The product (C) in the below mentioned reaction is:
${CH}_{3} - {CH}_{2} - {CH}_{2} - Br \to_{∆}^{{KOH}_{(alc)}} A \overset{HBr}{\to} B \to_{{KOH}_{(aq)}}^{∆} C$

Question

The product (C) in the below mentioned reaction is: CH 3 - CH 2 - CH 2 - Br → ∆ KOH alc A → HBr B → KOH aq ∆ C

Aicharya · Accepted Answer

When haloalkane or alkyl halide with a β hydrogen is heated with alcoholic solution of $KOH$ , elimination of a hydrogen atom from $β$ carbon and halogen atom from $α -$ carbon occurs as a result, alkene is formed as product. Since $β$ hydrogen atom is involved in the elimination reaction, it is often called $β$ elimination. An alkene is formed. This alkene then on reaction with $HBr$ undergoes addition reaction to give bromopropane, which on reaction with aqueous $KOH$ gives alcohol, propan $- 2 -$ ol.

The product (C) in the below mentioned reaction is: CH 3 - CH 2 - CH 2 - Br → ∆ KOH alc A → HBr B → KOH aq ∆ C

Solution