent for the hydroxyl group, while alcoholic potassium hydroxide favours removal of \( HBr \) to make an alkene. Recognising these two roles fixes the route: alcohol to haloalkane to alkene.

Step 2: Conversion of alcohol to haloalkane.
The \( -OH \) of propan-1-ol is swapped for bromine. Product [A] is the primary halide 1-bromopropane, \( CH_3-CH_2-CH_2-Br \).

Step 3: Elimination to the alkene.
With a base in ethanol and heating, a beta hydrogen and the halogen leave together. For 1-bromopropane only one alkene is possible, so no Saytzeff choice arises.
\( CH_3CH_2CH_2Br + KOH(alc.) \rightarrow CH_3CH=CH_2 + KBr + H_2O \)

Step 4: State the products.
\[\boxed{[A] = 1\text{-bromopropane};\ [B] = \text{propene}}\]