A) propane
B) butane
C) propene
D) ethane
Correct Answer: A
Solution :
\[\underset{propene}{\mathop{C{{H}_{3}}CH=C{{H}_{2}}}}\,\xrightarrow[\begin{smallmatrix} add\text{ }according\text{ }to \\ Markownikoffs \\ rule \end{smallmatrix}]{HBr}\]\[4C{{H}_{3}}-\underset{\begin{align} & | \\ & Br \\ & 2-bromopropane \\ & (A) \\ \end{align}}{\mathop{C}}\,H-C{{H}_{3}}\] \[\xrightarrow{\left( I \right)\text{ }Mg/ether}C{{H}_{3}}-\underset{\begin{align} & | \\ & MgBr \\ & isopropyi\text{ }magnesium \\ & bromide \\ & \left( Grignards\text{ }reagent \right) \\ \end{align}}{\mathop{C}}\,H-C{{H}_{3}}\] Note - Grignards reagent on reaction with any compound captaining active H-atom decompose to give alkane (when H-atom is attached to O, N or S or acidic in nature is known as active H-atom).You need to login to perform this action.
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