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In the reactions of alkene, which alkene and reagents would be necessary to produce

  • 2,3-dimethyl-2-butanol
  • 3-bromo-2-methylbutan-1-ol

For 2,3-dimethyl-2-butanol, I came to the conclusion that it would be the 2,3-dimethyl-1-butenol, assuming that a rearrangement would happen. I am right about this one?

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  • $\begingroup$ Welcome to Chemistry.SE! This is a Q&A site, but it is not a do-my-homework service. The members will be glad to help and point you in the right direction, but you have to show us your efforts, i.e. what you've done so far and what is it exactly that you don't understand. Otherwise, your question may be voted down and eventually closed. $\endgroup$
    – Don_S
    Commented Mar 2, 2017 at 19:47
  • $\begingroup$ Do you mean 3-bromo-2-methylbutan-1-ol for your second product? $\endgroup$
    – Waylander
    Commented Mar 2, 2017 at 19:50
  • $\begingroup$ Yes, exactly this $\endgroup$
    – user42041
    Commented Mar 2, 2017 at 19:51

1 Answer 1

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Let me try to give you some hints without answering your homework question directly ;-)

  1. Draw your products!
  2. You can still take rearrangements into account on a later stage if nothing else works out, but initially: think simple
  3. Would you agree that the most simple way to convert an alkene to an alkanol is the reaction with water under acidic conditions? You probably remember how this works:

    proton adds first, a cation is formed, then comes water as the nucleophile

  4. Will this work if you need to introduce $\ce{Br}$ and $\ce{OH}$ at adjacent positions?

    At 3., $\ce{H+}$ and (formally) $\ce{OH-}$ were added. The combination of $\ce{Br+}$ and $\ce{OH-}$ doesn't sound like a common reagent, but your teacher might have talked about the addition of halogens to alkenes :) With bromine, the $\ce{Br2}$ adds to the alkene and forms a three-membered ring, a bromonium ion. Then $\ce{Br-}$ comes along and adds too. Formally, this is like adding $\ce{Br+}$ and $\ce{Br-}$ to the adjacent carbon atoms of the $\ce{C=C}$ bond. Now imagine that there is not just bromine, but also a lot of water. The first step, the formation of the bomonium ion, will be the same, but now you have (lots of) water as a nucleophile that competes with $\ce{Br-}$!

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  • $\begingroup$ Thank you for your feedback! My teacher didn't talked about epoxides yet. What i know is some very basic things that i searched. May i ask you: When you say "Adjacent positions" what do you mean? Sorry if it is a dumb question, but english is not my main language, and this happens sometimes =D. $\endgroup$
    – user42041
    Commented Mar 2, 2017 at 20:21
  • $\begingroup$ @terra2322 No worries :) It is not a dumb question at all and reading buteno, I already thought that your first language might be spanish. Adjacent just means being neighbours, side to side. So, carbon 2 and carbon 3 are adjacent positions, while carrbon 1 and carbon 3 are not. $\endgroup$
    – Klaus-Dieter Warzecha
    Commented Mar 2, 2017 at 20:35
  • $\begingroup$ @terra2322 Sorry about the epoxide! I forgot about a much simpler reaction and have updated my answer. $\endgroup$
    – Klaus-Dieter Warzecha
    Commented Mar 2, 2017 at 20:38
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    $\begingroup$ Aaah, i see. I think i understand how this kind of exercises are done. $\endgroup$
    – user42041
    Commented Mar 2, 2017 at 20:56
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    $\begingroup$ I'm gonna keep doing it thinking like that. Thanks! $\endgroup$
    – user42041
    Commented Mar 2, 2017 at 20:56

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