Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: NewtoAtoms on July 20, 2008, 06:46:14 PM
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Hello Organic Chemists.
Please consider the following reaction:
CH3 Br CH3 CH3
l HBr l l l
H2C=C-C-CH2 ------> H3C-C-C=CH2 + H3C-C=C-CH2Br
l l l
CH3 CH3 CH3
2,3-dimethyl,1,3-budadiene 3-bromo-2,3-dimethyl-1-butene 1-bromo,2-3-dimethyl 2-butene
PRODUCT 1,2 addition PRODUCT 1,4 addition
I understand that for product 1,2 addition the H and Br are added to adjacent carbons.
However for the product 1,4 addition. How do I know where to place the double bond?!?!? Really I could place it pretty much anywhere as long as I respect the same amount of H atoms as in the beginning reagant.
Can anyone explain where I decide to place the double bond in the 1,4 addition product???
Thank you so much!!!
New to Organic Chemistry
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You can't just add the double bond anywhere, you have to follow resonance. Check out the following link.
http://www.chem.sc.edu/faculty/handy/334Notes/Chapter22.doc
Read through the first few pages and you'll see how the reaction procedes.
Remember, first you add hydrogen, to form a carbocation (which is allylic), then it resonates, and you can add to the other end of the molecule.
Also, you forgot a double bond in your diagram.