How will an alkene react with dilute, cold, neutral $\ce{KMnO4}$?
$\begingroup$
$\endgroup$
1
-
$\begingroup$ @YashasSamaga We prefer to not use MathJax in the title field, see here for details. $\endgroup$– Martin - マーチン ♦Commented Mar 31, 2017 at 5:29
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
The behaviour of $\ce{KMnO4}$ in a neutral medium is nearly the same as its behaviour in slightly alkaline medium.
In alkaline medium:
$\ce{3R-CH=CH-R + 2MnO4- + 2OH- -> R-CH(OH)-CH(OH)-R + MnO4^2-}$
In neutral medium:
$\ce{3R-CH=CH-R + 2MnO4- + 4H2O -> R-CH(OH)-CH(OH)-R + MnO2 + 2OH-}$
-
$\begingroup$ Wow, interesting. I wasn't expecting that. Thank you for your help! $\endgroup$ Commented Mar 30, 2017 at 2:54