Conversion of tertiary alcohols to alkyl bromides using HBr

by James

Description: Treatment of tertiary alcohols with HBr leads to formation of tertiary alkyl bromides.
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Notes: This proceeds through an SN1 mechanism. Note that alcohols which can form stable carbocations such as allylic and benzylic alcohols will also go through this pathway.

Examples: 

Notes: Note that in the third example, there will be a mixture of products (stereoisomers) because the reaction proceeds through an SN1 pathway involving a carbocation. In this case a mixture of enantiomers will form.

Mechanism: Protonation of the oxygen by HBr leads to an oxonium ion (Step 1, arrows A and B) turning OH into OH2(+) – a better leaving group. Loss of water (Step 2, arrow C) gives the carbocation, which is then attacked by bromide ion (Step 3, arrow D) to give the alkyl bromide.

Notes: If the alcohol were on a stereocenter we would obtain a mixture of stereoisomers.

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