Q. Propane cannot be prepared from which reaction?
A) $CH_{3} - CH = {CH}_{2}\underset{OH^{-}}{\overset{{\ \ B}_{2}H_{6}\ \ }{\rightarrow}}$
B) $CH_{3}CH_{2}{CH}_{2}I\underset{P}{\overset{\ \ \ \ \ HI\ \ \ \ \ }{\rightarrow}}$
C) $CH_{3}CH_{2}CH_{2}COONa\overset{\frac{NaOH}{CaO,\mathrm{\Delta}\ \ \ \ }}{\rightarrow}$
D) None of the above
Explanation
(a)
(a)$CH_{3} - CH = {CH}_{2}\overset{B_{2}H_{6}}{\rightarrow}$
$\left( CH_{3} - {CH}_{2} - {CH}_{2} \right)_{3}B\overset{{OH}^{-}}{\rightarrow}{CH}_{3}{CH}_{2}{CH}_{2}OH$
Hydroboration of alkenes followed by hydrolysis in basic medium yield alcohol.
(B)$CH_{3} - CH_{2} - CH_{2}I\overset{\frac{HI}{P}}{\rightarrow}CH_{3} - CH_{2} - CH_{3}$
propane
Reduction of alkyl halides yield alkane.
(c)$CH_{3}CH_{2}CH_{2}COONa + NaOH$
$\overset{CaO\ \ \ \ \ }{\rightarrow}\ {CH}_{3}{CH}_{2}{CH}_{3} + {Na}_{2}{CO}_{3}$
Propane
Decarboxylation of sodium salt of fatty acid yield alkane having one carbon atom less than parent acid salt.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.