Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: bmanth60 on December 16, 2006, 08:39:37 PM
-
With the elimination 1 reaction, i have an example:
http://www.chem.ualberta.ca/~tykwinsk/images/CHEM161.chapter7.pdf
Page 5, E1 elimination of alcohol with treatment of acid and heat.
How come when the elimination occurs, the hydrogen isn't the one that gets attacked? Instead, the CH3 is attacked?
-
Hint: Remember it's a bulky base
-
Um, even for the problem concerning E1 elimination of alcohols upon treatment with an acid and heat??
-
That reaction is incorrect. The correct product should be (H2C)2=(CH2)2 (2,3-dimethyl-but-2-ene). The reaction is obviously wrong because there are six carbons in the reactants but only five carbons in the product.
-
Yea thanks for clearing that up, I was thinking that was the case, but was doubtful of prof making mistake.
-
The methyl group is not supposed to be attacked. Elimination reactions require strong bases that react by removng a hydrogen proton (the hydrogen bond cleaves heterolytically).
Even by showing the methyl substituent being removed, the problem is not correct in doing so even at that, because only the carbon next to the carbon bearing the OH group can be attacked. This is the only way to form our double bond.
But yes the product does not have as many carbons as you began with, and since elimination reactions do not involve removal of carbons, this is a clear way to see that this reaction is not going to work that way.