Qustion about leaving groups and bases

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The discussion centers on the reactivity of a nucleophile, specifically the conjugate base of benzene, in the presence of a bromide and a primary alcohol. The primary question is whether the nucleophile would attack the carbon attached to the bromide or the proton on the alcohol. It is concluded that the nucleophile would likely pick up the proton from the alcohol, reverting to benzene, rather than successfully attacking the bromide. Attempts to protect the alcohol, such as converting it to a benzyl ether, are also deemed ineffective due to steric hindrance around the bromine. The potential for using AlCl3 catalysis in a Friedel-Crafts alkylation is mentioned, but this method is known for producing complex mixtures, and the presence of the alcohol complicates the reaction further. The discussion questions the necessity of synthesizing this specific compound, suggesting that alternative, more straightforward approaches may be preferable.
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If I had a nucleophile, say the conjugate base of benzene, and a molecule with a bromide, and a primary alcohol. Would the nucleophile attack the carbon bound to the bromide, or the proton on the alcohol?

If it's hard to follow I can try to make a picture. Basically I want to add Isobutanol to benzene using a bromide leaving group. (2-bromo-2-methyl-propanol). Possible? I don't know if the alcohol would act like an acid or would the bromide leave first.
 
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This doesn't work. First, as you suspect, the phenyl anion will pick up the proton from the alcohol and return to benzene. Two, if you protect the alcohol as e.g. a benzyl ether, it will still be too crowded around the bromine for this to work.
Maybe you could fish some of the correct structure out of you use AlCl3 catalysis instead but this reaction (Friedel-Crafts alkylation) is notorius for giving messy mixtures. And the alcohol group certainly doesn't help.
Why do you need this awkward compound? Can't you make something sensible?
 
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