Concept:Dehydrohalogenation (an elimination reaction) of an alkyl halide removes a hydrogen atom and a halogen atom from adjacent carbons to form an alkene.
Formula:$$ \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + \text{KOH}_{(\text{alc})} \rightarrow \text{CH}_3\text{CH}=\text{CH}_2 + \text{KBr} + \text{H}_2\text{O} $$
Solution:- Bromopropane is a 3-carbon alkyl halide.
- When subjected to an strong base (like alcoholic KOH), the bromine atom and a beta-hydrogen are eliminated.
- The 3-carbon skeleton remains intact, resulting in the formation of a double bond. The product is the 3-carbon alkene, propene.
Why other options are incorrect:- Butene and ethene have incorrect carbon chain lengths (4 and 2 carbons, respectively).
- Propane is a saturated alkane, which would be the product of reduction, not elimination.
Quality & Fidelity Assurance:
Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.