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N-butyl alcohol and bromide reaction

  1. Jul 21, 2011 #1
    1. The problem statement, all variables and given/known data

    n-butyl alcohol was reacted with potassium bromide in the presence of concentrated sulphuric acid and water.

    1) Is hydrobromic acid involved in this reaction? If yes, how so?
    2) Is this reaction SN1 or SN2? What is the mechanism?
    3) What is the purpose of the sulphuric acid and water?

    2. Relevant equations

    -

    3. The attempt at a solution

    1) I believe HBr is involved, even though it's not added. Perhaps the bromide from potassium bromide and the hydrogen ion from sulphuric form HBr.
    2) This is an SN2 reaction (substrate is a primary carbon). The hydroxyl group of the n-butyl alcohol undergoes protonation with HBr, while the nucleophilic Br- does a backside attack on the carbon atom which is attached to the alcohol group. The leaving group is the water molecule while n-butyl alcohol is converted into 1-bromobutane.
    3) Sulphuric acid serves as a catalyst? Or to provide the hydrogen ions for protonation.. or for the formation of HBr? Quite confused over this question. Water seemingly serves as a solvent, but it's a protic solvent, which means that it will slow the reaction down. Why is this so?


    Any help would be greatly appreciated. Thanks in advance.
     
  2. jcsd
  3. Jan 28, 2017 #2

    TeethWhitener

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    Science Advisor
    Gold Member

    Yes, in an acidic environment, your nucleophile will be neutral (in a basic environment, it'll be an anion). The sulfuric acid protonates the KBr to give HBr.

    Yes, it's SN2. The rest looks fine.

    The sulfuric acid pushes the hydroxyl equilibrium toward protonated hydroxyl, setting up water as a good leaving group. HBr is a pretty decent nucleophile in water, given that the hydrogen bonding to HBr is quite a bit weaker than in other acids like HCl or HF.
     
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