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In the following reactions, RCH(2)CH=C...

In the following reactions,
`RCH_(2)CH=CH_(2)+(ICl) to (A)`
Markownikoff's product (A) is:

A

`RCH_(2)underset(Cl)underset(|)CH-CH_2I`

B

`RCH_(2) - underset(I)underset(|)CH - CH_(2) CL`

C

`RCH- underset(I)underset(|)CH=CH_(2)`

D

`RCH = CH - CH_(2)I`

Text Solution

Verified by Experts

The correct Answer is:
A
Markownikoff.s addition : The negative part of the unsymmetrical reagent 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. i.e., `I^+ Cl^-`
SO, the product is
`RCH_(2)CH = CH_(2) + ICI rarr R -CH_(2)-underset(Cl)underset(|)CH-CH_(2)I`
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