traumanurse9 Posted March 2, 2009 Posted March 2, 2009 Hi guys, first time posting and I am sure I will be a frequent poster as I am really trying to master this subject material. My question is concerning solvolysis with ethanol: If I mix 2-bromo-2-methylbutane with heated ehtanol (which is a protic solvent correct?) I should produce two products correct?.......The ethanol oxygen replacing the bromine molecule attached to the 2nd carbon as well as the hydrogen attached to the oxygen atom being bonded to the displaced bromine atom? I just want to make sure I am understanding all this for Sn2 and Sn1 reactions. Scott
Kaeroll Posted March 2, 2009 Posted March 2, 2009 Hi Scott Your description is correct. The major products would be the ether derivative of the starting bromoalkane, along with hydrobromic acid in solution. As the bromoalkane is tertiary, it would most likely proceed via the Sn1 mechanism (or Dn + An if you prefer). Question for bonus points: can you think of a possible competing reaction? How likely is it that this will occur? Hope this helps. Kaeroll
vedmecum Posted March 14, 2009 Posted March 14, 2009 you are creating confusion in between two terms :- solvolysis and reaction as they differ a bit . and if i am not wrong then the competing reaction is elimination reaction just because of tertiary carbon . i have one problem which is blundering my CNS . when an alkyl halide is added to alc.KOH , there is an alkene formation. Alcohol is there but no ether formation takes place . how ? why ?
Kaeroll Posted April 14, 2009 Posted April 14, 2009 you are creating confusion in between two terms :- solvolysis and reaction as they differ a bit . and if i am not wrong then the competing reaction is elimination reaction just because of tertiary carbon . i have one problem which is blundering my CNS . when an alkyl halide is added to alc.KOH , there is an alkene formation. Alcohol is there but no ether formation takes place . how ? why ? Yes, elimination would likely compete. The product ratios depend strongly upon reagents and conditions though. I'm not sure about your second question. What alcohols and alkyl halides are you thinking of? Those conditions should allow the substitution reaction to occur. Again, the product ratios are dependant upon the reagents and conditions, but KOH should be a good enough base to effect the substitution.
UC Posted April 14, 2009 Posted April 14, 2009 (edited) Hi guys, first time posting and I am sure I will be a frequent poster as I am really trying to master this subject material. My question is concerning solvolysis with ethanol: If I mix 2-bromo-2-methylbutane with heated ehtanol (which is a protic solvent correct?) I should produce two products correct?.......The ethanol oxygen replacing the bromine molecule attached to the 2nd carbon as well as the hydrogen attached to the oxygen atom being bonded to the displaced bromine atom? I just want to make sure I am understanding all this for Sn2 and Sn1 reactions. Scott You have a tertiary alkyl bromide there. Bromide is a so-so leaving group and because it's tertiary, SN2 is never going to happen. Ethanol, being very weakly basic, will promote SN1 type reactions. Because the mixture is being heated, dehydrohalogenation via E1 is the favored reaction. If it were in the cold, SN1 substitution might show up more. Merged post follows: Consecutive posts mergedyou are creating confusion in between two terms :- solvolysis and reaction as they differ a bit . and if i am not wrong then the competing reaction is elimination reaction just because of tertiary carbon . i have one problem which is blundering my CNS . when an alkyl halide is added to alc.KOH , there is an alkene formation. Alcohol is there but no ether formation takes place . how ? why ? Depends what kind of alkyl halide you're using, and what kind of alcohol is being used. If the halide is primary and a bromide or especially an iodide, one could expect some amount of ether formation. If the alkyl halide is secondary or tertiary, E1 elimination will be the favored reaction path, especially if the mixture is heated strongly. Edited April 14, 2009 by UC Consecutive posts merged.
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