1.

In the following reaction, RCH_2CH=CH_2+I Cl to [A] Markownikoff's product [A] is

Answer»

`RCH_2undersetunderset(Cl)(|)CH-CH_2I`
`RCH_2-undersetunderset(I)|CH-CH_2Cl`
`RCH-undersetundersetI|CH=CH_2`
`RCH=CH-CH_2I`

Solution :Markownikoff's addition : The NEGATIVE part of the unsymmetrical REAGENT adds to a less hydrogenated (more substituted ) carbon atom of the double bond.
In IC L , Cl is more electronegative. So it will take negative CHARGE . i.e., `I^(+)Cl^(-)`
So, the PRODUCT is
`RCH_2CH=CH_2+IC l to R-CH_2-undersetunderset(Cl)(|)CH-CH_2I`


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