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Ochem: H20 and HBR

  1. Feb 19, 2011 #1
    1. The problem statement, all variables and given/known data
    2nd semester Organic Chemistry
    not a HW problem, I just wanted to get this right...

    (note: by primary/secondary/tertiary I mean carbon atom bonded to 1,2,3 carbons respectively
    if they are actually called 1st degree/2nd degree or anything else, please correct me)
    When I add H20 to primary Carbon. 1-bromobutane will it react?
    How about with 2-bromobutane (2ndary carbon)
    how about with t-butyl bromine (tertiary carbon)

    When I add HBr to 1-butanol, 2-butanol, and t-butylol
    will they react? or not react?

    2. Relevant equations
    mostly SN1, SN2, E1, E2 reactions, or NR (no reaction

    3. The attempt at a solution

    H2O + 1-bromobutane > NR

    H2O is a weak nucleophile and weak base, and we have primary carbon so NR

    H2O + 2-bromobutane > 2-butanol + (E) 2-butene and other minor butene

    H2O is a weak nucleophile and weak base, and we have secondary carbon so SN1+E1 reaction

    H2O + t-butyl bromine > t-butanol

    Classical SN1+E1 reaction, Br will leave and deprotinate H+ , leaving OH- which then attachs to t-butoxide

    HBr + 1-butanol > 1-bromobutane

    HBr is a strong acid, so we have Br(-1) ion, which attacks 1-butanol and forces OH group to leave
    going through SN2 reaction

    HBr + 2-butanol > 2-bromobutane + (E) 2-butene and other minor butene

    answer 1
    HBr is a strong acid, so we have Br(-1) ion, which attacks 2-butanol and goes through
    SN2 reaction.

    or... (tell me which one is right ^^)

    answer 2
    HBr is a strong acid, so H+ will be added to butanol so OH group will be OH2 (+1 charge)
    instead make it very good leaving group. Now Br- will attack (errr SN2?) and OH2+ will leave

    HBr + t-butanol > 1-methyl-2-propene

    tertiary carbon so no SN2, but E2 occurs so it forms alkene

    I really hate using HBr and H2O in these cases in synthesis due to their uncertainty in reaction,
    nevertheless I thought I should know these in case I can't use other reagents.

    Am I right? Instead of simple right or wrong, would you also add a note to each one how
    strong the reaction would be?
    like number 1 and 4 (2and5, and 3 and 6) can't both be right because they are opposite reaction.

    as for number 5, I got the wrong in test during synthesis, but I can't understand why because I saw this reaction in my notes as well.
    or could it because E2 is major in this reaction (and that notes were only explaining SN2 because E2 wasn't being covered at that time)
  2. jcsd
  3. Sep 14, 2017 #2


    User Avatar
    Science Advisor
    Gold Member

    Handy reference table:
    Most O-chem books will have something like this table. As for the reactions:
    1. This is fine. Water won't react (at least in the short term).
    2. I can see where you would get this from, but in reality 2-bromobutane isn't really reactive with water. If I had to say what a possible reaction would be, SN1 would be it.
    3. This is fine too. In reality, in neutral water, Br- is not basic. The OH- will come from autoionization of water.
    4. HBr protonates the OH group, giving an H2O+ group, which is an excellent leaving group. Then another HBr attacks the backside in an SN2 and the extra proton on the HBr is grabbed by a water molecule. One important feature here is that if you're in a strongly acidic environment (like with HBr), you aren't really ever going to see formal negative charges show up in reaction mechanisms, only positive ones. And vice versa in a strongly basic environment. If, for example, your first 3 reactions involved putting butyl bromide into NaOH, you wouldn't have reaction mechanisms with formal positive charges flying around.
    5. As mentioned before, since you're in a strongly acidic environment, the HBr will protonate the OH group. From here, secondary alkyl halides tend to see competition between SN1 and SN2 mechanisms. The elimination mechanisms are more important in basic media, but here E1 might compete to some extent (especially since the product is a gas in this case: it is at least strongly entropically favored). (Sometimes eliminations are also highly enthalpically favored, e.g., sulfuric acid + ethanol at high temperatures will give the dehydration product ethylene)
    6. Tertiary carbons only ever react through SN1 mechanisms. So the SN2/E2 question is irrelevant. You will probably see competition between SN1 and E1 reactions in this case, if I had to guess.
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