Q. The reaction conditions used for converting 1,2-dibromopropane to propylene are
A) $KOH,\ \frac{alcohol}{\mathrm{\Delta}}$
B) $KOH,\ \frac{water}{\mathrm{\Delta}}$
C) $Zn,\ \frac{alcohol}{\mathrm{\Delta}}$
D) $Na,\ \frac{alcohol}{\mathrm{\Delta}}$
Explanation
(c)
1, 2-dihalogen ($vicinal$) derivatives of the alkanes on reaction with zinc dust and methanol produces alkenes by loss of two halogen atoms (dehalogenation).
$CH_{3} - CH - CH_{2} + Zn\overset{\frac{Alcohol}{\mathrm{\Delta}}}{\rightarrow}CH_{3}CH = CH_{2}$
| | propylene
$Br\ \ \ \ \ \ \ Br$
1,2-dibromopropane
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.