Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: LanierC on November 30, 2016, 02:14:26 PM
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Why would bromine take place of an alcohol group? Specifically the reaction of n-butyl alcohol to n-butyl bromine. I know it is an Sn2 reaction, because a primary carbocation is unstable, but I've been told its electronegativity, and I'm curious if a higher electronegativity means its more likely to hold onto the electrons, or if it's for another reason.
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I question whether any reaction will occur. Is -OH a good leaving group? Why or why not? I don't think that electronegativity is the property on which to focus your attention.
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I question whether any reaction will occur. Is -OH a good leaving group? Why or why not? I don't think that electronegativity is the property on which to focus your attention.
I believe OH is a good leaving group because it is a polar protic solvent. It is a very slow reaction (this was done in lab, I'm writing the formal paper so I want to make sure I understand the full reaction). I also know that the more electronegative the leaving group is, the more polarized the carbon is, making it an easier target for the nucleophile to attack.
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Or could H2O be the leaving group?
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Or could H2O be the leaving group?
Good observation... water is a good leaving group because it is a weak base, correct? How would I correctly name the water group on the end of the butane chain?
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more electronegative atoms are may not be better leaving group. If you take the sample of halogen family, florine is not better leaving group compared to bromine and iodine even though florine is more electronegative than other members of same family. This can be explain in term of Pka value. If the conjugate base of a acid is a weaker base ( low Pka value) then it is better leaving group but if the conjugate base of acid is stronger base it is poor leaving group as in case of hydroxide ion.