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$$\ce{CH3CH=CH2 ->[Br2] CH3-CHBr-CH2Br}$$

In the reaction mechanism, why does the nucleophile attack at the 2nd carbon atom? Shouldn't it attack at the 1st one, so that in the transition state, the carbanion is relatively more stable?

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    $\begingroup$ Which site is the more stable carbocation in the transition state? $\endgroup$
    – Waylander
    Dec 14, 2019 at 12:13
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    $\begingroup$ There is no carbanion in this reaction mechanism. $\endgroup$ Dec 14, 2019 at 13:03

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There are no carbanion in this reaction. In bromine water, $\ce{Br2}$ is partially made of $\ce{Br+}$ and $\ce{Br-}$ ions. Then $\ce{Br+}$ attacks the double bond, producing a carbocation, with $\ce{Br}$ at the outer $\ce{C}$ atom, and the + charge on the central $\ce{C}$ atom. This is the most stable carbocation. Then it will react with the $\ce{Br-}$ ion which will be attached on the central $\ce{C}$ atom.

It also should be mentioned that a secondary reaction may happen where $\ce{H2O}$ reacts with the transition state carbocation. As a result, the molecule $\ce{CH3-CHOH-CH2Br}$ is produced.

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    $\begingroup$ Your explanaion is okay but mechanism is wrong. Did you hear about cyclic bromonium ion mechanism? $\endgroup$ Dec 14, 2019 at 13:04
  • $\begingroup$ Of course I know the bromonium ion mechanism. I have not mentioned it, because I think it might have been avoided here. $\endgroup$
    – Maurice
    Dec 14, 2019 at 16:24

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