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(CH(3))(2)C=CH(2)underset(H(2)O)overset(...

`(CH_(3))_(2)C=CH_(2)underset(H_(2)O)overset(Br_(2))rarr`
The major product of the reaction is

A

`(CH_(3))_(2)C(OH)CH_(2)Br`

B

`(CH_(3))_(2)C(Br)CH_(2)OH`

C

`(CH_(3))_(2)C(OH)CH_(2)OH`

D

`(CH_(3))_(2)C(Br)CH_(2)Br`

Text Solution

Verified by Experts

The correct Answer is:
A
If the alkene is unsymmetrical, the halogen ends up on the doubly bonded `C` with more `H's`. When the nucleophile is poor `(H_(2)O)` and the solvent has a high polarity `(H_(2)O)`, the `C` that can best be an incipient `R^(+)` is the more reactive site.
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