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When HBr reacts with ethene it proceeds by the attack of proton on Ethene to give Ethyl carbocation and then attack of nucleophilic bromide ion to give Bromoethane.

My question is why does the proton react with Ethene earlier than the bromide ion? Is a proton more reactive than bromide ion or is there some other reason?

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The mechanism of the addition of hydrogen bromide to ethene is called electrophilic addition. The region of highest electron density in ethene is the double bond, where the π electrons are. Hydrogen bromide acts as an electrophile toward the π electrons of the double bond. The proton of hydrogen bromide is positively polarized and thus electrophilic. (Obviously, the negatively charged bromide ion is not electrophilic.) Hence, the first step is an acid-base reaction in which hydrogen bromide donates a proton to ethene, forming a carbocation.

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    $\begingroup$ Also note that ethene is not usually reacted with $\ce{HBr(aq)}$. Ethene is usually reacted with $\ce{HBr(g)}$, in which case the acid is not ionized. In this case, the electron-rich (nucleophilic) part of ethene (the $\pi$ bond) reacts with the electron poor (electrophilic) part of $\ce{HBr}$ (the hydrogen atom). $\endgroup$
    – Ben Norris
    Nov 5, 2014 at 12:00

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