Card Set Information

2012-04-23 03:17:06

Show Answers:

  1. OH of 2' with CrO3 / KCr2O7
    • Turns into a Keytone. removes two H's.
    • 1 from alpha Carbon and 1 from the OH.
    • the elminiation becomes a bond between the c and o.
    • now it is a double bond.
  2. OH (IR)
    very broad; around 3600-3200
  3. OH of 1' with PCC/HCrO4
    takes 2 H's. one from OH and one from alpha carbon (there's 2 H's on the alpha) those will form a bond.
  4. Cleavage by O3 (alkenes)
    Breaks in Half (literally) and add an O at the end of the bonds.
  5. enantiopopic
    stereocenter when formed
  6. McPBA
    alkenes to epoxide. top and bottom side attack.

    **the groups stay in the same place (stereocenter wise, no changes in the 3D) since it is top and bottom attack, the only thing will change is the epoxide itself. that will go on top or bottom.(plain stereo)
  7. C=C triple bond (IR)
  8. Cleavage for alkynes by O3
    if the triple bond is at the end, the C-H bond turns into CO2

    • triple bond breaks into a double bond with O
    • and a bond with an OH.
  9. Alcohol (OH) of 1' with CrO3 /KCr2O7
    • Keep OH and the 2H's turn into C=O.
    • *Cr- Creep (keep OH)
  10. Sharpless
    Alkene with OH turns into epoxide with OH.

    • OH needs to be on the Right Hand corner.
    • (+)Det is below
    • (-)Det is above.

    Any product OH will be going away. All products OH will be going away. Below or above attacks should be in plain stereo. because OH is going away and what other product will be going towards. leaving plain to use.
  11. H2 and Pd-C
    Adding 2H's to any double or triple bonds. what for stereocenters.
  12. KMnO4/NaOH

    OsO4/NaH SO3
    Double bond breaks open and will turn into 2 OH bond.
  13. C=C (IR)
  14. N-H (IR)
  15. Cleavage of Alkenes by KMnO4
    Breaks in Half and at the end of the double bonds will be an O. so c=o
  16. (CH3)3COO3H/ OsO4
    anti OH
  17. Csp3-H (IR)
  18. Benzene (IR)
  19. C==N
  20. Csp2H (IR)
  21. LiALH4 /h2H
    If it is not reacting with epoxide then it will only remove the alkyl halide group. cl would become h
  22. C triple C KMNO4/h3O
    C=O and C-OH bond arm formed.
  23. Lindlar's Cat.
    Will only, ONLY break a triple bond into a double bond.
  24. LiALH4/h2O (epoxide)
    • H- leaves from LiALH4
    • attacks the less sub.
    • O is on more sub. H2o adds H to O. becomes OH.
  25. Diasteroctropic
    2 stereocent after formed
  26. CspH (IR)
  27. C=O (IR)
  28. 2' OH with PCC or CrO3
    2H's turns into C-O bond. which means C-O will be a C=O.