How to prepare 2-bromobutane from 1-butanol? [closed]








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So my answer (g then b) was incorrect. I thought that the bulky strong base of b would do E2 on 2-bromobutane, making it but-1-ene, then just add $\ce{HBr}$. What is wrong with my way of thinking?




The hydroxyl fragment isn't a good leaving group, unless the medium is acidic. Using just HBr, you can convert the hydroxyl group into an oxonium ion (the H2O fragment would be a good leaving group). Upon heterolysis of the C-O bond, a hydride shift occurs simultaneously to create a secondary carbocation. The exogenous bromide can act as a nucleophile towards the secondary carbocation to form 2-bromobutane.


 2

$\ce{(CH3)3COK}$ reacts with $\ce{R-X}$ to give a Hoffman alkene.
Elimination of the OH group usually requires protonation of oxygen, which requires an acidic medium.


A good solution to the given problem would be to add HBr without peroxide, which will give SN$\_2$ (no rearrangement, i.e. no cation shift) and give us $\ce{CH3CH2CH2CH2Br}$.


Then, add $\ce{(CH3)3COK/(CH3)3COH}$ to get but-1-ene. Then, add HBr to get 2-bromobutane.


Also a useful sidenote is that methanol, ethanol, n-propanol, and n-butanol all undergo SN$\_2$ when reacted with $\ce{HX}$. Other alcohols undergo SN$\_1$.


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