Why do alkanes need UV light to undergo a reaction with Bromine?

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Why do alkanes need UV light to undergo a reaction with Bromine?

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Answer 1 · 2018-05-16T15:58:29

Well, $C - H$ $C-H$ bonds ARE strong.....

Explanation:

...and in fact it is the bromine molecule, with a WEAKER $B r - B r$ $Br-Br$ bond, that undergoes photolysis to give TWO neutral bromine radicals...

$B r_{2} h ν - - \to 2 . B r$ $Br_2stackrel(hnu)rarr2dotBr$

$. B r + H_{2} C R_{2} \to B r - H + . C H R_{2}$ $dotBr+H_2CR_2 rarr Br-H + dotCHR_2$

$B r_{2} + . C H R \to . B r + B r C H R_{2}$ $Br_2 + dotCHRrarrdotBr+BrCHR_2$

You can look up $initiation, propagation, termination steps$ $"initiation, propagation, termination steps"$ of radical reactions in your text; this is worth doing. It is the propagation step that is the important step. The formation of a radical, and its subsequent reaction, creates ANOTHER radical, which can propagate the radical chain of reaction....

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Why do alkanes need UV light to undergo a reaction with Bromine?

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