Answers to Problem Set 4

 

4.2

  1. Phosphorus-for-Cl substitution (SN1 or SN2!), followed by attack of chloride on methyl ( SN2 ).
  2. Form the activated OBs ester; on ionization migrate the bond that is aligned and behind.
  3. Protonate the alkyne; LP from Cl attacks to form a cyclic intermediate. Note: trifluoroacetate is not nucleophilic enough to displace chloride unless it is in this bridging arrangement!
  4. First case:  Diazotization; ionization accompanied by alkyl migration to form resonance-stabilized oxocarbocation.  Second case:  The locked conformation (tBu) prevents alkyl migration as above; now hydride is anti-periplanar and migrates.
  5. A norbornyl-like rearrangement gives an oxocarbocation that on addition of water gives product as the hemiacetal form.
  6. Neighboring group participation by N gives a symmetric intermediate. Note that exo and endo isomers are diastereomeric, not enantiomeric!
  7. Neighboring group participation by the amide C=O oxygen forms a cyclic intermediate.

4.3.

  1. Fluorinated sulfonate a better LG.
  2. CH3 more electron rich; stabilizes carbocation.
  3. Both would require substantial deformation of the strained norbornyl ring system; the 7-tosylate has a sigma bond aligned in the back that sets up a rearrangement.
  4. Moving the tBu group provides relief of steric strain.
  5. The double bond cannot conjugate, and the sp2 center makes rehybridization more difficult.
  6. This is an SN2 reaction; which the sulfonyl makes the attached carbon more electrophilic, it also adds steric bulk. The aactual rate ratio is 38:2.
  7. Cyclopropylcarbinyl; the first example is meopentyl and undergoes a slightly slower but concommitant rearrangement to a tertiary cation.
  8. Presence of the alkene provides neighboring group participation & formation of the 2-norbornyl ion.
  9. Neighboring group participation; 5-member ring > 4-member ring.
  10. Neighboring group participation by the ortho acetate.


4.13. a.  Participation of the aromatic electrons is more costly than of the C=C pi electrons.

c. Two bond migrations compete in each case. One leads to a cyclopropylcarbinyl cation, differing for the two systems in being perpendicular or parallel at the transition state. This predominates for E because of the parallel orientation.

Look at models for the two compounds (different rotamers) and the two ions:

From anti isomer: Migration of C-1 to form the C-3 cation:

415c_1a.mol

 From the anti isomer: Migration of C-3 to form the C-2 cation:

415c_2a.mol
From the syn isomer: migration of C-1 to form the C-3 cation:

415c_1s.mol

 From the syn isomer: migration of C-3 to form the C-2 cation:

415c_2s.mol
The product from the anti isomer:

1_310.mol

 The product from the syn isomer:

2_310.mol

F would lead to a perpendicular cyclopropylcarbinyl cation, so it does the other migration, leading to a nonclassical cation delocalized into the bridgehead-bridgehead cyclopropane bond.


B.  Additional problems.

1.Polar solvent in red; nonpolar in black.

a.  Ionic reactants, neutral TS/product.  Stabilize reactants, increase activation barrier and slow rate in polar solvents.

b.  Polar reactants, TS and products.  Little change, either a slight increase or slight decrease in rate could be rationalized.

c.  Neutral reactants; polar TS/products.  The stabilization of developing charge will assist the formation of the transition state and accelerate the reaction.

2. 

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Last updated: 10/09/03