Look the following reaction:
CH2=CH2+Br2--->BrCH2-CH2Br .
Here,How the products BrCH2-CH2Br creates?
2006-08-22 07:37:49 · 4 answers · asked by star123 2 in Science & Mathematics ➔ Chemistry
This reaction can be driven by light.
Exposure to light causes the Br2 bromine molecule to split into two free radicals: Br. and Br.
The Br. has an un-paired electron which really wants to combine with something with a partial positive charge.
In the CH2=CH2 molecule, the electrons spend most of their time in the double bond between the carbons, leaving the hydrogen atoms with a partial positive charge.
An intermediate compound is formed
CH2=CH2 ---- Br.
The Bromine atom is very electronegative, meaning it pulls an electron away to complete it's unpaired electron.
This makes the remaining hydrogens on the other end of the intermediate compound even more positively charged.
So the other Br. free radical combines on the other end of the intermediate, resulting in the final product .
The final product has no double bond, so it is lower energy, and therefore more stable than the two starting reactants. This means the reaction is favorable.
But, it takes light to get it all started.
2006-08-23 09:08:27 · answer #1 · answered by Steve 2 · 1⤊ 0⤋
the reaction is correct.
That reaction happens because the total potential energy of BrCH2-CH2Br is smaller than the total potential energy of CH2=CH2 + Br2.
see quantum physics for precise calculation of both energies. But you can usually get a good approximation by summing the individual energies of each bond
2006-08-22 07:46:11 · answer #2 · answered by Anonymous · 0⤊ 0⤋
This is actually a really interesting mechanism. The double bond acts as the neucleophile, and the bromine acts intially as the electrophile. The double bond attacks one of the bromine molecules, and the other bromine molecule takes the electrons in the br-br covalent bond. This leave you with a transition state where both carbons in the double bond are bonded to the first bromine. The complete the reaction, the newly made bromide ion attacks the carbon atom in the C-C-Br bridge that is the LESS STABLE carbocation.
Sterochemically, the addition is always Trans!
2006-08-22 11:33:15 · answer #3 · answered by Duluth06ChE 3 · 1⤊ 0⤋
the relation between the two methylene groups is week so the prom which is halogen can break this relation and consist 1.2 dye promo ethane to get that u may need some heat
2006-08-22 08:07:24 · answer #4 · answered by Anonymous · 0⤊ 0⤋