Chem301 SN1, SN2, E1, E2

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Mattyj1388
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186455
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Chem301 SN1, SN2, E1, E2
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2012-12-15 14:19:41
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chem 301 USD Ch6
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general questions about SN1, SN2, E1, E2
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  1. Factors that affect rate constant
    • 1. Energy of activation
    • 2. Temperature
    • 3. Sterics (giant groups put in the way to block binding)
    • 4. Catalysts
  2. Identify the mechanisim(s) expected to operate when 1-bromobutane is treated with NaOH reagent. [SN1, SN2, E1, E2]
    SN2 and/or E2
  3. Identify the mechanisim(s) expected to operate when 1-bromobutane is treated with NaSH reagent. [SN1, SN2, E1, E2]
    SN2
  4. Identify the mechanisim(s) expected to operate when 1-bromobutane is treated with t-buOKreagent. [SN1, SN2, E1, E2]
    SN2, E2
  5. Identify the mechanisim(s) expected to operate when 1-bromobutane is treated with DBN reagent. [SN1, SN2, E1, E2]
    E2
  6. Identify the mechanisim(s) expected to operate when 1-bromobutane is treated with NaOMe reagent. [SN1, SN2, E1, E2]
    SN2, E2
  7. Identify the mechanisim(s) expected to operate when 2-bromopentane is treated with NaOEt reagent. [SN1, SN2, E1, E2]
    SN2, E2
  8. Identify the mechanisim(s) expected to operate when 2-bromopentane is treated with NaI/DMSO reagent. [SN1, SN2, E1, E2]
    SN2
  9. Identify the mechanisim(s) expected to operate when 2-bromopentane is treated with DBU reagent. [SN1, SN2, E1, E2]
    E2
  10. Identify the mechanisim(s) expected to operate when 2-bromopentane is treated with NaOH reagent. [SN1, SN2, E1, E2]
    SN2, E2
  11. Identify the mechanisim(s) expected to operate when 2-bromopentane is treated with t-BuOK reagent. [SN1, SN2, E1, E2]
    SN2, E2
  12. Identify the mechanisim(s) expected to operate when 2-bromo-2-methylpentane is treated with EtOH reagent. [SN1, SN2, E1, E2]
    SN1, E1
  13. Identify the mechanisim(s) expected to operate when 2-bromo-2-methylpentane is treated with NaH reagent. [SN1, SN2, E1, E2]
    E2
  14. Identify the mechanisim(s) expected to operate when 2-bromo-2-methylpentane is treated with NaI reagent. [SN1, SN2, E1, E2]
    SN1
  15. Identify the mechanisim(s) expected to operate when 2-bromo-2-methylpentane is treated with NaOEt reagent. [SN1, SN2, E1, E2]
    E2
  16. Identify the mechanisim(s) expected to operate when 2-bromo-2-methylpentane is treated with NaOH reagent. [SN1, SN2, E1, E2]
    E2
  17. When 1-chlorobutane is treated with ethanol, neither elimination process (E1 or E2) is observed at an appreciable rate; Explain why an E2 reaction does not occure.
    An E2 reaction does not readily occure because the base is weak.
  18. When 1-chlorobutane is treated with ethanol, neither elimination process (E1 or E2) is observed at an appreciable rate; Explain why an E1 reaction does not occure.
    An E1 reaction does not readily occure because the substrate is a primary.
  19. Modifying the reactants can have a profound effect on the rate of elimination. What modification would you suggest to enhance the rate of an E2 process?
    Replacing the weak base (EtOH) with a strong base (such as NaOEt) would greatly enhance the rate of an E2 process.
  20. What modification would suggest to enhance the rate of an E1 process?
    Replacing the primary substrate with a tirtiary substrate (such as 1-chloro-1,1-dimethylbutane) would greatly enhance the rates of an E1 process.

  21. 1, 2, 3, 4, 5 or more?
  22. When 1-methoxy-2-methylpropene is treated with HCl, the major product is 1-chloro-1-methyoxy-2-methylpropane. Although this reaction proceeds via an ionic mechanism, the HCl is ultimately positioned at the less substituted carbon. Draw a mechanism that is consistent with this outcome.
  23. When 1-methoxy-2-methylpropene is treated with HCl, the major product is 1-chloro-1-methyoxy-2-methylpropane. Although this reaction proceeds via an ionic mechanism, the HCl is ultimately positioned at the less substituted carbon.
    Draw the resonance-stabilized intermediate for the previous step.
    i.e.
  24. Draw step two of:
  25. Explain why the less substituted carbocation intermediate is more stable in this case.
    The less-substituted carbocation is more stable, because it is resonance-stabalized.
  26. What reactions would you expect to get using a Nucleophil (only) with a 1o, 2o, 3o? [
    • [nucleophile only i.e. Br-, Cl -, I -, HS -, H2S, RS -, RSH
    • 1o= SN2
    • 2o= SN2 + SN1
    • 3o= SN1
  27. What reactions would you expect to get using a Base (only) with a 1o, 2o, 3o?
    • [Base (only) i.e. H -, DBN, DBU]
    • 1o = E2
    • 2o = E2
    • 3o = E2
  28. What reactions would you expect to get using a Strong nuc / Strong Base with a 1o, 2o, 3o?
    • [Strong nuc / Strong Base i.e. OH -, MeO -, EtO -, t-BuO -]
    • 1o = SN2 (major) + E2 (minor)
    • 2o =  E2 (major) + SN2 (minor)
    • 3o = E2
  29. What reactions would you expect to get using a Weak nuc / Weak Base with a 1o, 2o, 3o?
    • [Weak Nuc / Weak Base i.e. H2O, MeOH, EtOH]
    • 1o = SN2 + E2 Not Practical
    • 2o = SN2 + SN1 + E2 + E1 Not Practical
    • 3o = SN1 + E1
  30. SN2 Regiochemical Outcome
    The nucleophile attacks the alpha position, where the leaving group is connected.
  31. SN2 Stereochemical Outcome
    The nucleophile replaces the leaving group with inversion of configuration.
  32. SN1 Regiochemical Outcome
    The nucleophile attacks the carbocation, which is where the leaving group was originally connected, unless a carbocation rearrangement took place.
  33. SN1 Stereochemical Outcome
    The nucleophile replaces the leaving group with recamization.
  34. E2 Regiochemical
    The Zaitsev product is generally favored over the Hofmann product, unless a sterically hindered base is used (Hofmann favored)
  35. E2 Stereochemical
    Process is both stereoselective and stereospecific. When applicable, a trans disubstituted alkene will be favored over a cis disubstituted alkene. When the Beta position of the substrate has only one proton, the stereoisomeric alkene resulting from anti-periplanar elimination will be obtained (exclusively, in most cases).
  36. E1 Regiochemical
    The Zaitsev product is always favored over the Hofmann product.
  37. E1 Stereochemical
    The process is stereoselective. When applicable, a trans disubstituted alkene will be favored over the cis disubstituted alkene
  38. Polar Aprotic
    • DMSO: dimethylsulfoxide
    • THF: tetrahydrofuran
    • DMF: dimethylformamide
    • DCM: dichloromethane
    • MeCN: acetonitrile
    • ethyl acetate
    • Acetone
  39. Polar Protic
    • H2O
    • MeOH
    • EtOH
    • n-propanol
    • IPA: isopropanol
    • n-butanol
    • formic acid
    • acetic acid
  40. SN2 Characteristics
    • One step
    • nucleophile attacks electrophile
    • Polar Aprotic solvent
    • 1o>2o>>>>>>>3o
    • rate: K [substrate] [base]
    • single enantiomer formed
    • no rearrangement (no carbocation)
    • Backside attack by nucleophile, pushing off LG, causes inversion at chiral carbon
    • Can go through SN2 twice (or any even number of times) to get the same direction for reactant and product.
    • Control!
  41. SN1 Charateristics
    • Two Steps (at least)
    • Polar Protic solvent
    • 3o>2o>>>>>>>1o
    • rate: K [electrophile or substrate]
    • forms racemic mixture
    • Carbocation intermediate (can be isolated)
    • L.G. leaves, forming a carbocation, nucleophile attacks to create a racemic mixture.
    • No stereoselectection
    • THINGS THAT DO NOT WORK FOR SN1
    • Allyl
    • Benzyl

  42. E2 Charateristics
    • Causes double bonds!
    • One step process: Concerted
    • typically 1o and 2o but can also do 3o
    • rate: k [substrate] [base]
    • Antiperiplanar
    • Typically caused by a strong base
    • can be Stereospecific
    • No carbocation!!
  43. E1 Charateristics
    • Causes double bonds!
    • Two steps: or step wise
    • Carbocation intermediate
    • LG then deprotonation
    • 3o, some 2o
    • rate: K [substrate]
    • weak or no base in solution
    • Most stable form (Zaitsev)
    • No atiperiplanar position needed
  44. Regiochemistry
    • When a double bond has different options to form and the differences are considerable differences.
    • Attacks carbocations
    • When using a big bulky base you get Hofmann product as major.

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