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Addition of alkene with water

  1. Jun 2, 2014 #1
    1. The problem statement, all variables and given/known data

    my question is at part 2 , since c form ch3ch2 is electron deficient, but after OH2 added in , with lone pair electron on O, the c should be no longer electron deficient anymore. am i right? why the + sign is moved to O now?

    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Jun 2, 2014 #2
    Please calculate the formal charge on oxygen. Remember that H2O has two lone electron pairs.
     
  4. Jun 3, 2014 #3
    do u mean now the O become electron deficient, since one lone pair of lectron is donated to C+, and now there's one lone pair left one the O atom?
     
  5. Jun 3, 2014 #4
    Why is oxygen electron deficient at this point? How many valence electrons should it have to be neutral? How many to be positive? How many to be negative? Do all atoms have the same requirements for their valence shells?

    Hint: this deals with the calculation of its Formal Charge…
     
  6. Jun 3, 2014 #5
    That is not what I said but it is another, less formal, way of looking at the same thing. Knowing how to calculate formal charges is good because books and people can make mistakes. Keeping track of your electrons is incredibly important in O Chem. Once you have seen an oxygen with three sigma bonds and a lone electron pair 100 times, you probably won't need to run the formal charge calculations because you know it is positive.

    I prefer to know why rather than rely on memorizing though. The memorization is simply a byproduct.
     
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