Q. In the following reaction, $RCH_{2}CH = CH_{2} + ICI \rightarrow \lbrack A\rbrack$ Markownikoff’s product $\lbrack A\rbrack$ is
A) $RCH_{2}CH - CH_{2}I$
|
Cl
B) $RCH_{2}CH - CH_{2}CI\$
|
$I$
C) $RCH_{2} - C = CH_{2}$
|
$I$
D) $RCH = CH - CH_{2}I$
Explanation
(a)
Markownikoff’s addition the negative part of the unsymmetrical reagents adds to a less hydrogenated (more substituted) carbon atom of the double bond. In ICl, Cl is more electronegative. So, it will take negative charge, $.e.,$ $I^{+}Cl^{-}$. So, the product is
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Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.